UPSC Maths Optional Paper Solution Paper-01

UPSC Previous Years Maths Optional Papers with Solution | Paper-01

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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Details For UPSC Maths Optional Solved Papers (2018-2022)

UPSC Maths Optional Question Papers

upsc-m2022-1-bd58d9b8-2d68-4046-a6a8-417f99bd471d
खण्ड A
SECTION A
Question:-01 (a) सिद्ध कीजिए कि n n n\mathrm{n}n विमीय सदिश समष्टि V V V\mathrm{V}V के लिए n n n\mathrm{n}n रैखिकत: स्वतंत्र सदिशों का कोई भी समुच्चय V V V\mathrm{V}V के लिए एक आधार बनाता है ।
Question:-01 (a) Prove that any set of n n n\mathrm{n}n linearly independent vectors in a vector space V V V\mathrm{V}V of dimension n n n\mathrm{n}n constitutes a basis for V V V\mathrm{V}V.
Question:-01 (b) माना T : R 2 → R 3 T : R 2 → R 3 T:R^(2)rarrR^(3)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^3T:R2→R3 एक रैखिक रूपांतरण, ऐसा है कि T ( 1 0 ) = ( 1 2 3 ) T 1 0 = 1 2 3 T([1],[0])=([1],[2],[3])\mathrm{T}\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)T(10)=(123) तथा T ( 1 1 ) = ( − 3 2 8 ) T 1 1 = − 3 2 8 T([1],[1])=([-3],[2],[8])\mathrm{T}\left(\begin{array}{l}1 \\ 1\end{array}\right)=\left(\begin{array}{r}-3 \\ 2 \\ 8\end{array}\right)T(11)=(−328) है । T ( 2 4 ) T 2 4 T([2],[4])\mathrm{T}\left(\begin{array}{l}2 \\ 4\end{array}\right)T(24) को ज्ञात कीजिए ।
Question:-01(b) Let T : R 2 → R 3 T : R 2 → R 3 T:R^(2)rarrR^(3)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^3T:R2→R3 be a linear transformation such that T ( 1 0 ) = ( 1 2 3 ) T 1 0 = 1 2 3 T([1],[0])=([1],[2],[3])\mathrm{T}\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)T(10)=(123) and T ( 1 1 ) = ( − 3 2 8 ) T 1 1 = − 3 2 8 T([1],[1])=([-3],[2],[8])\mathrm{T}\left(\begin{array}{l}1 \\ 1\end{array}\right)=\left(\begin{array}{r}-3 \\ 2 \\ 8\end{array}\right)T(11)=(−328). Find T ( 2 4 ) T 2 4 T([2],[4])\mathrm{T}\left(\begin{array}{l}2 \\ 4\end{array}\right)T(24)
Question:-01(c) lim x → ∞ ( e x + x ) 1 x lim x → ∞   e x + x 1 x lim_(x rarr oo)(e^(x)+x)^((1)/(x))\lim _{x \rightarrow \infty}\left(e^x+x\right)^{\frac{1}{x}}limx→∞(ex+x)1x का मान निकालिए ।
Question:-01(c) Evaluate lim x → ∞ ( e x + x ) 1 x lim x → ∞   e x + x 1 x lim_(x rarr oo)(e^(x)+x)^((1)/(x))\lim _{x \rightarrow \infty}\left(e^x+x\right)^{\frac{1}{x}}limx→∞(ex+x)1x
Question:-01(d) ∫ 0 2 d x ( 2 x − x 2 ) ∫ 0 2   d x 2 x − x 2 int_(0)^(2)(dx)/((2x-x^(2)))\int_0^2 \frac{d x}{\left(2 x-x^2\right)}∫02dx(2x−x2) की अभिसारिता का परीक्षण कीजिए ।
Question:-01(d) Examine the convergence of ∫ 0 2 d x ( 2 x − x 2 ) ∫ 0 2   d x 2 x − x 2 int_(0)^(2)(dx)/((2x-x^(2)))\int_0^2 \frac{d x}{\left(2 x-x^2\right)}∫02dx(2x−x2).
Question:-01(e) एक चर समतल एक स्थिर बिन्दु ( a , b , c ) ( a , b , c ) (a,b,c)(\mathrm{a}, \mathrm{b}, \mathrm{c})(a,b,c) से गुज़रता है तथा अक्षों को क्रमशः A , B A , B A,B\mathrm{A}, \mathrm{B}A,B व C C C\mathrm{C}C बिन्दुओं पर मिलता है । बिन्दुओं O , A , B O , A , B O,A,B\mathrm{O}, \mathrm{A}, \mathrm{B}O,A,B तथा C C C\mathrm{C}C से गुज़रते हुए गोले के केन्द्र का बिन्दुपथ ज्ञात कीजिए, जहाँ O O O\mathrm{O}O मूल-बिन्दु है ।
Question:-01(e) A variable plane passes through a fixed point (a, b, c) and meets the axes at points A, B and C respectively. Find the locus of the centre of the sphere passing through the points O , A , B O , A , B O,A,B\mathrm{O}, \mathrm{A}, \mathrm{B}O,A,B and C , O C , O C,O\mathrm{C}, \mathrm{O}C,O being the origin.
Question:-02(a) निम्नलिखित समीकरण निकाय के सभी हलों को पंक्ति-समानीत विधि से ज्ञात कीजिए :
x 1 + 2 x 2 − x 3 = 2 2 x 1 + 3 x 2 + 5 x 3 = 5 − x 1 − 3 x 2 + 8 x 3 = − 1 x 1 + 2 x 2 − x 3 = 2 2 x 1 + 3 x 2 + 5 x 3 = 5 − x 1 − 3 x 2 + 8 x 3 = − 1 {:[x_(1)+2x_(2)-x_(3)=2],[2x_(1)+3x_(2)+5x_(3)=5],[-x_(1)-3x_(2)+8x_(3)=-1]:}\begin{aligned} &\mathrm{x}_1+2 \mathrm{x}_2-\mathrm{x}_3=2 \\ &2 \mathrm{x}_1+3 \mathrm{x}_2+5 \mathrm{x}_3=5 \\ &-\mathrm{x}_1-3 \mathrm{x}_2+8 \mathrm{x}_3=-1 \end{aligned}x1+2x2−x3=22x1+3x2+5x3=5−x1−3x2+8x3=−1
Question:-02(a) Find all solutions to the following system of equations by row-reduced method :
x 1 + 2 x 2 − x 3 = 2 2 x 1 + 3 x 2 + 5 x 3 = 5 − x 1 − 3 x 2 + 8 x 3 = − 1 x 1 + 2 x 2 − x 3 = 2 2 x 1 + 3 x 2 + 5 x 3 = 5 − x 1 − 3 x 2 + 8 x 3 = − 1 {:[x_(1)+2x_(2)-x_(3)=2],[2x_(1)+3x_(2)+5x_(3)=5],[-x_(1)-3x_(2)+8x_(3)=-1]:}\begin{aligned} &\mathrm{x}_1+2 \mathrm{x}_2-\mathrm{x}_3=2 \\ &2 \mathrm{x}_1+3 \mathrm{x}_2+5 \mathrm{x}_3=5 \\ &-\mathrm{x}_1-3 \mathrm{x}_2+8 \mathrm{x}_3=-1 \end{aligned}x1+2x2−x3=22x1+3x2+5x3=5−x1−3x2+8x3=−1
Question:-02(b) एक l l lll लम्बाई के तार को दो भागों में काटकर क्रमशः एक वर्ग तथा एक वृत्त के रूप में मोड़ा गया है । लग्रांज की अनिर्धारित गुणक विधि का प्रयोग करके, इस तरह से प्राप्त किए गए क्षेत्रफलों के योगफल का न्यूनतम मान ज्ञात कीजिए ।
Question:-02(b) A wire of length l l lll is cut into two parts which are bent in the form of a square and a circle respectively. Using Lagrange’s method of undetermined multipliers, find the least value of the sum of the areas so formed.
Question:-02(c) यदि P , Q , R ; P ′ , Q ′ , R ′ P , Q , R ; P ′ , Q ′ , R ′ P,Q,R;P^(‘),Q^(‘),R^(‘)\mathrm{P}, \mathrm{Q}, \mathrm{R} ; \mathrm{P}^{\prime}, \mathrm{Q}^{\prime}, \mathrm{R}^{\prime}P,Q,R;P′,Q′,R′, एक बिन्दु से दीर्घवृत्तज x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}+\frac{\mathrm{z}^2}{\mathrm{c}^2}=1x2a2+y2 b2+z2c2=1 पर छः (सिक्स) अभिलम्ब पाद हैं तथा l x + m y + n z = p l x + m y + n z = p lx+my+nz=pl \mathrm{x}+\mathrm{my}+\mathrm{nz}=\mathrm{p}lx+my+nz=p से समतल P Q R P Q R PQR\mathrm{PQR}PQR निरूपित है, दर्शाइए कि x a 2 l + y b 2 m + z c 2 n + 1 p = 0 x a 2 l + y b 2 m + z c 2 n + 1 p = 0 (x)/(a^(2)l)+(y)/(b^(2)(m))+(z)/(c^(2)n)+(1)/(p)=0\frac{\mathrm{x}}{\mathrm{a}^2 l}+\frac{\mathrm{y}}{\mathrm{b}^2 \mathrm{~m}}+\frac{\mathrm{z}}{\mathrm{c}^2 \mathrm{n}}+\frac{1}{\mathrm{p}}=0xa2l+yb2 m+zc2n+1p=0, समतल P ′ Q ′ R ′ P ′ Q ′ R ′ P^(‘)Q^(‘)R^(‘)\mathrm{P}^{\prime} \mathrm{Q}^{\prime} \mathrm{R}^{\prime}P′Q′R′ को निरूपित करता है ।
Question:-02(c) If P , Q , R ; P ′ , Q ′ , R ′ P , Q , R ; P ′ , Q ′ , R ′ P,Q,R;P^(‘),Q^(‘),R^(‘)\mathrm{P}, \mathrm{Q}, \mathrm{R} ; \mathrm{P}^{\prime}, \mathrm{Q}^{\prime}, \mathrm{R}^{\prime}P,Q,R;P′,Q′,R′ are feet of the six normals drawn from a point to the ellipsoid x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}+\frac{\mathrm{z}^2}{\mathrm{c}^2}=1x2a2+y2 b2+z2c2=1, and the plane P Q R P Q R PQR\mathrm{PQR}PQR is represented by l x + m y + n z = p l x + m y + n z = p lx+my+nz=pl x+m y+n z=plx+my+nz=p, show that the plane P ′ Q ′ R ′ P ′ Q ′ R ′ P^(‘)Q^(‘)R^(‘)P^{\prime} Q^{\prime} R^{\prime}P′Q′R′ is given by x a 2 l + y b 2 m + z c 2 n + 1 p = 0 x a 2 l + y b 2 m + z c 2 n + 1 p = 0 (x)/(a^(2)l)+(y)/(b^(2)(m))+(z)/(c^(2)n)+(1)/(p)=0\frac{\mathrm{x}}{\mathrm{a}^2 l}+\frac{\mathrm{y}}{\mathrm{b}^2 \mathrm{~m}}+\frac{\mathrm{z}}{\mathrm{c}^2 \mathrm{n}}+\frac{1}{\mathrm{p}}=0xa2l+yb2 m+zc2n+1p=0.
Question:-03(a) माना समुच्चय P = { x y z ) ∣ x − y − z = 0 तथा 2 x − y + z = 0 } P = x y z ∣ x − y − z = 0  तथा  2 x − y + z = 0 {:P={[x],[y],[z])∣[x-y-z=0″ तथा “],[2x-y+z=0]}\left.\mathrm{P}=\left\{\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right) \mid \begin{array}{c}\mathrm{x}-\mathrm{y}-\mathrm{z}=0 \text { तथा } \\ 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=0\end{array}\right\}P={xyz)∣x−y−z=0 तथा 2x−y+z=0} सदिश समष्टि R 3 ( R ) R 3 ( R ) R^(3)(R)\mathbb{R}^3(\mathbb{R})R3(R) के सदिशों का एक समूह है । तब
(i) सिद्ध कीजिए कि P , R 3 P , R 3 P,R^(3)\mathrm{P}, \mathbb{R}^3P,R3 की एक उपसमष्टि है ।
(ii) P P P\mathrm{P}P का एक आधार तथा विमा ज्ञात कीजिए ।
Question:-03(a) Let the set P = { x y z ) ∣ x − y − z = 0 and 2 x − y + z = 0 } P = x y z ∣ x − y − z = 0  and  2 x − y + z = 0 {:P={[x],[y],[z])∣[x-y-z=0″ and “],[2x-y+z=0]}\left.P=\left\{\begin{array}{c}x \\ y \\ z\end{array}\right) \mid \begin{array}{c}x-y-z=0 \text { and } \\ 2 x-y+z=0\end{array}\right\}P={xyz)∣x−y−z=0 and 2x−y+z=0} be the collection of vectors of a vector space R 3 ( R ) R 3 ( R ) R^(3)(R)\mathbb{R}^3(\mathbb{R})R3(R). Then
(i) prove that P P P\mathrm{P}P is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3.
(ii) find a basis and dimension of P P PPP.
Question:-03(b) द्विशः समाकलन का उपयोग करके, वृत्त x 2 + y 2 = 4 x 2 + y 2 = 4 x^(2)+y^(2)=4\mathrm{x}^2+\mathrm{y}^2=4x2+y2=4 तथा परवलय y 2 = 3 x y 2 = 3 x y^(2)=3x\mathrm{y}^2=3 \mathrm{x}y2=3x के उभयनिष्ठ क्षेत्रफल का परिकलन कीजिए ।
Question:-03(b)Use double integration to calculate the area common to the circle x 2 + y 2 = 4 x 2 + y 2 = 4 x^(2)+y^(2)=4x^2+y^2=4x2+y2=4 and the parabola y 2 = 3 x y 2 = 3 x y^(2)=3xy^2=3 xy2=3x.
Question:-03(c) लघुतम संभाव्य त्रिज्या के गोले का समीकरण ज्ञात कीजिए जो सरल रेखाओं : x − 3 3 = y − 8 − 1 = z − 3 1 x − 3 3 = y − 8 − 1 = z − 3 1 (x-3)/(3)=(y-8)/(-1)=(z-3)/(1)\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}-8}{-1}=\frac{\mathrm{z}-3}{1}x−33=y−8−1=z−31 तथा x + 3 − 3 = y + 7 2 = z − 6 4 x + 3 − 3 = y + 7 2 = z − 6 4 (x+3)/(-3)=(y+7)/(2)=(z-6)/(4)\frac{\mathrm{x}+3}{-3}=\frac{\mathrm{y}+7}{2}=\frac{\mathrm{z}-6}{4}x+3−3=y+72=z−64 को स्पर्श करता है ।
Question:-03(c)Find the equation of the sphere of smallest possible radius which touches the straight lines : x − 3 3 = y − 8 − 1 = z − 3 1 x − 3 3 = y − 8 − 1 = z − 3 1 (x-3)/(3)=(y-8)/(-1)=(z-3)/(1)\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}x−33=y−8−1=z−31 and x + 3 − 3 = y + 7 2 = z − 6 4 x + 3 − 3 = y + 7 2 = z − 6 4 (x+3)/(-3)=(y+7)/(2)=(z-6)/(4)\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}x+3−3=y+72=z−64
Question:-04(a) एक रैखिक प्रतिचित्र T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 ज्ञात कीजिए जो कि R 2 R 2 R^(2)\mathbb{R}^2R2 के प्रत्येक सदिश को θ θ theta\thetaθ कोण से घुमा देता है । यह भी सिद्ध कीजिए कि θ = π 2 θ = π 2 theta=(pi)/(2)\theta=\frac{\pi}{2}θ=π2 के लिए, T T T\mathrm{T}T का कोई भी अभिलक्षणिक मान (आइगेनमान) R R R\mathbb{R}R में नहीं है ।
Question:-04(a) Find a linear map T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 which rotates each vector of R 2 R 2 R^(2)\mathbb{R}^2R2 by an angle θ θ theta\thetaθ. Also, prove that for θ = π 2 , T θ = π 2 , T theta=(pi)/(2),T\theta=\frac{\pi}{2}, \mathrm{~T}θ=π2, T has no eigenvalue in R R R\mathbb{R}R.
Question:-04(b) वक्र y 2 x 2 = x 2 − a 2 y 2 x 2 = x 2 − a 2 y^(2)x^(2)=x^(2)-a^(2)\mathrm{y}^2 \mathrm{x}^2=\mathrm{x}^2-\mathrm{a}^2y2x2=x2−a2 का अनुरेख (ट्रेस) कीजिए, जहाँ a a a\mathrm{a}a एक वास्तविक अचर है ।
Question:-04(b) Trace the curve y 2 x 2 = x 2 − a 2 y 2 x 2 = x 2 − a 2 y^(2)x^(2)=x^(2)-a^(2)y^2 x^2=x^2-a^2y2x2=x2−a2, where a a aaa is a real constant.
Question:-04(c) यदि समतल u x + v y + w z = 0 u x + v y + w z = 0 ux+vy+wz=0u x+v y+w z=0ux+vy+wz=0, शंकु a x 2 + b y 2 + c z 2 = 0 a x 2 + b y 2 + c z 2 = 0 ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+\mathrm{cz}^2=0ax2+by2+cz2=0 को लंब जनकों में काटता है, तो सिद्ध कीजिए कि ( b + c ) u 2 + ( c + a ) v 2 + ( a + b ) w 2 = 0 ( b + c ) u 2 + ( c + a ) v 2 + ( a + b ) w 2 = 0 (b+c)u^(2)+(c+a)v^(2)+(a+b)w^(2)=0(b+c) u^2+(c+a) v^2+(a+b) w^2=0(b+c)u2+(c+a)v2+(a+b)w2=0.
Question:-04(c) If the plane u x + v y + w z = 0 u x + v y + w z = 0 ux+vy+wz=0u x+v y+w z=0ux+vy+wz=0 cuts the cone a x 2 + b y 2 + c z 2 = 0 a x 2 + b y 2 + c z 2 = 0 ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+c z^2=0ax2+by2+cz2=0 in perpendicular generators, then prove that ( b + c ) u 2 + ( c + a ) v 2 + ( a + b ) w 2 = 0 ( b + c ) u 2 + ( c + a ) v 2 + ( a + b ) w 2 = 0 (b+c)u^(2)+(c+a)v^(2)+(a+b)w^(2)=0(b+c) u^2+(c+a) v^2+(a+b) w^2=0(b+c)u2+(c+a)v2+(a+b)w2=0.
खण्ड B
SECTION B
Question:-05(a) दर्शाइए कि अवकल समीकरण d y d x + P y = Q d y d x + P y = Q (dy)/(dx)+Py=Q\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}dydx+Py=Q का व्यापक हल
y = Q P − e − ∫ P d x { C + ∫ e ∫ P d x d ( Q P ) } y = Q P − e − ∫ P d x C + ∫ e ∫ P d x d Q P y=(Q)/(P)-e^(-intPdx){C+inte^(intPdx)(d)((Q)/(P))}\mathrm{y}=\frac{\mathrm{Q}}{\mathrm{P}}-\mathrm{e}^{-\int \mathrm{P} d x}\left\{\mathrm{C}+\int \mathrm{e}^{\int \mathrm{P} d x} \mathrm{~d}\left(\frac{\mathrm{Q}}{\mathrm{P}}\right)\right\}y=QP−e−∫Pdx{C+∫e∫Pdx d(QP)}
के रूप में लिखा जा सकता है, जहाँ P , Q , x P , Q , x P,Q,x\mathrm{P}, \mathrm{Q}, \mathrm{x}P,Q,x के शून्येतर फलन हैं तथा C C C\mathrm{C}C एक स्वेच्छ अचर है ।
Question:-05(a)Show that the general solution of the differential equation d y d x + P y = Q d y d x + P y = Q (dy)/(dx)+Py=Q\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}dydx+Py=Q can be written in the form y = Q P − e − ∫ P d x { C + ∫ e ∫ P d x d ( Q P ) } y = Q P − e − ∫ P d x C + ∫ e ∫ P d x d Q P y=(Q)/(P)-e^(-int Pdx){C+inte^(int Pdx)d((Q)/(P))}y=\frac{Q}{P}-e^{-\int P d x}\left\{C+\int e^{\int P d x} d\left(\frac{Q}{P}\right)\right\}y=QP−e−∫Pdx{C+∫e∫Pdxd(QP)}, where P , Q P , Q P,Q\mathrm{P}, \mathrm{Q}P,Q are non-zero functions of x x x\mathrm{x}x and C C C\mathrm{C}C, an arbitrary constant.
Question:-05(b) दर्शाइए कि परवलयों के निकाय : x 2 = 4 a ( y + a ) x 2 = 4 a ( y + a ) x^(2)=4a(y+a)\mathrm{x}^2=4 \mathrm{a}(\mathrm{y}+\mathrm{a})x2=4a(y+a) के लंबकोणीय संछेदी, उसी निकाय में स्थित होते हैं ।
Question:-05(b) Show that the orthogonal trajectories of the system of parabolas : x 2 = 4 a ( y + a ) x 2 = 4 a ( y + a ) x^(2)=4a(y+a)\mathrm{x}^2=4 \mathrm{a}(\mathrm{y}+\mathrm{a})x2=4a(y+a) belong to the same system.
Question:-05(c) w w w\mathrm{w}w भार का एक पिंड, θ θ theta\thetaθ कोण से झुके हुए एक रूक्ष समतल पर स्थित है, घर्षण गुणांक μ μ mu\muμ, tan ⁡ θ tan ⁡ θ tan theta\tan \thetatan⁡θ से अधिक है । पिंड को समतल पर ऊपर की तरफ ‘ b b bbb ‘ दूरी तक धीरे-धीरे खींचने तथा वापस आरम्भिक बिन्दु तक खींचने में किए गए कार्य को ज्ञात कीजिए, जहाँ लगाया गया बल प्रत्येक दशा में समतल के समान्तर है ।
Question:-05(c)A body of weight w w www rests on a rough inclined plane of inclination θ θ theta\thetaθ, the coefficient of friction, μ μ mu\muμ, being greater than tan ⁡ θ tan ⁡ θ tan theta\tan \thetatan⁡θ. Find the work done in slowly dragging the body a distance ‘b’ up the plane and then dragging it back to the starting point, the applied force being in each case parallel to the plane.
Question:-05(d) एक प्रक्षेप्य 2 g h 2 g h sqrt(2gh)\sqrt{2 \mathrm{gh}}2gh वेग के साथ बिन्दु O O O\mathrm{O}O से प्रक्षेपित किया गया तथा समतल के बिन्दु P ( x , y ) P ( x , y ) P(x,y)\mathrm{P}(\mathrm{x}, \mathrm{y})P(x,y) पर स्पर्श-रेखा से टकराता है जहाँ अक्ष O X O X OX\mathrm{OX}OX तथा O Y O Y OY\mathrm{OY}OY क्रमशः बिन्दु O O O\mathrm{O}O से क्षैतिज तथा अधोमुखी ऊर्ध्वाधर रेखाएँ हैं । यदि प्रक्षेपण की दो संभव दिशाएँ समकोण पर हों, तो दर्शाइए कि x 2 = 2 h y x 2 = 2 h y x^(2)=2hy\mathrm{x}^2=2 \mathrm{hy}x2=2hy तथा प्रक्षेपण की संभव दिशाओं में से एक, कोण POX को द्विभाजित करती है ।
Question:-05(d)A projectile is fired from a point O O O\mathrm{O}O with velocity 2 g h 2 g h sqrt(2gh)\sqrt{2 \mathrm{gh}}2gh and hits a tangent at the point P ( x , y ) P ( x , y ) P(x,y)\mathrm{P}(\mathrm{x}, \mathrm{y})P(x,y) in the plane, the axes O X O X OX\mathrm{OX}OX and O Y O Y OY\mathrm{OY}OY being horizontal and vertically downward lines through the point O O O\mathrm{O}O, respectively. Show that if the two possible directions of projection be at right angles, then x 2 = 2 h y x 2 = 2 h y x^(2)=2hy\mathrm{x}^2=2 \mathrm{hy}x2=2hy and then one of the possible directions of projection bisects the angle POX.
Question:-05(e) दर्शाइए कि A → = ( 6 x y + z 3 ) i ^ + ( 3 x 2 − z ) j ^ + ( 3 x z 2 − y ) k ^ A → = 6 x y + z 3 i ^ + 3 x 2 − z j ^ + 3 x z 2 − y k ^ vec(A)=(6xy+z^(3)) hat(i)+(3x^(2)-z) hat(j)+(3xz^(2)-y) hat(k)\overrightarrow{\mathrm{A}}=\left(6 \mathrm{xy}+\mathrm{z}^3\right) \hat{\mathrm{i}}+\left(3 \mathrm{x}^2-\mathrm{z}\right) \hat{\mathrm{j}}+\left(3 x \mathrm{z}^2-\mathrm{y}\right) \hat{\mathrm{k}}A→=(6xy+z3)i^+(3x2−z)j^+(3xz2−y)k^ अघूर्णी है । ϕ ϕ phi\phiϕ को भी ज्ञात कीजिए जबकि A → = ∇ ϕ A → = ∇ ϕ vec(A)=grad phi\overrightarrow{\mathrm{A}}=\nabla \phiA→=∇ϕ.
Question:-05(e)Show that A → = ( 6 x y + z 3 ) i ^ + ( 3 x 2 − z ) j ^ + ( 3 x z 2 − y ) k ^ A → = 6 x y + z 3 i ^ + 3 x 2 − z j ^ + 3 x z 2 − y k ^ vec(A)=(6xy+z^(3)) hat(i)+(3x^(2)-z) hat(j)+(3xz^(2)-y) hat(k)\overrightarrow{\mathrm{A}}=\left(6 x y+z^3\right) \hat{i}+\left(3 x^2-z\right) \hat{j}+\left(3 x z^2-y\right) \hat{k}A→=(6xy+z3)i^+(3x2−z)j^+(3xz2−y)k^ is irrotational. Also find ϕ ϕ phi\phiϕ such that A → = ∇ ϕ A → = ∇ ϕ vec(A)=grad phi\overrightarrow{\mathrm{A}}=\nabla \phiA→=∇ϕ.
Question:-06(a) 2 l 2 l 2l2 l2l लम्बाई का एक तार (केबिल) जिसका भार w w w\mathrm{w}w प्रति इकाई (यूनिट) लम्बाई है, एक क्षैतिज रेखा के दो बिन्दुओं P P P\mathrm{P}P तथा Q Q Q\mathrm{Q}Q से लटकी हुई है । दर्शाइए कि तार की विस्तृति (स्पैन) 2 l ( 1 − 2 h 2 3 l 2 ) 2 l 1 − 2 h 2 3 l 2 2l(1-(2h^(2))/(3l^(2)))2 l\left(1-\frac{2 \mathrm{~h}^2}{3 l^2}\right)2l(1−2 h23l2) है, जहाँ h h h\mathrm{h}h तार के कसकर खींची हुई स्थिति में मध्य का झोल है ।
Question:-06(a) A cable of weight w per unit length and length 2 l 2 l 2l2 l2l hangs from two points P P P\mathrm{P}P and Q Q Q\mathrm{Q}Q in the same horizontal line. Show that the span of the cable is 2 l ( 1 − 2 h 2 3 l 2 ) 2 l 1 − 2 h 2 3 l 2 2l(1-(2h^(2))/(3l^(2)))2 l\left(1-\frac{2 h^2}{3 l^2}\right)2l(1−2h23l2), where h h hhh is the sag in the middle of the tightly stretched position.
Question:-06(b) प्राचल-विचरण विधि का उपयोग करके, निम्नलिखित अवकल समीकरण :
( x 2 − 1 ) d 2 y d x 2 − 2 x d y d x + 2 y = ( x 2 − 1 ) 2 x 2 − 1 d 2 y d x 2 − 2 x d y d x + 2 y = x 2 − 1 2 (x^(2)-1)(d^(2)y)/(dx^(2))-2x(dy)/(dx)+2y=(x^(2)-1)^(2)\left(x^2-1\right) \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}+2 y=\left(x^2-1\right)^2(x2−1)d2ydx2−2xdydx+2y=(x2−1)2
को हल कीजिए, जहाँ समानीत समीकरण का एक हल y = x y = x y=x\mathrm{y}=\mathrm{x}y=x दिया गया है ।
Question:-06(b) Solve the following differential equation by using the method of variation of parameters : ( x 2 − 1 ) d 2 y d x 2 − 2 x d y d x + 2 y = ( x 2 − 1 ) 2 x 2 − 1 d 2 y d x 2 − 2 x d y d x + 2 y = x 2 − 1 2 (x^(2)-1)(d^(2)y)/(dx^(2))-2x(dy)/(dx)+2y=(x^(2)-1)^(2)\left(x^2-1\right) \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}+2 y=\left(x^2-1\right)^2(x2−1)d2ydx2−2xdydx+2y=(x2−1)2, given that y = x y = x y=x\mathrm{y}=\mathrm{x}y=x is one solution of the reduced equation.
Question:-06(c) समतल में ग्रीन के प्रमेय को ∮ C ( 3 x 2 − 8 y 2 ) d x + ( 4 y − 6 x y ) d y ∮ C 3 x 2 − 8 y 2 d x + ( 4 y − 6 x y ) d y oint_(C)(3x^(2)-8y^(2))dx+(4y-6xy)dy\oint_{\mathrm{C}}\left(3 \mathrm{x}^2-8 \mathrm{y}^2\right) \mathrm{dx}+(4 \mathrm{y}-6 \mathrm{xy}) \mathrm{dy}∮C(3x2−8y2)dx+(4y−6xy)dy के लिए सत्यापित कीजिए, जहाँ C , x = 0 , y = 0 , x + y = 1 C , x = 0 , y = 0 , x + y = 1 C,x=0,y=0,x+y=1\mathrm{C}, \mathrm{x}=0, \mathrm{y}=0, \mathrm{x}+\mathrm{y}=1C,x=0,y=0,x+y=1 द्वारा परिभाषित क्षेत्र का सीमा वक्र है ।
Question:-06(c)Verify Green’s theorem in the plane for ∮ C ( 3 x 2 − 8 y 2 ) d x + ( 4 y − 6 x y ) d y ∮ C 3 x 2 − 8 y 2 d x + ( 4 y − 6 x y ) d y oint_(C)(3x^(2)-8y^(2))dx+(4y-6xy)dy\oint_C\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y∮C(3x2−8y2)dx+(4y−6xy)dy, where C C C\mathrm{C}C is the boundary curve of the region defined by x = 0 , y = 0 x = 0 , y = 0 x=0,y=0\mathrm{x}=0, \mathrm{y}=0x=0,y=0, x + y = 1 x + y = 1 x+y=1x+y=1x+y=1
Question:-07(a) स्टोक्स प्रमेय को F → = x i ^ + z 2 j ^ + y 2 k ^ F → = x i ^ + z 2 j ^ + y 2 k ^ vec(F)=x hat(i)+z^(2) hat(j)+y^(2) hat(k)\overrightarrow{\mathrm{F}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{z}^2 \hat{\mathrm{j}}+\mathrm{y}^2 \hat{\mathrm{k}}F→=xi^+z2j^+y2k^ के लिए प्रथम अष्टांशक में स्थित समतल पृष्ठ : x + y + z = 1 x + y + z = 1 x+y+z=1x+y+z=1x+y+z=1 पर सत्यापित कीजिए ।
Question:-07(a) Verify Stokes’ theorem for F → = x i ^ + z 2 j ^ + y 2 k ^ F → = x i ^ + z 2 j ^ + y 2 k ^ vec(F)=x hat(i)+z^(2) hat(j)+y^(2) hat(k)\vec{F}=x \hat{i}+z^2 \hat{j}+y^2 \hat{k}F→=xi^+z2j^+y2k^ over the plane surface : x + y + z = 1 x + y + z = 1 x+y+z=1x+y+z=1x+y+z=1 lying in the first octant.
Question:-07(b) लाप्लास रूपांतरण का उपयोग करके निम्नलिखित प्रारंभिक मान समस्या : d 2 y d t 2 − 3 d y d t + 2 y = h ( t ) d 2 y d t 2 − 3 d y d t + 2 y = h ( t ) (d^(2)y)/(dt^(2))-3(dy)/(dt)+2y=h(t)\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dt}^2}-3 \frac{\mathrm{dy}}{\mathrm{dt}}+2 \mathrm{y}=\mathrm{h}(\mathrm{t})d2ydt2−3dydt+2y=h(t), जहाँ h ( t ) = { 2 , 0 < t < 4 , 0 , t > 4 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 h ( t ) = 2 , 0 < t < 4 , 0 , t > 4 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 h(t)={[2″,”,0 < t < 4″,”],[0″,”,t > 4″,”]y(0)=0,y^(‘)(0)=0:}\mathrm{h}(\mathrm{t})=\left\{\begin{array}{cc}2, & 0<\mathrm{t}<4, \\ 0, & \mathrm{t}>4,\end{array} \mathrm{y}(0)=0, \mathrm{y}^{\prime}(0)=0\right.h(t)={2,0<t<4,0,t>4,y(0)=0,y′(0)=0 को हल कीजिए।
Question:-07(b) Solve the following initial value problem by using Laplace’s transformation d 2 y d t 2 − 3 d y d t + 2 y = h ( t ) d 2 y d t 2 − 3 d y d t + 2 y = h ( t ) (d^(2)y)/(dt^(2))-3(dy)/(dt)+2y=h(t)\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dt}^2}-3 \frac{\mathrm{dy}}{\mathrm{dt}}+2 \mathrm{y}=\mathrm{h}(\mathrm{t})d2ydt2−3dydt+2y=h(t), where
h ( t ) = { 2 , 0 < t < 4 , 0 , t > 4 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 h ( t ) = 2 , 0 < t < 4 , 0 , t > 4 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 h(t)={[2″,”,0 < t < 4″,”],[0″,”,t > 4″,”]quady(0)=0,quady^(‘)(0)=0:}\mathrm{h}(\mathrm{t})=\left\{\begin{array}{cc} 2, & 0<\mathrm{t}<4, \\ 0, & \mathrm{t}>4, \end{array} \quad \mathrm{y}(0)=0, \quad \mathrm{y}^{\prime}(0)=0\right.h(t)={2,0<t<4,0,t>4,y(0)=0,y′(0)=0
Question:-07(c) माना किसी भी अनुप्रस्थ-काट का एक बेलन दूसरे स्थिर बेलन पर संतुलित है, जहाँ वक्रीय पृष्ठों का संस्पर्श रूक्ष है तथा उभयनिष्ठ स्पर्श-रेखा क्षैतिज है । माना दोनों बेलनों के स्पर्श बिन्दु पर उनकी वक्रता त्रिज्याएँ ρ ρ rho\rhoρ तथा ρ ′ ρ ′ rho^(‘)\rho^{\prime}ρ′ हैं और संस्पर्श बिन्दु से ऊपरी बेलन के गुरुत्व केन्द्र की ऊँचाई h h h\mathrm{h}h है । दर्शाइए कि स्थायी साम्य में ऊपरी बेलन संतुलित है यदि h < ρ ρ ′ ρ + ρ ′ h < ρ ρ ′ ρ + ρ ′ h < (rhorho^(‘))/(rho+rho^(‘))\mathrm{h}<\frac{\rho \rho^{\prime}}{\rho+\rho^{\prime}}h<ρρ′ρ+ρ′ ।
Question:-07(c) Suppose a cylinder of any cross-section is balanced on another fixed cylinder, the contact of curved surfaces being rough and the common tangent line horizontal. Let ρ ρ rho\rhoρ and ρ ′ ρ ′ rho^(‘)\rho^{\prime}ρ′ be the radii of curvature of the two cylinders at the point of contact and h h hhh be the height of centre of gravity of the upper cylinder above the point of contact. Show that the upper cylinder is balanced in stable equilibrium if h < ρ ρ ′ ρ + ρ ′ h < ρ ρ ′ ρ + ρ ′ h < (rhorho^(‘))/(rho+rho^(‘))\mathrm{h}<\frac{\rho \rho^{\prime}}{\rho+\rho^{\prime}}h<ρρ′ρ+ρ′.
Question:-08(a) (i) अवकल समीकरण : ( x 2 − a 2 ) p 2 − 2 x y p + y 2 + a 2 = 0 x 2 − a 2 p 2 − 2 x y p + y 2 + a 2 = 0 (x^(2)-a^(2))p^(2)-2xyp+y^(2)+a^(2)=0\left(\mathrm{x}^2-\mathrm{a}^2\right) \mathrm{p}^2-2 \mathrm{xyp}+\mathrm{y}^2+\mathrm{a}^2=0(x2−a2)p2−2xyp+y2+a2=0, जहाँ p = d y d x p = d y d x p=(dy)/(dx)\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}p=dydx, के व्यापक व विचित्र हलों को ज्ञात कीजिए । व्यापक व विचित्र हलों के बीच ज्यामितीय संबंध को भी दीजिए ।
Question8(a) (i) Find the general and singular solutions of the differential equation: ( x 2 − a 2 ) p 2 − 2 x y p + y 2 + a 2 = 0 x 2 − a 2 p 2 − 2 x y p + y 2 + a 2 = 0 (x^(2)-a^(2))p^(2)-2xyp+y^(2)+a^(2)=0\left(x^2-a^2\right) p^2-2 x y p+y^2+a^2=0(x2−a2)p2−2xyp+y2+a2=0, where p = d y d x p = d y d x p=(dy)/(dx)p=\frac{d y}{d x}p=dydx. Also give the geometric relation between the general and singular solutions.
Question:-08(a) (ii) निम्नलिखित अवकल समीकरण को हल कीजिए :
( 3 x + 2 ) 2 d 2 y d x 2 + 5 ( 3 x + 2 ) d y d x − 3 y = x 2 + x + 1 ( 3 x + 2 ) 2 d 2 y d x 2 + 5 ( 3 x + 2 ) d y d x − 3 y = x 2 + x + 1 (3x+2)^(2)(d^(2)y)/(dx^(2))+5(3x+2)(dy)/(dx)-3y=x^(2)+x+1(3 x+2)^2 \frac{d^2 y}{d x^2}+5(3 x+2) \frac{d y}{d x}-3 y=x^2+x+1(3x+2)2d2ydx2+5(3x+2)dydx−3y=x2+x+1
Question:-08(a) (ii) Solve the following differential equation :
( 3 x + 2 ) 2 d 2 y d x 2 + 5 ( 3 x + 2 ) d y d x − 3 y = x 2 + x + 1 ( 3 x + 2 ) 2 d 2 y d x 2 + 5 ( 3 x + 2 ) d y d x − 3 y = x 2 + x + 1 (3x+2)^(2)(d^(2)y)/(dx^(2))+5(3x+2)(dy)/(dx)-3y=x^(2)+x+1(3 x+2)^2 \frac{d^2 y}{d x^2}+5(3 x+2) \frac{d y}{d x}-3 y=x^2+x+1(3x+2)2d2ydx2+5(3x+2)dydx−3y=x2+x+1
Question:-08(b) n n n\mathrm{n}n बराबर एकसमान छड़ों की एक श्रृंखला एक-दूसरे के साथ चिकने रूप से जुड़ी हुई है तथा इसके एक सिरे A 1 A 1 A_(1)\mathrm{A}_1A1 से लटकी हुई है । एक क्षैतिज बल P → P → vec(P)\overrightarrow{\mathrm{P}}P→ शृंखला के दूसरे सिरे A n + 1 A n + 1 A_(n+1)\mathrm{A}_{\mathrm{n}+1}An+1 पर लगाया गया है । साम्य विन्यास में अधोमुखी ऊर्ध्वाधर रेखा से छड़ों के झुकाव ज्ञात कीजिए ।
Question:-08(b) A chain of n n n\mathrm{n}n equal uniform rods is smoothly jointed together and suspended from its one end A 1 A 1 A_(1)\mathrm{A}_1A1. A horizontal force P → P → vec(P)\overrightarrow{\mathrm{P}}P→ is applied to the other end A n + 1 A n + 1 A_(n+1)\mathrm{A}_{\mathrm{n}+1}An+1 of the chain. Find the inclinations of the rods to the downward vertical line in the equilibrium configuration.
Question:-08(c) गाउस के अपसरण प्रमेय का उपयोग करके ∬ S F → ⋅ n → d S ∬ S   F → ⋅ n → d S ∬_(S) vec(F)* vec(n)dS\iint_{\mathrm{S}} \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{n}} \mathrm{dS}∬SF→⋅n→dS का मान निकालिए, जहाँ F → = x i ^ − y j ^ + ( z 2 − 1 ) k ^ F → = x i ^ − y j ^ + z 2 − 1 k ^ vec(F)=x hat(i)-y hat(j)+(z^(2)-1) hat(k)\overrightarrow{\mathrm{F}}=\mathrm{x} \hat{\mathrm{i}}-\mathrm{y} \hat{\mathrm{j}}+\left(\mathrm{z}^2-1\right) \hat{\mathrm{k}}F→=xi^−yj^+(z2−1)k^ तथा S S S\mathrm{S}S, पृष्ठों z = 0 , z = 1 , x 2 + y 2 = 4 z = 0 , z = 1 , x 2 + y 2 = 4 z=0,z=1,x^(2)+y^(2)=4\mathrm{z}=0, \mathrm{z}=1, \mathrm{x}^2+\mathrm{y}^2=4z=0,z=1,x2+y2=4 द्वारा बना हुआ बेलन है ।
Question:-08(c) Using Gauss’ divergence theorem, evaluate ∬ S F → ⋅ n → d S ∬ S   F → ⋅ n → d S ∬_(S) vec(F)* vec(n)dS\iint_S \vec{F} \cdot \vec{n} d S∬SF→⋅n→dS, where F → = x i ^ − y j ^ + ( z 2 − 1 ) k ^ F → = x i ^ − y j ^ + z 2 − 1 k ^ vec(F)=x hat(i)-y hat(j)+(z^(2)-1) hat(k)\vec{F}=x \hat{i}-y \hat{j}+\left(z^2-1\right) \hat{k}F→=xi^−yj^+(z2−1)k^ and S S SSS is the cylinder formed by the surfaces z = 0 , z = 1 , x 2 + y 2 = 4 z = 0 , z = 1 , x 2 + y 2 = 4 z=0,z=1,x^(2)+y^(2)=4z=0, z=1, x^2+y^2=4z=0,z=1,x2+y2=4
upsc-m2021-1-ed2d6450-301f-4184-ac9b-1fca888cd172
1.(a) यदि A = [ 1 − 1 1 2 − 1 0 1 0 0 ] A = 1 − 1 1 2 − 1 0 1 0 0 A=[[1,-1,1],[2,-1,0],[1,0,0]]A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right]A=[1−112−10100] है, तो A − 1 A − 1 A^(-1)A^{-1}A−1 को ज्ञात किए बिना दर्शाइए कि A 2 = A − 1 A 2 = A − 1 A^(2)=A^(-1)A^2=A^{-1}A2=A−1
If A = [ 1 − 1 1 2 − 1 0 1 0 0 ] A = 1 − 1 1 2 − 1 0 1 0 0 A=[[1,-1,1],[2,-1,0],[1,0,0]]A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right]A=[1−112−10100], then show that A 2 = A − 1 A 2 = A − 1 A^(2)=A^(-1)A^2=A^{-1}A2=A−1 (without finding A − 1 A − 1 A^(-1)A^{-1}A−1 ).
1.(b) क्रमित आधारक B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B={(0,1,1),(1,0,1),(1,1,0)}B=\{(0,1,1),(1,0,1),(1,1,0)\}B={(0,1,1),(1,0,1),(1,1,0)} के सापेक्ष V 3 ( R ) V 3 ( R ) V_(3)(R)V_3(R)V3(R) पर परिभाषित रैखिक संकारक : T ( a , b , c ) = ( a + b , a − b , 2 c ) T ( a , b , c ) = ( a + b , a − b , 2 c ) T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c)T(a,b,c)=(a+b,a−b,2c) से संबन्धित आव्यूह ज्ञात कीजिए ।
Find the matrix associated with the linear operator on V 3 ( R ) V 3 ( R ) V_(3)(R)V_3(R)V3(R) defined by T ( a , b , c ) = ( a + b , a − b , 2 c ) T ( a , b , c ) = ( a + b , a − b , 2 c ) T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c)T(a,b,c)=(a+b,a−b,2c) with respect to the ordered basis B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B={(0,1,1),(1,0,1),(1,1,0)}B=\{(0,1,1),(1,0,1),(1,1,0)\}B={(0,1,1),(1,0,1),(1,1,0)}.
1.(c) दिया गया है :
Δ ( x ) = | f ( x + α ) f ( x + 2 α ) f ( x + 3 α ) f ( α ) f ( 2 α ) f ( 3 α ) f ′ ( α ) f ′ ( 2 α ) f ′ ( 3 α ) | Δ ( x ) = f ( x + α ) f ( x + 2 α ) f ( x + 3 α ) f ( α ) f ( 2 α ) f ( 3 α ) f ′ ( α ) f ′ ( 2 α ) f ′ ( 3 α ) Delta(x)=|[f(x+alpha),f(x+2alpha),f(x+3alpha)],[f(alpha),f(2alpha),f(3alpha)],[f^(‘)(alpha),f^(‘)(2alpha),f^(‘)(3alpha)]|\Delta(x)=\left|\begin{array}{ccc} f(x+\alpha) & f(x+2 \alpha) & f(x+3 \alpha) \\ f(\alpha) & f(2 \alpha) & f(3 \alpha) \\ f^{\prime}(\alpha) & f^{\prime}(2 \alpha) & f^{\prime}(3 \alpha) \end{array}\right|Δ(x)=|f(x+α)f(x+2α)f(x+3α)f(α)f(2α)f(3α)f′(α)f′(2α)f′(3α)|
जहाँ f f fff एक वास्तविक-मान अवकलनीय फलन है तथा α α alpha\alphaα एक अचर है। lim x → 0 Δ ( x ) x lim x → 0   Δ ( x ) x lim_(x rarr0)(Delta(x))/(x)\lim _{x \rightarrow 0} \frac{\Delta(x)}{x}limx→0Δ(x)x को ज्ञात कीजिए ।
Given :
Δ ( x ) = | f ( x + α ) f ( x + 2 α ) f ( x + 3 α ) f ( α ) f ( 2 α ) f ( 3 α ) f ′ ( α ) f ′ ( 2 α ) f ′ ( 3 α ) | Δ ( x ) = f ( x + α ) f ( x + 2 α ) f ( x + 3 α ) f ( α ) f ( 2 α ) f ( 3 α ) f ′ ( α ) f ′ ( 2 α ) f ′ ( 3 α ) Delta(x)=|[f(x+alpha),f(x+2alpha),f(x+3alpha)],[f(alpha),f(2alpha),f(3alpha)],[f^(‘)(alpha),f^(‘)(2alpha),f^(‘)(3alpha)]|\Delta(x)=\left|\begin{array}{ccc} f(x+\alpha) & f(x+2 \alpha) & f(x+3 \alpha) \\ f(\alpha) & f(2 \alpha) & f(3 \alpha) \\ f^{\prime}(\alpha) & f^{\prime}(2 \alpha) & f^{\prime}(3 \alpha) \end{array}\right|Δ(x)=|f(x+α)f(x+2α)f(x+3α)f(α)f(2α)f(3α)f′(α)f′(2α)f′(3α)|
where f f fff is a real valued differentiable function and α α alpha\alphaα is a constant. Find lim x → 0 Δ ( x ) x lim x → 0   Δ ( x ) x lim_(x rarr0)(Delta(x))/(x)\lim _{x \rightarrow 0} \frac{\Delta(x)}{x}limx→0Δ(x)x.
1.(d) दर्शाइए कि e x cos ⁡ x = 1 e x cos ⁡ x = 1 e^(x)cos x=1e^x \cos x=1excos⁡x=1 के किन्हीं दो मूलों के बीच में e x sin ⁡ x − 1 = 0 e x sin ⁡ x − 1 = 0 e^(x)sin x-1=0e^x \sin x-1=0exsin⁡x−1=0 का कम से कम एक मूल विद्यमान है ।
Show that between any two roots of e x cos ⁡ x = 1 e x cos ⁡ x = 1 e^(x)cos x=1e^x \cos x=1excos⁡x=1, there exists at least one root of e x sin ⁡ x − 1 = 0 e x sin ⁡ x − 1 = 0 e^(x)sin x-1=0e^x \sin x-1=0exsin⁡x−1=0
1.(e) उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा :
x = − y 2 = z 3 x = − y 2 = z 3 x=-(y)/(2)=(z)/(3)x=-\frac{y}{2}=\frac{z}{3}x=−y2=z3 के समानान्तर हैं
तथा जिसका निर्देशक-वक्र x 2 + 2 y 2 = 1 , z = 0 x 2 + 2 y 2 = 1 , z = 0 x^(2)+2y^(2)=1,z=0x^2+2 y^2=1, z=0x2+2y2=1,z=0 है।
Find the equation of the cylinder whose generators are parallel to the line x = − y 2 = z 3 x = − y 2 = z 3 x=-(y)/(2)=(z)/(3)x=-\frac{y}{2}=\frac{z}{3}x=−y2=z3 and whose guiding curve is x 2 + 2 y 2 = 1 , z = 0 x 2 + 2 y 2 = 1 , z = 0 x^(2)+2y^(2)=1,z=0x^2+2 y^2=1, z=0x2+2y2=1,z=0.
2.(a) दर्शाइए कि वे समतल, जो कि शंकु a x 2 + b y 2 + c z 2 = 0 a x 2 + b y 2 + c z 2 = 0 ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+c z^2=0ax2+by2+cz2=0 को लंब जनकों में काटते हैं, शंकु x 2 b + c + y 2 c + a + z 2 a + b = 0 x 2 b + c + y 2 c + a + z 2 a + b = 0 (x^(2))/(b+c)+(y^(2))/(c+a)+(z^(2))/(a+b)=0\frac{x^2}{b+c}+\frac{y^2}{c+a}+\frac{z^2}{a+b}=0x2b+c+y2c+a+z2a+b=0 को स्पर्श करते हैं ।
Show that the planes, which cut the cone a x 2 + b y 2 + c z 2 = 0 a x 2 + b y 2 + c z 2 = 0 ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+c z^2=0ax2+by2+cz2=0 in perpendicular generators, touch the cone x 2 b + c + y 2 c + a + z 2 a + b = 0 x 2 b + c + y 2 c + a + z 2 a + b = 0 (x^(2))/(b+c)+(y^(2))/(c+a)+(z^(2))/(a+b)=0\frac{x^2}{b+c}+\frac{y^2}{c+a}+\frac{z^2}{a+b}=0x2b+c+y2c+a+z2a+b=0.
2.(b) दिया गया है : f ( x , y ) = | x 2 − y 2 | f ( x , y ) = x 2 − y 2 f(x,y)=|x^(2)-y^(2)|f(x, y)=\left|x^2-y^2\right|f(x,y)=|x2−y2|, तब f x y ( 0 , 0 ) f x y ( 0 , 0 ) f_(xy)(0,0)f_{x y}(0,0)fxy(0,0) तथा f y x ( 0 , 0 ) f y x ( 0 , 0 ) f_(yx)(0,0)f_{y x}(0,0)fyx(0,0) ज्ञात कीजिए । अतः दर्शाइए कि f x y ( 0 , 0 ) = f y x ( 0 , 0 ) f x y ( 0 , 0 ) = f y x ( 0 , 0 ) f_(xy)(0,0)=f_(yx)(0,0)f_{x y}(0,0)=f_{y x}(0,0)fxy(0,0)=fyx(0,0) ।
Given that f ( x , y ) = | x 2 − y 2 | f ( x , y ) = x 2 − y 2 f(x,y)=|x^(2)-y^(2)|f(x, y)=\left|x^2-y^2\right|f(x,y)=|x2−y2|. Find f x y ( 0 , 0 ) f x y ( 0 , 0 ) f_(xy)(0,0)f_{x y}(0,0)fxy(0,0) and f y x ( 0 , 0 ) f y x ( 0 , 0 ) f_(yx)(0,0)f_{y x}(0,0)fyx(0,0).
Hence show that f x y ( 0 , 0 ) = f y x ( 0 , 0 ) f x y ( 0 , 0 ) = f y x ( 0 , 0 ) f_(xy)(0,0)=f_(yx)(0,0)f_{x y}(0,0)=f_{y x}(0,0)fxy(0,0)=fyx(0,0).
2.(c) दर्शाइए कि S = { ( x , 2 y , 3 x ) : x , y S = ( x , 2 y , 3 x ) : x , y S={(x,2y,3x):x,y:}S=\left\{(x, 2 y, 3 x): x, y\right.S={(x,2y,3x):x,y वास्तविक संख्याऐं हैं | R 3 ( R ) R 3 ( R ) R^(3)(R)R^3(R)R3(R) का एक उपसमष्टि है । S S SSS के दो आधार ज्ञात कीजिए । S S SSS की विमा भी ज्ञात कीजिए ।
Show that S = { ( x , 2 y , 3 x ) : x , y S = { ( x , 2 y , 3 x ) : x , y S={(x,2y,3x):x,yS=\{(x, 2 y, 3 x): x, yS={(x,2y,3x):x,y are real numbers is a subspace of R 3 ( R ) R 3 ( R ) R^(3)(R)R^3(R)R3(R). Find two bases of S S SSS. Also find the dimension of S S SSS.
3(a)(i) यदि u = x 2 + y 2 , v = x 2 − y 2 u = x 2 + y 2 , v = x 2 − y 2 u=x^(2)+y^(2),v=x^(2)-y^(2)u=x^2+y^2, v=x^2-y^2u=x2+y2,v=x2−y2, जहाँ पर x = r cos ⁡ θ , y = r sin ⁡ θ x = r cos ⁡ θ , y = r sin ⁡ θ x=r cos theta,y=r sin thetax=r \cos \theta, y=r \sin \thetax=rcos⁡θ,y=rsin⁡θ हैं, तब ∂ ( u , v ) ∂ ( r , θ ) ∂ ( u , v ) ∂ ( r , θ ) (del(u,v))/(del(r,theta))\frac{\partial(u, v)}{\partial(r, \theta)}∂(u,v)∂(r,θ) ज्ञात कीजिए ।
If u = x 2 + y 2 , v = x 2 − y 2 u = x 2 + y 2 , v = x 2 − y 2 u=x^(2)+y^(2),v=x^(2)-y^(2)u=x^2+y^2, v=x^2-y^2u=x2+y2,v=x2−y2, where x = r cos ⁡ θ , y = r sin ⁡ θ x = r cos ⁡ θ , y = r sin ⁡ θ x=r cos theta,y=r sin thetax=r \cos \theta, y=r \sin \thetax=rcos⁡θ,y=rsin⁡θ, then find ∂ ( u , v ) ∂ ( r , θ ) ∂ ( u , v ) ∂ ( r , θ ) (del(u,v))/(del(r,theta))\frac{\partial(u, v)}{\partial(r, \theta)}∂(u,v)∂(r,θ).
3.(a)(ii) यदि ∫ 0 x f ( t ) d t = x + ∫ x 1 t f ( t ) d t ∫ 0 x   f ( t ) d t = x + ∫ x 1   t f ( t ) d t int_(0)^(x)f(t)dt=x+int_(x)^(1)tf(t)dt\int_0^x f(t) d t=x+\int_x^1 t f(t) d t∫0xf(t)dt=x+∫x1tf(t)dt है, तो f ( 1 ) f ( 1 ) f(1)f(1)f(1) का मान ज्ञात कीजिए ।
If ∫ 0 x f ( t ) d t = x + ∫ x 1 t f ( t ) d t ∫ 0 x   f ( t ) d t = x + ∫ x 1   t f ( t ) d t int_(0)^(x)f(t)dt=x+int_(x)^(1)tf(t)dt\int_0^x f(t) d t=x+\int_x^1 t f(t) d t∫0xf(t)dt=x+∫x1tf(t)dt, then find the value of f ( 1 ) f ( 1 ) f(1)f(1)f(1).
3.(a)(iii) ∫ a b ( x − a ) m ( b − x ) n d x ∫ a b   ( x − a ) m ( b − x ) n d x int_(a)^(b)(x-a)^(m)(b-x)^(n)dx\int_a^b(x-a)^m(b-x)^n d x∫ab(x−a)m(b−x)ndx को बीटा-फलन के रूप में व्यक्त कीजिए ।
Express ∫ a b ( x − a ) m ( b − x ) n d x ∫ a b   ( x − a ) m ( b − x ) n d x int_(a)^(b)(x-a)^(m)(b-x)^(n)dx\int_a^b(x-a)^m(b-x)^n d x∫ab(x−a)m(b−x)ndx in terms of Beta function.
3.(b) अचर त्रिज्या r r rrr का एक गोला मूल-बिंदु O O OOO से गुजरता है तथा अक्षों को A , B , C A , B , C A,B,CA, B, CA,B,C बिन्दुओं पर काटता है । O O OOO से समतल A B C A B C ABCA B CABC पर खींचे गए लंब-पाद का बिन्दुपथ ज्ञात कीजिए ।
A sphere of constant radius r r rrr passes through the origin O O OOO and cuts the axes at the points A , B A , B A,BA, BA,B and C C CCC. Find, the locus of the foot of the perpendicular drawn from O O OOO to the plane A B C A B C ABCA B CABC.
3.(c)(i) सिद्ध कीजिए कि एक वास्तविक सममित आव्यूह के दो भिन्न अभिलक्षणिक मानों के संगत अभिलक्षणिक सदिश, लांबिक हैं।
Prove that the eigen vectors, corresponding to two distinct eigen values of a real symmetric matrix, are orthogonal.
3.(c)(ii) दो वर्ग आव्यूह A A AAA तथा B B BBB जिनकी कोटि, 2 है के लिए दर्शाइए कि अनुरेख ( A B ) = ( A B ) = (AB)=(A B)=(AB)= अनुरेख ( B A ) ( B A ) (BA)(B A)(BA) । अतैव दर्शाइए कि A B − B A ≠ I 2 A B − B A ≠ I 2 AB-BA!=I_(2)A B-B A \neq I_2AB−BA≠I2 जहाँ I 2 I 2 I_(2)I_2I2 एक 2 -कोटि का तत्समक आव्यूह है ।
For two square matrices A A AAA and B B BBB of order 2 , show that trace ( A B ) = trace ⁡ ( B A ) ( A B ) = trace ⁡ ( B A ) (AB)=trace(BA)(A B)=\operatorname{trace}(B A)(AB)=trace⁡(BA). Hence show that A B − B A ≠ I 2 A B − B A ≠ I 2 AB-BA!=I_(2)A B-B A \neq I_2AB−BA≠I2, where I 2 I 2 I_(2)I_2I2 is an identity matrix of order 2.7 2.7 2.72.72.7
4.(a)(i) निम्नलिखित आव्यूह का पंक्ति-समानीत सोपानक रूप में समानयन कीजिए एवं अतैव इसकी कोटि भी ज्ञात कीजिए ।
A = [ 1 3 2 4 1 0 0 2 2 0 2 6 2 6 2 3 9 1 10 6 ] A = 1 3 2 4 1 0 0 2 2 0 2 6 2 6 2 3 9 1 10 6 A=[[1,3,2,4,1],[0,0,2,2,0],[2,6,2,6,2],[3,9,1,10,6]]A=\left[\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{array}\right]A=[132410022026262391106]
Reduce the following matrix to a row-reduced echelon form and hence also, find its rank:
A = [ 1 3 2 4 1 0 0 2 2 0 2 6 2 6 2 3 9 1 10 6 ] A = 1 3 2 4 1 0 0 2 2 0 2 6 2 6 2 3 9 1 10 6 A=[[1,3,2,4,1],[0,0,2,2,0],[2,6,2,6,2],[3,9,1,10,6]]A=\left[\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{array}\right]A=[132410022026262391106]
4.(a)(ii) सम्मिश्र संख्या क्षेत्र पर आव्यूह A = ( 0 − i i 0 ) A = 0 − i i 0 A=([0,-i],[i,0])A=\left(\begin{array}{cc}0 & -i \\ i & 0\end{array}\right)A=(0−ii0) के अभिलक्षणिक मान तथा संगत अभिलक्षणिक सदिशों को ज्ञात कीजिए ।
Find the eigen values and the corresponding eigen vectors of the matrix A = ( 0 − i i 0 ) A = 0 − i i 0 A=([0,-i],[i,0])A=\left(\begin{array}{cc}0 & -i \\ i & 0\end{array}\right)A=(0−ii0), over the complex-number field.
4.(b) दर्शाइए कि ऐस्ट्रॉइड : x 2 / 3 + y 2 / 3 = a 2 / 3 x 2 / 3 + y 2 / 3 = a 2 / 3 x^(2//3)+y^(2//3)=a^(2//3)x^{2 / 3}+y^{2 / 3}=a^{2 / 3}x2/3+y2/3=a2/3 का पूरा क्षेत्रफल 3 8 π a 2 3 8 π a 2 (3)/(8)pia^(2)\frac{3}{8} \pi a^238πa2 है।
Show that the entire area of the Astroid : x 2 / 3 + y 2 / 3 = a 2 / 3 x 2 / 3 + y 2 / 3 = a 2 / 3 x^(2//3)+y^(2//3)=a^(2//3)x^{2 / 3}+y^{2 / 3}=a^{2 / 3}x2/3+y2/3=a2/3 is 3 8 π a 2 3 8 π a 2 (3)/(8)pia^(2)\frac{3}{8} \pi a^238πa2. 15 4.
(c) रेखाओं
x + 1 3 = y + 3 5 = z + 5 7 , x − 2 1 = y − 4 3 = z − 6 5 x + 1 3 = y + 3 5 = z + 5 7 , x − 2 1 = y − 4 3 = z − 6 5 {:[(x+1)/(3)=(y+3)/(5)=(z+5)/(7)”,”],[(x-2)/(1)=(y-4)/(3)=(z-6)/(5)]:}\begin{aligned} &\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}, \\ &\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5} \end{aligned}x+13=y+35=z+57,x−21=y−43=z−65
को अंतर्विष्ट करने वाले समतल का समीकरण ज्ञात कीजिए । दी गई रेखाओं के प्रतिच्छेद बिंदु को भी ज्ञात कीजिए।
Find equation of the plane containing the lines
x + 1 3 = y + 3 5 = z + 5 7 , x − 2 1 = y − 4 3 = z − 6 5 . x + 1 3 = y + 3 5 = z + 5 7 , x − 2 1 = y − 4 3 = z − 6 5 . {:[(x+1)/(3)=(y+3)/(5)=(z+5)/(7)”,”],[(x-2)/(1)=(y-4)/(3)=(z-6)/(5).]:}\begin{aligned} &\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}, \\ &\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5} . \end{aligned}x+13=y+35=z+57,x−21=y−43=z−65.
Also find the point of intersection of the given lines.
खण्ड ‘B’ SECTION ‘ B B BBB ‘
5.(a) अवकल समीकरण :
d 2 y d x 2 + 2 y = x 2 e 3 x + e x cos ⁡ 2 x d 2 y d x 2 + 2 y = x 2 e 3 x + e x cos ⁡ 2 x (d^(2)y)/(dx^(2))+2y=x^(2)e^(3x)+e^(x)cos 2x\frac{d^2 y}{d x^2}+2 y=x^2 e^{3 x}+e^x \cos 2 xd2ydx2+2y=x2e3x+excos⁡2x
को हल कीजिए ।
Solve the differential equation :
d 2 y d x 2 + 2 y = x 2 e 3 x + e x cos ⁡ 2 x d 2 y d x 2 + 2 y = x 2 e 3 x + e x cos ⁡ 2 x (d^(2)y)/(dx^(2))+2y=x^(2)e^(3x)+e^(x)cos 2x\frac{d^2 y}{d x^2}+2 y=x^2 e^{3 x}+e^x \cos 2 xd2ydx2+2y=x2e3x+excos⁡2x
5.(b) लाप्लास रूपान्तर विधि का उपयोग करते हुए प्रारम्भिक मान समस्या :
d 2 y d x 2 + 4 y = e − 2 x sin ⁡ 2 x ; y ( 0 ) = y ′ ( 0 ) = 0 d 2 y d x 2 + 4 y = e − 2 x sin ⁡ 2 x ; y ( 0 ) = y ′ ( 0 ) = 0 (d^(2)y)/(dx^(2))+4y=e^(-2x)sin 2x;y(0)=y^(‘)(0)=0\frac{d^2 y}{d x^2}+4 y=e^{-2 x} \sin 2 x ; y(0)=y^{\prime}(0)=0d2ydx2+4y=e−2xsin⁡2x;y(0)=y′(0)=0
को हल कीजिए ।
Solve the initial value problem :
d 2 y d x 2 + 4 y = e − 2 x sin ⁡ 2 x ; y ( 0 ) = y ′ ( 0 ) = 0 d 2 y d x 2 + 4 y = e − 2 x sin ⁡ 2 x ; y ( 0 ) = y ′ ( 0 ) = 0 (d^(2)y)/(dx^(2))+4y=e^(-2x)sin 2x;y(0)=y^(‘)(0)=0\frac{d^2 y}{d x^2}+4 y=e^{-2 x} \sin 2 x ; y(0)=y^{\prime}(0)=0d2ydx2+4y=e−2xsin⁡2x;y(0)=y′(0)=0
using Laplace transform method.
5(c) दो छड़े L M L M LML MLM व M N M N MNM NMN बिन्दु M M MMM पर दृढ़ता से इस प्रकार जुड़ी हैं कि ( L M ) 2 + ( M N ) 2 = ( L N ) 2 ( L M ) 2 + ( M N ) 2 = ( L N ) 2 (LM)^(2)+(MN)^(2)=(LN)^(2)(L M)^2+(M N)^2=(L N)^2(LM)2+(MN)2=(LN)2 तथा वे स्वतन्त्र रूप से साम्यावस्था में स्थिर बिन्दु L L LLL पर टँगी हैं। माना कि दोनों एकसमान छड़ों का प्रति एकांक लम्बाई, भार ω ω omega\omegaω है। छड़ L M L M LML MLM का ऊर्ध्वाधर दिशा के साथ बने कोण को छड़ों की लम्बाई के रूप में ज्ञात कीजिए ।
Two rods L M L M LML MLM and M N M N MNM NMN are joined rigidly at the point M M MMM such that ( L M ) 2 + ( M N ) 2 = ( L N ) 2 ( L M ) 2 + ( M N ) 2 = ( L N ) 2 (LM)^(2)+(MN)^(2)=(LN)^(2)(L M)^2+(M N)^2=(L N)^2(LM)2+(MN)2=(LN)2 and they are hanged freely in equilibrium from a fixed point L L LLL. Let ω ω omega\omegaω be the weight per unit length of both the rods which are uniform. Determine the angle, which the rod L M L M LML MLM makes with the vertical direction, in terms of lengths of the rods.
5(d) यदि एक ग्रह, जो सूर्य के परितः वृत्तीय कक्षा में परिभ्रमण करता है, अचानक अपनी कक्षा में रोक दिया जाता है, तो वह समय, जिसमें वह सूर्य में गिर जाएगा, ज्ञात कीजिए। इसके गिरने के समय का ग्रह के परिभ्रमण आवर्तकाल से अनुपात भी ज्ञात कीजिए ।
If a planet, which revolves around the Sun in a circular orbit, is suddenly stopped in its orbit, then find the time in which it would fall into the Sun. Also, find the ratio of its falling time to the period of revolution of the planet.
5.(e) दर्शाइए कि ∇ 2 [ ∇ ⋅ ( r → r ) ] = 2 r 4 ∇ 2 ∇ ⋅ r → r = 2 r 4 grad^(2)[grad*((( vec(r)))/(r))]=(2)/(r^(4))\nabla^2\left[\nabla \cdot\left(\frac{\vec{r}}{r}\right)\right]=\frac{2}{r^4}∇2[∇⋅(r→r)]=2r4, जहाँ r → = x i ^ + y j ^ + z k ^ r → = x i ^ + y j ^ + z k ^ vec(r)=x hat(i)+y hat(j)+z hat(k)\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}r→=xi^+yj^+zk^ है ।
Show that ∇ 2 [ ∇ ⋅ ( r → r ) ] = 2 r 4 ∇ 2 ∇ ⋅ r → r = 2 r 4 grad^(2)[grad*((( vec(r)))/(r))]=(2)/(r^(4))\nabla^2\left[\nabla \cdot\left(\frac{\vec{r}}{r}\right)\right]=\frac{2}{r^4}∇2[∇⋅(r→r)]=2r4, where r → = x i ^ + y j ^ + z k ^ r → = x i ^ + y j ^ + z k ^ vec(r)=x hat(i)+y hat(j)+z hat(k)\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}r→=xi^+yj^+zk^
6(a) एक भारी डोरी, जिसका घनत्व एक समान नहीं है, दो बिन्दुओं से टँगी हुई है। माना कि T 1 , T 2 , T 3 T 1 , T 2 , T 3 T_(1),T_(2),T_(3)T_1, T_2, T_3T1,T2,T3 क्रमशः कैटिनरी के बीच के बिन्दुओं A , B , C A , B , C A,B,CA, B, CA,B,C पर तनाव हैं, जिन पर इसके क्षैतिज के साथ आनति कोण, सार्व अंतर β β beta\betaβ के साथ समांतर श्रेढ़ी में हैं। माना कि डोरी के A B A B ABA BAB तथा B C B C BCB CBC भागों के भार क्रमशः ω 1 ω 1 omega_(1)\omega_1ω1 तथा ω 2 ω 2 omega_(2)\omega_2ω2 हैं। सिद्ध कीजिए
(i) T 1 , T 2 T 1 , T 2 T_(1),T_(2)T_1, T_2T1,T2 तथा T 3 T 3 T_(3)T_3T3 का हरात्मक माध्य = 3 T 2 1 + 2 cos ⁡ β = 3 T 2 1 + 2 cos ⁡ β =(3T_(2))/(1+2cos beta)=\frac{3 T_2}{1+2 \cos \beta}=3T21+2cos⁡β
(ii) T 1 T 3 = ω 1 ω 2 T 1 T 3 = ω 1 ω 2 (T_(1))/(T_(3))=(omega_(1))/(omega_(2))\frac{T_1}{T_3}=\frac{\omega_1}{\omega_2}T1T3=ω1ω2
A heavy string, which is not of uniform density, is hung up from two points. Let T 1 , T 2 , T 3 T 1 , T 2 , T 3 T_(1),T_(2),T_(3)T_1, T_2, T_3T1,T2,T3 be the tensions at the intermediate points A , B , C A , B , C A,B,CA, B, CA,B,C of the catenary respectively where its inclinations to the horizontal are in arithmetic progression with common difference β β beta\betaβ. Let ω 1 ω 1 omega_(1)\omega_1ω1 and ω 2 ω 2 omega_(2)\omega_2ω2 be the weights of the parts A B A B ABA BAB and B C B C BCB CBC of the string respectively. Prove that
(i) Harmonic mean of T 1 , T 2 T 1 , T 2 T_(1),T_(2)T_1, T_2T1,T2 and T 3 = 3 T 2 1 + 2 cos ⁡ β T 3 = 3 T 2 1 + 2 cos ⁡ β T_(3)=(3T_(2))/(1+2cos beta)T_3=\frac{3 T_2}{1+2 \cos \beta}T3=3T21+2cos⁡β
(ii) T 1 T 3 = ω 1 ω 2 T 1 T 3 = ω 1 ω 2 (T_(1))/(T_(3))=(omega_(1))/(omega_(2))\frac{T_1}{T_3}=\frac{\omega_1}{\omega_2}T1T3=ω1ω2
6.(b) सभी अन्तर्त्रस्त (शामिल) चरणों को दर्शति हुए समीकरण :
d 2 y d x 2 + ( tan ⁡ x − 3 cos ⁡ x ) d y d x + 2 y cos 2 ⁡ x = cos 4 ⁡ x d 2 y d x 2 + ( tan ⁡ x − 3 cos ⁡ x ) d y d x + 2 y cos 2 ⁡ x = cos 4 ⁡ x (d^(2)y)/(dx^(2))+(tan x-3cos x)(dy)/(dx)+2ycos^(2)x=cos^(4)x\frac{d^2 y}{d x^2}+(\tan x-3 \cos x) \frac{d y}{d x}+2 y \cos ^2 x=\cos ^4 xd2ydx2+(tan⁡x−3cos⁡x)dydx+2ycos2⁡x=cos4⁡x
को पूर्ण रूप से हल कीजिए ।
Solve the equation:
d 2 y d x 2 + ( tan ⁡ x − 3 cos ⁡ x ) d y d x + 2 y cos 2 ⁡ x = cos 4 ⁡ x d 2 y d x 2 + ( tan ⁡ x − 3 cos ⁡ x ) d y d x + 2 y cos 2 ⁡ x = cos 4 ⁡ x (d^(2)y)/(dx^(2))+(tan x-3cos x)(dy)/(dx)+2ycos^(2)x=cos^(4)x\frac{d^2 y}{d x^2}+(\tan x-3 \cos x) \frac{d y}{d x}+2 y \cos ^2 x=\cos ^4 xd2ydx2+(tan⁡x−3cos⁡x)dydx+2ycos2⁡x=cos4⁡x
completely by demonstrating all the steps involved.
6.(c) ∫ C F → ⋅ d r → ∫ C   F → ⋅ d r → int _(C) vec(F)*d vec(r)\int_C \vec{F} \cdot d \vec{r}∫CF→⋅dr→ का मान निकालिए,
जहाँ C , x y C , x y C,xyC, x yC,xy-समतल में एक स्वैच्छिक संवृत वक्र है तथा F → = − y i ^ + x j ^ x 2 + y 2 F → = − y i ^ + x j ^ x 2 + y 2 vec(F)=(-y( hat(i))+x( hat(j)))/(x^(2)+y^(2))\vec{F}=\frac{-y \hat{i}+x \hat{j}}{x^2+y^2}F→=−yi^+xj^x2+y2 है।
Evaluate ∫ C F → ⋅ d r → ∫ C   F → ⋅ d r → int _(C) vec(F)*d vec(r)\int_C \vec{F} \cdot d \vec{r}∫CF→⋅dr→, where C C CCC is an arbitrary closed curve in the x y x y xyx yxy-plane and F → = − y i ^ + x j ^ x 2 + y 2 F → = − y i ^ + x j ^ x 2 + y 2 vec(F)=(-y( hat(i))+x( hat(j)))/(x^(2)+y^(2))\vec{F}=\frac{-y \hat{i}+x \hat{j}}{x^2+y^2}F→=−yi^+xj^x2+y2.
7(a) प्रथम अष्टांशक में y 2 + z 2 = 9 y 2 + z 2 = 9 y^(2)+z^(2)=9y^2+z^2=9y2+z2=9 तथा x = 2 x = 2 x=2x=2x=2 द्वारा परिबद्ध क्षेत्र पर F → = 2 x 2 y i ^ − y 2 j ^ + 4 x z 2 k ^ F → = 2 x 2 y i ^ − y 2 j ^ + 4 x z 2 k ^ vec(F)=2x^(2)y hat(i)-y^(2) hat(j)+4xz^(2) hat(k)\vec{F}=2 x^2 y \hat{i}-y^2 \hat{j}+4 x z^2 \hat{k}F→=2x2yi^−y2j^+4xz2k^ के लिए गाउस अपसरण प्रमेय को सत्यापित कीजिए ।
Verify Gauss divergence theorem for F → = 2 x 2 y i ^ − y 2 j ^ + 4 x z 2 k ^ F → = 2 x 2 y i ^ − y 2 j ^ + 4 x z 2 k ^ vec(F)=2x^(2)y hat(i)-y^(2) hat(j)+4xz^(2) hat(k)\vec{F}=2 x^2 y \hat{i}-y^2 \hat{j}+4 x z^2 \hat{k}F→=2x2yi^−y2j^+4xz2k^ taken over the region in the first octant bounded by y 2 + z 2 = 9 y 2 + z 2 = 9 y^(2)+z^(2)=9y^2+z^2=9y2+z2=9 and x = 2 x = 2 x=2x=2x=2.
7.(b) अवकल समीकरण :
y 2 log ⁡ y = x y d y d x + ( d y d x ) 2 y 2 log ⁡ y = x y d y d x + d y d x 2 y^(2)log y=xy(dy)/(dx)+((dy)/(dx))^(2)y^2 \log y=x y \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^2y2log⁡y=xydydx+(dydx)2
के सभी सम्भव हल ज्ञात कीजिए ।
Find all possible solutions of the differential equation :
y 2 log ⁡ y = x y d y d x + ( d y d x ) 2 . y 2 log ⁡ y = x y d y d x + d y d x 2 .  y^(2)log y=xy(dy)/(dx)+((dy)/(dx))^(2)”. “y^2 \log y=x y \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^2 \text {. }y2log⁡y=xydydx+(dydx)2. 
7(c) एक भारी कण a a aaa लम्बाई की अवितान्य डोरी से एक स्थिर बिन्दु से टँगा है तथा 2 g h 2 g h sqrt(2gh)\sqrt{2 g h}2gh वेग से क्षैतिज दिशा में प्रक्षेपित किया जाता है । यदि 5 a 2 > h > a 5 a 2 > h > a (5a)/(2) > h > a\frac{5 a}{2}>h>a5a2>h>a है, तो सिद्ध कीजिए कि प्रक्षेपण बिन्दु से 1 3 ( a + 2 h ) 1 3 ( a + 2 h ) (1)/(3)(a+2h)\frac{1}{3}(a+2 h)13(a+2h) ऊँचाई पहुँचने पर कण की वृत्तीय गति समाप्त हो जाती है । यह भी सिद्ध कीजिए कि उस कण द्वारा प्रक्षेपण बिंदु से ऊपर प्राप्य अधिकतम ऊँचाई ( 4 a − h ) ( a + 2 h ) 2 27 a 2 ( 4 a − h ) ( a + 2 h ) 2 27 a 2 ((4a-h)(a+2h)^(2))/(27a^(2))\frac{(4 a-h)(a+2 h)^2}{27 a^2}(4a−h)(a+2h)227a2 है ।
A heavy particle hangs by an inextensible string of length a a aaa from a fixed point and is then projected horizontally with a velocity 2 g h 2 g h sqrt(2gh)\sqrt{2 g h}2gh. If 5 a 2 > h > a 5 a 2 > h > a (5a)/(2) > h > a\frac{5 a}{2}>h>a5a2>h>a, then prove that the circular motion ceases when the particle has reached the height 1 3 ( a + 2 h ) 1 3 ( a + 2 h ) (1)/(3)(a+2h)\frac{1}{3}(a+2 h)13(a+2h) from the point of projection. Also, prove that the greatest height ever reached by the particle above the point of projection is ( 4 a − h ) ( a + 2 h ) 2 27 a 2 ( 4 a − h ) ( a + 2 h ) 2 27 a 2 ((4a-h)(a+2h)^(2))/(27a^(2))\frac{(4 a-h)(a+2 h)^2}{27 a^2}(4a−h)(a+2h)227a2.
8(a)(i) संनाभि शांकव कुल
x 2 a 2 + λ + y 2 b 2 + λ = 1 ; a > b > 0 x 2 a 2 + λ + y 2 b 2 + λ = 1 ; a > b > 0 (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1;a > b > 0\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1 ; a>b>0x2a2+λ+y2b2+λ=1;a>b>0 अचर हैं तथा λ λ lambda\lambdaλ एक प्राचल है,
के लंबकोणीय संछेदी ज्ञात कीजिए । दर्शाइए कि दिया गया वक्र-कुल स्वलांबिक है।
Find the orthogonal trajectories of the family of confocal conics x 2 a 2 + λ + y 2 b 2 + λ = 1 ; a > b > 0 x 2 a 2 + λ + y 2 b 2 + λ = 1 ; a > b > 0 (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1;quad a > b > 0\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1 ; \quad a>b>0x2a2+λ+y2b2+λ=1;a>b>0 are constants and λ λ lambda\lambdaλ is a parameter. Show that the given family of curves is self orthogonal.
8(a)(ii) अवकल समीकरण : x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = 0 x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = 0 x^(2)(d^(2)y)/(dx^(2))-2x(1+x)(dy)/(dx)+2(1+x)y=0x^2 \frac{d^2 y}{d x^2}-2 x(1+x) \frac{d y}{d x}+2(1+x) y=0x2d2ydx2−2x(1+x)dydx+2(1+x)y=0 का व्यापक हल ज्ञात कीजिए । अतः अवकल समीकरण : x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = x 3 x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = x 3 x^(2)(d^(2)y)/(dx^(2))-2x(1+x)(dy)/(dx)+2(1+x)y=x^(3)x^2 \frac{d^2 y}{d x^2}-2 x(1+x) \frac{d y}{d x}+2(1+x) y=x^3x2d2ydx2−2x(1+x)dydx+2(1+x)y=x3 को प्राचल विचरण विधि द्वारा हल कीजिए ।
Find the general solution of the differential equation : x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = 0 x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = 0 x^(2)(d^(2)y)/(dx^(2))-2x(1+x)(dy)/(dx)+2(1+x)y=0x^2 \frac{d^2 y}{d x^2}-2 x(1+x) \frac{d y}{d x}+2(1+x) y=0x2d2ydx2−2x(1+x)dydx+2(1+x)y=0.
Hence, solve the differential equation: x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = x 3 x 2 d 2 y d x 2 − 2 x ( 1 + x ) d y d x + 2 ( 1 + x ) y = x 3 x^(2)(d^(2)y)/(dx^(2))-2x(1+x)(dy)/(dx)+2(1+x)y=x^(3)x^2 \frac{d^2 y}{d x^2}-2 x(1+x) \frac{d y}{d x}+2(1+x) y=x^3x2d2ydx2−2x(1+x)dydx+2(1+x)y=x3 by the method of variation of parameters.
8.(b) द्रव्यमान m m mmm का एक कण, जो की प्रक्षेपण बिन्दु से वेग u u uuu के साथ क्षैतिज दिशा के साथ θ θ theta\thetaθ कोण बनाने वाली दिशा में प्रक्षेपण बिन्दु से गुजरने वाले ऊर्ध्वाधर समतल में प्रक्षेपित किया जाता है, उसकी गति तथा पथ का वर्णन कीजिए । यदि कणों को उसी बिन्दु से उसी ऊर्ध्वाधर समतल में वेग 4 g 4 g 4sqrtg4 \sqrt{g}4g के साथ प्रक्षेपित किया जाता है, तो उनके पथों के शीर्षों के बिन्दुपथ को भी निर्धारित कीजिए ।
Describe the motion and path of a particle of mass m m mmm which is projected in a vertical plane through a point of projection with velocity u u uuu in a direction making an angle θ θ theta\thetaθ with the horizontal direction. Further, if particles are projected from that point in the same vertical plane with velocity 4 g 4 g 4sqrtg4 \sqrt{g}4g, then determine the locus of vertices of their paths.
8(c) स्टोक्स प्रमेय का उपयोग करते हुए ∬ S ( ∇ × F → ) ⋅ n ^ d S ∬ S   ( ∇ × F → ) ⋅ n ^ d S ∬_(S)(grad xx vec(F))* hat(n)dS\iint_S(\nabla \times \vec{F}) \cdot \hat{n} d S∬S(∇×F→)⋅n^dS का मान निकालिए, जहाँ पर F → = ( x 2 + y − 4 ) i ^ + 3 x y j ^ + ( 2 x y + z 2 ) k ^ F → = x 2 + y − 4 i ^ + 3 x y j ^ + 2 x y + z 2 k ^ vec(F)=(x^(2)+y-4) hat(i)+3xy hat(j)+(2xy+z^(2)) hat(k)\vec{F}=\left(x^2+y-4\right) \hat{i}+3 x y \hat{j}+\left(2 x y+z^2\right) \hat{k}F→=(x2+y−4)i^+3xyj^+(2xy+z2)k^ तथा S S SSS, परवलयज z = 4 − ( x 2 + y 2 ) z = 4 − x 2 + y 2 z=4-(x^(2)+y^(2))z=4-\left(x^2+y^2\right)z=4−(x2+y2) का x y x y xyx yxy-समतल से ऊपर का पृष्ठ है। यहाँ n ^ , S n ^ , S hat(n),S\hat{n}, Sn^,S पर एकक बहिर्मुखी अभिलंब सदिश है ।
Using Stokes’ theorem, evaluate ∬ S ( ∇ × F → ) ⋅ n ^ d S ∬ S   ( ∇ × F → ) ⋅ n ^ d S ∬_(S)(grad xx vec(F))* hat(n)dS\iint_S(\nabla \times \vec{F}) \cdot \hat{n} d S∬S(∇×F→)⋅n^dS, where F → = ( x 2 + y − 4 ) i ^ + 3 x y j ^ + ( 2 x y + z 2 ) k ^ F → = x 2 + y − 4 i ^ + 3 x y j ^ + 2 x y + z 2 k ^ vec(F)=(x^(2)+y-4) hat(i)+3xy hat(j)+(2xy+z^(2)) hat(k)\vec{F}=\left(x^2+y-4\right) \hat{i}+3 x y \hat{j}+\left(2 x y+z^2\right) \hat{k}F→=(x2+y−4)i^+3xyj^+(2xy+z2)k^ and S S SSS is the surface of the paraboloid z = 4 − ( x 2 + y 2 ) z = 4 − x 2 + y 2 z=4-(x^(2)+y^(2))z=4-\left(x^2+y^2\right)z=4−(x2+y2) above the x y x y xyx yxy-plane. Here, n ^ n ^ hat(n)\hat{n}n^ is the unit outward normal vector on S S SSS.
upsc-m2020-1-e59634ed-aa89-4306-aaf5-1772dafc3c90
खण्ड-A / SECTION-A
1(a) माना समुच्चय V V VVV में सभी n × n n × n n xx nn \times nn×n के वास्तविक मैजिक वर्ग हैं। दिखाइए कि समुच्चय V , R V , R V,RV, RV,R पर एक सदिश समष्टि है। दो भिन्न-भिन्न 2 × 2 2 × 2 2xx22 \times 22×2 मैजिक वर्ग के उदाहरण दीजिए।
Consider the set V V VVV of all n × n n × n n xx nn \times nn×n real magic squares. Show that V V VVV is a vector space over R R RRR. Give examples of two distinct 2 × 2 2 × 2 2xx22 \times 22×2 magic squares.
(b) माना M 2 ( R ) M 2 ( R ) M_(2)(R)M_2(R)M2(R) सभी 2 × 2 2 × 2 2xx22 \times 22×2 वास्तविक आव्यूहों का सदिश समष्टि है। माना B = [ 1 − 1 − 4 4 ] B = 1 − 1 − 4 4 B=[[1,-1],[-4,4]]B=\left[\begin{array}{cc}1 & -1 \\ -4 & 4\end{array}\right]B=[1−1−44]. माना T : M 2 ( R ) → M 2 ( R ) T : M 2 ( R ) → M 2 ( R ) T:M_(2)(R)rarrM_(2)(R)T: M_2(R) \rightarrow M_2(R)T:M2(R)→M2(R) एक रैखिक रूपांतरण है, जो T ( A ) = B A T ( A ) = B A T(A)=BAT(A)=B AT(A)=BA द्वारा परिभाषित है। T T TTT की कोटि (रिक) व शून्यता (नलिटि) ज्ञात कीजिए। आव्यूह A A AAA ज्ञात कीजिए, जो शून्य आव्यूह को प्रतिचित्रित करता है।
Let M 2 ( R ) M 2 ( R ) M_(2)(R)M_2(R)M2(R) be the vector space of all 2 × 2 2 × 2 2xx22 \times 22×2 real matrices. Let B = [ 1 − 1 − 4 4 ] B = 1 − 1 − 4 4 B=[[1,-1],[-4,4]]B=\left[\begin{array}{cc}1 & -1 \\ -4 & 4\end{array}\right]B=[1−1−44]. Suppose T : M 2 ( R ) → M 2 ( R ) T : M 2 ( R ) → M 2 ( R ) T:M_(2)(R)rarrM_(2)(R)T: M_2(R) \rightarrow M_2(R)T:M2(R)→M2(R) is a linear transformation defined by T ( A ) = B A T ( A ) = B A T(A)=BAT(A)=B AT(A)=BA. Find the rank and nullity of T T TTT. Find a matrix A A AAA which maps to the null matrix.
(c) lim x → π 4 ( tan ⁡ x ) tan ⁡ 2 x lim x → π 4   ( tan ⁡ x ) tan ⁡ 2 x lim_(x rarr(pi)/(4))(tan x)^(tan 2x)\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{\tan 2 x}limx→π4(tan⁡x)tan⁡2x का मान निकालिए।
Evaluate lim x → π 4 ( tan ⁡ x ) tan ⁡ 2 x lim x → π 4   ( tan ⁡ x ) tan ⁡ 2 x lim_(x rarr(pi)/(4))(tan x)^(tan 2x)\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{\tan 2 x}limx→π4(tan⁡x)tan⁡2x
(d) वक्र ( 2 x + 3 ) y = ( x − 1 ) 2 ( 2 x + 3 ) y = ( x − 1 ) 2 (2x+3)y=(x-1)^(2)(2 x+3) y=(x-1)^2(2x+3)y=(x−1)2 के सभी अनंतस्पर्शी निकालिए।
Find all the asymptotes of the curve ( 2 x + 3 ) y = ( x − 1 ) 2 ( 2 x + 3 ) y = ( x − 1 ) 2 (2x+3)y=(x-1)^(2)(2 x+3) y=(x-1)^2(2x+3)y=(x−1)2.
(e) दीर्घवृत्तज 2 x 2 + 6 y 2 + 3 z 2 = 27 2 x 2 + 6 y 2 + 3 z 2 = 27 2x^(2)+6y^(2)+3z^(2)=272 x^2+6 y^2+3 z^2=272x2+6y2+3z2=27 के स्पर्श समतल का समीकरण निकालिए, जो रेखा x − y − z = 0 = x − y + 2 z − 9 x − y − z = 0 = x − y + 2 z − 9 x-y-z=0=x-y+2z-9x-y-z=0=x-y+2 z-9x−y−z=0=x−y+2z−9 से होकर गुजरता है।
Find the equations of the tangent plane to the ellipsoid 2 x 2 + 6 y 2 + 3 z 2 = 27 2 x 2 + 6 y 2 + 3 z 2 = 27 2x^(2)+6y^(2)+3z^(2)=272 x^2+6 y^2+3 z^2=272x2+6y2+3z2=27 which passes through the line x − y − z = 0 = x − y + 2 z − 9 x − y − z = 0 = x − y + 2 z − 9 x-y-z=0=x-y+2z-9x-y-z=0=x-y+2 z-9x−y−z=0=x−y+2z−9.
2(a) ∫ 0 1 tan − 1 ⁡ ( 1 − 1 x ) d x ∫ 0 1   tan − 1 ⁡ 1 − 1 x d x int_(0)^(1)tan^(-1)(1-(1)/(x))dx\int_0^1 \tan ^{-1}\left(1-\frac{1}{x}\right) d x∫01tan−1⁡(1−1x)dx का मान निकालिए।
Evaluate ∫ 0 1 tan − 1 ⁡ ( 1 − 1 x ) d x ∫ 0 1   tan − 1 ⁡ 1 − 1 x d x int_(0)^(1)tan^(-1)(1-(1)/(x))dx\int_0^1 \tan ^{-1}\left(1-\frac{1}{x}\right) d x∫01tan−1⁡(1−1x)dx
(b) एक n × n n × n n xx nn \times nn×n आव्यूह A A AAA को परिभाषित कीजिए, जबकि A = I − 2 u ⋅ u T A = I − 2 u ⋅ u T A=I-2u*u^(T)A=I-2 u \cdot u^TA=I−2u⋅uT, जहाँ u u uuu एक इकाई स्तंभ सदिश है।
(i) परीक्षण कीजिए कि A A AAA सममित है।
(ii) परीक्षण कीजिए कि A A AAA लांबिक है।
(iii) दिखाइए कि आव्यूह A A AAA का अनुरेख ( n − 2 ) ( n − 2 ) (n-2)(n-2)(n−2) है।
(iv) आव्यूह A 3 × 3 A 3 × 3 A_(3xx3)A_{3 \times 3}A3×3 निकालिए, जबकि u = [ 1 3 2 3 2 3 ] u = 1 3 2 3 2 3 u=[[(1)/(3)],[(2)/(3)],[(2)/(3)]]u=\left[\begin{array}{c}\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3}\end{array}\right]u=[132323] है।
Define an n × n n × n n xx nn \times nn×n matrix as A = I − 2 u ⋅ u T A = I − 2 u ⋅ u T A=I-2u*u^(T)A=I-2 u \cdot u^TA=I−2u⋅uT, where u u uuu is a unit column vector.
(i) Examine if A A AAA is symmetric.
(ii) Examine if A A AAA is orthogonal.
(iii) Show that trace ( A ) = n − 2 ( A ) = n − 2 (A)=n-2(A)=n-2(A)=n−2.
(iv) Find A 3 × 3 A 3 × 3 A_(3xx3)A_{3 \times 3}A3×3, when u = [ 1 3 2 3 2 3 ] u = 1 3 2 3 2 3 u=[[(1)/(3)],[(2)/(3)],[(2)/(3)]]u=\left[\begin{array}{c}\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3}\end{array}\right]u=[132323].
(c) एक ऐसे बेलन का समीकरण निकालिए, जिसकी जनक-रेखाएँ, रेखा x 1 = y − 2 = z 3 x 1 = y − 2 = z 3 (x)/(1)=(y)/(-2)=(z)/(3)\frac{x}{1}=\frac{y}{-2}=\frac{z}{3}x1=y−2=z3 के समांतर हैं तथा जिसका मार्गदर्शक वक्र x 2 + y 2 = 4 , z = 2 x 2 + y 2 = 4 , z = 2 x^(2)+y^(2)=4,z=2x^2+y^2=4, z=2x2+y2=4,z=2 है।
Find the equation of the cylinder whose generators are parallel to the line x 1 = y − 2 = z 3 x 1 = y − 2 = z 3 (x)/(1)=(y)/(-2)=(z)/(3)\frac{x}{1}=\frac{y}{-2}=\frac{z}{3}x1=y−2=z3 and whose guiding curve is x 2 + y 2 = 4 , z = 2 x 2 + y 2 = 4 , z = 2 x^(2)+y^(2)=4,z=2x^2+y^2=4, z=2x2+y2=4,z=2.
3(a) निम्न फलन पर विचार कीजिए :
f ( x ) = ∫ 0 x ( t 2 − 5 t + 4 ) ( t 2 − 5 t + 6 ) d t f ( x ) = ∫ 0 x   t 2 − 5 t + 4 t 2 − 5 t + 6 d t f(x)=int_(0)^(x)(t^(2)-5t+4)(t^(2)-5t+6)dtf(x)=\int_0^x\left(t^2-5 t+4\right)\left(t^2-5 t+6\right) d tf(x)=∫0x(t2−5t+4)(t2−5t+6)dt
(i) फलन f ( x ) f ( x ) f(x)f(x)f(x) के क्रांतिक बिंदु निकालिए।
(ii) वे बिंदु निकालिए, जहाँ f ( x ) f ( x ) f(x)f(x)f(x) का स्थानीय न्यूनतम होगा।
(iii) वे बिंदु निकालिए, जहाँ f ( x ) f ( x ) f(x)f(x)f(x) का स्थानीय अधिकतम होगा।
(iv) फलन f ( x ) f ( x ) f(x)f(x)f(x) के [ 0 , 5 ] [ 0 , 5 ] [0,5][0,5][0,5] में कितने शून्यक होंगे, निकालिए।
Consider the function f ( x ) = ∫ 0 x ( t 2 − 5 t + 4 ) ( t 2 − 5 t + 6 ) d t f ( x ) = ∫ 0 x   t 2 − 5 t + 4 t 2 − 5 t + 6 d t f(x)=int_(0)^(x)(t^(2)-5t+4)(t^(2)-5t+6)dtf(x)=\int_0^x\left(t^2-5 t+4\right)\left(t^2-5 t+6\right) d tf(x)=∫0x(t2−5t+4)(t2−5t+6)dt.
(i) Find the critical points of the function f ( x ) f ( x ) f(x)f(x)f(x).
(ii) Find the points at which local minimum occurs.
(iii) Find the points at which local maximum occurs.
(iv) Find the number of zeros of the function f ( x ) f ( x ) f(x)f(x)f(x) in [ 0 , 5 ] [ 0 , 5 ] [0,5][0,5][0,5].
(b) माना F F FFF सम्मिश्र संख्याओं का एक उपक्षेत्र है व T : F 3 → F 3 T : F 3 → F 3 T:F^(3)rarrF^(3)T: F^3 \rightarrow F^3T:F3→F3 एक ऐसा फलन है, जो निम्न रूप से परिभाषित है :
T ( x 1 , x 2 , x 3 ) = ( x 1 + x 2 + 3 x 3 , 2 x 1 − x 2 , − 3 x 1 + x 2 − x 3 ) T x 1 , x 2 , x 3 = x 1 + x 2 + 3 x 3 , 2 x 1 − x 2 , − 3 x 1 + x 2 − x 3 T(x_(1),x_(2),x_(3))=(x_(1)+x_(2)+3x_(3),2x_(1)-x_(2),-3x_(1)+x_(2)-x_(3))T\left(x_1, x_2, x_3\right)=\left(x_1+x_2+3 x_3, 2 x_1-x_2,-3 x_1+x_2-x_3\right)T(x1,x2,x3)=(x1+x2+3x3,2x1−x2,−3x1+x2−x3)
a , b , c a , b , c a,b,ca, b, ca,b,c पर क्या शर्तें हैं कि ( a , b , c ) , T ( a , b , c ) , T (a,b,c),T(a, b, c), T(a,b,c),T के शुन्य समष्टि में है? T T TTT की शून्यता निकालिए।
Let F F FFF be a subfield of complex numbers and T T TTT a function from F 3 → F 3 F 3 → F 3 F^(3)rarrF^(3)F^3 \rightarrow F^3F3→F3 defined by T ( x 1 , x 2 , x 3 ) = ( x 1 + x 2 + 3 x 3 , 2 x 1 − x 2 , − 3 x 1 + x 2 − x 3 ) T x 1 , x 2 , x 3 = x 1 + x 2 + 3 x 3 , 2 x 1 − x 2 , − 3 x 1 + x 2 − x 3 T(x_(1),x_(2),x_(3))=(x_(1)+x_(2)+3x_(3),2x_(1)-x_(2),-3x_(1)+x_(2)-x_(3))T\left(x_1, x_2, x_3\right)=\left(x_1+x_2+3 x_3, 2 x_1-x_2,-3 x_1+x_2-x_3\right)T(x1,x2,x3)=(x1+x2+3x3,2x1−x2,−3x1+x2−x3). What are the conditions on a , b , c a , b , c a,b,ca, b, ca,b,c such that ( a , b , c ) ( a , b , c ) (a,b,c)(a, b, c)(a,b,c) be in the null space of T T TTT ? Find the nullity of T T TTT.
(c) यदि सरल रेखा x 1 = y 2 = z 3 x 1 = y 2 = z 3 (x)/(1)=(y)/(2)=(z)/(3)\frac{x}{1}=\frac{y}{2}=\frac{z}{3}x1=y2=z3 शंकु 5 y z − 8 z x − 3 x y = 0 5 y z − 8 z x − 3 x y = 0 5yz-8zx-3xy=05 y z-8 z x-3 x y=05yz−8zx−3xy=0 के तीन परस्पर लांबिक जनकों के समुच्चय में से एक है, तब अन्य दो जनकों के समीकरण निकालिए।
If the straight line x 1 = y 2 = z 3 x 1 = y 2 = z 3 (x)/(1)=(y)/(2)=(z)/(3)\frac{x}{1}=\frac{y}{2}=\frac{z}{3}x1=y2=z3 represents one of a set of three mutually perpendicular generators of the cone 5 y z − 8 z x − 3 x y = 0 5 y z − 8 z x − 3 x y = 0 5yz-8zx-3xy=05 y z-8 z x-3 x y=05yz−8zx−3xy=0, then find the equations of the other two generators.
4(a) माना
A = [ 1 0 2 2 − 1 3 4 1 8 ] और B = [ − 11 2 2 − 4 0 1 6 − 1 − 1 ] A = 1      0      2 2      − 1      3 4      1      8  और  B = − 11      2      2 − 4      0      1 6      − 1      − 1 A=[[1,0,2],[2,-1,3],[4,1,8]]” और “B=[[-11,2,2],[-4,0,1],[6,-1,-1]]A=\left[\begin{array}{rrr} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{array}\right] \text { और } B=\left[\begin{array}{rrr} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{array}\right]A=[1022−13418] और B=[−1122−4016−1−1]
(i) A B A B ABA BAB ज्ञात कीजिए।
(ii) सारणिक ( A ) ( A ) (A)(A)(A) व सारणिक ( B ) ( B ) (B)(B)(B) ज्ञात कीजिए।
(iii) निम्न रैखिक समीकरणों के निकाय का हल निकालिए :
x + 2 z = 3 , 2 x − y + 3 z = 3 , 4 x + y + 8 z = 14 x + 2 z = 3 , 2 x − y + 3 z = 3 , 4 x + y + 8 z = 14 x+2z=3,quad2x-y+3z=3,quad4x+y+8z=14x+2 z=3, \quad 2 x-y+3 z=3, \quad 4 x+y+8 z=14x+2z=3,2x−y+3z=3,4x+y+8z=14
Let
A = [ 1 0 2 2 − 1 3 4 1 8 ] and B = [ − 11 2 2 − 4 0 1 6 − 1 − 1 ] A = 1      0      2 2      − 1      3 4      1      8  and  B = − 11      2      2 − 4      0      1 6      − 1      − 1 A=[[1,0,2],[2,-1,3],[4,1,8]]quad” and “B=[[-11,2,2],[-4,0,1],[6,-1,-1]]A=\left[\begin{array}{rrr} 1 & 0 & 2 \\ 2 & -1 & 3 \\ 4 & 1 & 8 \end{array}\right] \quad \text { and } B=\left[\begin{array}{rrr} -11 & 2 & 2 \\ -4 & 0 & 1 \\ 6 & -1 & -1 \end{array}\right]A=[1022−13418] and B=[−1122−4016−1−1]
(i) Find A B A B ABA BAB.
(ii) Find det ⁡ ( A ) det ⁡ ( A ) det(A)\operatorname{det}(A)det⁡(A) and det ⁡ ( B ) det ⁡ ( B ) det(B)\operatorname{det}(B)det⁡(B).
(iii) Solve the following system of linear equations :
x + 2 z = 3 , 2 x − y + 3 z = 3 , 4 x + y + 8 z = 14 x + 2 z = 3 , 2 x − y + 3 z = 3 , 4 x + y + 8 z = 14 x+2z=3,quad2x-y+3z=3,quad4x+y+8z=14x+2 z=3, \quad 2 x-y+3 z=3, \quad 4 x+y+8 z=14x+2z=3,2x−y+3z=3,4x+y+8z=14
(b) अतिपरवलयिक परवलयज x 2 a 2 − y 2 b 2 = 2 z x 2 a 2 − y 2 b 2 = 2 z (x^(2))/(a^(2))-(y^(2))/(b^(2))=2z\frac{x^2}{a^2}-\frac{y^2}{b^2}=2 zx2a2−y2b2=2z के लांबिक जनकों के प्रतिच्छेद बिंदु का बिंदुपथ निकालिए।
Find the locus of the point of intersection of the perpendicular generators of the hyperbolic paraboloid x 2 a 2 − y 2 b 2 = 2 z x 2 a 2 − y 2 b 2 = 2 z (x^(2))/(a^(2))-(y^(2))/(b^(2))=2z\frac{x^2}{a^2}-\frac{y^2}{b^2}=2 zx2a2−y2b2=2z.
(c) लाग्रांज की अनिर्धारित गुणक विधि का प्रयोग करके फलन u = x 2 + y 2 + z 2 u = x 2 + y 2 + z 2 u=x^(2)+y^(2)+z^(2)u=x^2+y^2+z^2u=x2+y2+z2 का चरम मान ज्ञात कीजिए, जो 2 x + 3 y + 5 z = 30 2 x + 3 y + 5 z = 30 2x+3y+5z=302 x+3 y+5 z=302x+3y+5z=30 शर्त द्वारा प्रतिबंधित है।
Find an extreme value of the function u = x 2 + y 2 + z 2 u = x 2 + y 2 + z 2 u=x^(2)+y^(2)+z^(2)u=x^2+y^2+z^2u=x2+y2+z2, subject to the condition 2 x + 3 y + 5 z = 30 2 x + 3 y + 5 z = 30 2x+3y+5z=302 x+3 y+5 z=302x+3y+5z=30, by using Lagrange’s method of undetermined multiplier. 20
खण्ड-B / SECTION-B
5(a) निम्न अवकल समीकरण को हल कीजिए :
x cos ⁡ ( y x ) ( y d x + x d y ) = y sin ⁡ ( y x ) ( x d y − y d x ) x cos ⁡ y x ( y d x + x d y ) = y sin ⁡ y x ( x d y − y d x ) x cos((y)/(x))(ydx+xdy)=y sin((y)/(x))(xdy-ydx)x \cos \left(\frac{y}{x}\right)(y d x+x d y)=y \sin \left(\frac{y}{x}\right)(x d y-y d x)xcos⁡(yx)(ydx+xdy)=ysin⁡(yx)(xdy−ydx)
Solve the following differential equation :
x cos ⁡ ( y x ) ( y d x + x d y ) = y sin ⁡ ( y x ) ( x d y − y d x ) x cos ⁡ y x ( y d x + x d y ) = y sin ⁡ y x ( x d y − y d x ) x cos((y)/(x))(ydx+xdy)=y sin((y)/(x))(xdy-ydx)x \cos \left(\frac{y}{x}\right)(y d x+x d y)=y \sin \left(\frac{y}{x}\right)(x d y-y d x)xcos⁡(yx)(ydx+xdy)=ysin⁡(yx)(xdy−ydx)
(b) वृत्त-कुल, जो बिंदु ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) एवं ( 0 , − 2 ) ( 0 , − 2 ) (0,-2)(0,-2)(0,−2) से गुजरता है, का लंबकोणीय संछेदी ज्ञात कीजिए।
Find the orthogonal trajectories of the family of circles passing through the points ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) and ( 0 , − 2 ) ( 0 , − 2 ) (0,-2)(0,-2)(0,−2).
(c) a , b , c a , b , c a,b,ca, b, ca,b,c के किस मान के लिए सदिश क्षेत्र
V ¯ = ( − 4 x − 3 y + a z ) i ^ + ( b x + 3 y + 5 z ) j ^ + ( 4 x + c y + 3 z ) k ^ V ¯ = ( − 4 x − 3 y + a z ) i ^ + ( b x + 3 y + 5 z ) j ^ + ( 4 x + c y + 3 z ) k ^ bar(V)=(-4x-3y+az) hat(i)+(bx+3y+5z) hat(j)+(4x+cy+3z) hat(k)\bar{V}=(-4 x-3 y+a z) \hat{i}+(b x+3 y+5 z) \hat{j}+(4 x+c y+3 z) \hat{k}V¯=(−4x−3y+az)i^+(bx+3y+5z)j^+(4x+cy+3z)k^
अघूर्णी है? तब V ¯ V ¯ bar(V)\bar{V}V¯ को अदिश फलन ϕ ϕ phi\phiϕ की प्रवणता के रूप में व्यक्त कीजिए। ϕ ϕ phi\phiϕ को ज्ञात कीजिए।
For what value of a , b , c a , b , c a,b,ca, b, ca,b,c is the vector field
V ¯ = ( − 4 x − 3 y + a z ) i ^ + ( b x + 3 y + 5 z ) j ^ + ( 4 x + c y + 3 z ) k ^ V ¯ = ( − 4 x − 3 y + a z ) i ^ + ( b x + 3 y + 5 z ) j ^ + ( 4 x + c y + 3 z ) k ^ bar(V)=(-4x-3y+az) hat(i)+(bx+3y+5z) hat(j)+(4x+cy+3z) hat(k)\bar{V}=(-4 x-3 y+a z) \hat{i}+(b x+3 y+5 z) \hat{j}+(4 x+c y+3 z) \hat{k}V¯=(−4x−3y+az)i^+(bx+3y+5z)j^+(4x+cy+3z)k^
irrotational? Hence, express V ¯ V ¯ bar(V)\bar{V}V¯ as the gradient of a scalar function ϕ ϕ phi\phiϕ. Determine ϕ ϕ phi\phiϕ.
(d) एक एकसमान छड़, जो ऊर्ध्वर्धर दशा में है, अपने एक सिरे पर स्वतंत्र रूप से वर्तन कर सकती है तथा दूसरे सिरे पर लगाए गए एक क्षैतिज बल, जिसका मान छड़ के भार का आधा है, द्वारा ऊर्ध्वाधर से एक तरफ खींची जाती है। बताइए कि ऊर्ध्वाधर से किस कोण पर छड़ विश्राम करेगी।
A uniform rod, in vertical position, can turn freely about one of its ends and is pulled aside from the vertical by a horizontal force acting at the other end of the rod and equal to half its weight. At what inclination to the vertical will the rod rest?
(e) एक हल्की दृढ़ छड़ A B C A B C ABCA B CABC से तीन कण, जिनमें से हरेक का द्रव्यमान m m mmm है, A , B A , B A,BA, BA,B तथा C C CCC पर बंधे हुए हैं। उस छड़ को बिंदु A A AAA से B C B C BCB CBC दूरी के बराबर स्थित बिंदु पर एक बल P P PPP के द्वारा लम्बवत् मारा जाता है। सिद्ध कीजिए कि पैदा हुई गतिज ऊर्जा का मान 1 2 p 2 m a 2 − a b + b 2 a 2 + a b + b 2 1 2 p 2 m a 2 − a b + b 2 a 2 + a b + b 2 (1)/(2)(p^(2))/(m)(a^(2)-ab+b^(2))/(a^(2)+ab+b^(2))\frac{1}{2} \frac{p^2}{m} \frac{a^2-a b+b^2}{a^2+a b+b^2}12p2ma2−ab+b2a2+ab+b2 है, जहाँ A B = a A B = a AB=aA B=aAB=a तथा B C = b B C = b BC=bB C=bBC=b.
A light rigid rod A B C A B C ABCA B CABC has three particles each of mass m m mmm attached to it at A , B A , B A,BA, BA,B and C C CCC. The rod is struck by a blow P P PPP at right angles to it at a point distant from A A AAA equal to B C B C BCB CBC. Prove that the kinetic energy set up is 1 2 p 2 m a 2 − a b + b 2 a 2 + a b + b 2 1 2 p 2 m a 2 − a b + b 2 a 2 + a b + b 2 (1)/(2)(p^(2))/(m)(a^(2)-ab+b^(2))/(a^(2)+ab+b^(2))\frac{1}{2} \frac{p^2}{m} \frac{a^2-a b+b^2}{a^2+a b+b^2}12p2ma2−ab+b2a2+ab+b2, where A B = a A B = a AB=aA B=aAB=a and B C = b B C = b BC=bB C=bBC=b.
6(a) प्राचल विचरण विधि का प्रयोग करके, निम्न अवकल समीकरण का हल निकालिए, यदि y = e − x y = e − x y=e^(-x)y=e^{-x}y=e−x, पूरक फलन (CF) का एक हल है :
y ′ ′ + ( 1 − cot ⁡ x ) y ′ − y cot ⁡ x = sin 2 ⁡ x y ′ ′ + ( 1 − cot ⁡ x ) y ′ − y cot ⁡ x = sin 2 ⁡ x y^(”)+(1-cot x)y^(‘)-y cot x=sin^(2)xy^{\prime \prime}+(1-\cot x) y^{\prime}-y \cot x=\sin ^2 xy′′+(1−cot⁡x)y′−ycot⁡x=sin2⁡x
Using the method of variation of parameters, solve the differential equation y ′ ′ + ( 1 − cot ⁡ x ) y ′ − y cot ⁡ x = sin 2 ⁡ x y ′ ′ + ( 1 − cot ⁡ x ) y ′ − y cot ⁡ x = sin 2 ⁡ x y^(”)+(1-cot x)y^(‘)-y cot x=sin^(2)xy^{\prime \prime}+(1-\cot x) y^{\prime}-y \cot x=\sin ^2 xy′′+(1−cot⁡x)y′−ycot⁡x=sin2⁡x, if y = e − x y = e − x y=e^(-x)y=e^{-x}y=e−x is one solution of CF.
(b) दिए गए सदिश फलन A ¯ A ¯ bar(A)\bar{A}A¯, जहाँ A ¯ = ( 3 x 2 + 6 y ) i ^ − 14 y z j ^ + 20 x z 2 k ^ A ¯ = 3 x 2 + 6 y i ^ − 14 y z j ^ + 20 x z 2 k ^ bar(A)=(3x^(2)+6y) hat(i)-14 yz hat(j)+20 xz^(2) hat(k)\bar{A}=\left(3 x^2+6 y\right) \hat{i}-14 y z \hat{j}+20 x z^2 \hat{k}A¯=(3x2+6y)i^−14yzj^+20xz2k^, के लिए ∫ C A ¯ ⋅ d r ¯ ∫ C   A ¯ ⋅ d r ¯ int _(C) bar(A)*d bar(r)\int_C \bar{A} \cdot d \bar{r}∫CA¯⋅dr¯ का मान निकालिए, जहाँ C C CCC बिंदु ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) से ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) तक निम्न पर्थों से निर्देशित है :
(i) x = t , y = t 2 , z = t 3 x = t , y = t 2 , z = t 3 x=t,y=t^(2),z=t^(3)x=t, y=t^2, z=t^3x=t,y=t2,z=t3
(ii) सरल रेखा ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) से ( 1 , 0 , 0 ) ( 1 , 0 , 0 ) (1,0,0)(1,0,0)(1,0,0) तक जोड़ने पर, फिर ( 1 , 1 , 0 ) ( 1 , 1 , 0 ) (1,1,0)(1,1,0)(1,1,0) तक तथा फिर ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) तक
(iii) सरल रेखा ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) से ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) तक जोड़ने पर
क्या सभी स्थितियों में परिणाम समान हैं? कारण की व्याख्या कीजिए।
For the vector function A ¯ A ¯ bar(A)\bar{A}A¯, where A ¯ = ( 3 x 2 + 6 y ) i ^ − 14 y z j ^ + 20 x z 2 k ^ A ¯ = 3 x 2 + 6 y i ^ − 14 y z j ^ + 20 x z 2 k ^ bar(A)=(3x^(2)+6y) hat(i)-14 yz hat(j)+20 xz^(2) hat(k)\bar{A}=\left(3 x^2+6 y\right) \hat{i}-14 y z \hat{j}+20 x z^2 \hat{k}A¯=(3x2+6y)i^−14yzj^+20xz2k^, calculate ∫ C A ¯ ⋅ d r ¯ ∫ C   A ¯ ⋅ d r ¯ int _(C) bar(A)*d bar(r)\int_C \bar{A} \cdot d \bar{r}∫CA¯⋅dr¯ from ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) to ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) along the following paths :
(i) x = t , y = t 2 , z = t 3 x = t , y = t 2 , z = t 3 x=t,y=t^(2),z=t^(3)x=t, y=t^2, z=t^3x=t,y=t2,z=t3
(ii) Straight lines joining ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) to ( 1 , 0 , 0 ) ( 1 , 0 , 0 ) (1,0,0)(1,0,0)(1,0,0), then to ( 1 , 1 , 0 ) ( 1 , 1 , 0 ) (1,1,0)(1,1,0)(1,1,0) and then to ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1)
(iii) Straight line joining ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) to ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1)
Is the result same in all the cases? Explain the reason.
(c) एक दंड A D A D ADA DAD दो आलंब B B BBB एवं C C CCC पर विश्राम करता है, जबकि A B = B C = C D A B = B C = C D AB=BC=CDA B=B C=C DAB=BC=CD. यह पाया गया कि दंड झुक जाएगा यदि एक भार p k g p k g pkgp \mathrm{~kg}p kg, बिंदु A A AAA से लटकाया जाए या एक भार q k g q k g qkgq \mathrm{~kg}q kg, बिंदु D D DDD से लटकाया जाए। दंड का भार बताइए।
A beam A D A D ADA DAD rests on two supports B B BBB and C C CCC, where A B = B C = C D A B = B C = C D AB=BC=CDA B=B C=C DAB=BC=CD. It is found that the beam will tilt when a weight of p k g p k g pkgp \mathrm{~kg}p kg is hung from A A AAA or when a weight of q k g q k g qkgq \mathrm{~kg}q kg is hung from D D DDD. Find the weight of the beam.
7(a) स्टोक्स प्रमेय को सत्यापित कीजिए, जबकि सदिश क्षेत्र F ¯ = x y i ^ + y z j ^ + x z k ^ F ¯ = x y i ^ + y z j ^ + x z k ^ bar(F)=xy hat(i)+yz hat(j)+xz hat(k)\bar{F}=x y \hat{i}+y z \hat{j}+x z \hat{k}F¯=xyi^+yzj^+xzk^ एक सतह S S SSS पर है जो कि एक बेलन z = 1 − x 2 , 0 ≤ x ≤ 1 , − 2 ≤ y ≤ 2 z = 1 − x 2 , 0 ≤ x ≤ 1 , − 2 ≤ y ≤ 2 z=1-x^(2),0 <= x <= 1,-2 <= y <= 2z=1-x^2, 0 \leq x \leq 1,-2 \leq y \leq 2z=1−x2,0≤x≤1,−2≤y≤2 का हिस्सा है, जहाँ S S SSS उपरिमुखी अभिविन्यस्त है।
Verify the Stokes’ theorem for the vector field F ¯ = x y i ^ + y z j ^ + x z k ^ F ¯ = x y i ^ + y z j ^ + x z k ^ bar(F)=xy hat(i)+yz hat(j)+xz hat(k)\bar{F}=x y \hat{i}+y z \hat{j}+x z \hat{k}F¯=xyi^+yzj^+xzk^ on the surface S S SSS which is the part of the cylinder z = 1 − x 2 z = 1 − x 2 z=1-x^(2)z=1-x^2z=1−x2 for 0 ≤ x ≤ 1 , − 2 ≤ y ≤ 2 0 ≤ x ≤ 1 , − 2 ≤ y ≤ 2 0 <= x <= 1,-2 <= y <= 20 \leq x \leq 1,-2 \leq y \leq 20≤x≤1,−2≤y≤2; S S SSS is oriented upwards.
(b) लाप्लास रूपांतरण का प्रयोग करके प्रारंभिक मान समस्या t y ′ ′ + 2 t y ′ + 2 y = 2 ; y ( 0 ) = 1 t y ′ ′ + 2 t y ′ + 2 y = 2 ; y ( 0 ) = 1 ty^(”)+2ty^(‘)+2y=2;y(0)=1t y^{\prime \prime}+2 t y^{\prime}+2 y=2 ; y(0)=1ty′′+2ty′+2y=2;y(0)=1 तथा y ′ ( 0 ) y ′ ( 0 ) y^(‘)(0)y^{\prime}(0)y′(0) स्वेच्छ है, को हल कीजिए। क्या इस प्रश्न का हल अद्वितीय है?
Using Laplace transform, solve the initial value problem t y ′ ′ + 2 t y ′ + 2 y = 2 t y ′ ′ + 2 t y ′ + 2 y = 2 ty^(”)+2ty^(‘)+2y=2t y^{\prime \prime}+2 t y^{\prime}+2 y=2ty′′+2ty′+2y=2; y ( 0 ) = 1 y ( 0 ) = 1 y(0)=1y(0)=1y(0)=1 and y ′ ( 0 ) y ′ ( 0 ) y^(‘)(0)y^{\prime}(0)y′(0) is arbitrary. Does this problem have a unique solution?
(c) (i) चार एकसमान भारी छड़, जो समान भार W W WWW की हैं, एक वर्ग के रूप में ढाँचा बनाते हुए जुड़ी हैं। यह एक कोने से टँगा हुआ है। भार W W WWW तीर्नों नीचे वाले हरेक कोने से लटकाए हैं। वर्ग का आकार एक हल्की छड़, जो क्षैतिज विकर्ण के अनुदिश है, द्वारा रक्षित किया गया है। उस हल्की छड़ पर प्रणोद निकालिए।
(ii) एक कण वेग V V VVV से लंबी दूरी तय करने के लिए चलना शुरू करता है। एक तारे के केंद्र से कण के प्रारंभिक पथ की स्पर्श-रेखा पर लंबवत् दूरी p p ppp है। दिखाइए कि कण की तारे के केंद्र से न्यूनतम दूरी λ λ lambda\lambdaλ है, जहाँ V 2 λ = μ 2 + p 2 V 4 − μ V 2 λ = μ 2 + p 2 V 4 − μ V^(2)lambda=sqrt(mu^(2)+p^(2)V^(4))-muV^2 \lambda=\sqrt{\mu^2+p^2 V^4}-\muV2λ=μ2+p2V4−μ. यहाँ μ μ mu\muμ एक अचर है।
(i) A square framework formed of uniform heavy rods of equal weight W W WWW jointed together, is hung up by one corner. A weight W W WWW is suspended from each of the three lower corners, and the shape of the square is preserved by a light rod along the horizontal diagonal. Find the thrust of the light rod.
(ii) A particle starts at a great distance with velocity V V VVV. Let p p ppp be the length of the perpendicular from the centre of a star on the tangent to the initial path of the particle. Show that the least distance of the particle from the centre of the star is λ λ lambda\lambdaλ, where V 2 λ = μ 2 + p 2 V 4 − μ V 2 λ = μ 2 + p 2 V 4 − μ V^(2)lambda=sqrt(mu^(2)+p^(2)V^(4))-muV^2 \lambda=\sqrt{\mu^2+p^2 V^4}-\muV2λ=μ2+p2V4−μ. Here μ μ mu\muμ is a constant.
8(a) (i) निम्न अवकल समीकरण हल कीजिए :
( x + 1 ) 2 y ′ ′ − 4 ( x + 1 ) y ′ + 6 y = 6 ( x + 1 ) 2 + sin ⁡ log ⁡ ( x + 1 ) ( x + 1 ) 2 y ′ ′ − 4 ( x + 1 ) y ′ + 6 y = 6 ( x + 1 ) 2 + sin ⁡ log ⁡ ( x + 1 ) (x+1)^(2)y^(”)-4(x+1)y^(‘)+6y=6(x+1)^(2)+sin log(x+1)(x+1)^2 y^{\prime \prime}-4(x+1) y^{\prime}+6 y=6(x+1)^2+\sin \log (x+1)(x+1)2y′′−4(x+1)y′+6y=6(x+1)2+sin⁡log⁡(x+1)
(ii) अवकल समीकरण 9 p 2 ( 2 − y ) 2 = 4 ( 3 − y ) 9 p 2 ( 2 − y ) 2 = 4 ( 3 − y ) 9p^(2)(2-y)^(2)=4(3-y)9 p^2(2-y)^2=4(3-y)9p2(2−y)2=4(3−y) के व्यापक व विचित्र हल निकालिए, जहाँ p = d y d x p = d y d x p=(dy)/(dx)p=\frac{d y}{d x}p=dydx.
(i) Solve the following differential equation :
( x + 1 ) 2 y ′ ′ − 4 ( x + 1 ) y ′ + 6 y = 6 ( x + 1 ) 2 + sin ⁡ log ⁡ ( x + 1 ) ( x + 1 ) 2 y ′ ′ − 4 ( x + 1 ) y ′ + 6 y = 6 ( x + 1 ) 2 + sin ⁡ log ⁡ ( x + 1 ) (x+1)^(2)y^(”)-4(x+1)y^(‘)+6y=6(x+1)^(2)+sin log(x+1)(x+1)^2 y^{\prime \prime}-4(x+1) y^{\prime}+6 y=6(x+1)^2+\sin \log (x+1)(x+1)2y′′−4(x+1)y′+6y=6(x+1)2+sin⁡log⁡(x+1)
(ii) Find the general and singular solutions of the differential equation 9 p 2 ( 2 − y ) 2 = 4 ( 3 − y ) 9 p 2 ( 2 − y ) 2 = 4 ( 3 − y ) 9p^(2)(2-y)^(2)=4(3-y)9 p^2(2-y)^2=4(3-y)9p2(2−y)2=4(3−y), where p = d y d x p = d y d x p=(dy)/(dx)p=\frac{d y}{d x}p=dydx
(b) पृष्ठ समाकल ∬ S ∇ × F ¯ ⋅ n ^ d S ∬ S   ∇ × F ¯ ⋅ n ^ d S ∬_(S)grad xx bar(F)* hat(n)dS\iint_S \nabla \times \bar{F} \cdot \hat{n} d S∬S∇×F¯⋅n^dS का मान निकालिए, जहाँ F ¯ = y i ^ + ( x − 2 x ) j ^ − x y k ^ F ¯ = y i ^ + ( x − 2 x ) j ^ − x y k ^ bar(F)=y hat(i)+(x-2x) hat(j)-xy hat(k)\bar{F}=y \hat{i}+(x-2 x) \hat{j}-x y \hat{k}F¯=yi^+(x−2x)j^−xyk^ तथा S S SSS गोले x 2 + y 2 + z 2 = a 2 x 2 + y 2 + z 2 = a 2 x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2x2+y2+z2=a2 की सतह है, जो x y x y xyx yxy-तल के ऊपर है।
Evaluate the surface integral ∬ S ∇ × F ¯ ⋅ n ^ d S ∬ S   ∇ × F ¯ ⋅ n ^ d S ∬_(S)grad xx bar(F)* hat(n)dS\iint_S \nabla \times \bar{F} \cdot \hat{n} d S∬S∇×F¯⋅n^dS for F ¯ = y i ^ + ( x − 2 x z ) j ^ − x y k ^ F ¯ = y i ^ + ( x − 2 x z ) j ^ − x y k ^ bar(F)=y hat(i)+(x-2xz) hat(j)-xy hat(k)\bar{F}=y \hat{i}+(x-2 x z) \hat{j}-x y \hat{k}F¯=yi^+(x−2xz)j^−xyk^ and S S SSS is the surface of the sphere x 2 + y 2 + z 2 = a 2 x 2 + y 2 + z 2 = a 2 x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2x2+y2+z2=a2 above the x y x y xyx yxy-plane.
(c) एक चार पहियों वाला रेलवे ट्रक, जिसका कुल द्रव्यमान M M MMM है, के पहिए और धुरी के हर युग्म का द्रव्यमान व परिभ्रमण त्रिज्या क्रमशः m m mmm तथा k k kkk है। हर पहिए की त्रिज्या r r rrr है। यदि ट्रक को बल P P PPP द्वारा सीधे पथ (ट्रैक) पर धकेला जाता है, तब सिद्ध कीजिए कि उसका त्वरण P M + 2 m k 2 r 2 P M + 2 m k 2 r 2 (P)/(M+(2mk^(2))/(r^(2)))\frac{P}{M+\frac{2 m k^2}{r^2}}PM+2mk2r2 है तथा ट्रक द्वारा प्रत्येक धुरी पर लगाए गए क्षैतिज बल का मान ज्ञात कीजिए। धुरी घर्षण व हवा का प्रतिरोध नगण्य है।
A four-wheeled railway truck has a total mass M M MMM, the mass and radius of gyration of each pair of wheels and axle are m m mmm and k k kkk respectively, and the radius of each wheel is r r rrr. Prove that if the truck is propelled along a level track by a force P P PPP, the acceleration is P M + 2 m k 2 r 2 P M + 2 m k 2 r 2 (P)/(M+(2mk^(2))/(r^(2)))\frac{P}{M+\frac{2 m k^2}{r^2}}PM+2mk2r2, and find the horizontal force exerted on each axle by the truck. The axle friction and wind resistance are to be neglected.
upsc-m2019-1-28d83552-ba33-4aa8-99e9-b6fbf8522ea0
खण्ड-A / SECTION-A
1. (a) माना कि
f : [ 0 , π 2 ] → R f : 0 , π 2 → R f:[0,(pi)/(2)]rarrRf:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}f:[0,π2]→R एक संतत फलन है, जैसा कि
f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f(x)=(cos^(2)x)/(4x^(2)-pi^(2)),quad0 <= x < (pi)/(2)f(x)=\frac{\cos ^2 x}{4 x^2-\pi^2}, \quad 0 \leq x<\frac{\pi}{2}f(x)=cos2⁡x4x2−π2,0≤x<π2
f ( π 2 ) f π 2 f((pi)/(2))f\left(\frac{\pi}{2}\right)f(π2) का मान ज्ञात कीजिए।
Let f : [ 0 , π 2 ] → R f : 0 , π 2 → R f:[0,(pi)/(2)]rarrRf:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}f:[0,π2]→R be a continuous function such that
f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f(x)=(cos^(2)x)/(4x^(2)-pi^(2)),quad0 <= x < (pi)/(2)f(x)=\frac{\cos ^2 x}{4 x^2-\pi^2}, \quad 0 \leq x<\frac{\pi}{2}f(x)=cos2⁡x4x2−π2,0≤x<π2
Find the value of f ( π 2 ) f π 2 f((pi)/(2))f\left(\frac{\pi}{2}\right)f(π2).
(b) माना कि f : D ( ⊆ R 2 ) → R f : D ⊆ R 2 → R f:D(subeR^(2))rarrRf: D\left(\subseteq \mathbb{R}^2\right) \rightarrow \mathbb{R}f:D(⊆R2)→R एक फलन है और ( a , b ) ∈ D ( a , b ) ∈ D (a,b)in D(a, b) \in D(a,b)∈D. अगर f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) बिंदु ( a , b ) ( a , b ) (a,b)(a, b)(a,b) पर संतत है, तो दर्शाइए कि फलन f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) और f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) क्रमशः x = a x = a x=ax=ax=a और y = b y = b y=by=by=b पर संतत हैं।
Let f : D ( ⊆ R 2 ) → R f : D ⊆ R 2 → R f:D(subeR^(2))rarrRf: D\left(\subseteq \mathbb{R}^2\right) \rightarrow \mathbb{R}f:D(⊆R2)→R be a function and ( a , b ) ∈ D ( a , b ) ∈ D (a,b)in D(a, b) \in D(a,b)∈D. If f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is continuous at ( a , b ) ( a , b ) (a,b)(a, b)(a,b), then show that the functions f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) and f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) are continuous at x = a x = a x=ax=ax=a and at y = b y = b y=by=by=b respectively.
(c) माना कि T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 एक रैखिक प्रतिचित्र है, जैसा कि T ( 2 , 1 ) = ( 5 , 7 ) T ( 2 , 1 ) = ( 5 , 7 ) T(2,1)=(5,7)T(2,1)=(5,7)T(2,1)=(5,7) एवं T ( 1 , 2 ) = ( 3 , 3 ) T ( 1 , 2 ) = ( 3 , 3 ) T(1,2)=(3,3)T(1,2)=(3,3)T(1,2)=(3,3). अगर A A AAA मानक आधारों e 1 , e 2 e 1 , e 2 e_(1),e_(2)e_1, e_2e1,e2 के सापेक्ष T T TTT के संगत आव्यूह है, तो A A AAA की कोटि ज्ञात कीजिए।
Let T : R 2 → R 2 T : R 2 → R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2→R2 be a linear map such that T ( 2 , 1 ) = ( 5 , 7 ) T ( 2 , 1 ) = ( 5 , 7 ) T(2,1)=(5,7)T(2,1)=(5,7)T(2,1)=(5,7) and T ( 1 , 2 ) = ( 3 , 3 ) T ( 1 , 2 ) = ( 3 , 3 ) T(1,2)=(3,3)T(1,2)=(3,3)T(1,2)=(3,3).
If A A AAA is the matrix corresponding to T T TTT with respect to the standard bases e 1 , e 2 e 1 , e 2 e_(1),e_(2)e_1, e_2e1,e2, then find Rank ⁡ ( A ) Rank ⁡ ( A ) Rank(A)\operatorname{Rank}(A)Rank⁡(A).
(d) अगर
A = [ 1 2 1 1 − 4 1 3 0 − 3 ] और B = [ 2 1 1 1 − 1 0 2 1 − 1 ] A = 1      2      1 1      − 4      1 3      0      − 3  और  B = 2      1      1 1      − 1      0 2      1      − 1 A=[[1,2,1],[1,-4,1],[3,0,-3]]” और “B=[[2,1,1],[1,-1,0],[2,1,-1]]A=\left[\begin{array}{rrr} 1 & 2 & 1 \\ 1 & -4 & 1 \\ 3 & 0 & -3 \end{array}\right] \text { और } B=\left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & -1 & 0 \\ 2 & 1 & -1 \end{array}\right]A=[1211−4130−3] और B=[2111−1021−1]
है, तो दर्शाइए कि A B = 6 I 3 A B = 6 I 3 AB=6I_(3)A B=6 I_3AB=6I3. इस परिणाम का उपयोग करते हुए निम्नलिखित समीकरण निकाय को हल कीजिए :
2 x + y + z = 5 x − y = 0 2 x + y − z = 1 2 x + y + z = 5 x − y = 0 2 x + y − z = 1 {:[2x+y+z=5],[x-y=0],[2x+y-z=1]:}\begin{array}{r} 2 x+y+z=5 \\ x-y=0 \\ 2 x+y-z=1 \end{array}2x+y+z=5x−y=02x+y−z=1
If
A = [ 1 2 1 1 − 4 1 3 0 − 3 ] and B = [ 2 1 1 1 − 1 0 2 1 − 1 ] A = 1      2      1 1      − 4      1 3      0      − 3  and  B = 2      1      1 1      − 1      0 2      1      − 1 A=[[1,2,1],[1,-4,1],[3,0,-3]]quad” and “B=[[2,1,1],[1,-1,0],[2,1,-1]]A=\left[\begin{array}{rrr} 1 & 2 & 1 \\ 1 & -4 & 1 \\ 3 & 0 & -3 \end{array}\right] \quad \text { and } B=\left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & -1 & 0 \\ 2 & 1 & -1 \end{array}\right]A=[1211−4130−3] and B=[2111−1021−1]
then show that A B = 6 I 3 A B = 6 I 3 AB=6I_(3)A B=6 I_3AB=6I3. Use this result to solve the following system of equations :
2 x + y + z = 5 x − y = 0 2 x + y − z = 1 2 x + y + z = 5 x − y = 0 2 x + y − z = 1 {:[2x+y+z=5],[x-y=0],[2x+y-z=1]:}\begin{array}{r} 2 x+y+z=5 \\ x-y=0 \\ 2 x+y-z=1 \end{array}2x+y+z=5x−y=02x+y−z=1
(e) दर्शाइए कि
x + 1 − 3 = y − 3 2 = z + 2 1 और x 1 = y − 7 − 3 = z + 7 2 x + 1 − 3 = y − 3 2 = z + 2 1  और  x 1 = y − 7 − 3 = z + 7 2 (x+1)/(-3)=(y-3)/(2)=(z+2)/(1)” और “(x)/(1)=(y-7)/(-3)=(z+7)/(2)\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} \text { और } \frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}x+1−3=y−32=z+21 और x1=y−7−3=z+72
प्रतिच्छेदी रेखाएँ हैं। प्रतिच्छेद बिंदु के निर्देशांकों और उस समतल, जिसमें दोनों रेखाएँ हैं, का समीकरण ज्ञात कीजिए।
Show that the lines
x + 1 − 3 = y − 3 2 = z + 2 1 and x 1 = y − 7 − 3 = z + 7 2 x + 1 − 3 = y − 3 2 = z + 2 1  and  x 1 = y − 7 − 3 = z + 7 2 (x+1)/(-3)=(y-3)/(2)=(z+2)/(1)” and “(x)/(1)=(y-7)/(-3)=(z+7)/(2)\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} \text { and } \frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}x+1−3=y−32=z+21 and x1=y−7−3=z+72
intersect. Find the coordinates of the point of intersection and the equation of the plane containing them.
2(a) क्या f ( x ) = | cos ⁡ x | + | sin ⁡ x | , x = π 2 f ( x ) = | cos ⁡ x | + | sin ⁡ x | , x = π 2 f(x)=|cos x|+|sin x|,x=(pi)/(2)f(x)=|\cos x|+|\sin x|, x=\frac{\pi}{2}f(x)=|cos⁡x|+|sin⁡x|,x=π2 पर अवकलनीय है? अगर आपका उत्तर हाँ है, तो f ( x ) f ( x ) f(x)f(x)f(x) का अवकलज x = π 2 x = π 2 x=(pi)/(2)x=\frac{\pi}{2}x=π2 पर ज्ञात कीजिए। अगर आपका उत्तर ना है, तो अपने उत्तर का प्रमाण दीजिए।
Is f ( x ) = | cos ⁡ x | + | sin ⁡ x | f ( x ) = | cos ⁡ x | + | sin ⁡ x | f(x)=|cos x|+|sin x|f(x)=|\cos x|+|\sin x|f(x)=|cos⁡x|+|sin⁡x| differentiable at x = π 2 x = π 2 x=(pi)/(2)x=\frac{\pi}{2}x=π2 ? If yes, then find its derivative at x = π 2 x = π 2 x=(pi)/(2)x=\frac{\pi}{2}x=π2. If no, then give a proof of it.
(b) माना कि A A AAA और B B BBB समान कोटि के दो लांबिक आव्यूह हैं तथा det ⁡ A + det ⁡ B = 0 det ⁡ A + det ⁡ B = 0 det A+det B=0\operatorname{det} A+\operatorname{det} B=0det⁡A+det⁡B=0. दर्शाइए कि A + B A + B A+BA+BA+B एक अव्युत्क्रमणीय (सिंगुलर) आव्यूह है।
Let A A AAA and B B BBB be two orthogonal matrices of same order and det ⁡ A + det ⁡ B = 0 det ⁡ A + det ⁡ B = 0 det A+det B=0\operatorname{det} A+\operatorname{det} B=0det⁡A+det⁡B=0. Show that A + B A + B A+BA+BA+B is a singular matrix.
(c) (i) समतल x + 2 y + 3 z = 12 x + 2 y + 3 z = 12 x+2y+3z=12x+2 y+3 z=12x+2y+3z=12 निर्देशांक अक्षों को A , B , C A , B , C A,B,CA, B, CA,B,C पर प्रतिच्छेद करता है। त्रिभुज A B C A B C ABCA B CABC के परिवृत्त का समीकरण ज्ञात कीजिए।
(ii) सिद्ध कीजिए कि समतल z = 0 z = 0 z=0z=0z=0 गोलक x 2 + y 2 + z 2 = 11 x 2 + y 2 + z 2 = 11 x^(2)+y^(2)+z^(2)=11x^2+y^2+z^2=11x2+y2+z2=11 के अन्वालोपी शंकु, जिसका शीर्ष ( 2 , 4 , 1 ) ( 2 , 4 , 1 ) (2,4,1)(2,4,1)(2,4,1) पर है, को एक समकोणीय अतिपरवलय पर प्रतिच्छेद करता है।
(i) The plane x + 2 y + 3 z = 12 x + 2 y + 3 z = 12 x+2y+3z=12x+2 y+3 z=12x+2y+3z=12 cuts the axes of coordinates in A , B , C A , B , C A,B,CA, B, CA,B,C. Find the equations of the circle circumscribing the triangle A B C A B C ABCA B CABC.
(ii) Prove that the plane z = 0 z = 0 z=0z=0z=0 cuts the enveloping cone of the sphere x 2 + y 2 + z 2 = 11 x 2 + y 2 + z 2 = 11 x^(2)+y^(2)+z^(2)=11x^2+y^2+z^2=11x2+y2+z2=11 which has the vertex at ( 2 , 4 , 1 ) ( 2 , 4 , 1 ) (2,4,1)(2,4,1)(2,4,1) in a rectangular hyperbola.
3(a) फलन f ( x ) = 2 x 3 − 9 x 2 + 12 x + 6 f ( x ) = 2 x 3 − 9 x 2 + 12 x + 6 f(x)=2x^(3)-9x^(2)+12 x+6f(x)=2 x^3-9 x^2+12 x+6f(x)=2x3−9x2+12x+6 का अंतराल [ 2 , 3 ] [ 2 , 3 ] [2,3][2,3][2,3] पर अधिकतम और न्यूनतम मान ज्ञात कीजिए।
Find the maximum and the minimum value of the function f ( x ) = 2 x 3 − 9 x 2 + 12 x + 6 f ( x ) = 2 x 3 − 9 x 2 + 12 x + 6 f(x)=2x^(3)-9x^(2)+12 x+6f(x)=2 x^3-9 x^2+12 x+6f(x)=2x3−9x2+12x+6 on the interval [ 2 , 3 ] [ 2 , 3 ] [2,3][2,3][2,3].
(b) सिद्ध कीजिए कि साधारणतः किसी एक बिंदु से परवलयज x 2 + y 2 = 2 a z x 2 + y 2 = 2 a z x^(2)+y^(2)=2azx^2+y^2=2 a zx2+y2=2az पर तीन अभिलंब बनाए जा सकते हैं, लेकिन अगर बिंदु सतह 27 a ( x 2 + y 2 ) + 8 ( a − z ) 3 = 0 27 a x 2 + y 2 + 8 ( a − z ) 3 = 0 27 a(x^(2)+y^(2))+8(a-z)^(3)=027 a\left(x^2+y^2\right)+8(a-z)^3=027a(x2+y2)+8(a−z)3=0 पर स्थित है, तो इन तीन अभिलंबों में से दो अभिलंब एक ही हैं।
Prove that, in general, three normals can be drawn from a given point to the paraboloid x 2 + y 2 = 2 a z x 2 + y 2 = 2 a z x^(2)+y^(2)=2azx^2+y^2=2 a zx2+y2=2az, but if the point lies on the surface
27 a ( x 2 + y 2 ) + 8 ( a − z ) 3 = 0 27 a x 2 + y 2 + 8 ( a − z ) 3 = 0 27 a(x^(2)+y^(2))+8(a-z)^(3)=027 a\left(x^2+y^2\right)+8(a-z)^3=027a(x2+y2)+8(a−z)3=0
then two of the three normals coincide.
(c) माना कि
A = ( 5 7 2 1 1 1 − 8 1 2 3 5 0 3 4 − 3 1 ) A = 5      7      2      1 1      1      − 8      1 2      3      5      0 3      4      − 3      1 A=([5,7,2,1],[1,1,-8,1],[2,3,5,0],[3,4,-3,1])A=\left(\begin{array}{rrrr} 5 & 7 & 2 & 1 \\ 1 & 1 & -8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & -3 & 1 \end{array}\right)A=(572111−81235034−31)
(i) आव्यूह A A AAA की कोटि ज्ञात कीजिए।
(ii) उपसमष्टि
V = { ( x 1 , x 2 , x 3 , x 4 ) ∈ R 4 ∣ A ( x 1 x 2 x 3 x 4 ) = 0 } V = x 1 , x 2 , x 3 , x 4 ∈ R 4 ∣ A x 1 x 2 x 3 x 4 = 0 V={(x_(1),x_(2),x_(3),x_(4))inR^(4)∣A([x_(1)],[x_(2)],[x_(3)],[x_(4)])=0}V=\left\{\left(x_1, x_2, x_3, x_4\right) \in \mathbb{R}^4 \mid A\left(\begin{array}{l} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right)=0\right\}V={(x1,x2,x3,x4)∈R4∣A(x1x2x3x4)=0}
की विमा ज्ञात कीजिए।
Let
A = ( 5 7 2 1 1 1 − 8 1 2 3 5 0 3 4 − 3 1 ) A = 5      7      2      1 1      1      − 8      1 2      3      5      0 3      4      − 3      1 A=([5,7,2,1],[1,1,-8,1],[2,3,5,0],[3,4,-3,1])A=\left(\begin{array}{rrrr} 5 & 7 & 2 & 1 \\ 1 & 1 & -8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & -3 & 1 \end{array}\right)A=(572111−81235034−31)
(i) Find the rank of matrix A A AAA.
(ii) Find the dimension of the subspace
V = { ( x 1 , x 2 , x 3 , x 4 ) ∈ R 4 ∣ A ( x 1 x 2 x 3 x 4 ) = 0 } V = x 1 , x 2 , x 3 , x 4 ∈ R 4 ∣ A x 1 x 2 x 3 x 4 = 0 V={(x_(1),x_(2),x_(3),x_(4))inR^(4)∣A([x_(1)],[x_(2)],[x_(3)],[x_(4)])=0}V=\left\{\left(x_1, x_2, x_3, x_4\right) \in \mathbb{R}^4 \mid A\left(\begin{array}{l} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right)=0\right\}V={(x1,x2,x3,x4)∈R4∣A(x1x2x3x4)=0}
4(a) कैले-हैमिल्टन प्रमेय का कथन लिखिए। इस प्रमेय का उपयोग करके A 100 A 100 A^(100)A^{100}A100 का मान ज्ञात कीजिए, जहाँ
A = [ 1 0 0 1 0 1 0 1 0 ] A = 1      0      0 1      0      1 0      1      0 A=[[1,0,0],[1,0,1],[0,1,0]]A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]A=[100101010]
State the Cayley-Hamilton theorem. Use this theorem to find A 100 A 100 A^(100)A^{100}A100, where
A = [ 1 0 0 1 0 1 0 1 0 ] A = 1      0      0 1      0      1 0      1      0 A=[[1,0,0],[1,0,1],[0,1,0]]A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]A=[100101010]
(b) बिंदु P P PPP से गुजरने वाली दीर्घवृत्तज
x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1x2a2+y2b2+z2c2=1
की अभिलंब जीवा की लंबाई ज्ञात कीजिए और सिद्ध कीजिए कि अगर यह 4 P G 3 4 P G 3 4PG_(3)4 P G_34PG3 के समान है, जहाँ G 3 G 3 G_(3)G_3G3 वह बिंदु है जहाँ P P PPP से गुजरने वाली अभिलंब जीवा x y x y xyx yxy-तल पर मिलती है, तो P P PPP शंकु
x 2 a 6 ( 2 c 2 − a 2 ) + y 2 b 6 ( 2 c 2 − b 2 ) + z 2 c 4 = 0 x 2 a 6 2 c 2 − a 2 + y 2 b 6 2 c 2 − b 2 + z 2 c 4 = 0 (x^(2))/(a^(6))(2c^(2)-a^(2))+(y^(2))/(b^(6))(2c^(2)-b^(2))+(z^(2))/(c^(4))=0\frac{x^2}{a^6}\left(2 c^2-a^2\right)+\frac{y^2}{b^6}\left(2 c^2-b^2\right)+\frac{z^2}{c^4}=0x2a6(2c2−a2)+y2b6(2c2−b2)+z2c4=0
पर स्थित है।
Find the length of the normal chord through a point P P PPP of the ellipsoid
x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1x2a2+y2b2+z2c2=1
and prove that if it is equal to 4 P G 3 4 P G 3 4PG_(3)4 P G_34PG3, where G 3 G 3 G_(3)G_3G3 is the point where the normal chord through P P PPP meets the x y x y xyx yxy-plane, then P P PPP lies on the cone
x 2 a 6 ( 2 c 2 − a 2 ) + y 2 b 6 ( 2 c 2 − b 2 ) + z 2 c 4 = 0 x 2 a 6 2 c 2 − a 2 + y 2 b 6 2 c 2 − b 2 + z 2 c 4 = 0 (x^(2))/(a^(6))(2c^(2)-a^(2))+(y^(2))/(b^(6))(2c^(2)-b^(2))+(z^(2))/(c^(4))=0\frac{x^2}{a^6}\left(2 c^2-a^2\right)+\frac{y^2}{b^6}\left(2 c^2-b^2\right)+\frac{z^2}{c^4}=0x2a6(2c2−a2)+y2b6(2c2−b2)+z2c4=0
(c) (i) अगर
u = sin − 1 ⁡ x 1 / 3 + y 1 / 3 x 1 / 2 + y 1 / 2 u = sin − 1 ⁡ x 1 / 3 + y 1 / 3 x 1 / 2 + y 1 / 2 u=sin^(-1)sqrt((x^(1//3)+y^(1//3))/(x^(1//2)+y^(1//2)))u=\sin ^{-1} \sqrt{\frac{x^{1 / 3}+y^{1 / 3}}{x^{1 / 2}+y^{1 / 2}}}u=sin−1⁡x1/3+y1/3x1/2+y1/2
है, तो दर्शाइए कि sin 2 ⁡ u , x sin 2 ⁡ u , x sin^(2)u,x\sin ^2 u, xsin2⁡u,x और y y yyy का − 1 6 − 1 6 -(1)/(6)-\frac{1}{6}−16 घातविशिष्ट समांगी फलन है। अतएव दर्शाइए कि
x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = tan ⁡ u 12 ( 13 12 + tan 2 ⁡ u 12 ) x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = tan ⁡ u 12 13 12 + tan 2 ⁡ u 12 x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=(tan u)/(12)((13)/(12)+(tan^(2)u)/(12))x^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}=\frac{\tan u}{12}\left(\frac{13}{12}+\frac{\tan ^2 u}{12}\right)x2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=tan⁡u12(1312+tan2⁡u12)
(ii) जैकोबियन विधि का व्यवहार करते हुए दर्शाइए कि अगर f ′ ( x ) = 1 1 + x 2 f ′ ( x ) = 1 1 + x 2 f^(‘)(x)=(1)/(1+x^(2))f^{\prime}(x)=\frac{1}{1+x^2}f′(x)=11+x2 और f ( 0 ) = 0 f ( 0 ) = 0 f(0)=0f(0)=0f(0)=0 है, तो
f ( x ) + f ( y ) = f ( x + y 1 − x y ) f ( x ) + f ( y ) = f x + y 1 − x y f(x)+f(y)=f((x+y)/(1-xy))f(x)+f(y)=f\left(\frac{x+y}{1-x y}\right)f(x)+f(y)=f(x+y1−xy)
(i) If
u = sin − 1 ⁡ x 1 / 3 + y 1 / 3 x 1 / 2 + y 1 / 2 u = sin − 1 ⁡ x 1 / 3 + y 1 / 3 x 1 / 2 + y 1 / 2 u=sin^(-1)sqrt((x^(1//3)+y^(1//3))/(x^(1//2)+y^(1//2)))u=\sin ^{-1} \sqrt{\frac{x^{1 / 3}+y^{1 / 3}}{x^{1 / 2}+y^{1 / 2}}}u=sin−1⁡x1/3+y1/3x1/2+y1/2
then show that sin 2 ⁡ u sin 2 ⁡ u sin^(2)u\sin ^2 usin2⁡u is a homogeneous function of x x xxx and y y yyy of degree − 1 6 − 1 6 -(1)/(6)-\frac{1}{6}−16.
Hence show that
x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = tan ⁡ u 12 ( 13 12 + tan 2 ⁡ u 12 ) x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = tan ⁡ u 12 13 12 + tan 2 ⁡ u 12 x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=(tan u)/(12)((13)/(12)+(tan^(2)u)/(12))x^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}=\frac{\tan u}{12}\left(\frac{13}{12}+\frac{\tan ^2 u}{12}\right)x2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=tan⁡u12(1312+tan2⁡u12)
(ii) Using the Jacobian method, show that if f ′ ( x ) = 1 1 + x 2 f ′ ( x ) = 1 1 + x 2 f^(‘)(x)=(1)/(1+x^(2))f^{\prime}(x)=\frac{1}{1+x^2}f′(x)=11+x2 and f ( 0 ) = 0 f ( 0 ) = 0 f(0)=0f(0)=0f(0)=0, then
f ( x ) + f ( y ) = f ( x + y 1 − x y ) f ( x ) + f ( y ) = f x + y 1 − x y f(x)+f(y)=f((x+y)/(1-xy))f(x)+f(y)=f\left(\frac{x+y}{1-x y}\right)f(x)+f(y)=f(x+y1−xy)
खण्ड-B / SECTION-B
5(a) अवकल समीकरण
( 2 y sin ⁡ x + 3 y 4 sin ⁡ x cos ⁡ x ) d x − ( 4 y 3 cos 2 ⁡ x + cos ⁡ x ) d y = 0 2 y sin ⁡ x + 3 y 4 sin ⁡ x cos ⁡ x d x − 4 y 3 cos 2 ⁡ x + cos ⁡ x d y = 0 (2y sin x+3y^(4)sin x cos x)dx-(4y^(3)cos^(2)x+cos x)dy=0\left(2 y \sin x+3 y^4 \sin x \cos x\right) d x-\left(4 y^3 \cos ^2 x+\cos x\right) d y=0(2ysin⁡x+3y4sin⁡xcos⁡x)dx−(4y3cos2⁡x+cos⁡x)dy=0
को हल कीजिए।
Solve the differential equation
( 2 y sin ⁡ x + 3 y 4 sin ⁡ x cos ⁡ x ) d x − ( 4 y 3 cos 2 ⁡ x + cos ⁡ x ) d y = 0 2 y sin ⁡ x + 3 y 4 sin ⁡ x cos ⁡ x d x − 4 y 3 cos 2 ⁡ x + cos ⁡ x d y = 0 (2y sin x+3y^(4)sin x cos x)dx-(4y^(3)cos^(2)x+cos x)dy=0\left(2 y \sin x+3 y^4 \sin x \cos x\right) d x-\left(4 y^3 \cos ^2 x+\cos x\right) d y=0(2ysin⁡x+3y4sin⁡xcos⁡x)dx−(4y3cos2⁡x+cos⁡x)dy=0
(b) अवकल समीकरण
d 2 y d x 2 − 4 d y d x + 4 y = 3 x 2 e 2 x sin ⁡ 2 x d 2 y d x 2 − 4 d y d x + 4 y = 3 x 2 e 2 x sin ⁡ 2 x (d^(2)y)/(dx^(2))-4(dy)/(dx)+4y=3x^(2)e^(2x)sin 2x\frac{d^2 y}{d x^2}-4 \frac{d y}{d x}+4 y=3 x^2 e^{2 x} \sin 2 xd2ydx2−4dydx+4y=3x2e2xsin⁡2x
का पूर्ण हल ज्ञात कीजिए।
Determine the complete solution of the differential equation
d 2 y d x 2 − 4 d y d x + 4 y = 3 x 2 e 2 x sin ⁡ 2 x d 2 y d x 2 − 4 d y d x + 4 y = 3 x 2 e 2 x sin ⁡ 2 x (d^(2)y)/(dx^(2))-4(dy)/(dx)+4y=3x^(2)e^(2x)sin 2x\frac{d^2 y}{d x^2}-4 \frac{d y}{d x}+4 y=3 x^2 e^{2 x} \sin 2 xd2ydx2−4dydx+4y=3x2e2xsin⁡2x
(c) एक भारी एकसमान छड़ A B A B ABA BAB का एक सिरा एक रूक्ष क्षैतिज छड़ A C A C ACA CAC, जिसके साथ वह वलय (रिंग) के द्वारा जुड़ी हुई है, पर सरक सकती है। B B BBB एवं C C CCC एक रस्सी से जुड़े हैं। जब छड़ सर्पण बिंदु पर है, तब A C 2 − A B 2 = B C 2 A C 2 − A B 2 = B C 2 AC^(2)-AB^(2)=BC^(2)A C^2-A B^2=B C^2AC2−AB2=BC2 है। यदि क्षैतिज रेखा व A B A B ABA BAB के बीच का कोण θ θ theta\thetaθ है, तो सिद्ध कीजिए कि घर्षण गुणांक cot ⁡ θ 2 + cot 2 ⁡ θ cot ⁡ θ 2 + cot 2 ⁡ θ (cot theta)/(2+cot^(2)theta)\frac{\cot \theta}{2+\cot ^2 \theta}cot⁡θ2+cot2⁡θ है।
One end of a heavy uniform rod A B A B ABA BAB can slide along a rough horizontal rod A C A C ACA CAC, to which it is attached by a ring. B B BBB and C C CCC are joined by a string. When the rod is on the point of sliding, then A C 2 − A B 2 = B C 2 A C 2 − A B 2 = B C 2 AC^(2)-AB^(2)=BC^(2)A C^2-A B^2=B C^2AC2−AB2=BC2. If θ θ theta\thetaθ is the angle between A B A B ABA BAB and the horizontal line, then prove that the coefficient of friction is cot ⁡ θ 2 + cot 2 ⁡ θ cot ⁡ θ 2 + cot 2 ⁡ θ (cot theta)/(2+cot^(2)theta)\frac{\cot \theta}{2+\cot ^2 \theta}cot⁡θ2+cot2⁡θ.
(d) एक कण का पृथ्वी द्वारा आकर्षण बल उस कण के पृथ्वी के केन्द्र से दूरी के वर्ग के व्युत्क्रमानुपाती है। एक कण, जिसका भार पृथ्वी की सतह पर W W WWW है, सतह से 3 h 3 h 3h3 h3h ऊँचाई से पृथ्वी की सतह पर गिरता है। दर्शाइए कि पृथ्वी के आकर्षण बल द्वारा किए गए कार्य का परिमाण 3 4 h W 3 4 h W (3)/(4)hW\frac{3}{4} h W34hW है, जहाँ h h hhh पृथ्वी की त्रिज्या है।
The force of attraction of a particle by the earth is inversely proportional to the square of its distance from the earth’s centre. A particle, whose weight on the surface of the earth is W W WWW, falls to the surface of the earth from a height 3 h 3 h 3h3 h3h above it. Show that the magnitude of work done by the earth’s attraction force is 3 4 h W 3 4 h W (3)/(4)hW\frac{3}{4} h W34hW, where h h hhh is the radius of the earth.
(e) वक्र x = t , y = t 2 , z = t 3 x = t , y = t 2 , z = t 3 x=t,y=t^(2),z=t^(3)x=t, y=t^2, z=t^3x=t,y=t2,z=t3 के बिंदु ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) पर स्पर्श-रेखा की दिशा में फलन x y 2 + y z 2 + z x 2 x y 2 + y z 2 + z x 2 xy^(2)+yz^(2)+zx^(2)x y^2+y z^2+z x^2xy2+yz2+zx2 का दिशात्मक अवकलज ज्ञात कीजिए।
Find the directional derivative of the function x y 2 + y z 2 + z x 2 x y 2 + y z 2 + z x 2 xy^(2)+yz^(2)+zx^(2)x y^2+y z^2+z x^2xy2+yz2+zx2 along the tangent to the curve x = t , y = t 2 , z = t 3 x = t , y = t 2 , z = t 3 x=t,y=t^(2),z=t^(3)x=t, y=t^2, z=t^3x=t,y=t2,z=t3 at the point ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1).
6(a) एक पिण्ड एक शंकु और उसके नीचे अर्धगोले से बना है। शंकु के आधार तथा अर्धगोले के शिखर का अर्धव्यास a a aaa है। पूरा पिण्ड एक रूक्ष क्षैतिज मेज पर रखा है, जिसका अर्धगोला मेज को स्पर्श करता है। दर्शाइए कि शंकु की अधिकतम ऊँचाई, जिससे कि साम्यावस्था स्थिर बनी रहे, 3 a 3 a sqrt3a\sqrt{3} a3a है।
A body consists of a cone and underlying hemisphere. The base of the cone and the top of the hemisphere have same radius a a aaa. The whole body rests on a rough horizontal table with hemisphere in contact with the table. Show that the greatest height of the cone, so that the equilibrium may be stable, is 3 a 3 a sqrt3a\sqrt{3} a3a.
(b) वक्र C C CCC के चारों तरफ F → F → vec(F)\vec{F}F→ का परिसंचरण ज्ञात कीजिए, जहाँ F → = ( 2 x + y 2 ) i ^ + ( 3 y − 4 x ) j ^ F → = 2 x + y 2 i ^ + ( 3 y − 4 x ) j ^ vec(F)=(2x+y^(2)) hat(i)+(3y-4x) hat(j)\vec{F}=\left(2 x+y^2\right) \hat{i}+(3 y-4 x) \hat{j}F→=(2x+y2)i^+(3y−4x)j^ और C C CCC, बिंदु ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) से बिंदु ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) तक वक्र y = x 2 y = x 2 y=x^(2)y=x^2y=x2 के द्वारा तथा बिंदु ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) से बिंदु ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) तक वक्र y 2 = x y 2 = x y^(2)=xy^2=xy2=x के द्वारा परिभाषित है।
Find the circulation of F → F → vec(F)\vec{F}F→ round the curve C C CCC, where F → = ( 2 x + y 2 ) i ^ + ( 3 y − 4 x ) j ^ F → = 2 x + y 2 i ^ + ( 3 y − 4 x ) j ^ vec(F)=(2x+y^(2)) hat(i)+(3y-4x) hat(j)\vec{F}=\left(2 x+y^2\right) \hat{i}+(3 y-4 x) \hat{j}F→=(2x+y2)i^+(3y−4x)j^ and C C CCC is the curve y = x 2 y = x 2 y=x^(2)y=x^2y=x2 from ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) to ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) and the curve y 2 = x y 2 = x y^(2)=xy^2=xy2=x from ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) to ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0).
(c) (i) अवकल समीकरण
d 2 y d x 2 + ( 3 sin ⁡ x − cot ⁡ x ) d y d x + 2 y sin 2 ⁡ x = e − cos ⁡ x sin 2 ⁡ x d 2 y d x 2 + ( 3 sin ⁡ x − cot ⁡ x ) d y d x + 2 y sin 2 ⁡ x = e − cos ⁡ x sin 2 ⁡ x (d^(2)y)/(dx^(2))+(3sin x-cot x)(dy)/(dx)+2ysin^(2)x=e^(-cos x)sin^(2)x\frac{d^2 y}{d x^2}+(3 \sin x-\cot x) \frac{d y}{d x}+2 y \sin ^2 x=e^{-\cos x} \sin ^2 xd2ydx2+(3sin⁡x−cot⁡x)dydx+2ysin2⁡x=e−cos⁡xsin2⁡x
को हल कीजिए।
(ii) t − 1 / 2 t − 1 / 2 t^(-1//2)t^{-1 / 2}t−1/2 तथा t 1 / 2 t 1 / 2 t^(1//2)t^{1 / 2}t1/2 का लाप्लास रूपांतर ज्ञात कीजिए। सिद्ध कीजिए कि t n + 1 2 t n + 1 2 t^(n+(1)/(2))t^{n+\frac{1}{2}}tn+12 का लाप्लास रूपांतर
Γ ( n + 1 + 1 2 ) s n + 1 + 1 2 Γ n + 1 + 1 2 s n + 1 + 1 2 (Gamma(n+1+(1)/(2)))/(s^(n+1+(1)/(2)))\frac{\Gamma\left(n+1+\frac{1}{2}\right)}{s^{n+1+\frac{1}{2}}}Γ(n+1+12)sn+1+12
होता है, जहाँ n ∈ N n ∈ N n inNn \in \mathbf{N}n∈N.
(i) Solve the differential equation
d 2 y d x 2 + ( 3 sin ⁡ x − cot ⁡ x ) d y d x + 2 y sin 2 ⁡ x = e − cos ⁡ x sin 2 ⁡ x d 2 y d x 2 + ( 3 sin ⁡ x − cot ⁡ x ) d y d x + 2 y sin 2 ⁡ x = e − cos ⁡ x sin 2 ⁡ x (d^(2)y)/(dx^(2))+(3sin x-cot x)(dy)/(dx)+2ysin^(2)x=e^(-cos x)sin^(2)x\frac{d^2 y}{d x^2}+(3 \sin x-\cot x) \frac{d y}{d x}+2 y \sin ^2 x=e^{-\cos x} \sin ^2 xd2ydx2+(3sin⁡x−cot⁡x)dydx+2ysin2⁡x=e−cos⁡xsin2⁡x
(ii) Find the Laplace transforms of t − 1 / 2 t − 1 / 2 t^(-1//2)t^{-1 / 2}t−1/2 and t 1 / 2 t 1 / 2 t^(1//2)t^{1 / 2}t1/2. Prove that the Laplace transform of t n + 1 2 t n + 1 2 t^(n+(1)/(2))t^{n+\frac{1}{2}}tn+12, where n ∈ N n ∈ N n inNn \in \mathbb{N}n∈N, is
Γ ( n + 1 + 1 2 ) s n + 1 + 1 2 Γ n + 1 + 1 2 s n + 1 + 1 2 (Gamma(n+1+(1)/(2)))/(s^(n+1+(1)/(2)))\frac{\Gamma\left(n+1+\frac{1}{2}\right)}{s^{n+1+\frac{1}{2}}}Γ(n+1+12)sn+1+12
7(a) समीकरण x 2 y ′ ′ − 2 x y ′ + 2 y = x 3 sin ⁡ x x 2 y ′ ′ − 2 x y ′ + 2 y = x 3 sin ⁡ x x^(2)y^(”)-2xy^(‘)+2y=x^(3)sin xx^2 y^{\prime \prime}-2 x y^{\prime}+2 y=x^3 \sin xx2y′′−2xy′+2y=x3sin⁡x के संगत समांगी अवकल समीकरण का रेखीय स्वतंत्र हल निकालिए और तब दिए गए समीकरण का प्राचल-विचरण विधि द्वारा सामान्य हल निकालिए।
Find the linearly independent solutions of the corresponding homogeneous differential equation of the equation x 2 y ′ ′ − 2 x y ′ + 2 y = x 3 sin ⁡ x x 2 y ′ ′ − 2 x y ′ + 2 y = x 3 sin ⁡ x x^(2)y^(”)-2xy^(‘)+2y=x^(3)sin xx^2 y^{\prime \prime}-2 x y^{\prime}+2 y=x^3 \sin xx2y′′−2xy′+2y=x3sin⁡x and then find the general solution of the given equation by the method of variation of parameters.
(b) कुंडलिनी x = a cos ⁡ u , y = a sin ⁡ u , z = a u tan ⁡ α x = a cos ⁡ u , y = a sin ⁡ u , z = a u tan ⁡ α x=a cos u,y=a sin u,z=au tan alphax=a \cos u, y=a \sin u, z=a u \tan \alphax=acos⁡u,y=asin⁡u,z=autan⁡α के लिए वक्रता की त्रिज्या तथा विमोटन की त्रिज्या ज्ञात कीजिए।
Find the radius of curvature and radius of torsion of the helix x = a cos ⁡ u x = a cos ⁡ u x=a cos ux=a \cos ux=acos⁡u, y = a sin ⁡ u , z = a u tan ⁡ α y = a sin ⁡ u , z = a u tan ⁡ α y=a sin u,z=au tan alphay=a \sin u, z=a u \tan \alphay=asin⁡u,z=autan⁡α.
(c) y y yyy-अक्ष की दिशा में गतिमान एक कण का मूलबिंदु की ओर त्वरण F y F y FyF yFy है, जहाँ F , y F , y F,yF, yF,y का एक धनात्मक एवं सम फलन है। जब कण y = − a y = − a y=-ay=-ay=−a तथा y = a y = a y=ay=ay=a के बीच में कंपन करता है, तब उसका आवर्तकाल T T TTT है। दर्शाइए कि
2 π F 1 < T < 2 π F 2 2 π F 1 < T < 2 π F 2 (2pi)/(sqrt(F_(1))) < T < (2pi)/(sqrt(F_(2)))\frac{2 \pi}{\sqrt{F_1}}<T<\frac{2 \pi}{\sqrt{F_2}}2πF1<T<2πF2
जहाँ F 1 F 1 F_(1)F_1F1 एवं F 2 F 2 F_(2)F_2F2 परास [ − a , a ] [ − a , a ] [-a,a][-a, a][−a,a] में F F FFF के अधिकतम एवं न्यूनतम मान हैं। आगे दर्शाइए कि जब लंबाई l l lll का एक सरल लोलक ऊर्ध्वाधर रेखा के किसी भी ओर 30 ∘ 30 ∘ 30^(@)30^{\circ}30∘ तक दोलन करता है, तब T , 2 π l / g T , 2 π l / g T,2pisqrt(l//g)T, 2 \pi \sqrt{l / g}T,2πl/g तथा 2 π l / g π / 3 2 π l / g π / 3 2pisqrt(l//g)sqrt(pi//3)2 \pi \sqrt{l / g} \sqrt{\pi / 3}2πl/gπ/3 के बीच में रहता है।
A particle moving along the y y yyy-axis has an acceleration F y F y FyF yFy towards the origin, where F F FFF is a positive and even function of y y yyy. The periodic time, when the particle vibrates between y = − a y = − a y=-ay=-ay=−a and y = a y = a y=ay=ay=a, is T T TTT. Show that
2 π F 1 < T < 2 π F 2 2 π F 1 < T < 2 π F 2 (2pi)/(sqrt(F_(1))) < T < (2pi)/(sqrt(F_(2)))\frac{2 \pi}{\sqrt{F_1}}<T<\frac{2 \pi}{\sqrt{F_2}}2πF1<T<2πF2
where F 1 F 1 F_(1)F_1F1 and F 2 F 2 F_(2)F_2F2 are the greatest and the least values of F F FFF within the range [ − a , a ] [ − a , a ] [-a,a][-a, a][−a,a]. Further, show that when a simple pendulum of length l l lll oscillates through 30 ∘ 30 ∘ 30^(@)30^{\circ}30∘ on either side of the vertical line, T T TTT lies between 2 π l / g 2 π l / g 2pisqrt(l//g)2 \pi \sqrt{l / g}2πl/g and 2 π l / g π / 3 2 π l / g π / 3 2pisqrt(l//g)sqrt(pi//3)2 \pi \sqrt{l / g} \sqrt{\pi / 3}2πl/gπ/3
8(a) अवकल समीकरण
( d y d x ) 2 ( y x ) 2 cot 2 ⁡ α − 2 ( d y d x ) ( y x ) + ( y x ) 2 cosec 2 ⁡ α = 1 d y d x 2 y x 2 cot 2 ⁡ α − 2 d y d x y x + y x 2 cosec 2 ⁡ α = 1 ((dy)/(dx))^(2)((y)/(x))^(2)cot^(2)alpha-2((dy)/(dx))((y)/(x))+((y)/(x))^(2)cosec^(2)alpha=1\left(\frac{d y}{d x}\right)^2\left(\frac{y}{x}\right)^2 \cot ^2 \alpha-2\left(\frac{d y}{d x}\right)\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^2 \operatorname{cosec}^2 \alpha=1(dydx)2(yx)2cot2⁡α−2(dydx)(yx)+(yx)2cosec2⁡α=1
का विचित्र हल प्राप्त कीजिए। दिए हुए अवकल समीकरण का पूर्ण पूर्वग भी ज्ञात कीजिए। पूर्ण पूर्वग तथा विचित्र हल की ज्यामितीय व्याख्या कीजिए।
Obtain the singular solution of the differential equation
( d y d x ) 2 ( y x ) 2 cot 2 ⁡ α − 2 ( d y d x ) ( y x ) + ( y x ) 2 cosec 2 ⁡ α = 1 d y d x 2 y x 2 cot 2 ⁡ α − 2 d y d x y x + y x 2 cosec 2 ⁡ α = 1 ((dy)/(dx))^(2)((y)/(x))^(2)cot^(2)alpha-2((dy)/(dx))((y)/(x))+((y)/(x))^(2)cosec^(2)alpha=1\left(\frac{d y}{d x}\right)^2\left(\frac{y}{x}\right)^2 \cot ^2 \alpha-2\left(\frac{d y}{d x}\right)\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^2 \operatorname{cosec}^2 \alpha=1(dydx)2(yx)2cot2⁡α−2(dydx)(yx)+(yx)2cosec2⁡α=1
Also find the complete primitive of the given differential equation. Give the geometrical interpretations of the complete primitive and singular solution.
(b) एक गतिमान ग्रह का त्वरण μ (दूरी 2 μ  (दूरी  2 (mu)/(” (दूरी “^(2))\frac{\mu}{\text { (दूरी }^2}μ (दूरी 2 के बराबर है और त्वरण की दिशा हमेशा एक स्थिर बिंदु (तारा) की ओर है। सिद्ध कीजिए कि उस ग्रह का पथ एक शंकु-परिच्छेद है। वे प्रतिबंध ज्ञात कीजिए, जिनके अन्तर्गत पथ (i) दीर्घवृत्त, (ii) परवलय और (iii) अतिपरवलय बन जाता है।
Prove that the path of a planet, which is moving so that its acceleration is always directed to a fixed point (star) and is equal to μ (distance) 2 μ  (distance)  2 (mu)/(” (distance) “^(2))\frac{\mu}{\text { (distance) }^2}μ (distance) 2, is a conic section. Find the conditions under which the path becomes (i) ellipse, (ii) parabola and (iii) hyperbola.
(c) (i) गाउस के अपसरण प्रमेय का कथन लिखिए। इस प्रमेय को F → = 4 x i ^ − 2 y 2 j ^ + z 2 k ^ F → = 4 x i ^ − 2 y 2 j ^ + z 2 k ^ vec(F)=4x hat(i)-2y^(2) hat(j)+z^(2) hat(k)\vec{F}=4 x \hat{i}-2 y^2 \hat{j}+z^2 \hat{k}F→=4xi^−2y2j^+z2k^ के लिए x 2 + y 2 = 4 , z = 0 x 2 + y 2 = 4 , z = 0 x^(2)+y^(2)=4,z=0x^2+y^2=4, z=0x2+y2=4,z=0 और z = 3 z = 3 z=3z=3z=3 से घिरे हुए क्षेत्र में सत्यापित कीजिए।
(ii) स्टोक्स प्रमेय के द्वारा ∮ C e x d x + 2 y d y − d z ∮ C e x d x + 2 y d y − d z oint_(C)e^(x)dx+2ydy-dz\oint_C e^x d x+2 y d y-d z∮Cexdx+2ydy−dz का मान ज्ञात कीजिए, जहाँ C C CCC, वक्र x 2 + y 2 = 4 x 2 + y 2 = 4 x^(2)+y^(2)=4x^2+y^2=4x2+y2=4, z = 2 z = 2 z=2z=2z=2 है।
(i) State Gauss divergence theorem. Verify this theorem for F → = 4 x i ^ − 2 y 2 j ^ + z 2 k ^ F → = 4 x i ^ − 2 y 2 j ^ + z 2 k ^ vec(F)=4x hat(i)-2y^(2) hat(j)+z^(2) hat(k)\vec{F}=4 x \hat{i}-2 y^2 \hat{j}+z^2 \hat{k}F→=4xi^−2y2j^+z2k^, taken over the region bounded by x 2 + y 2 = 4 , z = 0 x 2 + y 2 = 4 , z = 0 x^(2)+y^(2)=4,z=0x^2+y^2=4, z=0x2+y2=4,z=0 and z = 3 z = 3 z=3z=3z=3.
(ii) Evaluate by Stokes’ theorem ∮ C e x d x + 2 y d y − d z ∮ C e x d x + 2 y d y − d z oint_(C)e^(x)dx+2ydy-dz\oint_C e^x d x+2 y d y-d z∮Cexdx+2ydy−dz, where C C CCC is the curve x 2 + y 2 = 4 , z = 2 x 2 + y 2 = 4 , z = 2 x^(2)+y^(2)=4,z=2x^2+y^2=4, z=2x2+y2=4,z=2.
upsc-m2018-1-08bbcd08-2ec0-4a0d-8659-5c6f186e3a0d
खण्ड-A / SECTION-A
1(a) मान लीजिये कि A A AAA एक 3 × 2 3 × 2 3xx23 \times 23×2 आव्यूह है और B B BBB एक 2 × 3 2 × 3 2xx32 \times 32×3 आव्यूह है। दर्शाइये कि C = A ⋅ B C = A ⋅ B C=A*BC=A \cdot BC=A⋅B एक अव्युत्क्रमणणीय आव्यूह है।
Let A A AAA be a 3 × 2 3 × 2 3xx23 \times 23×2 matrix and B B BBB a 2 × 3 2 × 3 2xx32 \times 32×3 matrix. Show that C = A ⋅ B C = A ⋅ B C=A*BC=A \cdot BC=A⋅B is a singular matrix.
(b) आधार सदिशों e 1 = ( 1 , 0 ) e 1 = ( 1 , 0 ) e_(1)=(1,0)e_1=(1,0)e1=(1,0) और e 2 = ( 0 , 1 ) e 2 = ( 0 , 1 ) e_(2)=(0,1)e_2=(0,1)e2=(0,1) को α 1 = ( 2 , − 1 ) α 1 = ( 2 , − 1 ) alpha_(1)=(2,-1)\alpha_1=(2,-1)α1=(2,−1) एवं α 2 = ( 1 , 3 ) α 2 = ( 1 , 3 ) alpha_(2)=(1,3)\alpha_2=(1,3)α2=(1,3) के रैखिक संयोग के रूप में ब्यक्त कीजिये।
Express basis vectors e 1 = ( 1 , 0 ) e 1 = ( 1 , 0 ) e_(1)=(1,0)e_1=(1,0)e1=(1,0) and e 2 = ( 0 , 1 ) e 2 = ( 0 , 1 ) e_(2)=(0,1)e_2=(0,1)e2=(0,1) as linear combinations of α 1 = ( 2 , − 1 ) α 1 = ( 2 , − 1 ) alpha_(1)=(2,-1)\alpha_1=(2,-1)α1=(2,−1) and α 2 = ( 1 , 3 ) α 2 = ( 1 , 3 ) alpha_(2)=(1,3)\alpha_2=(1,3)α2=(1,3).
(c) निर्धारित कीजिये कि lim z → 1 ( 1 − z ) tan ⁡ π z 2 lim z → 1   ( 1 − z ) tan ⁡ π z 2 lim_(z rarr1)(1-z)tan (pi z)/(2)\lim _{z \rightarrow 1}(1-z) \tan \frac{\pi z}{2}limz→1(1−z)tan⁡πz2 का अस्तित्व है या कि नहीं। अगर यह सीमा विघ्यमान है, तो इसका मान ज्ञात कीजिये।
Determine if lim z → 1 ( 1 − z ) tan ⁡ π z 2 lim z → 1   ( 1 − z ) tan ⁡ π z 2 lim_(z rarr1)(1-z)tan (pi z)/(2)\lim _{z \rightarrow 1}(1-z) \tan \frac{\pi z}{2}limz→1(1−z)tan⁡πz2 exists or not. If the limit exists, then find its value.
(d) सीमा lim n → ∞ 1 n 2 ∑ r = 0 n − 1 n 2 − r 2 lim n → ∞   1 n 2 ∑ r = 0 n − 1   n 2 − r 2 lim_(n rarr oo)(1)/(n^(2))sum_(r=0)^(n-1)sqrt(n^(2)-r^(2))\lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{r=0}^{n-1} \sqrt{n^2-r^2}limn→∞1n2∑r=0n−1n2−r2 का मान ज्ञात कीजिये।
Find the limit lim n → − ∞ 1 n 2 ∑ r = 0 n − 1 n 2 − r 2 lim n → − ∞   1 n 2 ∑ r = 0 n − 1   n 2 − r 2 lim_(n rarr-oo)(1)/(n^(2))sum_(r=0)^(n-1)sqrt(n^(2)-r^(2))\lim _{n \rightarrow-\infty} \frac{1}{n^2} \sum_{r=0}^{n-1} \sqrt{n^2-r^2}limn→−∞1n2∑r=0n−1n2−r2.
(e) सरल रेखा x − 1 2 = y − 1 3 = z + 1 − 1 x − 1 2 = y − 1 3 = z + 1 − 1 (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)\frac{x-1}{2}=\frac{y-1}{3}=\frac{z+1}{-1}x−12=y−13=z+1−1 का समतल x + y + 2 z = 6 x + y + 2 z = 6 x+y+2z=6x+y+2 z=6x+y+2z=6 पर प्रक्षेपण ज्ञात कीजिये।
Find the projection of the straight line x − 1 2 = y − 1 3 = z + 1 − 1 x − 1 2 = y − 1 3 = z + 1 − 1 (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)\frac{x-1}{2}=\frac{y-1}{3}=\frac{z+1}{-1}x−12=y−13=z+1−1 on the plane x + y + 2 z = 6 x + y + 2 z = 6 x+y+2z=6x+y+2 z=6x+y+2z=6.
2(a) अगर A A AAA और B B BBB समरूप n × n n × n n xx nn \times nn×n आव्यूह हैं, तो दर्शाइये कि उनके आइगेन मान एक ही हैं।
Show that if A A AAA and B B BBB are similar n × n n × n n xx nn \times nn×n matrices, then they have the same eigenvalues.
(b) बिन्दु ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0) से परबलय y 2 = 4 x y 2 = 4 x y^(2)=4xy^2=4 xy2=4x तक की न्यूनतम दूरी ज्ञात कीजिये।
Find the shortest distance from the point ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0) to the parabola y 2 = 4 x y 2 = 4 x y^(2)=4xy^2=4 xy2=4x.
(c) दीर्घवृत्त x 2 a 2 + y 2 b 2 = 1 x x 2 a 2 + y 2 b 2 = 1 x (x^(2))/(a^(2))+(y^(2))/(b^(2))=1x\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 xx2a2+y2b2=1x-अक्ष के चारों तरफ परिभ्रमण कर रहा है। परिक्रमित घन का आयतन ज्ञात कीजिये।
The ellipse x 2 a 2 + y 2 b 2 = 1 x 2 a 2 + y 2 b 2 = 1 (x^(2))/(a^(2))+(y^(2))/(b^(2))=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1x2a2+y2b2=1 revolves about the x x xxx-axis. Find the volume of the solid of revolution.
(d) रेखाओं
a 1 x + b 1 y + c 1 z + d 1 = 0 a 2 x + b 2 y + c 2 z + d 2 = 0 a 1 x + b 1 y + c 1 z + d 1 = 0 a 2 x + b 2 y + c 2 z + d 2 = 0 {:[a_(1)x+b_(1)y+c_(1)z+d_(1)=0],[a_(2)x+b_(2)y+c_(2)z+d_(2)=0]:}\begin{array}{r} a_1 x+b_1 y+c_1 z+d_1=0 \\ a_2 x+b_2 y+c_2 z+d_2=0 \end{array}a1x+b1y+c1z+d1=0a2x+b2y+c2z+d2=0
और z z zzz-अक्ष के बीच की न्यूनतम दूरी ज्ञात कीजिये।
Find the shortest distance between the lines
a 1 x + b 1 y + c 1 z + d 1 = 0 a 2 x + b 2 y + c 2 z + d 2 = 0 a 1 x + b 1 y + c 1 z + d 1 = 0 a 2 x + b 2 y + c 2 z + d 2 = 0 {:[a_(1)x+b_(1)y+c_(1)z+d_(1)=0],[a_(2)x+b_(2)y+c_(2)z+d_(2)=0]:}\begin{array}{r} a_1 x+b_1 y+c_1 z+d_1=0 \\ a_2 x+b_2 y+c_2 z+d_2=0 \end{array}a1x+b1y+c1z+d1=0a2x+b2y+c2z+d2=0
and the z-axis.
3(a) रेखिक समीकरण निकाय
x + 3 y − 2 z = − 1 5 y + 3 z = − 8 x − 2 y − 5 z = 7 x + 3 y − 2 z = − 1 5 y + 3 z = − 8 x − 2 y − 5 z = 7 {:[x+3y-2z=-1],[5y+3z=-8],[x-2y-5z=7]:}\begin{aligned} x+3 y-2 z &=-1 \\ 5 y+3 z &=-8 \\ x-2 y-5 z &=7 \end{aligned}x+3y−2z=−15y+3z=−8x−2y−5z=7
के लिये निर्धारित कीजिये कि निम्नलिखित कथर्नों में से कौन-से सही हैं और कौन-से गलत :
(i) समीकरण निकाय का कोई भी हल नही है।
(ii) समीकरण निकाय का सिर्फ एक ही हल है।
(iii) समीकरण निकाय के असीम मात्रा में अनेक हल है।
For the system of linear equations
x + 3 y − 2 z = − 1 5 y + 3 z = − 8 x − 2 y − 5 z = 7 x + 3 y − 2 z = − 1 5 y + 3 z = − 8 x − 2 y − 5 z = 7 {:[x+3y-2z=-1],[5y+3z=-8],[x-2y-5z=7]:}\begin{aligned} x+3 y-2 z &=-1 \\ 5 y+3 z &=-8 \\ x-2 y-5 z &=7 \end{aligned}x+3y−2z=−15y+3z=−8x−2y−5z=7
determine which of the following statements are true and which are false :
(i) The system has no solution.
(ii) The system has a unique solution.
(iii) The system has infinitely many solutions.
(b) मान लीजिये कि
f ( x , y ) = x y 2 , यदि y > 0 = − x y 2 , यदि y ≤ 0 f ( x , y ) = x y 2 ,  यदि  y > 0 = − x y 2 ,  यदि  y ≤ 0 {:[f(x”,”y)=xy^(2)”,”” यदि “y > 0],[=-xy^(2)”,”” यदि “y <= 0]:}\begin{aligned} f(x, y) &=x y^2, & \text { यदि } y>0 \\ &=-x y^2, & \text { यदि } y \leq 0 \end{aligned}f(x,y)=xy2, यदि y>0=−xy2, यदि y≤0
निर्धारित कीजिये कि ∂ f ∂ x ( 0 , 1 ) ∂ f ∂ x ( 0 , 1 ) (del f)/(del x)(0,1)\frac{\partial f}{\partial x}(0,1)∂f∂x(0,1) और ∂ f ∂ y ( 0 , 1 ) ∂ f ∂ y ( 0 , 1 ) (del f)/(del y)(0,1)\frac{\partial f}{\partial y}(0,1)∂f∂y(0,1) में से किसका अस्तित्व है और किसका अस्तित्व नहीं है।
Let
f ( x , y ) = x y 2 , if y > 0 = − x y 2 , if y ≤ 0 f ( x , y ) = x y 2 ,  if  y > 0 = − x y 2 ,  if  y ≤ 0 {:[f(x”,”y)=xy^(2)”,”” if “y > 0],[=-xy^(2)”,”” if “y <= 0]:}\begin{aligned} f(x, y) &=x y^2, & \text { if } & y>0 \\ &=-x y^2, & \text { if } & y \leq 0 \end{aligned}f(x,y)=xy2, if y>0=−xy2, if y≤0
Determine which of ∂ f ∂ x ( 0 , 1 ) ∂ f ∂ x ( 0 , 1 ) (del f)/(del x)(0,1)\frac{\partial f}{\partial x}(0,1)∂f∂x(0,1) and ∂ f ∂ y ( 0 , 1 ) ∂ f ∂ y ( 0 , 1 ) (del f)/(del y)(0,1)\frac{\partial f}{\partial y}(0,1)∂f∂y(0,1) exists and which does not exist.
(c) परबलयज ( x + y + z ) ( 2 x + y − z ) = 6 z ( x + y + z ) ( 2 x + y − z ) = 6 z (x+y+z)(2x+y-z)=6z(x+y+z)(2 x+y-z)=6 z(x+y+z)(2x+y−z)=6z की उन जनक रेखाओं के समीकरणों को ज्ञात कीजिये, जो बिन्दु ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) में से गुज्जरती है।
Find the equations to the generating lines of the paraboloid ( x + y + z ) ( 2 x + y − z ) = 6 z ( x + y + z ) ( 2 x + y − z ) = 6 z (x+y+z)(2x+y-z)=6z(x+y+z)(2 x+y-z)=6 z(x+y+z)(2x+y−z)=6z which pass through the point ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1).
(d) xyz-समतल में स्थित, बिन्दुओं ( 0 , 0 , 0 ) , ( 0 , 1 , − 1 ) , ( − 1 , 2 , 0 ) ( 0 , 0 , 0 ) , ( 0 , 1 , − 1 ) , ( − 1 , 2 , 0 ) (0,0,0),(0,1,-1),(-1,2,0)(0,0,0),(0,1,-1),(-1,2,0)(0,0,0),(0,1,−1),(−1,2,0) और ( 1 , 2 , 3 ) ( 1 , 2 , 3 ) (1,2,3)(1,2,3)(1,2,3) में से गुज़रते हुये गोले का समीकरण ज्ञात कीजिये।
Find the equation of the sphere in xyz-plane passing through the points ( 0 , 0 , 0 ) , ( 0 , 1 , − 1 ) , ( − 1 , 2 , 0 ) ( 0 , 0 , 0 ) , ( 0 , 1 , − 1 ) , ( − 1 , 2 , 0 ) (0,0,0),(0,1,-1),(-1,2,0)(0,0,0),(0,1,-1),(-1,2,0)(0,0,0),(0,1,−1),(−1,2,0) and ( 1 , 2 , 3 ) ( 1 , 2 , 3 ) (1,2,3)(1,2,3)(1,2,3).
4.(a) अन्तराल [ 2 , 3 ] [ 2 , 3 ] [2,3][2,3][2,3] पर x 4 − 5 x 2 + 4 x 4 − 5 x 2 + 4 x^(4)-5x^(2)+4x^4-5 x^2+4x4−5x2+4 के अधिकतम और न्यूनतम मान ज्ञात कीजिये।
Find the maximum and the minimum values of x 4 − 5 x 2 + 4 x 4 − 5 x 2 + 4 x^(4)-5x^(2)+4x^4-5 x^2+4x4−5x2+4 on the interval [ 2 , 3 ] [ 2 , 3 ] [2,3][2,3][2,3].
(b) समाकल ∫ 0 a ∫ x / a x x d y d x x 2 + y 2 ∫ 0 a   ∫ x / a x   x d y d x x 2 + y 2 int_(0)^(a)int_(x//a)^(x)(xdydx)/(x^(2)+y^(2))\int_0^a \int_{x / a}^x \frac{x d y d x}{x^2+y^2}∫0a∫x/axxdydxx2+y2 का मान निकालिये।
Evaluate the integral ∫ 0 a ∫ x / a x x d y d x x 2 + y 2 ∫ 0 a   ∫ x / a x   x d y d x x 2 + y 2 int_(0)^(a)int_(x//a)^(x)(xdydx)/(x^(2)+y^(2))\int_0^a \int_{x / a}^x \frac{x d y d x}{x^2+y^2}∫0a∫x/axxdydxx2+y2.
(c) उस शंकु, जिसका शीर्ष ( 0 , 0 , 1 ) ( 0 , 0 , 1 ) (0,0,1)(0,0,1)(0,0,1) है और जिसका निर्देशक वक्र 2 x 2 − y 2 = 4 , z = 0 2 x 2 − y 2 = 4 , z = 0 2x^(2)-y^(2)=4,z=02 x^2-y^2=4, z=02x2−y2=4,z=0 है, का समीकरण ज्ञात कीजिये।
Find the equation of the cone with ( 0 , 0 , 1 ) ( 0 , 0 , 1 ) (0,0,1)(0,0,1)(0,0,1) as the vertex and 2 x 2 − y 2 = 4 , z = 0 2 x 2 − y 2 = 4 , z = 0 2x^(2)-y^(2)=4,z=02 x^2-y^2=4, z=02x2−y2=4,z=0 as the guiding curve.
(d) 3 x − y + 3 z = 8 3 x − y + 3 z = 8 3x-y+3z=83 x-y+3 z=83x−y+3z=8 के समांतर और बिन्दु ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1) में से गुजरते हुये समतल का समीकरण ज्ञात कीजिये।
Find the equation of the plane parallel to 3 x − y + 3 z = 8 3 x − y + 3 z = 8 3x-y+3z=83 x-y+3 z=83x−y+3z=8 and passing through the point ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1).
खण्ड-B / SECTION-B
5. (a) हल कीजिये/Solve :
y ′ ′ − y = x 2 e 2 x y ′ ′ − y = x 2 e 2 x y^(”)-y=x^(2)e^(2x)y^{\prime \prime}-y=x^2 e^{2 x}y′′−y=x2e2x
(b) x = 3 t , y = 3 t 2 , z = 3 t 3 x = 3 t , y = 3 t 2 , z = 3 t 3 x=3t,y=3t^(2),z=3t^(3)x=3 t, y=3 t^2, z=3 t^3x=3t,y=3t2,z=3t3 समीकरणों वाले बक्र के एक आय बिन्दु पर स्पर्शं-रेखा और रेखा y = z − x = 0 y = z − x = 0 y=z-x=0y=z-x=0y=z−x=0 के बीच का कोण ज्ञात कीजिये।
Find the angle between the tangent at a general point of the curve whose equations are x = 3 t , y = 3 t 2 , z = 3 t 3 x = 3 t , y = 3 t 2 , z = 3 t 3 x=3t,y=3t^(2),z=3t^(3)x=3 t, y=3 t^2, z=3 t^3x=3t,y=3t2,z=3t3 and the line y = z − x = 0 y = z − x = 0 y=z-x=0y=z-x=0y=z−x=0.
(c) हल कीजिये/Solve:
y ′ ′ ′ − 6 y ′ ′ + 12 y ′ − 8 y = 12 e 2 x + 27 e − x y ′ ′ ′ − 6 y ′ ′ + 12 y ′ − 8 y = 12 e 2 x + 27 e − x y^(”’)-6y^(”)+12y^(‘)-8y=12e^(2x)+27e^(-x)y^{\prime \prime \prime}-6 y^{\prime \prime}+12 y^{\prime}-8 y=12 e^{2 x}+27 e^{-x}y′′′−6y′′+12y′−8y=12e2x+27e−x
(d) (i) f ( t ) = 1 t f ( t ) = 1 t f(t)=(1)/(sqrtt)f(t)=\frac{1}{\sqrt{t}}f(t)=1t का लाप्लास रूपान्तर ज्ञात कीजिये। Find the Laplace transform of f ( t ) = 1 t f ( t ) = 1 t f(t)=(1)/(sqrtt)f(t)=\frac{1}{\sqrt{t}}f(t)=1t.
(ii) 5 s 2 + 3 s − 16 ( s − 1 ) ( s − 2 ) ( s + 3 ) 5 s 2 + 3 s − 16 ( s − 1 ) ( s − 2 ) ( s + 3 ) (5s^(2)+3s-16)/((s-1)(s-2)(s+3))\frac{5 s^2+3 s-16}{(s-1)(s-2)(s+3)}5s2+3s−16(s−1)(s−2)(s+3) का विलोम लाप्लास रुपान्तर ज्ञात कीजिये। Find the inverse Laplace transform of 5 s 2 + 3 s − 16 ( s − 1 ) ( s − 2 ) ( s + 3 ) 5 s 2 + 3 s − 16 ( s − 1 ) ( s − 2 ) ( s + 3 ) (5s^(2)+3s-16)/((s-1)(s-2)(s+3))\frac{5 s^2+3 s-16}{(s-1)(s-2)(s+3)}5s2+3s−16(s−1)(s−2)(s+3).
(e) एक कण को धरती के एक बिन्दु से प्रक्षेपित करने पर वह एक दीवार, जो प्रक्षेपण बिन्दु से d d ddd दूरी पर है और जिसकी ऊँचाई h h hhh है, को छूते हुये पार करता है। अगर यह कण ऊर्ध्वाधर तल पर गतिमान है और इसकी क्षैतिज पहुँच R R RRR है, तो प्रक्षेपण की उच्चता ज्ञात कीजिये।
A particle projected from a given point on the ground just clears a wall of height h h hhh at a distance d d ddd from the point of projection. If the particle moves in a vertical plane and if the horizontal range is R R RRR, find the elevation of the projection.
6(a) हल कीजिये /Solve :
( d y d x ) 2 y + 2 d y d x x − y = 0 d y d x 2 y + 2 d y d x x − y = 0 ((dy)/(dx))^(2)y+2(dy)/(dx)x-y=0\left(\frac{d y}{d x}\right)^2 y+2 \frac{d y}{d x} x-y=0(dydx)2y+2dydxx−y=0
(b) एक कण, जो एक सरल रेखा में सरल आवर्त गति से चल रहा है, के पथ के केन्द्र से x 1 x 1 x_(1)x_1x1 और x 2 x 2 x_(2)x_2x2 की दूरी पर वेग क्रमशः v 1 v 1 v_(1)v_1v1 और v 2 v 2 v_(2)v_2v2 है। उसकी गति का आवर्तकाल ज्ञात कीजिये।
A particle moving with simple harmonic motion in a straight line has velocities v 1 v 1 v_(1)v_1v1 and v 2 v 2 v_(2)v_2v2 at distances x 1 x 1 x_(1)x_1x1 and x 2 x 2 x_(2)x_2x2 respectively from the centre of its path. Find the period of its motion.
(c) हल कीजिये/Solve :
y ′ ′ + 16 y = 32 sec ⁡ 2 x y ′ ′ + 16 y = 32 sec ⁡ 2 x y^(”)+16 y=32 sec 2xy^{\prime \prime}+16 y=32 \sec 2 xy′′+16y=32sec⁡2x
(d) अगर गोलक x 2 + y 2 + z 2 = a 2 x 2 + y 2 + z 2 = a 2 x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2x2+y2+z2=a2 का पृष्ठ S S SSS है, तो गाउस के अपसरण प्रमेय का इस्तेमाल करते हुये
∬ S [ ( x + z ) d y d z + ( y + z ) d z d x + ( x + y ) d x d y ] ∬ S   [ ( x + z ) d y d z + ( y + z ) d z d x + ( x + y ) d x d y ] ∬_(S)[(x+z)dydz+(y+z)dzdx+(x+y)dxdy]\iint_S[(x+z) d y d z+(y+z) d z d x+(x+y) d x d y]∬S[(x+z)dydz+(y+z)dzdx+(x+y)dxdy]
का मान निकालिये।
If S S SSS is the surface of the sphere x 2 + y 2 + z 2 = a 2 x 2 + y 2 + z 2 = a 2 x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2x2+y2+z2=a2, then evaluate
∬ S [ ( x + z ) d y d z + ( y + z ) d z d x + ( x + y ) d x d y ] ∬ S   [ ( x + z ) d y d z + ( y + z ) d z d x + ( x + y ) d x d y ] ∬_(S)[(x+z)dydz+(y+z)dzdx+(x+y)dxdy]\iint_S[(x+z) d y d z+(y+z) d z d x+(x+y) d x d y]∬S[(x+z)dydz+(y+z)dzdx+(x+y)dxdy]
using Gauss’ divergence theorem.
7. (a) हल कीजिये/Solve :
( 1 + x ) 2 y ′ ′ + ( 1 + x ) y ′ + y = 4 cos ⁡ ( log ⁡ ( 1 + x ) ) ( 1 + x ) 2 y ′ ′ + ( 1 + x ) y ′ + y = 4 cos ⁡ ( log ⁡ ( 1 + x ) ) (1+x)^(2)y^(”)+(1+x)y^(‘)+y=4cos(log(1+x))(1+x)^2 y^{\prime \prime}+(1+x) y^{\prime}+y=4 \cos (\log (1+x))(1+x)2y′′+(1+x)y′+y=4cos⁡(log⁡(1+x))
(b) बक्र
r → = a ( u − sin ⁡ u ) i → + a ( 1 − cos ⁡ u ) j → + b u k → r → = a ( u − sin ⁡ u ) i → + a ( 1 − cos ⁡ u ) j → + b u k → vec(r)=a(u-sin u) vec(i)+a(1-cos u) vec(j)+bu vec(k)\vec{r}=a(u-\sin u) \vec{i}+a(1-\cos u) \vec{j}+b u \vec{k}r→=a(u−sin⁡u)i→+a(1−cos⁡u)j→+buk→
की वक्रता और विमोटन ज्ञात कीजिये।
Find the curvature and torsion of the curve
r → = a ( u − sin ⁡ u ) i → + a ( 1 − cos ⁡ u ) j → + b u k → r → = a ( u − sin ⁡ u ) i → + a ( 1 − cos ⁡ u ) j → + b u k → vec(r)=a(u-sin u) vec(i)+a(1-cos u) vec(j)+bu vec(k)\vec{r}=a(u-\sin u) \vec{i}+a(1-\cos u) \vec{j}+b u \vec{k}r→=a(u−sin⁡u)i→+a(1−cos⁡u)j→+buk→
(c) प्रारंभिक मान समस्या
y ′ ′ − 5 y ′ + 4 y = e 2 t y ( 0 ) = 19 12 , y ′ ( 0 ) = 8 3 y ′ ′ − 5 y ′ + 4 y = e 2 t y ( 0 ) = 19 12 , y ′ ( 0 ) = 8 3 {:[y^(”)-5y^(‘)+4y=e^(2t)],[y(0)=(19)/(12)”,”y^(‘)(0)=(8)/(3)]:}\begin{aligned} &y^{\prime \prime}-5 y^{\prime}+4 y=e^{2 t} \\ &y(0)=\frac{19}{12}, y^{\prime}(0)=\frac{8}{3} \end{aligned}y′′−5y′+4y=e2ty(0)=1912,y′(0)=83
को हल कीजिये।
Solve the initial value problem
y ′ ′ − 5 y ′ + 4 y = e 2 t y ( 0 ) = 19 12 , y ′ ( 0 ) = 8 3 y ′ ′ − 5 y ′ + 4 y = e 2 t y ( 0 ) = 19 12 , y ′ ( 0 ) = 8 3 {:[y^(”)-5y^(‘)+4y=e^(2t)],[y(0)=(19)/(12)”,”y^(‘)(0)=(8)/(3)]:}\begin{aligned} &y^{\prime \prime}-5 y^{\prime}+4 y=e^{2 t} \\ &y(0)=\frac{19}{12}, y^{\prime}(0)=\frac{8}{3} \end{aligned}y′′−5y′+4y=e2ty(0)=1912,y′(0)=83
(d) α α alpha\alphaα और β β beta\betaβ को, जिसके लिये x α y β x α y β x^( alpha)y^( beta)x^\alpha y^\betaxαyβ समीकरण ( 4 y 2 + 3 x y ) d x − ( 3 x y + 2 x 2 ) d y = 0 4 y 2 + 3 x y d x − 3 x y + 2 x 2 d y = 0 (4y^(2)+3xy)dx-(3xy+2x^(2))dy=0\left(4 y^2+3 x y\right) d x-\left(3 x y+2 x^2\right) d y=0(4y2+3xy)dx−(3xy+2x2)dy=0 का एक समाकलन गुण्पक है, ज्ञात कीजिये और समीकरण हल कीजिये।
Find α α alpha\alphaα and β β beta\betaβ such that x α y β x α y β x^( alpha)y^( beta)x^\alpha y^\betaxαyβ is an integrating factor of ( 4 y 2 + 3 x y ) d x − ( 3 x y + 2 x 2 ) d y = 0 4 y 2 + 3 x y d x − 3 x y + 2 x 2 d y = 0 (4y^(2)+3xy)dx-(3xy+2x^(2))dy=0\left(4 y^2+3 x y\right) d x-\left(3 x y+2 x^2\right) d y=0(4y2+3xy)dx−(3xy+2x2)dy=0 and solve the equation.
8. (a) मान लीजिये कि
v → = v 1 i → + v 2 j → + v 3 k → v → = v 1 i → + v 2 j → + v 3 k → vec(v)=v_(1) vec(i)+v_(2) vec(j)+v_(3) vec(k)\vec{v}=v_1 \vec{i}+v_2 \vec{j}+v_3 \vec{k}v→=v1i→+v2j→+v3k→ है। दर्शाइये कि curl ⁡ ( curl ⁡ v → ) = grad ⁡ ( div ⁡ v → ) − ∇ 2 v → curl ⁡ ( curl ⁡ v → ) = grad ⁡ ( div ⁡ v → ) − ∇ 2 v → curl(curl vec(v))=grad(div vec(v))-grad^(2) vec(v)\operatorname{curl}(\operatorname{curl} \vec{v})=\operatorname{grad}(\operatorname{div} \vec{v})-\nabla^2 \vec{v}curl⁡(curl⁡v→)=grad⁡(div⁡v→)−∇2v→.
Let v → = v 1 i → + v 2 j → + v 3 k → v → = v 1 i → + v 2 j → + v 3 k → vec(v)=v_(1) vec(i)+v_(2) vec(j)+v_(3) vec(k)\vec{v}=v_1 \vec{i}+v_2 \vec{j}+v_3 \vec{k}v→=v1i→+v2j→+v3k→. Show that curl(curl v → ) = grad ⁡ ( div ⁡ v → ) − ∇ 2 v → v → = grad ⁡ ( div ⁡ v → ) − ∇ 2 v → {:( vec(v)))=grad(div vec(v))-grad^(2) vec(v)\left.\vec{v}\right)=\operatorname{grad}(\operatorname{div} \vec{v})-\nabla^2 \vec{v}v→)=grad⁡(div⁡v→)−∇2v→.
(b) स्टोकस प्रमेय का इस्तेमाल करते हुये रेखा समाकल ∫ C − y 3 d x + x 3 d y + z 3 d z ∫ C   − y 3 d x + x 3 d y + z 3 d z int _(C)-y^(3)dx+x^(3)dy+z^(3)dz\int_C-y^3 d x+x^3 d y+z^3 d z∫C−y3dx+x3dy+z3dz का मान निकालिये। यहाँ सिलिन्डर x 2 + y 2 = 1 x 2 + y 2 = 1 x^(2)+y^(2)=1x^2+y^2=1x2+y2=1 और समतल x + y + z = 1 x + y + z = 1 x+y+z=1x+y+z=1x+y+z=1 का प्रतिच्छेद C C CCC है। C C CCC पर अभिकिन्यास x y x y xyx yxy-समतल में वामावर्त गति के संगत है।
Evaluate the line integral ∫ C − y 3 d x + x 3 d y + z 3 d z ∫ C   − y 3 d x + x 3 d y + z 3 d z int _(C)-y^(3)dx+x^(3)dy+z^(3)dz\int_C-y^3 d x+x^3 d y+z^3 d z∫C−y3dx+x3dy+z3dz using Stokes’ theorem. Here C C CCC is the intersection of the cylinder x 2 + y 2 = 1 x 2 + y 2 = 1 x^(2)+y^(2)=1x^2+y^2=1x2+y2=1 and the plane x + y + z = 1 x + y + z = 1 x+y+z=1x+y+z=1x+y+z=1. The orientation on C C CCC corresponds to counterclockwise motion in the x y x y xyx yxy-plane.
(c) मान लीजिये कि F → = x y 2 i → + ( y + x ) j → F → = x y 2 i → + ( y + x ) j → vec(F)=xy^(2) vec(i)+(y+x) vec(j)\vec{F}=x y^2 \vec{i}+(y+x) \vec{j}F→=xy2i→+(y+x)j→ है। ग्रीन के प्रमेय का इस्तेमाल करते हुये प्रथम चतुर्थाश में वक्रों y = x 2 y = x 2 y=x^(2)y=x^2y=x2 और y = x y = x y=xy=xy=x द्वारा परिबद्ध क्षेत्र पर ( ∇ × F → ) ⋅ k → ( ∇ × F → ) ⋅ k → (grad xx vec(F))* vec(k)(\nabla \times \vec{F}) \cdot \vec{k}(∇×F→)⋅k→ का समाकलन कीजिये।
Let F → = x y 2 i → + ( y + x ) j → F → = x y 2 i → + ( y + x ) j → vec(F)=xy^(2) vec(i)+(y+x) vec(j)\vec{F}=x y^2 \vec{i}+(y+x) \vec{j}F→=xy2i→+(y+x)j→. Integrate ( ∇ × F → ) ⋅ k → ( ∇ × F → ) ⋅ k → (grad xx vec(F))* vec(k)(\nabla \times \vec{F}) \cdot \vec{k}(∇×F→)⋅k→ over the region in the first quadrant bounded by the curves y = x 2 y = x 2 y=x^(2)y=x^2y=x2 and y = x y = x y=xy=xy=x using Green’s theorem.
(d) f ( y ) f ( y ) f(y)f(y)f(y), जिसके लिये समीकरण ( 2 x e y + 3 y 2 ) d y + ( 3 x 2 + f ( y ) ) d x = 0 2 x e y + 3 y 2 d y + 3 x 2 + f ( y ) d x = 0 (2xe^(y)+3y^(2))dy+(3x^(2)+f(y))dx=0\left(2 x e^y+3 y^2\right) d y+\left(3 x^2+f(y)\right) d x=0(2xey+3y2)dy+(3x2+f(y))dx=0 सथातथ्य है, ज्ञात कीजिये और हाल निकालिये।
Find f ( y ) f ( y ) f(y)f(y)f(y) such that ( 2 x e y + 3 y 2 ) d y + ( 3 x 2 + f ( y ) ) d x = 0 2 x e y + 3 y 2 d y + 3 x 2 + f ( y ) d x = 0 (2xe^(y)+3y^(2))dy+(3x^(2)+f(y))dx=0\left(2 x e^y+3 y^2\right) d y+\left(3 x^2+f(y)\right) d x=0(2xey+3y2)dy+(3x2+f(y))dx=0 is exact and hence solve.
\(b=c\:cos\:A+a\:cos\:C\)

UPSC Maths Optional Solved Papers (2018-2022)

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खण्ड A
SECTION A
Question:-01 (a) सिद्ध कीजिए कि n n n\mathrm{n}n विमीय सदिश समष्टि V V V\mathrm{V}V के लिए n n n\mathrm{n}n रैखिकत: स्वतंत्र सदिशों का कोई भी समुच्चय V V V\mathrm{V}V के लिए एक आधार बनाता है ।
Question:-01 (a) Prove that any set of n n n\mathrm{n}n linearly independent vectors in a vector space V V V\mathrm{V}V of dimension n n n\mathrm{n}n constitutes a basis for V V V\mathrm{V}V.
Answer:
To prove that any set of n n nnn linearly independent vectors in a vector space V V VVV of dimension n n nnn constitutes a basis for V V VVV, we need to show two things:
  1. The set spans V V VVV.
  2. The set is linearly independent.

Given:

  • A vector space V V VVV of dimension n n nnn.
  • A set S = { v 1 , v 2 , … , v n } S = { v 1 , v 2 , … , v n } S={v_(1),v_(2),dots,v_(n)}S = \{ \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \}S={v1,v2,…,vn} of n n nnn linearly independent vectors in V V VVV.

Proof:

Step 1: S S SSS is Linearly Independent (Given)

We are given that S S SSS is a set of n n nnn linearly independent vectors. Therefore, no vector in S S SSS can be written as a linear combination of the other vectors in S S SSS.

Step 2: S S SSS Spans V V VVV

To show that S S SSS spans V V VVV, we need to show that any vector w w w\mathbf{w}w in V V VVV can be written as a linear combination of vectors in S S SSS.
Let’s consider another basis B B BBB for V V VVV. Since V V VVV has dimension n n nnn, B B BBB contains exactly n n nnn vectors. We can write w w w\mathbf{w}w as a linear combination of vectors in B B BBB:
w = c 1 b 1 + c 2 b 2 + … + c n b n w = c 1 b 1 + c 2 b 2 + … + c n b n w=c_(1)b_(1)+c_(2)b_(2)+dots+c_(n)b_(n)\mathbf{w} = c_1 \mathbf{b}_1 + c_2 \mathbf{b}_2 + \ldots + c_n \mathbf{b}_nw=c1b1+c2b2+…+cnbn
Now, each b i b i b_(i)\mathbf{b}_ibi can also be written as a linear combination of vectors in S S SSS because S S SSS and B B BBB are both bases for V V VVV:
b i = a i 1 v 1 + a i 2 v 2 + … + a i n v n b i = a i 1 v 1 + a i 2 v 2 + … + a i n v n b_(i)=a_(i1)v_(1)+a_(i2)v_(2)+dots+a_(in)v_(n)\mathbf{b}_i = a_{i1} \mathbf{v}_1 + a_{i2} \mathbf{v}_2 + \ldots + a_{in} \mathbf{v}_nbi=ai1v1+ai2v2+…+ainvn
Substituting this into the equation for w w w\mathbf{w}w, we get:
w = ∑ i = 1 n c i ( a i 1 v 1 + a i 2 v 2 + … + a i n v n ) w = ∑ i = 1 n   c i ( a i 1 v 1 + a i 2 v 2 + … + a i n v n ) w=sum_(i=1)^(n)c_(i)(a_(i1)v_(1)+a_(i2)v_(2)+dots+a_(in)v_(n))\mathbf{w} = \sum_{i=1}^{n} c_i (a_{i1} \mathbf{v}_1 + a_{i2} \mathbf{v}_2 + \ldots + a_{in} \mathbf{v}_n)w=∑i=1nci(ai1v1+ai2v2+…+ainvn)
Simplifying, we find that w w w\mathbf{w}w can indeed be written as a linear combination of vectors in S S SSS, proving that S S SSS spans V V VVV.

Conclusion

Since S S SSS is both linearly independent and spans V V VVV, it constitutes a basis for V V VVV.
Thus, we have proven that any set of n n nnn linearly independent vectors in a vector space V V VVV of dimension n n nnn constitutes a basis for V V VVV.
Question:-01 (b) माना T : R 2 → R 3 T : R 2 → R 3 T:R^(2)rarrR^(3)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^3T:R2→R3 एक रैखिक रूपांतरण, ऐसा है कि T ( 1 0 ) = ( 1 2 3 ) T 1 0 = 1 2 3 T([1],[0])=([1],[2],[3])\mathrm{T}\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)T(10)=(123) तथा T ( 1 1 ) = ( − 3 2 8 ) T 1 1 = − 3 2 8 T([1],[1])=([-3],[2],[8])\mathrm{T}\left(\begin{array}{l}1 \\ 1\end{array}\right)=\left(\begin{array}{r}-3 \\ 2 \\ 8\end{array}\right)T(11)=(−328) है । T ( 2 4 ) T 2 4 T([2],[4])\mathrm{T}\left(\begin{array}{l}2 \\ 4\end{array}\right)T(24) को ज्ञात कीजिए ।
Question:-01(b) Let T : R 2 → R 3 T : R 2 → R 3 T:R^(2)rarrR^(3)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^3T:R2→R3 be a linear transformation such that T ( 1 0 ) = ( 1 2 3 ) T 1 0 = 1 2 3 T([1],[0])=([1],[2],[3])\mathrm{T}\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)T(10)=(123) and T ( 1 1 ) = ( − 3 2 8 ) T 1 1 = − 3 2 8 T([1],[1])=([-3],[2],[8])\mathrm{T}\left(\begin{array}{l}1 \\ 1\end{array}\right)=\left(\begin{array}{r}-3 \\ 2 \\ 8\end{array}\right)T(11)=(−328). Find T ( 2 4 ) T 2 4 T([2],[4])\mathrm{T}\left(\begin{array}{l}2 \\ 4\end{array}\right)T(24)
Answer:
Introduction: In this problem, we are given a linear transformation T : R 2 → R 3 T : R 2 → R 3 T:R^(2)rarrR^(3)T: \mathbb{R}^2 \rightarrow \mathbb{R}^3T:R2→R3 and are asked to find the image of a vector under this transformation. We are given the images of two vectors under T T T\mathrm{T}T and will use this information along with the linearity property of the transformation to find the desired image.
Assumptions: We assume that T T T\mathrm{T}T is a linear transformation.
Definition:
  1. Linear Transformation: A function T : V → W T : V → W T:V rarr WT: V \rightarrow WT:V→W between two vector spaces V V VVV and W W WWW is called a linear transformation if for all vectors u , v ∈ V u , v ∈ V u,v in Vu, v \in Vu,v∈V and scalars c c ccc, the following conditions hold:
    a. T ( u + v ) = T ( u ) + T ( v ) T ( u + v ) = T ( u ) + T ( v ) T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v)
    b. T ( c u ) = c T ( u ) T ( c u ) = c T ( u ) T(cu)=cT(u)T(c u)=c T(u)T(cu)=cT(u)
    Method/Approach: To solve this problem, we will first express the given vector as a linear combination of the two given vectors. Then, we will use the linearity property of the transformation to find the image of the given vector under the transformation T T T\mathrm{T}T.
    Work/Calculations: The calculations have already been provided in the original answer. I’ll restate them here for clarity:
  2. Represent the vector ( 2 4 ) 2 4 ([2],[4])\left(\begin{array}{c}2 \\ 4\end{array}\right)(24) as a linear combination of the given vectors:
( 2 4 ) = a ( 1 0 ) + b ( 1 1 ) 2 4 = a 1 0 + b 1 1 ([2],[4])=a([1],[0])+b([1],[1])\left(\begin{array}{l} 2 \\ 4 \end{array}\right)=a\left(\begin{array}{l} 1 \\ 0 \end{array}\right)+b\left(\begin{array}{l} 1 \\ 1 \end{array}\right)(24)=a(10)+b(11)
  1. Solve for a a aaa and b b bbb :
2 = a + b 4 = 0 + b 2 = a + b 4 = 0 + b {:[2=a+b],[4=0+b]:}\begin{aligned} & 2=a+b \\ & 4=0+b \end{aligned}2=a+b4=0+b
  1. Find a = − 2 a = − 2 a=-2a=-2a=−2 and b = 4 b = 4 b=4b=4b=4 and write the linear combination:
( 2 4 ) = − 2 ( 1 0 ) + 4 ( 1 1 ) 2 4 = − 2 1 0 + 4 1 1 ([2],[4])=-2([1],[0])+4([1],[1])\left(\begin{array}{l} 2 \\ 4 \end{array}\right)=-2\left(\begin{array}{l} 1 \\ 0 \end{array}\right)+4\left(\begin{array}{l} 1 \\ 1 \end{array}\right)(24)=−2(10)+4(11)
  1. Apply the linear transformation T T T\mathrm{T}T :
T ( 2 4 ) = − 2 T ( 1 0 ) + 4 T ( 1 1 ) T 2 4 = − 2 T 1 0 + 4 T 1 1 T([2],[4])=-2T([1],[0])+4T([1],[1])\mathrm{T}\left(\begin{array}{l} 2 \\ 4 \end{array}\right)=-2 \mathrm{~T}\left(\begin{array}{l} 1 \\ 0 \end{array}\right)+4 \mathrm{~T}\left(\begin{array}{l} 1 \\ 1 \end{array}\right)T(24)=−2 T(10)+4 T(11)
  1. Use the given values for the transformation, perform scalar multiplication and vector addition:
T ( 2 4 ) = ( − 2 − 4 − 6 ) + ( − 12 8 32 ) = ( − 14 4 26 ) Thus, T ( 2 4 ) = ( − 14 4 26 ) . T 2 4 = − 2 − 4 − 6 + − 12 8 32 = − 14 4 26  Thus,  T 2 4 = − 14 4 26 . {:[T([2],[4])=([-2],[-4],[-6])+([-12],[8],[32])=([-14],[4],[26])],[” Thus, “T([2],[4])=([-14],[4],[26]).]:}\begin{aligned} & T\left(\begin{array}{l} 2 \\ 4 \end{array}\right)=\left(\begin{array}{l} -2 \\ -4 \\ -6 \end{array}\right)+\left(\begin{array}{c} -12 \\ 8 \\ 32 \end{array}\right)=\left(\begin{array}{c} -14 \\ 4 \\ 26 \end{array}\right) \\ & \text { Thus, } \mathrm{T}\left(\begin{array}{l} 2 \\ 4 \end{array}\right)=\left(\begin{array}{c} -14 \\ 4 \\ 26 \end{array}\right) . \end{aligned}T(24)=(−2−4−6)+(−12832)=(−14426) Thus, T(24)=(−14426).
Conclusion: In this problem, we found the image of the vector ( 2 4 ) 2 4 ([2],[4])\left(\begin{array}{c}2 \\ 4\end{array}\right)(24) under the linear transformation T : R 2 → R 3 T : R 2 → R 3 T:R^(2)rarrR^(3)T: \mathbb{R}^2 \rightarrow \mathbb{R}^3T:R2→R3. We expressed the given vector as a linear combination of the two given vectors and used the linearity property of the transformation to find the image.
The image of the given vector under the transformation T T TTT is ( − 14 4 26 ) − 14 4 26 ([-14],[4],[26])\left(\begin{array}{c}-14 \\ 4 \\ 26\end{array}\right)(−14426).
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1.(a) यदि A = [ 1 − 1 1 2 − 1 0 1 0 0 ] A = 1 − 1 1 2 − 1 0 1 0 0 A=[[1,-1,1],[2,-1,0],[1,0,0]]A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right]A=[1−112−10100] है, तो A − 1 A − 1 A^(-1)A^{-1}A−1 को ज्ञात किए बिना दर्शाइए कि A 2 = A − 1 A 2 = A − 1 A^(2)=A^(-1)A^2=A^{-1}A2=A−1
If A = [ 1 − 1 1 2 − 1 0 1 0 0 ] A = 1 − 1 1 2 − 1 0 1 0 0 A=[[1,-1,1],[2,-1,0],[1,0,0]]A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right]A=[1−112−10100], then show that A 2 = A − 1 A 2 = A − 1 A^(2)=A^(-1)A^2=A^{-1}A2=A−1 (without finding A − 1 A − 1 A^(-1)A^{-1}A−1 ).
Answer:
Introduction
We are given a 3×3 matrix A A AAA and asked to prove that the square of this matrix, A 2 A 2 A^(2)A^2A2, is equal to its inverse, A − 1 A − 1 A^(-1)A^{-1}A−1, without explicitly finding the inverse of A A AAA.
Assumptions
The problem assumes that the matrix A A AAA is invertible, i.e., it has an inverse A − 1 A − 1 A^(-1)A^{-1}A−1.
Definition
The inverse of a matrix A A AAA, denoted A − 1 A − 1 A^(-1)A^{-1}A−1, is a unique matrix such that when it is multiplied by A A AAA, the result is the identity matrix I I III. The identity matrix is a special square matrix with ones on the diagonal and zeros elsewhere.
Method/Approach
We will use the definition of the inverse of a matrix to solve this problem. If A 2 A 2 A^(2)A^2A2 is indeed the inverse of A A AAA, then the product A A 2 A A 2 AA^(2)AA^2AA2 should be the identity matrix I I III. We will calculate A 2 A 2 A^(2)A^2A2 and A A 2 A A 2 AA^(2)AA^2AA2 to verify this.
Work/Calculations
First, let’s calculate A 2 A 2 A^(2)A^2A2:
A 2 = A ⋅ A = [ 1 − 1 1 2 − 1 0 1 0 0 ] ⋅ [ 1 − 1 1 2 − 1 0 1 0 0 ] A 2 = A ⋅ A = 1 − 1 1 2 − 1 0 1 0 0 ⋅ 1 − 1 1 2 − 1 0 1 0 0 A^(2)=A*A=[[1,-1,1],[2,-1,0],[1,0,0]]*[[1,-1,1],[2,-1,0],[1,0,0]]A^2 = A \cdot A = \left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right] \cdot \left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right]A2=A⋅A=[1−112−10100]⋅[1−112−10100]
The calculation yields:
A 2 = [ 0 0 1 0 − 1 2 1 − 1 1 ] A 2 = 0 0 1 0 − 1 2 1 − 1 1 A^(2)=[[0,0,1],[0,-1,2],[1,-1,1]]A^2 = \left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{array}\right]A2=[0010−121−11]
Next, we calculate A A 2 A A 2 AA^(2)AA^2AA2:
A A 2 = A ⋅ A 2 = [ 1 − 1 1 2 − 1 0 1 0 0 ] ⋅ [ 0 0 1 0 − 1 2 1 − 1 1 ] A A 2 = A ⋅ A 2 = 1 − 1 1 2 − 1 0 1 0 0 ⋅ 0 0 1 0 − 1 2 1 − 1 1 AA^(2)=A*A^(2)=[[1,-1,1],[2,-1,0],[1,0,0]]*[[0,0,1],[0,-1,2],[1,-1,1]]AA^2 = A \cdot A^2 = \left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right] \cdot \left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1\end{array}\right]AA2=A⋅A2=[1−112−10100]⋅[0010−121−11]
The calculation yields:
A A 2 = [ 1 0 0 0 1 0 0 0 1 ] A A 2 = 1 0 0 0 1 0 0 0 1 AA^(2)=[[1,0,0],[0,1,0],[0,0,1]]AA^2 = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]AA2=[100010001]
Conclusion
The product of A A AAA and A 2 A 2 A^(2)A^2A2 is the identity matrix I I III. Therefore, we have shown that A 2 A 2 A^(2)A^2A2 is indeed the inverse of A A AAA, i.e., A 2 = A − 1 A 2 = A − 1 A^(2)=A^(-1)A^2 = A^{-1}A2=A−1, without explicitly finding A − 1 A − 1 A^(-1)A^{-1}A−1.
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1.(b) क्रमित आधारक B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B={(0,1,1),(1,0,1),(1,1,0)}B=\{(0,1,1),(1,0,1),(1,1,0)\}B={(0,1,1),(1,0,1),(1,1,0)} के सापेक्ष V 3 ( R ) V 3 ( R ) V_(3)(R)V_3(R)V3(R) पर परिभाषित रैखिक संकारक : T ( a , b , c ) = ( a + b , a − b , 2 c ) T ( a , b , c ) = ( a + b , a − b , 2 c ) T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c)T(a,b,c)=(a+b,a−b,2c) से संबन्धित आव्यूह ज्ञात कीजिए ।
Find the matrix associated with the linear operator on V 3 ( R ) V 3 ( R ) V_(3)(R)V_3(R)V3(R) defined by T ( a , b , c ) = ( a + b , a − b , 2 c ) T ( a , b , c ) = ( a + b , a − b , 2 c ) T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c)T(a,b,c)=(a+b,a−b,2c) with respect to the ordered basis B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B={(0,1,1),(1,0,1),(1,1,0)}B=\{(0,1,1),(1,0,1),(1,1,0)\}B={(0,1,1),(1,0,1),(1,1,0)}.
Answer:

Introduction

We are tasked with finding the matrix representation of the linear operator T : V 3 ( R ) → V 3 ( R ) T : V 3 ( R ) → V 3 ( R ) T:V_(3)(R)rarrV_(3)(R)T: V_3(\mathbb{R}) \to V_3(\mathbb{R})T:V3(R)→V3(R) defined by T ( a , b , c ) = ( a + b , a − b , 2 c ) T ( a , b , c ) = ( a + b , a − b , 2 c ) T(a,b,c)=(a+b,a-b,2c)T(a, b, c) = (a+b, a-b, 2c)T(a,b,c)=(a+b,a−b,2c). The matrix representation will be with respect to the ordered basis B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B={(0,1,1),(1,0,1),(1,1,0)}B = \{(0,1,1), (1,0,1), (1,1,0)\}B={(0,1,1),(1,0,1),(1,1,0)}.

Assumptions

  • T T TTT is a linear operator.
  • V 3 ( R ) V 3 ( R ) V_(3)(R)V_3(\mathbb{R})V3(R) is a vector space over the real numbers R R R\mathbb{R}R.
  • The ordered basis B B BBB consists of vectors in V 3 ( R ) V 3 ( R ) V_(3)(R)V_3(\mathbb{R})V3(R).

Method/Approach

To find the matrix representation of T T TTT with respect to the basis B B BBB, we will:
  1. Apply T T TTT to each vector in the basis B B BBB.
  2. Express the resulting vectors in terms of the basis B B BBB.
  3. Use the coefficients as columns of the matrix representation of T T TTT.

Work/Calculations

Step 1: Apply T T TTT to each vector in B B BBB

Let’s apply T T TTT to the vectors ( 0 , 1 , 1 ) ( 0 , 1 , 1 ) (0,1,1)(0,1,1)(0,1,1), ( 1 , 0 , 1 ) ( 1 , 0 , 1 ) (1,0,1)(1,0,1)(1,0,1), and ( 1 , 1 , 0 ) ( 1 , 1 , 0 ) (1,1,0)(1,1,0)(1,1,0).
  1. T ( 0 , 1 , 1 ) = ( 0 + 1 , 0 − 1 , 2 × 1 ) T ( 0 , 1 , 1 ) = ( 0 + 1 , 0 − 1 , 2 × 1 ) T(0,1,1)=(0+1,0-1,2xx1)T(0,1,1) = (0+1, 0-1, 2 \times 1)T(0,1,1)=(0+1,0−1,2×1)
  2. T ( 1 , 0 , 1 ) = ( 1 + 0 , 1 − 0 , 2 × 1 ) T ( 1 , 0 , 1 ) = ( 1 + 0 , 1 − 0 , 2 × 1 ) T(1,0,1)=(1+0,1-0,2xx1)T(1,0,1) = (1+0, 1-0, 2 \times 1)T(1,0,1)=(1+0,1−0,2×1)
  3. T ( 1 , 1 , 0 ) = ( 1 + 1 , 1 − 1 , 2 × 0 ) T ( 1 , 1 , 0 ) = ( 1 + 1 , 1 − 1 , 2 × 0 ) T(1,1,0)=(1+1,1-1,2xx0)T(1,1,0) = (1+1, 1-1, 2 \times 0)T(1,1,0)=(1+1,1−1,2×0)
After calculating, we get the following transformed vectors:
  1. T ( 0 , 1 , 1 ) = ( 1 , − 1 , 2 ) T ( 0 , 1 , 1 ) = ( 1 , − 1 , 2 ) T(0,1,1)=(1,-1,2)T(0,1,1) = (1, -1, 2)T(0,1,1)=(1,−1,2)
  2. T ( 1 , 0 , 1 ) = ( 1 , 1 , 2 ) T ( 1 , 0 , 1 ) = ( 1 , 1 , 2 ) T(1,0,1)=(1,1,2)T(1,0,1) = (1, 1, 2)T(1,0,1)=(1,1,2)
  3. T ( 1 , 1 , 0 ) = ( 2 , 0 , 0 ) T ( 1 , 1 , 0 ) = ( 2 , 0 , 0 ) T(1,1,0)=(2,0,0)T(1,1,0) = (2, 0, 0)T(1,1,0)=(2,0,0)

Step 2: Express the resulting vectors in terms of B B BBB

We need to express each of these vectors as a linear combination of the basis vectors in B B BBB.
  1. ( 1 , − 1 , 2 ) = a 1 ( 0 , 1 , 1 ) + a 2 ( 1 , 0 , 1 ) + a 3 ( 1 , 1 , 0 ) ( 1 , − 1 , 2 ) = a 1 ( 0 , 1 , 1 ) + a 2 ( 1 , 0 , 1 ) + a 3 ( 1 , 1 , 0 ) (1,-1,2)=a_(1)(0,1,1)+a_(2)(1,0,1)+a_(3)(1,1,0)(1, -1, 2) = a_1 (0,1,1) + a_2 (1,0,1) + a_3 (1,1,0)(1,−1,2)=a1(0,1,1)+a2(1,0,1)+a3(1,1,0)
  2. ( 1 , 1 , 2 ) = b 1 ( 0 , 1 , 1 ) + b 2 ( 1 , 0 , 1 ) + b 3 ( 1 , 1 , 0 ) ( 1 , 1 , 2 ) = b 1 ( 0 , 1 , 1 ) + b 2 ( 1 , 0 , 1 ) + b 3 ( 1 , 1 , 0 ) (1,1,2)=b_(1)(0,1,1)+b_(2)(1,0,1)+b_(3)(1,1,0)(1, 1, 2) = b_1 (0,1,1) + b_2 (1,0,1) + b_3 (1,1,0)(1,1,2)=b1(0,1,1)+b2(1,0,1)+b3(1,1,0)
  3. ( 2 , 0 , 0 ) = c 1 ( 0 , 1 , 1 ) + c 2 ( 1 , 0 , 1 ) + c 3 ( 1 , 1 , 0 ) ( 2 , 0 , 0 ) = c 1 ( 0 , 1 , 1 ) + c 2 ( 1 , 0 , 1 ) + c 3 ( 1 , 1 , 0 ) (2,0,0)=c_(1)(0,1,1)+c_(2)(1,0,1)+c_(3)(1,1,0)(2, 0, 0) = c_1 (0,1,1) + c_2 (1,0,1) + c_3 (1,1,0)(2,0,0)=c1(0,1,1)+c2(1,0,1)+c3(1,1,0)
After calculating, we find the coefficients as follows:
  1. a 1 = 0 , a 2 = 2 , a 3 = − 1 a 1 = 0 , a 2 = 2 , a 3 = − 1 a_(1)=0,a_(2)=2,a_(3)=-1a_1 = 0, a_2 = 2, a_3 = -1a1=0,a2=2,a3=−1
  2. b 1 = 1 , b 2 = 1 , b 3 = 0 b 1 = 1 , b 2 = 1 , b 3 = 0 b_(1)=1,b_(2)=1,b_(3)=0b_1 = 1, b_2 = 1, b_3 = 0b1=1,b2=1,b3=0
  3. c 1 = − 1 , c 2 = 1 , c 3 = 1 c 1 = − 1 , c 2 = 1 , c 3 = 1 c_(1)=-1,c_(2)=1,c_(3)=1c_1 = -1, c_2 = 1, c_3 = 1c1=−1,c2=1,c3=1
Thus, the vectors ( 1 , − 1 , 2 ) ( 1 , − 1 , 2 ) (1,-1,2)(1, -1, 2)(1,−1,2), ( 1 , 1 , 2 ) ( 1 , 1 , 2 ) (1,1,2)(1, 1, 2)(1,1,2), and ( 2 , 0 , 0 ) ( 2 , 0 , 0 ) (2,0,0)(2, 0, 0)(2,0,0) can be expressed in terms of the basis B B BBB as:
  1. ( 1 , − 1 , 2 ) = 0 ⋅ ( 0 , 1 , 1 ) + 2 ⋅ ( 1 , 0 , 1 ) + ( − 1 ) ⋅ ( 1 , 1 , 0 ) ( 1 , − 1 , 2 ) = 0 ⋅ ( 0 , 1 , 1 ) + 2 ⋅ ( 1 , 0 , 1 ) + ( − 1 ) ⋅ ( 1 , 1 , 0 ) (1,-1,2)=0*(0,1,1)+2*(1,0,1)+(-1)*(1,1,0)(1, -1, 2) = 0 \cdot (0,1,1) + 2 \cdot (1,0,1) + (-1) \cdot (1,1,0)(1,−1,2)=0⋅(0,1,1)+2⋅(1,0,1)+(−1)⋅(1,1,0)
  2. ( 1 , 1 , 2 ) = 1 ⋅ ( 0 , 1 , 1 ) + 1 ⋅ ( 1 , 0 , 1 ) + 0 ⋅ ( 1 , 1 , 0 ) ( 1 , 1 , 2 ) = 1 ⋅ ( 0 , 1 , 1 ) + 1 ⋅ ( 1 , 0 , 1 ) + 0 ⋅ ( 1 , 1 , 0 ) (1,1,2)=1*(0,1,1)+1*(1,0,1)+0*(1,1,0)(1, 1, 2) = 1 \cdot (0,1,1) + 1 \cdot (1,0,1) + 0 \cdot (1,1,0)(1,1,2)=1⋅(0,1,1)+1⋅(1,0,1)+0⋅(1,1,0)
  3. ( 2 , 0 , 0 ) = − 1 ⋅ ( 0 , 1 , 1 ) + 1 ⋅ ( 1 , 0 , 1 ) + 1 ⋅ ( 1 , 1 , 0 ) ( 2 , 0 , 0 ) = − 1 ⋅ ( 0 , 1 , 1 ) + 1 ⋅ ( 1 , 0 , 1 ) + 1 ⋅ ( 1 , 1 , 0 ) (2,0,0)=-1*(0,1,1)+1*(1,0,1)+1*(1,1,0)(2, 0, 0) = -1 \cdot (0,1,1) + 1 \cdot (1,0,1) + 1 \cdot (1,1,0)(2,0,0)=−1⋅(0,1,1)+1⋅(1,0,1)+1⋅(1,1,0)

Step 3: Form the Matrix Representation

The matrix representation [ T ] B [ T ] B [T]_(B)[T]_B[T]B of T T TTT with respect to the basis B B BBB is formed by using these coefficients as columns:
[ T ] B = ( 0 1 − 1 2 1 1 − 1 0 1 ) [ T ] B = 0 1 − 1 2 1 1 − 1 0 1 [T]_(B)=([0,1,-1],[2,1,1],[-1,0,1])[T]_B = \begin{pmatrix} 0 & 1 & -1 \\ 2 & 1 & 1 \\ -1 & 0 & 1 \end{pmatrix}[T]B=(01−1211−101)

Conclusion

The matrix representation of the linear operator T T TTT with respect to the ordered basis B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B = { ( 0 , 1 , 1 ) , ( 1 , 0 , 1 ) , ( 1 , 1 , 0 ) } B={(0,1,1),(1,0,1),(1,1,0)}B = \{(0,1,1), (1,0,1), (1,1,0)\}B={(0,1,1),(1,0,1),(1,1,0)} is:
[ T ] B = ( 0 1 − 1 2 1 1 − 1 0 1 ) [ T ] B = 0 1 − 1 2 1 1 − 1 0 1 [T]_(B)=([0,1,-1],[2,1,1],[-1,0,1])[T]_B = \begin{pmatrix} 0 & 1 & -1 \\ 2 & 1 & 1 \\ -1 & 0 & 1 \end{pmatrix}[T]B=(01−1211−101)
untitled-document-15-c4a14609-db5c-41e1-99f0-81aab3fdac8f
खण्ड-A / SECTION-A
  1. (a) माना समुच्चय V V VVV में सभी n × n n × n n xx nn \times nn×n के वास्तविक मैजिक वर्ग हैं। दिखाइए कि समुच्चय V , R V , R V,RV, RV,R पर एक सदिश समष्टि है। दो भिन्न-भिन्न 2 × 2 2 × 2 2xx22 \times 22×2 मैजिक वर्ग के उदाहरण दीजिए।
Consider the set V V VVV of all n × n n × n n xx nn \times nn×n real magic squares. Show that V V VVV is a vector space over R R RRR. Give examples of two distinct 2 × 2 2 × 2 2xx22 \times 22×2 magic squares.
Answer:

Introduction

The problem asks us to prove that the set V V VVV of all n × n n × n n xx nn \times nn×n real magic squares is a vector space over R R R\mathbb{R}R. A magic square is a square grid of numbers such that the sums of the numbers in each row, each column, and both main diagonals are the same. We will use the properties of vector spaces to prove this.
To make the proof more explicit, let’s assume X , Y , X , Y , X,Y,X, Y,X,Y, and Z Z ZZZ are n × n n × n n xx nn \times nn×n magic squares with general entries as follows:
X = [ x 11 x 12 x 13 … x 1 n x 21 x 22 x 23 … x 2 n ⋮ ⋮ ⋮ ⋱ ⋮ x n 1 x n 2 x n 3 … x n n ] X = x 11 x 12 x 13 … x 1 n x 21 x 22 x 23 … x 2 n ⋮ ⋮ ⋮ ⋱ ⋮ x n 1 x n 2 x n 3 … x n n X=[[x_(11),x_(12),x_(13),dots,x_(1n)],[x_(21),x_(22),x_(23),dots,x_(2n)],[vdots,vdots,vdots,ddots,vdots],[x_(n1),x_(n2),x_(n3),dots,x_(nn)]]X=\left[\begin{array}{ccccc} x_{11} & x_{12} & x_{13} & \ldots & x_{1n} \\ x_{21} & x_{22} & x_{23} & \ldots & x_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & x_{n3} & \ldots & x_{nn} \end{array}\right]X=[x11x12x13…x1nx21x22x23…x2n⋮⋮⋮⋱⋮xn1xn2xn3…xnn]
Y = [ y 11 y 12 y 13 … y 1 n y 21 y 22 y 23 … y 2 n ⋮ ⋮ ⋮ ⋱ ⋮ y n 1 y n 2 y n 3 … y n n ] Y = y 11 y 12 y 13 … y 1 n y 21 y 22 y 23 … y 2 n ⋮ ⋮ ⋮ ⋱ ⋮ y n 1 y n 2 y n 3 … y n n Y=[[y_(11),y_(12),y_(13),dots,y_(1n)],[y_(21),y_(22),y_(23),dots,y_(2n)],[vdots,vdots,vdots,ddots,vdots],[y_(n1),y_(n2),y_(n3),dots,y_(nn)]]Y=\left[\begin{array}{ccccc} y_{11} & y_{12} & y_{13} & \ldots & y_{1n} \\ y_{21} & y_{22} & y_{23} & \ldots & y_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ y_{n1} & y_{n2} & y_{n3} & \ldots & y_{nn} \end{array}\right]Y=[y11y12y13…y1ny21y22y23…y2n⋮⋮⋮⋱⋮yn1yn2yn3…ynn]
Z = [ z 11 z 12 z 13 … z 1 n z 21 z 22 z 23 … z 2 n ⋮ ⋮ ⋮ ⋱ ⋮ z n 1 z n 2 z n 3 … z n n ] Z = z 11 z 12 z 13 … z 1 n z 21 z 22 z 23 … z 2 n ⋮ ⋮ ⋮ ⋱ ⋮ z n 1 z n 2 z n 3 … z n n Z=[[z_(11),z_(12),z_(13),dots,z_(1n)],[z_(21),z_(22),z_(23),dots,z_(2n)],[vdots,vdots,vdots,ddots,vdots],[z_(n1),z_(n2),z_(n3),dots,z_(nn)]]Z=\left[\begin{array}{ccccc} z_{11} & z_{12} & z_{13} & \ldots & z_{1n} \\ z_{21} & z_{22} & z_{23} & \ldots & z_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ z_{n1} & z_{n2} & z_{n3} & \ldots & z_{nn} \end{array}\right]Z=[z11z12z13…z1nz21z22z23…z2n⋮⋮⋮⋱⋮zn1zn2zn3…znn]
And let a a aaa and b b bbb be real numbers.

Work/Calculations

Property 1: Closure under Addition and Scalar Multiplication

  1. X + Y ∈ MS ⁡ ( n ) X + Y ∈ MS ⁡ ( n ) X+Y in MS(n)X + Y \in \operatorname{MS}(n)X+Y∈MS⁡(n)
    Let’s substitute the values:
    ( X + Y ) = [ x i j + y i j ] ( X + Y ) = [ x i j + y i j ] (X+Y)=[x_(ij)+y_(ij)](X + Y) = [x_{ij} + y_{ij}](X+Y)=[xij+yij]
    The sum of each row, column, and diagonal in X X XXX and Y Y YYY is the same constant k k kkk. Therefore, the sum of each corresponding row, column, and diagonal in X + Y X + Y X+YX + YX+Y will be 2 k 2 k 2k2k2k, which means X + Y X + Y X+YX + YX+Y is also a magic square.
  2. a X ∈ MS ⁡ ( n ) a X ∈ MS ⁡ ( n ) aX in MS(n)aX \in \operatorname{MS}(n)aX∈MS⁡(n)
    Let’s substitute the values:
    a X = [ a x i j ] a X = [ a x i j ] aX=[ax_(ij)]aX = [ax_{ij}]aX=[axij]
    If we multiply X X XXX by a scalar a a aaa, each row, column, and diagonal sum becomes a k a k akakak, which means a X a X aXaXaX is also a magic square.
After calculating, we find that both X + Y X + Y X+YX + YX+Y and a X a X aXaXaX are in MS ⁡ ( n ) MS ⁡ ( n ) MS(n)\operatorname{MS}(n)MS⁡(n), proving closure under addition and scalar multiplication.

Property 2: Commutativity of Addition

X + Y = Y + X X + Y = Y + X X+Y=Y+XX + Y = Y + XX+Y=Y+X
This is straightforward because matrix addition is commutative.

Property 3: Associativity of Addition

X + ( Y + Z ) = ( X + Y ) + Z X + ( Y + Z ) = ( X + Y ) + Z X+(Y+Z)=(X+Y)+ZX + (Y + Z) = (X + Y) + ZX+(Y+Z)=(X+Y)+Z
Matrix addition is associative, so this property holds.

Property 4: Existence of Zero Vector

Let 0 0 0\mathbf{0}0 be the n × n n × n n xx nn \times nn×n magic square where every entry is zero. Then,
X + 0 = 0 + X = X X + 0 = 0 + X = X X+0=0+X=XX + \mathbf{0} = \mathbf{0} + X = XX+0=0+X=X

Property 5: Existence of Additive Inverse

Let X ′ = − X X ′ = − X X^(‘)=-XX’ = -XX′=−X. Then,
X + X ′ = X ′ + X = 0 X + X ′ = X ′ + X = 0 X+X^(‘)=X^(‘)+X=0X + X’ = X’ + X = \mathbf{0}X+X′=X′+X=0

Property 6: Distributive Law 1

a ( X + Y ) = a X + a Y a ( X + Y ) = a X + a Y a(X+Y)=aX+aYa(X + Y) = aX + aYa(X+Y)=aX+aY
This is a property of matrices, so it holds.

Property 7: Distributive Law 2

( a + b ) X = a X + b X ( a + b ) X = a X + b X (a+b)X=aX+bX(a + b)X = aX + bX(a+b)X=aX+bX
This is also a property of matrices.

Property 8: Associativity of Scalar Multiplication

( a b ) X = a ( b X ) ( a b ) X = a ( b X ) (ab)X=a(bX)(ab)X = a(bX)(ab)X=a(bX)
This is true for matrices.

Property 9: Multiplication by Identity

1 X = X 1 X = X 1X=X1X = X1X=X
This is true for any matrix X X XXX.
We have shown that the set V V VVV of all n × n n × n n xx nn \times nn×n real magic squares satisfies all the properties required for it to be a vector space over R R R\mathbb{R}R. Therefore, V V VVV is indeed a vector space over R R R\mathbb{R}R.
For a 2 × 2 2 × 2 2xx22 \times 22×2 matrix to be a magic square, the sum of each row, each column, and both diagonals must be the same. Let’s consider a general 2 × 2 2 × 2 2xx22 \times 22×2 magic square M M MMM with entries a , b , c , a , b , c , a,b,c,a, b, c,a,b,c, and d d ddd:
M = [ a b c d ] M = a b c d M=[[a,b],[c,d]]M = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right]M=[abcd]
For M M MMM to be a magic square, the following conditions must be met:
  1. The sum of each row must be the same: a + b = c + d a + b = c + d a+b=c+da + b = c + da+b=c+d
  2. The sum of each column must be the same: a + c = b + d a + c = b + d a+c=b+da + c = b + da+c=b+d
  3. The sum of the diagonals must be the same: a + d = b + c a + d = b + c a+d=b+ca + d = b + ca+d=b+c

Example 1

Let’s choose a = 1 , b = 1 , c = 1 , a = 1 , b = 1 , c = 1 , a=1,b=1,c=1,a = 1, b = 1, c = 1,a=1,b=1,c=1, and d = 1 d = 1 d=1d = 1d=1. All the sums are 1 + 1 = 2 1 + 1 = 2 1+1=21 + 1 = 21+1=2, so it’s a magic square.
M 1 = [ 1 1 1 1 ] M 1 = 1 1 1 1 M_(1)=[[1,1],[1,1]]M_1 = \left[\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right]M1=[1111]

Example 2

Let’s choose a = 2 , b = 2 , c = 2 , a = 2 , b = 2 , c = 2 , a=2,b=2,c=2,a = 2, b = 2, c = 2,a=2,b=2,c=2, and d = 2 d = 2 d=2d = 2d=2. All the sums are 2 + 2 = 4 2 + 2 = 4 2+2=42 + 2 = 42+2=4, so it’s a magic square.
M 2 = [ 2 2 2 2 ] M 2 = 2 2 2 2 M_(2)=[[2,2],[2,2]]M_2 = \left[\begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array}\right]M2=[2222]

Conclusion

Therefore, we conclude that the set V V VVV of all n × n n × n n xx nn \times nn×n real magic squares is a valid vector space over R R R\mathbb{R}R. This conclusion is underpinned by the rigorous application of vector space properties and the concrete examples provided for 2 × 2 2 × 2 2xx22 \times 22×2 magic squares, affirming the validity of this mathematical concept.
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(b) माना M 2 ( R ) M 2 ( R ) M_(2)(R)M_2(R)M2(R) सभी 2 × 2 2 × 2 2xx22 \times 22×2 वास्तविक आव्यूहों का सदिश समष्टि है। माना B = [ 1 − 1 − 4 4 ] B = 1 − 1 − 4 4 B=[[1,-1],[-4,4]]B=\left[\begin{array}{cc}1 & -1 \\ -4 & 4\end{array}\right]B=[1−1−44]. माना T : M 2 ( R ) → M 2 ( R ) T : M 2 ( R ) → M 2 ( R ) T:M_(2)(R)rarrM_(2)(R)T: M_2(R) \rightarrow M_2(R)T:M2(R)→M2(R) एक रैखिक रूपांतरण है, जो T ( A ) = B A T ( A ) = B A T(A)=BAT(A)=B AT(A)=BA द्वारा परिभाषित है। T T TTT की कोटि (रिक) व शून्यता (नलिटि) ज्ञात कीजिए। आव्यूह A A AAA ज्ञात कीजिए, जो शून्य आव्यूह को प्रतिचित्रित करता है।
Let M 2 ( R ) M 2 ( R ) M_(2)(R)M_2(R)M2(R) be the vector space of all 2 × 2 2 × 2 2xx22 \times 22×2 real matrices. Let B = [ 1 − 1 − 4 4 ] B = 1 − 1 − 4 4 B=[[1,-1],[-4,4]]B=\left[\begin{array}{cc}1 & -1 \\ -4 & 4\end{array}\right]B=[1−1−44]. Suppose T : M 2 ( R ) → M 2 ( R ) T : M 2 ( R ) → M 2 ( R ) T:M_(2)(R)rarrM_(2)(R)T: M_2(R) \rightarrow M_2(R)T:M2(R)→M2(R) is a linear transformation defined by T ( A ) = B A T ( A ) = B A T(A)=BAT(A)=B AT(A)=BA. Find the rank and nullity of T T TTT. Find a matrix A A AAA which maps to the null matrix.
Answer:

Introduction

We are given a vector space M 2 ( R ) M 2 ( R ) M_(2)(R)M_2(\mathbb{R})M2(R) of all 2 × 2 2 × 2 2xx22 \times 22×2 real matrices and a specific matrix B B BBB. A linear transformation T : M 2 ( R ) → M 2 ( R ) T : M 2 ( R ) → M 2 ( R ) T:M_(2)(R)rarrM_(2)(R)T: M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})T:M2(R)→M2(R) is defined as T ( A ) = B A T ( A ) = B A T(A)=BAT(A) = BAT(A)=BA. We are asked to find the rank and nullity of T T TTT and to find a matrix A A AAA that maps to the null matrix under T T TTT.

Work/Calculations

Finding a Matrix A A AAA that Maps to the Null Matrix

To find a matrix A A AAA that maps to the null matrix under T T TTT, we need to find A A AAA such that B A = 0 B A = 0 BA=0BA = 0BA=0.
Let A = [ a b c d ] A = a b c d A=[[a,b],[c,d]]A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right]A=[abcd].
Then B A B A BABABA is:
B A = [ 1 − 1 − 4 4 ] [ a b c d ] = [ a − c b − d − 4 a + 4 c − 4 b + 4 d ] B A = 1 − 1 − 4 4 a b c d = a − c b − d − 4 a + 4 c − 4 b + 4 d BA=[[1,-1],[-4,4]][[a,b],[c,d]]=[[a-c,b-d],[-4a+4c,-4b+4d]]BA = \left[\begin{array}{cc} 1 & -1 \\ -4 & 4 \end{array}\right] \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] = \left[\begin{array}{cc} a – c & b – d \\ -4a + 4c & -4b + 4d \end{array}\right]BA=[1−1−44][abcd]=[a−cb−d−4a+4c−4b+4d]
For B A B A BABABA to be the null matrix, we need a − c = 0 a − c = 0 a-c=0a – c = 0a−c=0, b − d = 0 b − d = 0 b-d=0b – d = 0b−d=0, − 4 a + 4 c = 0 − 4 a + 4 c = 0 -4a+4c=0-4a + 4c = 0−4a+4c=0, and − 4 b + 4 d = 0 − 4 b + 4 d = 0 -4b+4d=0-4b + 4d = 0−4b+4d=0.
Solving these equations, we find a = c a = c a=ca = ca=c and b = d b = d b=db = db=d.
Therefore, any matrix A A AAA of the form [ a b a b ] a b a b [[a,b],[a,b]]\left[\begin{array}{cc} a & b \\ a & b \end{array}\right][abab] will map to the null matrix under T T TTT.

Nullity of T T TTT

The nullity of T T TTT is the dimension of the null space of T T TTT, denoted as N ( T ) N ( T ) N(T)N(T)N(T). The null space consists of all matrices A A AAA such that B A = 0 B A = 0 BA=0BA = 0BA=0.
Any matrix A A AAA of the form [ a b a b ] a b a b [[a,b],[a,b]]\left[\begin{array}{cc} a & b \\ a & b \end{array}\right][abab] will map to the null matrix under T T TTT.
We can express this as:
[ x 1 x 2 x 3 x 4 ] = [ p q p q ] = p [ 1 0 1 0 ] + q [ 0 1 0 1 ] x 1      x 2 x 3      x 4 = p      q p      q = p 1      0 1      0 + q 0      1 0      1 [[x_(1),x_(2)],[x_(3),x_(4)]]=[[p,q],[p,q]]=p[[1,0],[1,0]]+q[[0,1],[0,1]]\left[\begin{array}{ll}x_1 & x_2 \\ x_3 & x_4\end{array}\right]=\left[\begin{array}{ll}p & q \\ p & q\end{array}\right]=p\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right]+q\left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right][x1x2x3x4]=[pqpq]=p[1010]+q[0101]
Thus, the null space is spanned by the matrices [ 1 0 1 0 ] 1      0 1      0 [[1,0],[1,0]]\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right][1010] and [ 0 1 0 1 ] 0      1 0      1 [[0,1],[0,1]]\left[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\right][0101], and its dimension is 2. Hence, Nullity ( T ) = 2 Nullity ( T ) = 2 “Nullity”(T)=2\text{Nullity}(T) = 2Nullity(T)=2.

Rank of T T TTT

By the rank-nullity theorem, we have:
ρ ( T ) + N ( T ) = 4 ⟹ ρ ( T ) = 4 − 2 = 2 ρ ( T ) + N ( T ) = 4 ⟹ ρ ( T ) = 4 − 2 = 2 rho(T)+N(T)=4Longrightarrowrho(T)=4-2=2\rho(T) + N(T) = 4 \implies \rho(T) = 4 – 2 = 2ρ(T)+N(T)=4⟹ρ(T)=4−2=2
So, the rank of T T TTT is 2.

Finding Matrix A A AAA that Maps to the Null Matrix

Any matrix A A AAA of the form [ a b a b ] a b a b [[a,b],[a,b]]\left[\begin{array}{cc} a & b \\ a & b \end{array}\right][abab] will map to the null matrix under T T TTT.
Lets take an example :
To find out if A A AAA maps to the null matrix under T T TTT, we need to compute B A B A BABABA.
Given B = [ 1 − 1 − 4 4 ] B = 1 − 1 − 4 4 B=[[1,-1],[-4,4]]B = \left[\begin{array}{cc}1 & -1 \\ -4 & 4\end{array}\right]B=[1−1−44] and A = [ 2 3 2 3 ] A = 2      3 2      3 A=[[2,3],[2,3]]A = \left[\begin{array}{ll}2 & 3 \\ 2 & 3\end{array}\right]A=[2323],
B A = [ 1 − 1 − 4 4 ] [ 2 3 2 3 ] = [ 0 0 0 0 ] B A = 1 − 1 − 4 4 2      3 2      3 = 0      0 0      0 BA=[[1,-1],[-4,4]][[2,3],[2,3]]=[[0,0],[0,0]]BA = \left[\begin{array}{cc}1 & -1 \\ -4 & 4\end{array}\right] \left[\begin{array}{ll}2 & 3 \\ 2 & 3\end{array}\right] = \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]BA=[1−1−44][2323]=[0000]
After calculating, we find that B A B A BABABA is indeed the null matrix.

Conclusion

  • The rank of T T TTT is 1.
  • The nullity of T T TTT is 1.
  • A matrix A A AAA that maps to the null matrix under T T TTT is of the form [ a b a b ] a b a b [[a,b],[a,b]]\left[\begin{array}{cc} a & b \\ a & b \end{array}\right][abab] Example is A = [ 2 3 2 3 ] A = 2      3 2      3 A=[[2,3],[2,3]]A = \left[\begin{array}{ll}2 & 3 \\ 2 & 3\end{array}\right]A=[2323] .
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खण्ड-A / SECTION-A
  1. (a) माना कि f : [ 0 , π 2 ] → R f : 0 , π 2 → R f:[0,(pi)/(2)]rarrRf:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}f:[0,π2]→R एक संतत फलन है, जैसा कि
f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f(x)=(cos^(2)x)/(4x^(2)-pi^(2)),quad0 <= x < (pi)/(2)f(x)=\frac{\cos ^2 x}{4 x^2-\pi^2}, \quad 0 \leq x<\frac{\pi}{2}f(x)=cos2⁡x4x2−π2,0≤x<π2
f ( π 2 ) f π 2 f((pi)/(2))f\left(\frac{\pi}{2}\right)f(π2) का मान ज्ञात कीजिए।
Let f : [ 0 , π 2 ] → R f : 0 , π 2 → R f:[0,(pi)/(2)]rarrRf:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}f:[0,π2]→R be a continuous function such that
f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 , 0 ≤ x < π 2 f(x)=(cos^(2)x)/(4x^(2)-pi^(2)),quad0 <= x < (pi)/(2)f(x)=\frac{\cos ^2 x}{4 x^2-\pi^2}, \quad 0 \leq x<\frac{\pi}{2}f(x)=cos2⁡x4x2−π2,0≤x<π2
Find the value of f ( π 2 ) f π 2 f((pi)/(2))f\left(\frac{\pi}{2}\right)f(π2).
Answer:

Introduction

The problem asks us to find the value of f ( π 2 ) f π 2 f((pi)/(2))f\left(\frac{\pi}{2}\right)f(π2) for a given function f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 f(x)=(cos^(2)x)/(4x^(2)-pi^(2))f(x) = \frac{\cos^2 x}{4x^2 – \pi^2}f(x)=cos2⁡x4x2−π2 defined on the interval [ 0 , π 2 ] 0 , π 2 [0,(pi)/(2)]\left[0, \frac{\pi}{2}\right][0,π2]. The function is continuous except at x = π 2 x = π 2 x=(pi)/(2)x = \frac{\pi}{2}x=π2 because the denominator becomes zero at that point. To find f ( π 2 ) f π 2 f((pi)/(2))f\left(\frac{\pi}{2}\right)f(π2), we’ll need to evaluate the limit of f ( x ) f ( x ) f(x)f(x)f(x) as x x xxx approaches π 2 π 2 (pi)/(2)\frac{\pi}{2}π2.

Work/Calculations

Step 1: Define the Function and the Limit

The function f ( x ) f ( x ) f(x)f(x)f(x) is given as:
f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 f ( x ) = cos 2 ⁡ x 4 x 2 − π 2 f(x)=(cos^(2)x)/(4x^(2)-pi^(2))f(x) = \frac{\cos^2 x}{4x^2 – \pi^2}f(x)=cos2⁡x4x2−π2
We need to find:
f ( π 2 ) = lim x → π 2 f ( x ) f π 2 = lim x → π 2   f ( x ) f((pi)/(2))=lim_(x rarr(pi)/(2))f(x)f\left(\frac{\pi}{2}\right) = \lim_{{x \to \frac{\pi}{2}}} f(x)f(π2)=limx→π2f(x)

Step 2: Simplify the Function

To find the limit, we can use L’Hôpital’s Rule, which states that if lim x → a f ( x ) g ( x ) lim x → a   f ( x ) g ( x ) lim_(x rarr a)(f(x))/(g(x))\lim_{{x \to a}} \frac{f(x)}{g(x)}limx→af(x)g(x) is an indeterminate form 0 0 0 0 (0)/(0)\frac{0}{0}00 or ∞ ∞ ∞ ∞ (oo )/(oo)\frac{\infty}{\infty}∞∞, then:
lim x → a f ( x ) g ( x ) = lim x → a f ′ ( x ) g ′ ( x ) lim x → a   f ( x ) g ( x ) = lim x → a   f ′ ( x ) g ′ ( x ) lim_(x rarr a)(f(x))/(g(x))=lim_(x rarr a)(f^(‘)(x))/(g^(‘)(x))\lim_{{x \to a}} \frac{f(x)}{g(x)} = \lim_{{x \to a}} \frac{f'(x)}{g'(x)}limx→af(x)g(x)=limx→af′(x)g′(x)
provided the limits on the right-hand side exist.
First, let’s check if the function f ( x ) f ( x ) f(x)f(x)f(x) is in indeterminate form at x = π 2 x = π 2 x=(pi)/(2)x = \frac{\pi}{2}x=π2:
Numerator at x = π 2 : cos 2 ⁡ ( π 2 ) = 0 Numerator at  x = π 2 : cos 2 ⁡ π 2 = 0 “Numerator at “x=(pi)/(2):cos^(2)((pi)/(2))=0\text{Numerator at } x = \frac{\pi}{2}: \cos^2 \left(\frac{\pi}{2}\right) = 0Numerator at x=π2:cos2⁡(π2)=0
Denominator at x = π 2 : 4 ( π 2 ) 2 − π 2 = 0 Denominator at  x = π 2 : 4 π 2 2 − π 2 = 0 “Denominator at “x=(pi)/(2):4((pi)/(2))^(2)-pi^(2)=0\text{Denominator at } x = \frac{\pi}{2}: 4 \left(\frac{\pi}{2}\right)^2 – \pi^2 = 0Denominator at x=π2:4(π2)2−π2=0
Both the numerator and the denominator are zero, so we have an indeterminate form 0 0 0 0 (0)/(0)\frac{0}{0}00.

Step 3: Apply L’Hôpital’s Rule

Let’s differentiate the numerator and the denominator with respect to x x xxx.
  • f ′ ( x ) f ′ ( x ) f^(‘)(x)f'(x)f′(x) (Derivative of cos 2 ⁡ x cos 2 ⁡ x cos^(2)x\cos^2 xcos2⁡x) = − 2 cos ⁡ ( x ) sin ⁡ ( x ) − 2 cos ⁡ ( x ) sin ⁡ ( x ) -2cos(x)sin(x)-2 \cos(x) \sin(x)−2cos⁡(x)sin⁡(x)
  • g ′ ( x ) g ′ ( x ) g^(‘)(x)g'(x)g′(x) (Derivative of 4 x 2 − π 2 4 x 2 − π 2 4x^(2)-pi^(2)4x^2 – \pi^24x2−π2) = 8 x 8 x 8x8x8x
Now, we can find the limit:
lim x → π 2 f ′ ( x ) g ′ ( x ) = lim x → π 2 − 2 cos ⁡ ( x ) sin ⁡ ( x ) 8 x lim x → π 2   f ′ ( x ) g ′ ( x ) = lim x → π 2   − 2 cos ⁡ ( x ) sin ⁡ ( x ) 8 x lim_(x rarr(pi)/(2))(f^(‘)(x))/(g^(‘)(x))=lim_(x rarr(pi)/(2))(-2cos(x)sin(x))/(8x)\lim_{{x \to \frac{\pi}{2}}} \frac{f'(x)}{g'(x)} = \lim_{{x \to \frac{\pi}{2}}} \frac{-2 \cos(x) \sin(x)}{8x}limx→π2f′(x)g′(x)=limx→π2−2cos⁡(x)sin⁡(x)8x
Let’s substitute the values into the formula and calculate the limit.
After substituting the values, we get:
lim x → π 2 − 2 cos ⁡ ( x ) sin ⁡ ( x ) 8 x = − 2 cos ⁡ ( π 2 ) sin ⁡ ( π 2 ) 8 × π 2 lim x → π 2   − 2 cos ⁡ ( x ) sin ⁡ ( x ) 8 x = − 2 cos ⁡ π 2 sin ⁡ π 2 8 × π 2 lim_(x rarr(pi)/(2))(-2cos(x)sin(x))/(8x)=(-2cos((pi)/(2))sin((pi)/(2)))/(8xx(pi)/(2))\lim_{{x \to \frac{\pi}{2}}} \frac{-2 \cos(x) \sin(x)}{8x} = \frac{-2 \cos\left(\frac{\pi}{2}\right) \sin\left(\frac{\pi}{2}\right)}{8 \times \frac{\pi}{2}}limx→π2−2cos⁡(x)sin⁡(x)8x=−2cos⁡(π2)sin⁡(π2)8×π2
After calculating, we get:
lim x → π 2 − 2 cos ⁡ ( x ) sin ⁡ ( x ) 8 x = 0 lim x → π 2   − 2 cos ⁡ ( x ) sin ⁡ ( x ) 8 x = 0 lim_(x rarr(pi)/(2))(-2cos(x)sin(x))/(8x)=0\lim_{{x \to \frac{\pi}{2}}} \frac{-2 \cos(x) \sin(x)}{8x} = 0limx→π2−2cos⁡(x)sin⁡(x)8x=0

Conclusion

The value of f ( π 2 ) f π 2 f((pi)/(2))f\left(\frac{\pi}{2}\right)f(π2) is 0 0 000. We used L’Hôpital’s Rule to evaluate the limit of the function f ( x ) f ( x ) f(x)f(x)f(x) as x x xxx approaches π 2 π 2 (pi)/(2)\frac{\pi}{2}π2, and found that the limit is 0 0 000. Therefore, f ( π 2 ) = 0 f π 2 = 0 f((pi)/(2))=0f\left(\frac{\pi}{2}\right) = 0f(π2)=0.
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(b) माना कि f : D ( ⊆ R 2 ) → R f : D ⊆ R 2 → R f:D(subeR^(2))rarrRf: D\left(\subseteq \mathbb{R}^2\right) \rightarrow \mathbb{R}f:D(⊆R2)→R एक फलन है और ( a , b ) ∈ D ( a , b ) ∈ D (a,b)in D(a, b) \in D(a,b)∈D. अगर f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) बिंदु ( a , b ) ( a , b ) (a,b)(a, b)(a,b) पर संतत है, तो दर्शाइए कि फलन f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) और f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) क्रमशः x = a x = a x=ax=ax=a और y = b y = b y=by=by=b पर संतत हैं।
Let f : D ( ⊆ R 2 ) → R f : D ⊆ R 2 → R f:D(subeR^(2))rarrRf: D\left(\subseteq \mathbb{R}^2\right) \rightarrow \mathbb{R}f:D(⊆R2)→R be a function and ( a , b ) ∈ D ( a , b ) ∈ D (a,b)in D(a, b) \in D(a,b)∈D. If f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is continuous at ( a , b ) ( a , b ) (a,b)(a, b)(a,b), then show that the functions f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) and f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) are continuous at x = a x = a x=ax=ax=a and at y = b y = b y=by=by=b respectively.
Answer:

Introduction

The problem asks us to prove that if a function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is continuous at a point ( a , b ) ( a , b ) (a,b)(a, b)(a,b), then the functions f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) and f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) are continuous at x = a x = a x=ax = ax=a and y = b y = b y=by = by=b respectively. To prove this, we will use the definition of continuity and manipulate the mathematical expressions accordingly.

Work/Calculations

Step 1: Definition of Continuity

A function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is said to be continuous at ( a , b ) ( a , b ) (a,b)(a, b)(a,b) if for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that:
( x − a ) 2 + ( y − b ) 2 < δ ⟹ | f ( x , y ) − f ( a , b ) | < ϵ ( x − a ) 2 + ( y − b ) 2 < δ ⟹ | f ( x , y ) − f ( a , b ) | < ϵ sqrt((x-a)^(2)+(y-b)^(2)) < deltaLongrightarrow|f(x,y)-f(a,b)| < epsilon\sqrt{(x-a)^2 + (y-b)^2} < \delta \implies |f(x, y) – f(a, b)| < \epsilon(x−a)2+(y−b)2<δ⟹|f(x,y)−f(a,b)|<ϵ

Step 2: Prove Continuity for f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) at x = a x = a x=ax = ax=a

We need to show that for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that:
| x − a | < δ ⟹ | f ( x , b ) − f ( a , b ) | < ϵ | x − a | < δ ⟹ | f ( x , b ) − f ( a , b ) | < ϵ |x-a| < deltaLongrightarrow|f(x,b)-f(a,b)| < epsilon|x – a| < \delta \implies |f(x, b) – f(a, b)| < \epsilon|x−a|<δ⟹|f(x,b)−f(a,b)|<ϵ
To prove this, let’s consider ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0. Since f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is continuous at ( a , b ) ( a , b ) (a,b)(a, b)(a,b), there exists δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that:
( x − a ) 2 + ( y − b ) 2 < δ ⟹ | f ( x , y ) − f ( a , b ) | < ϵ ( x − a ) 2 + ( y − b ) 2 < δ ⟹ | f ( x , y ) − f ( a , b ) | < ϵ sqrt((x-a)^(2)+(y-b)^(2)) < deltaLongrightarrow|f(x,y)-f(a,b)| < epsilon\sqrt{(x-a)^2 + (y-b)^2} < \delta \implies |f(x, y) – f(a, b)| < \epsilon(x−a)2+(y−b)2<δ⟹|f(x,y)−f(a,b)|<ϵ
Now, consider | x − a | < δ | x − a | < δ |x-a| < delta|x – a| < \delta|x−a|<δ. We can rewrite this as ( x − a ) 2 + ( b − b ) 2 < δ ( x − a ) 2 + ( b − b ) 2 < δ sqrt((x-a)^(2)+(b-b)^(2)) < delta\sqrt{(x-a)^2 + (b-b)^2} < \delta(x−a)2+(b−b)2<δ.
By the continuity of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) at ( a , b ) ( a , b ) (a,b)(a, b)(a,b), this implies:
| f ( x , b ) − f ( a , b ) | < ϵ | f ( x , b ) − f ( a , b ) | < ϵ |f(x,b)-f(a,b)| < epsilon|f(x, b) – f(a, b)| < \epsilon|f(x,b)−f(a,b)|<ϵ
Thus, we have shown that f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) is continuous at x = a x = a x=ax = ax=a.

Step 3: Prove Continuity for f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) at y = b y = b y=by = by=b

Similarly, we need to show that for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that:
| y − b | < δ ⟹ | f ( a , y ) − f ( a , b ) | < ϵ | y − b | < δ ⟹ | f ( a , y ) − f ( a , b ) | < ϵ |y-b| < deltaLongrightarrow|f(a,y)-f(a,b)| < epsilon|y – b| < \delta \implies |f(a, y) – f(a, b)| < \epsilon|y−b|<δ⟹|f(a,y)−f(a,b)|<ϵ
Again, consider ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0. By the continuity of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) at ( a , b ) ( a , b ) (a,b)(a, b)(a,b), there exists δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that:
( x − a ) 2 + ( y − b ) 2 < δ ⟹ | f ( x , y ) − f ( a , b ) | < ϵ ( x − a ) 2 + ( y − b ) 2 < δ ⟹ | f ( x , y ) − f ( a , b ) | < ϵ sqrt((x-a)^(2)+(y-b)^(2)) < deltaLongrightarrow|f(x,y)-f(a,b)| < epsilon\sqrt{(x-a)^2 + (y-b)^2} < \delta \implies |f(x, y) – f(a, b)| < \epsilon(x−a)2+(y−b)2<δ⟹|f(x,y)−f(a,b)|<ϵ
Now, consider | y − b | < δ | y − b | < δ |y-b| < delta|y – b| < \delta|y−b|<δ. We can rewrite this as ( a − a ) 2 + ( y − b ) 2 < δ ( a − a ) 2 + ( y − b ) 2 < δ sqrt((a-a)^(2)+(y-b)^(2)) < delta\sqrt{(a-a)^2 + (y-b)^2} < \delta(a−a)2+(y−b)2<δ.
By the continuity of f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) at ( a , b ) ( a , b ) (a,b)(a, b)(a,b), this implies:
| f ( a , y ) − f ( a , b ) | < ϵ | f ( a , y ) − f ( a , b ) | < ϵ |f(a,y)-f(a,b)| < epsilon|f(a, y) – f(a, b)| < \epsilon|f(a,y)−f(a,b)|<ϵ
Thus, we have shown that f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) is continuous at y = b y = b y=by = by=b.

Conclusion

We have successfully proven that if f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is continuous at ( a , b ) ( a , b ) (a,b)(a, b)(a,b), then f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) is continuous at x = a x = a x=ax = ax=a and f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) is continuous at y = b y = b y=by = by=b. We used the definition of continuity to establish these results.
untitled-document-15-c4a14609-db5c-41e1-99f0-81aab3fdac8f
खण्ड-A / SECTION-A
  1. (a) मान लीजिये कि A A AAA एक 3 × 2 3 × 2 3xx23 \times 23×2 आव्यूह है और B B BBB एक 2 × 3 2 × 3 2xx32 \times 32×3 आव्यूह है। दर्शाइये कि C = A ⋅ B C = A ⋅ B C=A*BC=A \cdot BC=A⋅B एक अव्युत्क्रमणणीय आव्यूह है।
Let A A AAA be a 3 × 2 3 × 2 3xx23 \times 23×2 matrix and B B BBB a 2 × 3 2 × 3 2xx32 \times 32×3 matrix. Show that C = A ⋅ B C = A ⋅ B C=A*BC=A \cdot BC=A⋅B is a singular matrix.
Answer:

Introduction

The problem asks us to show that the product C = A ⋅ B C = A ⋅ B C=A*BC = A \cdot BC=A⋅B is a singular matrix, given that A A AAA is a 3 × 2 3 × 2 3xx23 \times 23×2 matrix and B B BBB is a 2 × 3 2 × 3 2xx32 \times 32×3 matrix. A singular matrix is one that does not have an inverse, which means its determinant is zero.

Work/Calculations

Step 1: Dimensions of C C CCC

First, let’s find the dimensions of the resulting matrix C C CCC when A A AAA and B B BBB are multiplied.
The dimensions of C C CCC can be determined by the outer dimensions of A A AAA and B B BBB. In this case, A A AAA is 3 × 2 3 × 2 3xx23 \times 23×2 and B B BBB is 2 × 3 2 × 3 2xx32 \times 32×3, so C C CCC will be 3 × 3 3 × 3 3xx33 \times 33×3.

Step 2: Rank of C C CCC

The rank of C C CCC is limited by the smaller of the two ranks of A A AAA and B B BBB. Since A A AAA is 3 × 2 3 × 2 3xx23 \times 23×2, its rank can be at most 2. Similarly, B B BBB is 2 × 3 2 × 3 2xx32 \times 32×3, so its rank can also be at most 2.
Therefore, the rank of C C CCC can be at most 2.
Rank ( C ) ≤ min ( Rank ( A ) , Rank ( B ) ) ≤ 2 Rank ( C ) ≤ min ( Rank ( A ) , Rank ( B ) ) ≤ 2 “Rank”(C) <= min(“Rank”(A),”Rank”(B)) <= 2\text{Rank}(C) \leq \min(\text{Rank}(A), \text{Rank}(B)) \leq 2Rank(C)≤min(Rank(A),Rank(B))≤2

Step 3: Determinant of C C CCC

For a 3 × 3 3 × 3 3xx33 \times 33×3 matrix to be invertible (non-singular), its rank must be 3. However, we’ve established that the rank of C C CCC can be at most 2. Therefore, C C CCC must be singular.
To confirm, the determinant of a singular matrix is zero:
Det ( C ) = 0 Det ( C ) = 0 “Det”(C)=0\text{Det}(C) = 0Det(C)=0

Conclusion

We have shown that the matrix C = A ⋅ B C = A ⋅ B C=A*BC = A \cdot BC=A⋅B will be a 3 × 3 3 × 3 3xx33 \times 33×3 matrix with a rank of at most 2. Since the rank is less than 3, C C CCC is a singular matrix, and its determinant is zero. Therefore, C = A ⋅ B C = A ⋅ B C=A*BC = A \cdot BC=A⋅B is indeed a singular matrix.
Page Break
(b) आधार सदिशों e 1 = ( 1 , 0 ) e 1 = ( 1 , 0 ) e_(1)=(1,0)e_1=(1,0)e1=(1,0) और e 2 = ( 0 , 1 ) e 2 = ( 0 , 1 ) e_(2)=(0,1)e_2=(0,1)e2=(0,1) को α 1 = ( 2 , − 1 ) α 1 = ( 2 , − 1 ) alpha_(1)=(2,-1)\alpha_1=(2,-1)α1=(2,−1) एवं α 2 = ( 1 , 3 ) α 2 = ( 1 , 3 ) alpha_(2)=(1,3)\alpha_2=(1,3)α2=(1,3) के रैखिक संयोग के रूप में ब्यक्त कीजिये।
Express basis vectors e 1 = ( 1 , 0 ) e 1 = ( 1 , 0 ) e_(1)=(1,0)e_1=(1,0)e1=(1,0) and e 2 = ( 0 , 1 ) e 2 = ( 0 , 1 ) e_(2)=(0,1)e_2=(0,1)e2=(0,1) as linear combinations of α 1 = ( 2 , − 1 ) α 1 = ( 2 , − 1 ) alpha_(1)=(2,-1)\alpha_1=(2,-1)α1=(2,−1) and α 2 = ( 1 , 3 ) α 2 = ( 1 , 3 ) alpha_(2)=(1,3)\alpha_2=(1,3)α2=(1,3).
Answer:

Introduction

The problem asks us to express the basis vectors e 1 = ( 1 , 0 ) e 1 = ( 1 , 0 ) e_(1)=(1,0)e_1 = (1, 0)e1=(1,0) and e 2 = ( 0 , 1 ) e 2 = ( 0 , 1 ) e_(2)=(0,1)e_2 = (0, 1)e2=(0,1) as linear combinations of the vectors α 1 = ( 2 , − 1 ) α 1 = ( 2 , − 1 ) alpha_(1)=(2,-1)\alpha_1 = (2, -1)α1=(2,−1) and α 2 = ( 1 , 3 ) α 2 = ( 1 , 3 ) alpha_(2)=(1,3)\alpha_2 = (1, 3)α2=(1,3). In other words, we want to find constants c 1 c 1 c_(1)c_1c1 and c 2 c 2 c_(2)c_2c2 such that:
e 1 = c 1 α 1 + c 2 α 2 e 1 = c 1 α 1 + c 2 α 2 e_(1)=c_(1)alpha_(1)+c_(2)alpha_(2)e_1 = c_1 \alpha_1 + c_2 \alpha_2e1=c1α1+c2α2
e 2 = d 1 α 1 + d 2 α 2 e 2 = d 1 α 1 + d 2 α 2 e_(2)=d_(1)alpha_(1)+d_(2)alpha_(2)e_2 = d_1 \alpha_1 + d_2 \alpha_2e2=d1α1+d2α2

Work/Calculations

Step 1: Setting up the Equations for e 1 e 1 e_(1)e_1e1

The equation for e 1 e 1 e_(1)e_1e1 can be written as:
( 1 , 0 ) = c 1 ( 2 , − 1 ) + c 2 ( 1 , 3 ) ( 1 , 0 ) = c 1 ( 2 , − 1 ) + c 2 ( 1 , 3 ) (1,0)=c_(1)(2,-1)+c_(2)(1,3)(1, 0) = c_1 (2, -1) + c_2 (1, 3)(1,0)=c1(2,−1)+c2(1,3)
Breaking it down into components, we get:
1 = 2 c 1 + c 2 1 = 2 c 1 + c 2 1=2c_(1)+c_(2)1 = 2c_1 + c_21=2c1+c2
0 = − c 1 + 3 c 2 0 = − c 1 + 3 c 2 0=-c_(1)+3c_(2)0 = -c_1 + 3c_20=−c1+3c2

Step 2: Solving for c 1 c 1 c_(1)c_1c1 and c 2 c 2 c_(2)c_2c2

After Calculating, we get:
c 1 = 3 7 , c 2 = 1 7 c 1 = 3 7 , c 2 = 1 7 c_(1)=(3)/(7),quadc_(2)=(1)/(7)c_1 = \frac{3}{7}, \quad c_2 = \frac{1}{7}c1=37,c2=17

Step 3: Setting up the Equations for e 2 e 2 e_(2)e_2e2

The equation for e 2 e 2 e_(2)e_2e2 can be written as:
( 0 , 1 ) = d 1 ( 2 , − 1 ) + d 2 ( 1 , 3 ) ( 0 , 1 ) = d 1 ( 2 , − 1 ) + d 2 ( 1 , 3 ) (0,1)=d_(1)(2,-1)+d_(2)(1,3)(0, 1) = d_1 (2, -1) + d_2 (1, 3)(0,1)=d1(2,−1)+d2(1,3)
Breaking it down into components, we get:
0 = 2 d 1 + d 2 0 = 2 d 1 + d 2 0=2d_(1)+d_(2)0 = 2d_1 + d_20=2d1+d2
1 = − d 1 + 3 d 2 1 = − d 1 + 3 d 2 1=-d_(1)+3d_(2)1 = -d_1 + 3d_21=−d1+3d2

Step 4: Solving for d 1 d 1 d_(1)d_1d1 and d 2 d 2 d_(2)d_2d2

After Calculating, we get:
d 1 = − 1 7 , d 2 = 2 7 d 1 = − 1 7 , d 2 = 2 7 d_(1)=-(1)/(7),quadd_(2)=(2)/(7)d_1 = -\frac{1}{7}, \quad d_2 = \frac{2}{7}d1=−17,d2=27

Conclusion

The basis vector e 1 = ( 1 , 0 ) e 1 = ( 1 , 0 ) e_(1)=(1,0)e_1 = (1, 0)e1=(1,0) can be expressed as a linear combination of α 1 α 1 alpha_(1)\alpha_1α1 and α 2 α 2 alpha_(2)\alpha_2α2 as follows:
e 1 = 3 7 α 1 + 1 7 α 2 e 1 = 3 7 α 1 + 1 7 α 2 e_(1)=(3)/(7)alpha_(1)+(1)/(7)alpha_(2)e_1 = \frac{3}{7} \alpha_1 + \frac{1}{7} \alpha_2e1=37α1+17α2
Similarly, the basis vector e 2 = ( 0 , 1 ) e 2 = ( 0 , 1 ) e_(2)=(0,1)e_2 = (0, 1)e2=(0,1) can be expressed as:
e 2 = − 1 7 α 1 + 2 7 α 2 e 2 = − 1 7 α 1 + 2 7 α 2 e_(2)=-(1)/(7)alpha_(1)+(2)/(7)alpha_(2)e_2 = -\frac{1}{7} \alpha_1 + \frac{2}{7} \alpha_2e2=−17α1+27α2
Thus, both e 1 e 1 e_(1)e_1e1 and e 2 e 2 e_(2)e_2e2 can be represented as linear combinations of α 1 α 1 alpha_(1)\alpha_1α1 and α 2 α 2 alpha_(2)\alpha_2α2.
\(sin^2\left(\frac{\theta }{2}\right)=\frac{1-cos\:\theta }{2}\)
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