Details For UPSC Maths Optional Solved Papers (2018-2022)
UPSC Maths Optional Question Papers
UPSC Maths Optional Paper – 01 2022
upsc-m2022-1-bd58d9b8-2d68-4046-a6a8-417f99bd471d
खण्ड A
SECTION A
Question:-01 (a) सिद्ध कीजिए कि nmathrm{n} विमीय सदिश समष्टि Vmathrm{V} के लिए nmathrm{n} रैखिकत: स्वतंत्र सदिशों का कोई भी समुच्चय Vmathrm{V} के लिए एक आधार बनाता है ।
Question:-01 (a) Prove that any set of nmathrm{n} linearly independent vectors in a vector space Vmathrm{V} of dimension nmathrm{n} constitutes a basis for Vmathrm{V}.
Question:-01 (b) माना T:R^(2)rarrR^(3)mathrm{T}: mathbb{R}^2 rightarrow mathbb{R}^3 एक रैखिक रूपांतरण, ऐसा है कि T([1],[0])=([1],[2],[3])mathrm{T}left(begin{array}{l}1 \ 0end{array}right)=left(begin{array}{l}1 \ 2 \ 3end{array}right) तथा T([1],[1])=([-3],[2],[8])mathrm{T}left(begin{array}{l}1 \ 1end{array}right)=left(begin{array}{r}-3 \ 2 \ 8end{array}right) है । T([2],[4])mathrm{T}left(begin{array}{l}2 \ 4end{array}right) को ज्ञात कीजिए ।
Question:-01(b) Let T:R^(2)rarrR^(3)mathrm{T}: mathbb{R}^2 rightarrow mathbb{R}^3 be a linear transformation such that T([1],[0])=([1],[2],[3])mathrm{T}left(begin{array}{l}1 \ 0end{array}right)=left(begin{array}{l}1 \ 2 \ 3end{array}right) and T([1],[1])=([-3],[2],[8])mathrm{T}left(begin{array}{l}1 \ 1end{array}right)=left(begin{array}{r}-3 \ 2 \ 8end{array}right). Find T([2],[4])mathrm{T}left(begin{array}{l}2 \ 4end{array}right)
Question:-01(c) lim_(x rarr oo)(e^(x)+x)^((1)/(x))lim _{x rightarrow infty}left(e^x+xright)^{frac{1}{x}} का मान निकालिए ।
Question:-01(c) Evaluate lim_(x rarr oo)(e^(x)+x)^((1)/(x))lim _{x rightarrow infty}left(e^x+xright)^{frac{1}{x}}
Question:-01(d) int_(0)^(2)(dx)/((2x-x^(2)))int_0^2 frac{d x}{left(2 x-x^2right)} की अभिसारिता का परीक्षण कीजिए ।
Question:-01(d) Examine the convergence of int_(0)^(2)(dx)/((2x-x^(2)))int_0^2 frac{d x}{left(2 x-x^2right)}.
Question:-01(e) एक चर समतल एक स्थिर बिन्दु (a,b,c)(mathrm{a}, mathrm{b}, mathrm{c}) से गुज़रता है तथा अक्षों को क्रमशः A,Bmathrm{A}, mathrm{B} व Cmathrm{C} बिन्दुओं पर मिलता है । बिन्दुओं O,A,Bmathrm{O}, mathrm{A}, mathrm{B} तथा Cmathrm{C} से गुज़रते हुए गोले के केन्द्र का बिन्दुपथ ज्ञात कीजिए, जहाँ Omathrm{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,Bmathrm{O}, mathrm{A}, mathrm{B} and C,Omathrm{C}, mathrm{O} being the origin.
Question:-02(a) निम्नलिखित समीकरण निकाय के सभी हलों को पंक्ति-समानीत विधि से ज्ञात कीजिए : {:[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}
Question:-02(a) Find all solutions to the following system of equations by row-reduced method : {:[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}
Question:-02(b) एक ll 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 ll 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^(‘)mathrm{P}, mathrm{Q}, mathrm{R} ; mathrm{P}^{prime}, mathrm{Q}^{prime}, mathrm{R}^{prime}, एक बिन्दु से दीर्घवृत्तज (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1frac{mathrm{x}^2}{mathrm{a}^2}+frac{mathrm{y}^2}{mathrm{~b}^2}+frac{mathrm{z}^2}{mathrm{c}^2}=1 पर छः (सिक्स) अभिलम्ब पाद हैं तथा lx+my+nz=pl mathrm{x}+mathrm{my}+mathrm{nz}=mathrm{p} से समतल PQRmathrm{PQR} निरूपित है, दर्शाइए कि (x)/(a^(2)l)+(y)/(b^(2)(m))+(z)/(c^(2)n)+(1)/(p)=0frac{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}}=0, समतल P^(‘)Q^(‘)R^(‘)mathrm{P}^{prime} mathrm{Q}^{prime} mathrm{R}^{prime} को निरूपित करता है ।
Question:-02(c) If P,Q,R;P^(‘),Q^(‘),R^(‘)mathrm{P}, mathrm{Q}, mathrm{R} ; mathrm{P}^{prime}, mathrm{Q}^{prime}, mathrm{R}^{prime} 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))=1frac{mathrm{x}^2}{mathrm{a}^2}+frac{mathrm{y}^2}{mathrm{~b}^2}+frac{mathrm{z}^2}{mathrm{c}^2}=1, and the plane PQRmathrm{PQR} is represented by lx+my+nz=pl x+m y+n z=p, show that the plane P^(‘)Q^(‘)R^(‘)P^{prime} Q^{prime} R^{prime} is given by (x)/(a^(2)l)+(y)/(b^(2)(m))+(z)/(c^(2)n)+(1)/(p)=0frac{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}}=0.
Question:-03(a) माना समुच्चय {: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}=0end{array}right} सदिश समष्टि R^(3)(R)mathbb{R}^3(mathbb{R}) के सदिशों का एक समूह है । तब
(i) सिद्ध कीजिए कि P,R^(3)mathrm{P}, mathbb{R}^3 की एक उपसमष्टि है ।
(ii) Pmathrm{P} का एक आधार तथा विमा ज्ञात कीजिए ।
Question:-03(a) Let the set {:P={[x],[y],[z])∣[x-y-z=0″ and “],[2x-y+z=0]}left.P=left{begin{array}{c}x \ y \ zend{array}right) mid begin{array}{c}x-y-z=0 text { and } \ 2 x-y+z=0end{array}right} be the collection of vectors of a vector space R^(3)(R)mathbb{R}^3(mathbb{R}). Then
(i) prove that Pmathrm{P} is a subspace of R^(3)mathbb{R}^3.
(ii) find a basis and dimension of PP.
Question:-03(b) द्विशः समाकलन का उपयोग करके, वृत्त x^(2)+y^(2)=4mathrm{x}^2+mathrm{y}^2=4 तथा परवलय y^(2)=3xmathrm{y}^2=3 mathrm{x} के उभयनिष्ठ क्षेत्रफल का परिकलन कीजिए ।
Question:-03(b)Use double integration to calculate the area common to the circle x^(2)+y^(2)=4x^2+y^2=4 and the parabola y^(2)=3xy^2=3 x.
Question:-03(c) लघुतम संभाव्य त्रिज्या के गोले का समीकरण ज्ञात कीजिए जो सरल रेखाओं : (x-3)/(3)=(y-8)/(-1)=(z-3)/(1)frac{mathrm{x}-3}{3}=frac{mathrm{y}-8}{-1}=frac{mathrm{z}-3}{1} तथा (x+3)/(-3)=(y+7)/(2)=(z-6)/(4)frac{mathrm{x}+3}{-3}=frac{mathrm{y}+7}{2}=frac{mathrm{z}-6}{4} 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)frac{x-3}{3}=frac{y-8}{-1}=frac{z-3}{1} and (x+3)/(-3)=(y+7)/(2)=(z-6)/(4)frac{x+3}{-3}=frac{y+7}{2}=frac{z-6}{4}
Question:-04(a) एक रैखिक प्रतिचित्र T:R^(2)rarrR^(2)mathrm{T}: mathbb{R}^2 rightarrow mathbb{R}^2 ज्ञात कीजिए जो कि R^(2)mathbb{R}^2 के प्रत्येक सदिश को thetatheta कोण से घुमा देता है । यह भी सिद्ध कीजिए कि theta=(pi)/(2)theta=frac{pi}{2} के लिए, Tmathrm{T} का कोई भी अभिलक्षणिक मान (आइगेनमान) Rmathbb{R} में नहीं है ।
Question:-04(a) Find a linear map T:R^(2)rarrR^(2)mathrm{T}: mathbb{R}^2 rightarrow mathbb{R}^2 which rotates each vector of R^(2)mathbb{R}^2 by an angle thetatheta. Also, prove that for theta=(pi)/(2),Ttheta=frac{pi}{2}, mathrm{~T} has no eigenvalue in Rmathbb{R}.
Question:-04(b) वक्र y^(2)x^(2)=x^(2)-a^(2)mathrm{y}^2 mathrm{x}^2=mathrm{x}^2-mathrm{a}^2 का अनुरेख (ट्रेस) कीजिए, जहाँ amathrm{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, where aa is a real constant.
Question:-04(c) यदि समतल ux+vy+wz=0u x+v y+w z=0, शंकु ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+mathrm{cz}^2=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.
Question:-04(c) If the plane ux+vy+wz=0u x+v y+w z=0 cuts the cone ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+c z^2=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.
खण्ड B
SECTION B
Question:-05(a) दर्शाइए कि अवकल समीकरण (dy)/(dx)+Py=Qfrac{mathrm{dy}}{mathrm{dx}}+mathrm{Py}=mathrm{Q} का व्यापक हल 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}
के रूप में लिखा जा सकता है, जहाँ P,Q,xmathrm{P}, mathrm{Q}, mathrm{x} के शून्येतर फलन हैं तथा Cmathrm{C} एक स्वेच्छ अचर है ।
Question:-05(a)Show that the general solution of the differential equation (dy)/(dx)+Py=Qfrac{mathrm{dy}}{mathrm{dx}}+mathrm{Py}=mathrm{Q} can be written in the form 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} dleft(frac{Q}{P}right)right}, where P,Qmathrm{P}, mathrm{Q} are non-zero functions of xmathrm{x} and Cmathrm{C}, an arbitrary constant.
Question:-05(b) दर्शाइए कि परवलयों के निकाय : x^(2)=4a(y+a)mathrm{x}^2=4 mathrm{a}(mathrm{y}+mathrm{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)=4a(y+a)mathrm{x}^2=4 mathrm{a}(mathrm{y}+mathrm{a}) belong to the same system.
Question:-05(c) wmathrm{w} भार का एक पिंड, thetatheta कोण से झुके हुए एक रूक्ष समतल पर स्थित है, घर्षण गुणांक mumu, tan thetatan theta से अधिक है । पिंड को समतल पर ऊपर की तरफ ‘ bb ‘ 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 ww rests on a rough inclined plane of inclination thetatheta, the coefficient of friction, mumu, being greater than tan thetatan theta. 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 sqrt(2gh)sqrt{2 mathrm{gh}} वेग के साथ बिन्दु Omathrm{O} से प्रक्षेपित किया गया तथा समतल के बिन्दु P(x,y)mathrm{P}(mathrm{x}, mathrm{y}) पर स्पर्श-रेखा से टकराता है जहाँ अक्ष OXmathrm{OX} तथा OYmathrm{OY} क्रमशः बिन्दु Omathrm{O} से क्षैतिज तथा अधोमुखी ऊर्ध्वाधर रेखाएँ हैं । यदि प्रक्षेपण की दो संभव दिशाएँ समकोण पर हों, तो दर्शाइए कि x^(2)=2hymathrm{x}^2=2 mathrm{hy} तथा प्रक्षेपण की संभव दिशाओं में से एक, कोण POX को द्विभाजित करती है ।
Question:-05(d)A projectile is fired from a point Omathrm{O} with velocity sqrt(2gh)sqrt{2 mathrm{gh}} and hits a tangent at the point P(x,y)mathrm{P}(mathrm{x}, mathrm{y}) in the plane, the axes OXmathrm{OX} and OYmathrm{OY} being horizontal and vertically downward lines through the point Omathrm{O}, respectively. Show that if the two possible directions of projection be at right angles, then x^(2)=2hymathrm{x}^2=2 mathrm{hy} and then one of the possible directions of projection bisects the angle POX.
Question:-05(e) दर्शाइए कि 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}^3right) 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}} अघूर्णी है । phiphi को भी ज्ञात कीजिए जबकि vec(A)=grad phioverrightarrow{mathrm{A}}=nabla phi.
Question:-05(e)Show that 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^3right) hat{i}+left(3 x^2-zright) hat{j}+left(3 x z^2-yright) hat{k} is irrotational. Also find phiphi such that vec(A)=grad phioverrightarrow{mathrm{A}}=nabla phi.
Question:-06(a) 2l2 l लम्बाई का एक तार (केबिल) जिसका भार wmathrm{w} प्रति इकाई (यूनिट) लम्बाई है, एक क्षैतिज रेखा के दो बिन्दुओं Pmathrm{P} तथा Qmathrm{Q} से लटकी हुई है । दर्शाइए कि तार की विस्तृति (स्पैन) 2l(1-(2h^(2))/(3l^(2)))2 lleft(1-frac{2 mathrm{~h}^2}{3 l^2}right) है, जहाँ hmathrm{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 2l2 l hangs from two points Pmathrm{P} and Qmathrm{Q} in the same horizontal line. Show that the span of the cable is 2l(1-(2h^(2))/(3l^(2)))2 lleft(1-frac{2 h^2}{3 l^2}right), where hh 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)/(dx^(2))-2x(dy)/(dx)+2y=(x^(2)-1)^(2)left(x^2-1right) frac{d^2 y}{d x^2}-2 x frac{d y}{d x}+2 y=left(x^2-1right)^2
को हल कीजिए, जहाँ समानीत समीकरण का एक हल y=xmathrm{y}=mathrm{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)/(dx^(2))-2x(dy)/(dx)+2y=(x^(2)-1)^(2)left(x^2-1right) frac{d^2 y}{d x^2}-2 x frac{d y}{d x}+2 y=left(x^2-1right)^2, given that y=xmathrm{y}=mathrm{x} is one solution of the reduced equation.
Question:-06(c) समतल में ग्रीन के प्रमेय को oint_(C)(3x^(2)-8y^(2))dx+(4y-6xy)dyoint_{mathrm{C}}left(3 mathrm{x}^2-8 mathrm{y}^2right) mathrm{dx}+(4 mathrm{y}-6 mathrm{xy}) mathrm{dy} के लिए सत्यापित कीजिए, जहाँ C,x=0,y=0,x+y=1mathrm{C}, mathrm{x}=0, mathrm{y}=0, mathrm{x}+mathrm{y}=1 द्वारा परिभाषित क्षेत्र का सीमा वक्र है ।
Question:-06(c)Verify Green’s theorem in the plane for oint_(C)(3x^(2)-8y^(2))dx+(4y-6xy)dyoint_Cleft(3 x^2-8 y^2right) d x+(4 y-6 x y) d y, where Cmathrm{C} is the boundary curve of the region defined by x=0,y=0mathrm{x}=0, mathrm{y}=0, x+y=1x+y=1
Question:-07(a) स्टोक्स प्रमेय को 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}} के लिए प्रथम अष्टांशक में स्थित समतल पृष्ठ : x+y+z=1x+y+z=1 पर सत्यापित कीजिए ।
Question:-07(a) Verify Stokes’ theorem for 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} over the plane surface : x+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)/(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}), जहाँ 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)=0right. को हल कीजिए।
Question:-07(b) Solve the following initial value problem by using Laplace’s transformation (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}), where 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)=0right.
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 rhorho तथा rho^(‘)rho^{prime} हैं और संस्पर्श बिन्दु से ऊपरी बेलन के गुरुत्व केन्द्र की ऊँचाई hmathrm{h} है । दर्शाइए कि स्थायी साम्य में ऊपरी बेलन संतुलित है यदि h < (rhorho^(‘))/(rho+rho^(‘))mathrm{h}<frac{rho rho^{prime}}{rho+rho^{prime}} ।
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 rhorho and rho^(‘)rho^{prime} be the radii of curvature of the two cylinders at the point of contact and hh 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 < (rhorho^(‘))/(rho+rho^(‘))mathrm{h}<frac{rho rho^{prime}}{rho+rho^{prime}}.
Question:-08(a) (i) अवकल समीकरण : (x^(2)-a^(2))p^(2)-2xyp+y^(2)+a^(2)=0left(mathrm{x}^2-mathrm{a}^2right) mathrm{p}^2-2 mathrm{xyp}+mathrm{y}^2+mathrm{a}^2=0, जहाँ p=(dy)/(dx)mathrm{p}=frac{mathrm{dy}}{mathrm{dx}}, 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)-2xyp+y^(2)+a^(2)=0left(x^2-a^2right) p^2-2 x y p+y^2+a^2=0, where p=(dy)/(dx)p=frac{d y}{d x}. Also give the geometric relation between the general and singular solutions.
Question:-08(a) (ii) निम्नलिखित अवकल समीकरण को हल कीजिए : (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
Question:-08(a) (ii) Solve the following differential equation : (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
Question:-08(b) nmathrm{n} बराबर एकसमान छड़ों की एक श्रृंखला एक-दूसरे के साथ चिकने रूप से जुड़ी हुई है तथा इसके एक सिरे A_(1)mathrm{A}_1 से लटकी हुई है । एक क्षैतिज बल vec(P)overrightarrow{mathrm{P}} शृंखला के दूसरे सिरे A_(n+1)mathrm{A}_{mathrm{n}+1} पर लगाया गया है । साम्य विन्यास में अधोमुखी ऊर्ध्वाधर रेखा से छड़ों के झुकाव ज्ञात कीजिए ।
Question:-08(b) A chain of nmathrm{n} equal uniform rods is smoothly jointed together and suspended from its one end A_(1)mathrm{A}_1. A horizontal force vec(P)overrightarrow{mathrm{P}} is applied to the other end A_(n+1)mathrm{A}_{mathrm{n}+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) vec(F)* vec(n)dSiint_{mathrm{S}} overrightarrow{mathrm{F}} cdot overrightarrow{mathrm{n}} mathrm{dS} का मान निकालिए, जहाँ 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}^2-1right) hat{mathrm{k}} तथा Smathrm{S}, पृष्ठों z=0,z=1,x^(2)+y^(2)=4mathrm{z}=0, mathrm{z}=1, mathrm{x}^2+mathrm{y}^2=4 द्वारा बना हुआ बेलन है ।
Question:-08(c) Using Gauss’ divergence theorem, evaluate ∬_(S) vec(F)* vec(n)dSiint_S vec{F} cdot vec{n} d S, where vec(F)=x hat(i)-y hat(j)+(z^(2)-1) hat(k)vec{F}=x hat{i}-y hat{j}+left(z^2-1right) hat{k} and SS is the cylinder formed by the surfaces z=0,z=1,x^(2)+y^(2)=4z=0, z=1, x^2+y^2=4
UPSC Maths Optional Paper – 01 2021
upsc-m2021-1-ed2d6450-301f-4184-ac9b-1fca888cd172
1.(a) यदि A=[[1,-1,1],[2,-1,0],[1,0,0]]A=left[begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0end{array}right] है, तो A^(-1)A^{-1} को ज्ञात किए बिना दर्शाइए कि A^(2)=A^(-1)A^2=A^{-1}
If A=[[1,-1,1],[2,-1,0],[1,0,0]]A=left[begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0end{array}right], then show that A^(2)=A^(-1)A^2=A^{-1} (without finding 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)} के सापेक्ष V_(3)(R)V_3(R) पर परिभाषित रैखिक संकारक : T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c) से संबन्धित आव्यूह ज्ञात कीजिए ।
Find the matrix associated with the linear operator on V_(3)(R)V_3(R) defined by T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c) 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)}.
1.(c) दिया गया है : 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|
जहाँ ff एक वास्तविक-मान अवकलनीय फलन है तथा alphaalpha एक अचर है। lim_(x rarr0)(Delta(x))/(x)lim _{x rightarrow 0} frac{Delta(x)}{x} को ज्ञात कीजिए ।
Given : 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|
where ff is a real valued differentiable function and alphaalpha is a constant. Find lim_(x rarr0)(Delta(x))/(x)lim _{x rightarrow 0} frac{Delta(x)}{x}.
1.(d) दर्शाइए कि e^(x)cos x=1e^x cos x=1 के किन्हीं दो मूलों के बीच में e^(x)sin x-1=0e^x sin x-1=0 का कम से कम एक मूल विद्यमान है ।
Show that between any two roots of e^(x)cos x=1e^x cos x=1, there exists at least one root of e^(x)sin x-1=0e^x sin x-1=0
1.(e) उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा : x=-(y)/(2)=(z)/(3)x=-frac{y}{2}=frac{z}{3} के समानान्तर हैं
तथा जिसका निर्देशक-वक्र x^(2)+2y^(2)=1,z=0x^2+2 y^2=1, z=0 है।
Find the equation of the cylinder whose generators are parallel to the line x=-(y)/(2)=(z)/(3)x=-frac{y}{2}=frac{z}{3} and whose guiding curve is x^(2)+2y^(2)=1,z=0x^2+2 y^2=1, z=0.
2.(a) दर्शाइए कि वे समतल, जो कि शंकु ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+c z^2=0 को लंब जनकों में काटते हैं, शंकु (x^(2))/(b+c)+(y^(2))/(c+a)+(z^(2))/(a+b)=0frac{x^2}{b+c}+frac{y^2}{c+a}+frac{z^2}{a+b}=0 को स्पर्श करते हैं ।
Show that the planes, which cut the cone ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+c z^2=0 in perpendicular generators, touch the cone (x^(2))/(b+c)+(y^(2))/(c+a)+(z^(2))/(a+b)=0frac{x^2}{b+c}+frac{y^2}{c+a}+frac{z^2}{a+b}=0.
2.(b) दिया गया है : f(x,y)=|x^(2)-y^(2)|f(x, y)=left|x^2-y^2right|, तब f_(xy)(0,0)f_{x y}(0,0) तथा f_(yx)(0,0)f_{y x}(0,0) ज्ञात कीजिए । अतः दर्शाइए कि f_(xy)(0,0)=f_(yx)(0,0)f_{x y}(0,0)=f_{y x}(0,0) ।
Given that f(x,y)=|x^(2)-y^(2)|f(x, y)=left|x^2-y^2right|. Find f_(xy)(0,0)f_{x y}(0,0) and f_(yx)(0,0)f_{y x}(0,0).
Hence show that f_(xy)(0,0)=f_(yx)(0,0)f_{x y}(0,0)=f_{y x}(0,0).
2.(c) दर्शाइए कि S={(x,2y,3x):x,y:}S=left{(x, 2 y, 3 x): x, yright. वास्तविक संख्याऐं हैं | R^(3)(R)R^3(R) का एक उपसमष्टि है । SS के दो आधार ज्ञात कीजिए । SS की विमा भी ज्ञात कीजिए ।
Show that S={(x,2y,3x):x,yS={(x, 2 y, 3 x): x, y are real numbers is a subspace of R^(3)(R)R^3(R). Find two bases of SS. Also find the dimension of SS.
3(a)(i) यदि u=x^(2)+y^(2),v=x^(2)-y^(2)u=x^2+y^2, v=x^2-y^2, जहाँ पर x=r cos theta,y=r sin thetax=r cos theta, y=r sin theta हैं, तब (del(u,v))/(del(r,theta))frac{partial(u, v)}{partial(r, theta)} ज्ञात कीजिए ।
If u=x^(2)+y^(2),v=x^(2)-y^(2)u=x^2+y^2, v=x^2-y^2, where x=r cos theta,y=r sin thetax=r cos theta, y=r sin theta, then find (del(u,v))/(del(r,theta))frac{partial(u, v)}{partial(r, theta)}.
3.(a)(ii) यदि int_(0)^(x)f(t)dt=x+int_(x)^(1)tf(t)dtint_0^x f(t) d t=x+int_x^1 t f(t) d t है, तो f(1)f(1) का मान ज्ञात कीजिए ।
If int_(0)^(x)f(t)dt=x+int_(x)^(1)tf(t)dtint_0^x f(t) d t=x+int_x^1 t f(t) d t, then find the value of f(1)f(1).
3.(a)(iii) int_(a)^(b)(x-a)^(m)(b-x)^(n)dxint_a^b(x-a)^m(b-x)^n d x को बीटा-फलन के रूप में व्यक्त कीजिए ।
Express int_(a)^(b)(x-a)^(m)(b-x)^(n)dxint_a^b(x-a)^m(b-x)^n d x in terms of Beta function.
3.(b) अचर त्रिज्या rr का एक गोला मूल-बिंदु OO से गुजरता है तथा अक्षों को A,B,CA, B, C बिन्दुओं पर काटता है । OO से समतल ABCA B C पर खींचे गए लंब-पाद का बिन्दुपथ ज्ञात कीजिए ।
A sphere of constant radius rr passes through the origin OO and cuts the axes at the points A,BA, B and CC. Find, the locus of the foot of the perpendicular drawn from OO to the plane ABCA B C.
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 AA तथा BB जिनकी कोटि, 2 है के लिए दर्शाइए कि अनुरेख (AB)=(A B)= अनुरेख (BA)(B A) । अतैव दर्शाइए कि AB-BA!=I_(2)A B-B A neq I_2 जहाँ I_(2)I_2 एक 2 -कोटि का तत्समक आव्यूह है ।
For two square matrices AA and BB of order 2 , show that trace (AB)=trace(BA)(A B)=operatorname{trace}(B A). Hence show that AB-BA!=I_(2)A B-B A neq I_2, where I_(2)I_2 is an identity matrix of order 2.72.7
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=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]
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=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]
4.(a)(ii) सम्मिश्र संख्या क्षेत्र पर आव्यूह A=([0,-i],[i,0])A=left(begin{array}{cc}0 & -i \ i & 0end{array}right) 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=left(begin{array}{cc}0 & -i \ i & 0end{array}right), 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} का पूरा क्षेत्रफल (3)/(8)pia^(2)frac{3}{8} pi a^2 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} is (3)/(8)pia^(2)frac{3}{8} pi a^2. 15 4.
(c) रेखाओं {:[(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}
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).]:}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}
Also find the point of intersection of the given lines.
खण्ड ‘B’ SECTION ‘ BB ‘
5.(a) अवकल समीकरण : (d^(2)y)/(dx^(2))+2y=x^(2)e^(3x)+e^(x)cos 2xfrac{d^2 y}{d x^2}+2 y=x^2 e^{3 x}+e^x cos 2 x
को हल कीजिए ।
Solve the differential equation : (d^(2)y)/(dx^(2))+2y=x^(2)e^(3x)+e^(x)cos 2xfrac{d^2 y}{d x^2}+2 y=x^2 e^{3 x}+e^x cos 2 x
5.(b) लाप्लास रूपान्तर विधि का उपयोग करते हुए प्रारम्भिक मान समस्या : (d^(2)y)/(dx^(2))+4y=e^(-2x)sin 2x;y(0)=y^(‘)(0)=0frac{d^2 y}{d x^2}+4 y=e^{-2 x} sin 2 x ; y(0)=y^{prime}(0)=0
को हल कीजिए ।
Solve the initial value problem : (d^(2)y)/(dx^(2))+4y=e^(-2x)sin 2x;y(0)=y^(‘)(0)=0frac{d^2 y}{d x^2}+4 y=e^{-2 x} sin 2 x ; y(0)=y^{prime}(0)=0
using Laplace transform method.
5(c) दो छड़े LML M व MNM N बिन्दु MM पर दृढ़ता से इस प्रकार जुड़ी हैं कि (LM)^(2)+(MN)^(2)=(LN)^(2)(L M)^2+(M N)^2=(L N)^2 तथा वे स्वतन्त्र रूप से साम्यावस्था में स्थिर बिन्दु LL पर टँगी हैं। माना कि दोनों एकसमान छड़ों का प्रति एकांक लम्बाई, भार omegaomega है। छड़ LML M का ऊर्ध्वाधर दिशा के साथ बने कोण को छड़ों की लम्बाई के रूप में ज्ञात कीजिए ।
Two rods LML M and MNM N are joined rigidly at the point MM such that (LM)^(2)+(MN)^(2)=(LN)^(2)(L M)^2+(M N)^2=(L N)^2 and they are hanged freely in equilibrium from a fixed point LL. Let omegaomega be the weight per unit length of both the rods which are uniform. Determine the angle, which the rod LML M 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 grad^(2)[grad*((( vec(r)))/(r))]=(2)/(r^(4))nabla^2left[nabla cdotleft(frac{vec{r}}{r}right)right]=frac{2}{r^4}, जहाँ vec(r)=x hat(i)+y hat(j)+z hat(k)vec{r}=x hat{i}+y hat{j}+z hat{k} है ।
Show that grad^(2)[grad*((( vec(r)))/(r))]=(2)/(r^(4))nabla^2left[nabla cdotleft(frac{vec{r}}{r}right)right]=frac{2}{r^4}, where vec(r)=x hat(i)+y hat(j)+z hat(k)vec{r}=x hat{i}+y hat{j}+z hat{k}
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 क्रमशः कैटिनरी के बीच के बिन्दुओं A,B,CA, B, C पर तनाव हैं, जिन पर इसके क्षैतिज के साथ आनति कोण, सार्व अंतर betabeta के साथ समांतर श्रेढ़ी में हैं। माना कि डोरी के ABA B तथा BCB C भागों के भार क्रमशः omega_(1)omega_1 तथा omega_(2)omega_2 हैं। सिद्ध कीजिए
(i) T_(1),T_(2)T_1, T_2 तथा T_(3)T_3 का हरात्मक माध्य =(3T_(2))/(1+2cos beta)=frac{3 T_2}{1+2 cos beta}
(ii) (T_(1))/(T_(3))=(omega_(1))/(omega_(2))frac{T_1}{T_3}=frac{omega_1}{omega_2}
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 be the tensions at the intermediate points A,B,CA, B, C of the catenary respectively where its inclinations to the horizontal are in arithmetic progression with common difference betabeta. Let omega_(1)omega_1 and omega_(2)omega_2 be the weights of the parts ABA B and BCB C of the string respectively. Prove that
(i) Harmonic mean of T_(1),T_(2)T_1, T_2 and T_(3)=(3T_(2))/(1+2cos beta)T_3=frac{3 T_2}{1+2 cos beta}
(ii) (T_(1))/(T_(3))=(omega_(1))/(omega_(2))frac{T_1}{T_3}=frac{omega_1}{omega_2}
6.(b) सभी अन्तर्त्रस्त (शामिल) चरणों को दर्शति हुए समीकरण : (d^(2)y)/(dx^(2))+(tan x-3cos x)(dy)/(dx)+2ycos^(2)x=cos^(4)xfrac{d^2 y}{d x^2}+(tan x-3 cos x) frac{d y}{d x}+2 y cos ^2 x=cos ^4 x
को पूर्ण रूप से हल कीजिए ।
Solve the equation: (d^(2)y)/(dx^(2))+(tan x-3cos x)(dy)/(dx)+2ycos^(2)x=cos^(4)xfrac{d^2 y}{d x^2}+(tan x-3 cos x) frac{d y}{d x}+2 y cos ^2 x=cos ^4 x
completely by demonstrating all the steps involved.
6.(c) int _(C) vec(F)*d vec(r)int_C vec{F} cdot d vec{r} का मान निकालिए,
जहाँ C,xyC, x y-समतल में एक स्वैच्छिक संवृत वक्र है तथा 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} है।
Evaluate int _(C) vec(F)*d vec(r)int_C vec{F} cdot d vec{r}, where CC is an arbitrary closed curve in the xyx y-plane and 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}.
7(a) प्रथम अष्टांशक में y^(2)+z^(2)=9y^2+z^2=9 तथा x=2x=2 द्वारा परिबद्ध क्षेत्र पर 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} के लिए गाउस अपसरण प्रमेय को सत्यापित कीजिए ।
Verify Gauss divergence theorem for 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} taken over the region in the first octant bounded by y^(2)+z^(2)=9y^2+z^2=9 and x=2x=2.
7.(b) अवकल समीकरण : 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
के सभी सम्भव हल ज्ञात कीजिए ।
Find all possible solutions of the differential equation : 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 {. }
7(c) एक भारी कण aa लम्बाई की अवितान्य डोरी से एक स्थिर बिन्दु से टँगा है तथा sqrt(2gh)sqrt{2 g h} वेग से क्षैतिज दिशा में प्रक्षेपित किया जाता है । यदि (5a)/(2) > h > afrac{5 a}{2}>h>a है, तो सिद्ध कीजिए कि प्रक्षेपण बिन्दु से (1)/(3)(a+2h)frac{1}{3}(a+2 h) 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 ((4a-h)(a+2h)^(2))/(27a^(2))frac{(4 a-h)(a+2 h)^2}{27 a^2} है ।
A heavy particle hangs by an inextensible string of length aa from a fixed point and is then projected horizontally with a velocity sqrt(2gh)sqrt{2 g h}. If (5a)/(2) > h > afrac{5 a}{2}>h>a, then prove that the circular motion ceases when the particle has reached the height (1)/(3)(a+2h)frac{1}{3}(a+2 h) from the point of projection. Also, prove that the greatest height ever reached by the particle above the point of projection is ((4a-h)(a+2h)^(2))/(27a^(2))frac{(4 a-h)(a+2 h)^2}{27 a^2}.
8(a)(i) संनाभि शांकव कुल (x^(2))/(a^(2)+lambda)+(y^(2))/(b^(2)+lambda)=1;a > b > 0frac{x^2}{a^2+lambda}+frac{y^2}{b^2+lambda}=1 ; a>b>0 अचर हैं तथा lambdalambda 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)+lambda)+(y^(2))/(b^(2)+lambda)=1;quad a > b > 0frac{x^2}{a^2+lambda}+frac{y^2}{b^2+lambda}=1 ; quad a>b>0 are constants and lambdalambda is a parameter. Show that the given family of curves is self orthogonal.
8(a)(ii) अवकल समीकरण : 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=0 का व्यापक हल ज्ञात कीजिए । अतः अवकल समीकरण : 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^3 को प्राचल विचरण विधि द्वारा हल कीजिए ।
Find the general solution of the differential equation : 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=0.
Hence, solve the differential equation: 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^3 by the method of variation of parameters.
8.(b) द्रव्यमान mm का एक कण, जो की प्रक्षेपण बिन्दु से वेग uu के साथ क्षैतिज दिशा के साथ thetatheta 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 4sqrtg4 sqrt{g} 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 mm which is projected in a vertical plane through a point of projection with velocity uu in a direction making an angle thetatheta with the horizontal direction. Further, if particles are projected from that point in the same vertical plane with velocity 4sqrtg4 sqrt{g}, then determine the locus of vertices of their paths.
8(c) स्टोक्स प्रमेय का उपयोग करते हुए ∬_(S)(grad xx vec(F))* hat(n)dSiint_S(nabla times vec{F}) cdot hat{n} d S का मान निकालिए, जहाँ पर vec(F)=(x^(2)+y-4) hat(i)+3xy hat(j)+(2xy+z^(2)) hat(k)vec{F}=left(x^2+y-4right) hat{i}+3 x y hat{j}+left(2 x y+z^2right) hat{k} तथा SS, परवलयज z=4-(x^(2)+y^(2))z=4-left(x^2+y^2right) का xyx y-समतल से ऊपर का पृष्ठ है। यहाँ hat(n),Shat{n}, S पर एकक बहिर्मुखी अभिलंब सदिश है ।
Using Stokes’ theorem, evaluate ∬_(S)(grad xx vec(F))* hat(n)dSiint_S(nabla times vec{F}) cdot hat{n} d S, where vec(F)=(x^(2)+y-4) hat(i)+3xy hat(j)+(2xy+z^(2)) hat(k)vec{F}=left(x^2+y-4right) hat{i}+3 x y hat{j}+left(2 x y+z^2right) hat{k} and SS is the surface of the paraboloid z=4-(x^(2)+y^(2))z=4-left(x^2+y^2right) above the xyx y-plane. Here, hat(n)hat{n} is the unit outward normal vector on SS.
UPSC Maths Optional Paper – 01 2020
upsc-m2020-1-e59634ed-aa89-4306-aaf5-1772dafc3c90
खण्ड-A / SECTION-A
1(a) माना समुच्चय VV में सभी n xx nn times n के वास्तविक मैजिक वर्ग हैं। दिखाइए कि समुच्चय V,RV, R पर एक सदिश समष्टि है। दो भिन्न-भिन्न 2xx22 times 2 मैजिक वर्ग के उदाहरण दीजिए।
Consider the set VV of all n xx nn times n real magic squares. Show that VV is a vector space over RR. Give examples of two distinct 2xx22 times 2 magic squares.
(b) माना M_(2)(R)M_2(R) सभी 2xx22 times 2 वास्तविक आव्यूहों का सदिश समष्टि है। माना B=[[1,-1],[-4,4]]B=left[begin{array}{cc}1 & -1 \ -4 & 4end{array}right]. माना T:M_(2)(R)rarrM_(2)(R)T: M_2(R) rightarrow M_2(R) एक रैखिक रूपांतरण है, जो T(A)=BAT(A)=B A द्वारा परिभाषित है। TT की कोटि (रिक) व शून्यता (नलिटि) ज्ञात कीजिए। आव्यूह AA ज्ञात कीजिए, जो शून्य आव्यूह को प्रतिचित्रित करता है।
Let M_(2)(R)M_2(R) be the vector space of all 2xx22 times 2 real matrices. Let B=[[1,-1],[-4,4]]B=left[begin{array}{cc}1 & -1 \ -4 & 4end{array}right]. Suppose T:M_(2)(R)rarrM_(2)(R)T: M_2(R) rightarrow M_2(R) is a linear transformation defined by T(A)=BAT(A)=B A. Find the rank and nullity of TT. Find a matrix AA which maps to the null matrix.
(c) lim_(x rarr(pi)/(4))(tan x)^(tan 2x)lim _{x rightarrow frac{pi}{4}}(tan x)^{tan 2 x} का मान निकालिए।
Evaluate lim_(x rarr(pi)/(4))(tan x)^(tan 2x)lim _{x rightarrow frac{pi}{4}}(tan x)^{tan 2 x}
(d) वक्र (2x+3)y=(x-1)^(2)(2 x+3) y=(x-1)^2 के सभी अनंतस्पर्शी निकालिए।
Find all the asymptotes of the curve (2x+3)y=(x-1)^(2)(2 x+3) y=(x-1)^2.
(e) दीर्घवृत्तज 2x^(2)+6y^(2)+3z^(2)=272 x^2+6 y^2+3 z^2=27 के स्पर्श समतल का समीकरण निकालिए, जो रेखा x-y-z=0=x-y+2z-9x-y-z=0=x-y+2 z-9 से होकर गुजरता है।
Find the equations of the tangent plane to the ellipsoid 2x^(2)+6y^(2)+3z^(2)=272 x^2+6 y^2+3 z^2=27 which passes through the line x-y-z=0=x-y+2z-9x-y-z=0=x-y+2 z-9.
2(a) int_(0)^(1)tan^(-1)(1-(1)/(x))dxint_0^1 tan ^{-1}left(1-frac{1}{x}right) d x का मान निकालिए।
Evaluate int_(0)^(1)tan^(-1)(1-(1)/(x))dxint_0^1 tan ^{-1}left(1-frac{1}{x}right) d x
(b) एक n xx nn times n आव्यूह AA को परिभाषित कीजिए, जबकि A=I-2u*u^(T)A=I-2 u cdot u^T, जहाँ uu एक इकाई स्तंभ सदिश है।
(i) परीक्षण कीजिए कि AA सममित है।
(ii) परीक्षण कीजिए कि AA लांबिक है।
(iii) दिखाइए कि आव्यूह AA का अनुरेख (n-2)(n-2) है।
(iv) आव्यूह A_(3xx3)A_{3 times 3} निकालिए, जबकि u=[[(1)/(3)],[(2)/(3)],[(2)/(3)]]u=left[begin{array}{c}frac{1}{3} \ frac{2}{3} \ frac{2}{3}end{array}right] है।
Define an n xx nn times n matrix as A=I-2u*u^(T)A=I-2 u cdot u^T, where uu is a unit column vector.
(i) Examine if AA is symmetric.
(ii) Examine if AA is orthogonal.
(iii) Show that trace (A)=n-2(A)=n-2.
(iv) Find A_(3xx3)A_{3 times 3}, when u=[[(1)/(3)],[(2)/(3)],[(2)/(3)]]u=left[begin{array}{c}frac{1}{3} \ frac{2}{3} \ frac{2}{3}end{array}right].
(c) एक ऐसे बेलन का समीकरण निकालिए, जिसकी जनक-रेखाएँ, रेखा (x)/(1)=(y)/(-2)=(z)/(3)frac{x}{1}=frac{y}{-2}=frac{z}{3} के समांतर हैं तथा जिसका मार्गदर्शक वक्र x^(2)+y^(2)=4,z=2x^2+y^2=4, z=2 है।
Find the equation of the cylinder whose generators are parallel to the line (x)/(1)=(y)/(-2)=(z)/(3)frac{x}{1}=frac{y}{-2}=frac{z}{3} and whose guiding curve is x^(2)+y^(2)=4,z=2x^2+y^2=4, z=2.
3(a) निम्न फलन पर विचार कीजिए : f(x)=int_(0)^(x)(t^(2)-5t+4)(t^(2)-5t+6)dtf(x)=int_0^xleft(t^2-5 t+4right)left(t^2-5 t+6right) d t
(i) फलन f(x)f(x) के क्रांतिक बिंदु निकालिए।
(ii) वे बिंदु निकालिए, जहाँ f(x)f(x) का स्थानीय न्यूनतम होगा।
(iii) वे बिंदु निकालिए, जहाँ f(x)f(x) का स्थानीय अधिकतम होगा।
(iv) फलन f(x)f(x) के [0,5][0,5] में कितने शून्यक होंगे, निकालिए।
Consider the function f(x)=int_(0)^(x)(t^(2)-5t+4)(t^(2)-5t+6)dtf(x)=int_0^xleft(t^2-5 t+4right)left(t^2-5 t+6right) d t.
(i) Find the critical points of the function 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) in [0,5][0,5].
(b) माना FF सम्मिश्र संख्याओं का एक उपक्षेत्र है व T:F^(3)rarrF^(3)T: F^3 rightarrow F^3 एक ऐसा फलन है, जो निम्न रूप से परिभाषित है : T(x_(1),x_(2),x_(3))=(x_(1)+x_(2)+3x_(3),2x_(1)-x_(2),-3x_(1)+x_(2)-x_(3))Tleft(x_1, x_2, x_3right)=left(x_1+x_2+3 x_3, 2 x_1-x_2,-3 x_1+x_2-x_3right) a,b,ca, b, c पर क्या शर्तें हैं कि (a,b,c),T(a, b, c), T के शुन्य समष्टि में है? TT की शून्यता निकालिए।
Let FF be a subfield of complex numbers and TT a function from F^(3)rarrF^(3)F^3 rightarrow F^3 defined by T(x_(1),x_(2),x_(3))=(x_(1)+x_(2)+3x_(3),2x_(1)-x_(2),-3x_(1)+x_(2)-x_(3))Tleft(x_1, x_2, x_3right)=left(x_1+x_2+3 x_3, 2 x_1-x_2,-3 x_1+x_2-x_3right). What are the conditions on a,b,ca, b, c such that (a,b,c)(a, b, c) be in the null space of TT ? Find the nullity of TT.
(c) यदि सरल रेखा (x)/(1)=(y)/(2)=(z)/(3)frac{x}{1}=frac{y}{2}=frac{z}{3} शंकु 5yz-8zx-3xy=05 y z-8 z x-3 x y=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)frac{x}{1}=frac{y}{2}=frac{z}{3} represents one of a set of three mutually perpendicular generators of the cone 5yz-8zx-3xy=05 y z-8 z x-3 x y=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=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]
(i) ABA B ज्ञात कीजिए।
(ii) सारणिक (A)(A) व सारणिक (B)(B) ज्ञात कीजिए।
(iii) निम्न रैखिक समीकरणों के निकाय का हल निकालिए : 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=14
Let 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]
(i) Find ABA B.
(ii) Find det(A)operatorname{det}(A) and det(B)operatorname{det}(B).
(iii) Solve the following system of linear equations : 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=14
(b) अतिपरवलयिक परवलयज (x^(2))/(a^(2))-(y^(2))/(b^(2))=2zfrac{x^2}{a^2}-frac{y^2}{b^2}=2 z 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))=2zfrac{x^2}{a^2}-frac{y^2}{b^2}=2 z.
(c) लाग्रांज की अनिर्धारित गुणक विधि का प्रयोग करके फलन u=x^(2)+y^(2)+z^(2)u=x^2+y^2+z^2 का चरम मान ज्ञात कीजिए, जो 2x+3y+5z=302 x+3 y+5 z=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, subject to the condition 2x+3y+5z=302 x+3 y+5 z=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))(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)
Solve the following differential equation : 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)
(b) वृत्त-कुल, जो बिंदु (0,2)(0,2) एवं (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) and (0,-2)(0,-2).
(c) a,b,ca, b, c के किस मान के लिए सदिश क्षेत्र 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}
अघूर्णी है? तब bar(V)bar{V} को अदिश फलन phiphi की प्रवणता के रूप में व्यक्त कीजिए। phiphi को ज्ञात कीजिए।
For what value of a,b,ca, b, c is the vector field 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}
irrotational? Hence, express bar(V)bar{V} as the gradient of a scalar function phiphi. Determine phiphi.
(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 ABCA B C से तीन कण, जिनमें से हरेक का द्रव्यमान mm है, A,BA, B तथा CC पर बंधे हुए हैं। उस छड़ को बिंदु AA से BCB C दूरी के बराबर स्थित बिंदु पर एक बल PP के द्वारा लम्बवत् मारा जाता है। सिद्ध कीजिए कि पैदा हुई गतिज ऊर्जा का मान (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} है, जहाँ AB=aA B=a तथा BC=bB C=b.
A light rigid rod ABCA B C has three particles each of mass mm attached to it at A,BA, B and CC. The rod is struck by a blow PP at right angles to it at a point distant from AA equal to BCB C. Prove that the kinetic energy set up is (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}, where AB=aA B=a and BC=bB C=b.
6(a) प्राचल विचरण विधि का प्रयोग करके, निम्न अवकल समीकरण का हल निकालिए, यदि y=e^(-x)y=e^{-x}, पूरक फलन (CF) का एक हल है : y^(”)+(1-cot x)y^(‘)-y cot x=sin^(2)xy^{prime prime}+(1-cot x) y^{prime}-y cot x=sin ^2 x
Using the method of variation of parameters, solve the differential equation y^(”)+(1-cot x)y^(‘)-y cot x=sin^(2)xy^{prime prime}+(1-cot x) y^{prime}-y cot x=sin ^2 x, if y=e^(-x)y=e^{-x} is one solution of CF.
(b) दिए गए सदिश फलन bar(A)bar{A}, जहाँ bar(A)=(3x^(2)+6y) hat(i)-14 yz hat(j)+20 xz^(2) hat(k)bar{A}=left(3 x^2+6 yright) hat{i}-14 y z hat{j}+20 x z^2 hat{k}, के लिए int _(C) bar(A)*d bar(r)int_C bar{A} cdot d bar{r} का मान निकालिए, जहाँ CC बिंदु (0,0,0)(0,0,0) से (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
(ii) सरल रेखा (0,0,0)(0,0,0) से (1,0,0)(1,0,0) तक जोड़ने पर, फिर (1,1,0)(1,1,0) तक तथा फिर (1,1,1)(1,1,1) तक
(iii) सरल रेखा (0,0,0)(0,0,0) से (1,1,1)(1,1,1) तक जोड़ने पर
क्या सभी स्थितियों में परिणाम समान हैं? कारण की व्याख्या कीजिए।
For the vector function bar(A)bar{A}, where bar(A)=(3x^(2)+6y) hat(i)-14 yz hat(j)+20 xz^(2) hat(k)bar{A}=left(3 x^2+6 yright) hat{i}-14 y z hat{j}+20 x z^2 hat{k}, calculate int _(C) bar(A)*d bar(r)int_C bar{A} cdot d bar{r} from (0,0,0)(0,0,0) to (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
(ii) Straight lines joining (0,0,0)(0,0,0) to (1,0,0)(1,0,0), then to (1,1,0)(1,1,0) and then to (1,1,1)(1,1,1)
(iii) Straight line joining (0,0,0)(0,0,0) to (1,1,1)(1,1,1)
Is the result same in all the cases? Explain the reason.
(c) एक दंड ADA D दो आलंब BB एवं CC पर विश्राम करता है, जबकि AB=BC=CDA B=B C=C D. यह पाया गया कि दंड झुक जाएगा यदि एक भार pkgp mathrm{~kg}, बिंदु AA से लटकाया जाए या एक भार qkgq mathrm{~kg}, बिंदु DD से लटकाया जाए। दंड का भार बताइए।
A beam ADA D rests on two supports BB and CC, where AB=BC=CDA B=B C=C D. It is found that the beam will tilt when a weight of pkgp mathrm{~kg} is hung from AA or when a weight of qkgq mathrm{~kg} is hung from DD. Find the weight of the beam.
7(a) स्टोक्स प्रमेय को सत्यापित कीजिए, जबकि सदिश क्षेत्र 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} एक सतह SS पर है जो कि एक बेलन z=1-x^(2),0 <= x <= 1,-2 <= y <= 2z=1-x^2, 0 leq x leq 1,-2 leq y leq 2 का हिस्सा है, जहाँ SS उपरिमुखी अभिविन्यस्त है।
Verify the Stokes’ theorem for the vector field 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} on the surface SS which is the part of the cylinder z=1-x^(2)z=1-x^2 for 0 <= x <= 1,-2 <= y <= 20 leq x leq 1,-2 leq y leq 2; SS is oriented upwards.
(b) लाप्लास रूपांतरण का प्रयोग करके प्रारंभिक मान समस्या ty^(”)+2ty^(‘)+2y=2;y(0)=1t y^{prime prime}+2 t y^{prime}+2 y=2 ; y(0)=1 तथा y^(‘)(0)y^{prime}(0) स्वेच्छ है, को हल कीजिए। क्या इस प्रश्न का हल अद्वितीय है?
Using Laplace transform, solve the initial value problem ty^(”)+2ty^(‘)+2y=2t y^{prime prime}+2 t y^{prime}+2 y=2; y(0)=1y(0)=1 and y^(‘)(0)y^{prime}(0) is arbitrary. Does this problem have a unique solution?
(c) (i) चार एकसमान भारी छड़, जो समान भार WW की हैं, एक वर्ग के रूप में ढाँचा बनाते हुए जुड़ी हैं। यह एक कोने से टँगा हुआ है। भार WW 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 VV 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 pp है। दिखाइए कि कण की तारे के केंद्र से न्यूनतम दूरी lambdalambda है, जहाँ V^(2)lambda=sqrt(mu^(2)+p^(2)V^(4))-muV^2 lambda=sqrt{mu^2+p^2 V^4}-mu. यहाँ mumu एक अचर है।
(i) A square framework formed of uniform heavy rods of equal weight WW jointed together, is hung up by one corner. A weight WW 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 VV. Let pp 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 lambdalambda, where V^(2)lambda=sqrt(mu^(2)+p^(2)V^(4))-muV^2 lambda=sqrt{mu^2+p^2 V^4}-mu. Here mumu is a constant.
8(a) (i) निम्न अवकल समीकरण हल कीजिए : (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)
(ii) अवकल समीकरण 9p^(2)(2-y)^(2)=4(3-y)9 p^2(2-y)^2=4(3-y) के व्यापक व विचित्र हल निकालिए, जहाँ p=(dy)/(dx)p=frac{d y}{d x}.
(i) Solve the following differential equation : (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)
(ii) Find the general and singular solutions of the differential equation 9p^(2)(2-y)^(2)=4(3-y)9 p^2(2-y)^2=4(3-y), where p=(dy)/(dx)p=frac{d y}{d x}
(b) पृष्ठ समाकल ∬_(S)grad xx bar(F)* hat(n)dSiint_S nabla times bar{F} cdot hat{n} d S का मान निकालिए, जहाँ 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} तथा SS गोले x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2 की सतह है, जो xyx y-तल के ऊपर है।
Evaluate the surface integral ∬_(S)grad xx bar(F)* hat(n)dSiint_S nabla times bar{F} cdot hat{n} d S for 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} and SS is the surface of the sphere x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2 above the xyx y-plane.
(c) एक चार पहियों वाला रेलवे ट्रक, जिसका कुल द्रव्यमान MM है, के पहिए और धुरी के हर युग्म का द्रव्यमान व परिभ्रमण त्रिज्या क्रमशः mm तथा kk है। हर पहिए की त्रिज्या rr है। यदि ट्रक को बल PP द्वारा सीधे पथ (ट्रैक) पर धकेला जाता है, तब सिद्ध कीजिए कि उसका त्वरण (P)/(M+(2mk^(2))/(r^(2)))frac{P}{M+frac{2 m k^2}{r^2}} 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 MM, the mass and radius of gyration of each pair of wheels and axle are mm and kk respectively, and the radius of each wheel is rr. Prove that if the truck is propelled along a level track by a force PP, the acceleration is (P)/(M+(2mk^(2))/(r^(2)))frac{P}{M+frac{2 m k^2}{r^2}}, and find the horizontal force exerted on each axle by the truck. The axle friction and wind resistance are to be neglected.
UPSC Maths Optional Paper – 01 2019
upsc-m2019-1-28d83552-ba33-4aa8-99e9-b6fbf8522ea0
खण्ड-A / SECTION-A
1. (a) माना कि f:[0,(pi)/(2)]rarrRf:left[0, frac{pi}{2}right] rightarrow mathbb{R} एक संतत फलन है, जैसा कि 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((pi)/(2))fleft(frac{pi}{2}right) का मान ज्ञात कीजिए।
Let f:[0,(pi)/(2)]rarrRf:left[0, frac{pi}{2}right] rightarrow mathbb{R} be a continuous function such that 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}
Find the value of f((pi)/(2))fleft(frac{pi}{2}right).
(b) माना कि f:D(subeR^(2))rarrRf: Dleft(subseteq mathbb{R}^2right) rightarrow mathbb{R} एक फलन है और (a,b)in D(a, b) in D. अगर f(x,y)f(x, y) बिंदु (a,b)(a, b) पर संतत है, तो दर्शाइए कि फलन f(x,b)f(x, b) और f(a,y)f(a, y) क्रमशः x=ax=a और y=by=b पर संतत हैं।
Let f:D(subeR^(2))rarrRf: Dleft(subseteq mathbb{R}^2right) rightarrow mathbb{R} be a function and (a,b)in D(a, b) in D. If f(x,y)f(x, y) is continuous at (a,b)(a, b), then show that the functions f(x,b)f(x, b) and f(a,y)f(a, y) are continuous at x=ax=a and at y=by=b respectively.
(c) माना कि T:R^(2)rarrR^(2)T: mathbb{R}^2 rightarrow mathbb{R}^2 एक रैखिक प्रतिचित्र है, जैसा कि T(2,1)=(5,7)T(2,1)=(5,7) एवं T(1,2)=(3,3)T(1,2)=(3,3). अगर AA मानक आधारों e_(1),e_(2)e_1, e_2 के सापेक्ष TT के संगत आव्यूह है, तो AA की कोटि ज्ञात कीजिए।
Let T:R^(2)rarrR^(2)T: mathbb{R}^2 rightarrow mathbb{R}^2 be a linear map such that T(2,1)=(5,7)T(2,1)=(5,7) and T(1,2)=(3,3)T(1,2)=(3,3).
If AA is the matrix corresponding to TT with respect to the standard bases e_(1),e_(2)e_1, e_2, then find Rank(A)operatorname{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=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]
है, तो दर्शाइए कि AB=6I_(3)A B=6 I_3. इस परिणाम का उपयोग करते हुए निम्नलिखित समीकरण निकाय को हल कीजिए : {:[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}
If 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]
then show that AB=6I_(3)A B=6 I_3. Use this result to solve the following system of equations : {:[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}
(e) दर्शाइए कि (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}
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)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}
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=(pi)/(2)f(x)=|cos x|+|sin x|, x=frac{pi}{2} पर अवकलनीय है? अगर आपका उत्तर हाँ है, तो f(x)f(x) का अवकलज x=(pi)/(2)x=frac{pi}{2} 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| differentiable at x=(pi)/(2)x=frac{pi}{2} ? If yes, then find its derivative at x=(pi)/(2)x=frac{pi}{2}. If no, then give a proof of it.
(b) माना कि AA और BB समान कोटि के दो लांबिक आव्यूह हैं तथा det A+det B=0operatorname{det} A+operatorname{det} B=0. दर्शाइए कि A+BA+B एक अव्युत्क्रमणीय (सिंगुलर) आव्यूह है।
Let AA and BB be two orthogonal matrices of same order and det A+det B=0operatorname{det} A+operatorname{det} B=0. Show that A+BA+B is a singular matrix.
(c) (i) समतल x+2y+3z=12x+2 y+3 z=12 निर्देशांक अक्षों को A,B,CA, B, C पर प्रतिच्छेद करता है। त्रिभुज ABCA B C के परिवृत्त का समीकरण ज्ञात कीजिए।
(ii) सिद्ध कीजिए कि समतल z=0z=0 गोलक x^(2)+y^(2)+z^(2)=11x^2+y^2+z^2=11 के अन्वालोपी शंकु, जिसका शीर्ष (2,4,1)(2,4,1) पर है, को एक समकोणीय अतिपरवलय पर प्रतिच्छेद करता है।
(i) The plane x+2y+3z=12x+2 y+3 z=12 cuts the axes of coordinates in A,B,CA, B, C. Find the equations of the circle circumscribing the triangle ABCA B C.
(ii) Prove that the plane z=0z=0 cuts the enveloping cone of the sphere x^(2)+y^(2)+z^(2)=11x^2+y^2+z^2=11 which has the vertex at (2,4,1)(2,4,1) in a rectangular hyperbola.
3(a) फलन f(x)=2x^(3)-9x^(2)+12 x+6f(x)=2 x^3-9 x^2+12 x+6 का अंतराल [2,3][2,3] पर अधिकतम और न्यूनतम मान ज्ञात कीजिए।
Find the maximum and the minimum value of the function f(x)=2x^(3)-9x^(2)+12 x+6f(x)=2 x^3-9 x^2+12 x+6 on the interval [2,3][2,3].
(b) सिद्ध कीजिए कि साधारणतः किसी एक बिंदु से परवलयज x^(2)+y^(2)=2azx^2+y^2=2 a z पर तीन अभिलंब बनाए जा सकते हैं, लेकिन अगर बिंदु सतह 27 a(x^(2)+y^(2))+8(a-z)^(3)=027 aleft(x^2+y^2right)+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)=2azx^2+y^2=2 a z, but if the point lies on the surface 27 a(x^(2)+y^(2))+8(a-z)^(3)=027 aleft(x^2+y^2right)+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=left(begin{array}{rrrr}
5 & 7 & 2 & 1 \
1 & 1 & -8 & 1 \
2 & 3 & 5 & 0 \
3 & 4 & -3 & 1
end{array}right)
(i) आव्यूह AA की कोटि ज्ञात कीजिए।
(ii) उपसमष्टि 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_4right) in mathbb{R}^4 mid Aleft(begin{array}{l}
x_1 \
x_2 \
x_3 \
x_4
end{array}right)=0right}
की विमा ज्ञात कीजिए।
Let 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)
(i) Find the rank of matrix AA.
(ii) Find the dimension of the subspace 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_4right) in mathbb{R}^4 mid Aleft(begin{array}{l}
x_1 \
x_2 \
x_3 \
x_4
end{array}right)=0right}
4(a) कैले-हैमिल्टन प्रमेय का कथन लिखिए। इस प्रमेय का उपयोग करके A^(100)A^{100} का मान ज्ञात कीजिए, जहाँ 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]
State the Cayley-Hamilton theorem. Use this theorem to find A^(100)A^{100}, where 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]
(b) बिंदु PP से गुजरने वाली दीर्घवृत्तज (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1
की अभिलंब जीवा की लंबाई ज्ञात कीजिए और सिद्ध कीजिए कि अगर यह 4PG_(3)4 P G_3 के समान है, जहाँ G_(3)G_3 वह बिंदु है जहाँ PP से गुजरने वाली अभिलंब जीवा xyx y-तल पर मिलती है, तो PP शंकु (x^(2))/(a^(6))(2c^(2)-a^(2))+(y^(2))/(b^(6))(2c^(2)-b^(2))+(z^(2))/(c^(4))=0frac{x^2}{a^6}left(2 c^2-a^2right)+frac{y^2}{b^6}left(2 c^2-b^2right)+frac{z^2}{c^4}=0
पर स्थित है।
Find the length of the normal chord through a point PP of the ellipsoid (x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1
and prove that if it is equal to 4PG_(3)4 P G_3, where G_(3)G_3 is the point where the normal chord through PP meets the xyx y-plane, then PP lies on the cone (x^(2))/(a^(6))(2c^(2)-a^(2))+(y^(2))/(b^(6))(2c^(2)-b^(2))+(z^(2))/(c^(4))=0frac{x^2}{a^6}left(2 c^2-a^2right)+frac{y^2}{b^6}left(2 c^2-b^2right)+frac{z^2}{c^4}=0
(c) (i) अगर 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}}}
है, तो दर्शाइए कि sin^(2)u,xsin ^2 u, x और yy का -(1)/(6)-frac{1}{6} घातविशिष्ट समांगी फलन है। अतएव दर्शाइए कि 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)
(ii) जैकोबियन विधि का व्यवहार करते हुए दर्शाइए कि अगर f^(‘)(x)=(1)/(1+x^(2))f^{prime}(x)=frac{1}{1+x^2} और f(0)=0f(0)=0 है, तो f(x)+f(y)=f((x+y)/(1-xy))f(x)+f(y)=fleft(frac{x+y}{1-x y}right)
(i) If 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}}}
then show that sin^(2)usin ^2 u is a homogeneous function of xx and yy of degree -(1)/(6)-frac{1}{6}.
Hence show that 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)
(ii) Using the Jacobian method, show that if f^(‘)(x)=(1)/(1+x^(2))f^{prime}(x)=frac{1}{1+x^2} and f(0)=0f(0)=0, then f(x)+f(y)=f((x+y)/(1-xy))f(x)+f(y)=fleft(frac{x+y}{1-x y}right)
खण्ड-B / SECTION-B
5(a) अवकल समीकरण (2y sin x+3y^(4)sin x cos x)dx-(4y^(3)cos^(2)x+cos x)dy=0left(2 y sin x+3 y^4 sin x cos xright) d x-left(4 y^3 cos ^2 x+cos xright) d y=0
को हल कीजिए।
Solve the differential equation (2y sin x+3y^(4)sin x cos x)dx-(4y^(3)cos^(2)x+cos x)dy=0left(2 y sin x+3 y^4 sin x cos xright) d x-left(4 y^3 cos ^2 x+cos xright) d y=0
(b) अवकल समीकरण (d^(2)y)/(dx^(2))-4(dy)/(dx)+4y=3x^(2)e^(2x)sin 2xfrac{d^2 y}{d x^2}-4 frac{d y}{d x}+4 y=3 x^2 e^{2 x} sin 2 x
का पूर्ण हल ज्ञात कीजिए।
Determine the complete solution of the differential equation (d^(2)y)/(dx^(2))-4(dy)/(dx)+4y=3x^(2)e^(2x)sin 2xfrac{d^2 y}{d x^2}-4 frac{d y}{d x}+4 y=3 x^2 e^{2 x} sin 2 x
(c) एक भारी एकसमान छड़ ABA B का एक सिरा एक रूक्ष क्षैतिज छड़ ACA C, जिसके साथ वह वलय (रिंग) के द्वारा जुड़ी हुई है, पर सरक सकती है। BB एवं CC एक रस्सी से जुड़े हैं। जब छड़ सर्पण बिंदु पर है, तब AC^(2)-AB^(2)=BC^(2)A C^2-A B^2=B C^2 है। यदि क्षैतिज रेखा व ABA B के बीच का कोण thetatheta है, तो सिद्ध कीजिए कि घर्षण गुणांक (cot theta)/(2+cot^(2)theta)frac{cot theta}{2+cot ^2 theta} है।
One end of a heavy uniform rod ABA B can slide along a rough horizontal rod ACA C, to which it is attached by a ring. BB and CC are joined by a string. When the rod is on the point of sliding, then AC^(2)-AB^(2)=BC^(2)A C^2-A B^2=B C^2. If thetatheta is the angle between ABA B and the horizontal line, then prove that the coefficient of friction is (cot theta)/(2+cot^(2)theta)frac{cot theta}{2+cot ^2 theta}.
(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 WW है, सतह से 3h3 h ऊँचाई से पृथ्वी की सतह पर गिरता है। दर्शाइए कि पृथ्वी के आकर्षण बल द्वारा किए गए कार्य का परिमाण (3)/(4)hWfrac{3}{4} h W है, जहाँ hh 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 WW, falls to the surface of the earth from a height 3h3 h above it. Show that the magnitude of work done by the earth’s attraction force is (3)/(4)hWfrac{3}{4} h W, where hh is the radius of the earth.
(e) वक्र x=t,y=t^(2),z=t^(3)x=t, y=t^2, z=t^3 के बिंदु (1,1,1)(1,1,1) पर स्पर्श-रेखा की दिशा में फलन xy^(2)+yz^(2)+zx^(2)x y^2+y z^2+z x^2 का दिशात्मक अवकलज ज्ञात कीजिए।
Find the directional derivative of the function xy^(2)+yz^(2)+zx^(2)x y^2+y z^2+z x^2 along the tangent to the curve x=t,y=t^(2),z=t^(3)x=t, y=t^2, z=t^3 at the point (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 aa 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, sqrt3asqrt{3} a है।
A body consists of a cone and underlying hemisphere. The base of the cone and the top of the hemisphere have same radius aa. 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 sqrt3asqrt{3} a.
(b) वक्र CC के चारों तरफ vec(F)vec{F} का परिसंचरण ज्ञात कीजिए, जहाँ vec(F)=(2x+y^(2)) hat(i)+(3y-4x) hat(j)vec{F}=left(2 x+y^2right) hat{i}+(3 y-4 x) hat{j} और CC, बिंदु (0,0)(0,0) से बिंदु (1,1)(1,1) तक वक्र y=x^(2)y=x^2 के द्वारा तथा बिंदु (1,1)(1,1) से बिंदु (0,0)(0,0) तक वक्र y^(2)=xy^2=x के द्वारा परिभाषित है।
Find the circulation of vec(F)vec{F} round the curve CC, where vec(F)=(2x+y^(2)) hat(i)+(3y-4x) hat(j)vec{F}=left(2 x+y^2right) hat{i}+(3 y-4 x) hat{j} and CC is the curve y=x^(2)y=x^2 from (0,0)(0,0) to (1,1)(1,1) and the curve y^(2)=xy^2=x from (1,1)(1,1) to (0,0)(0,0).
(c) (i) अवकल समीकरण (d^(2)y)/(dx^(2))+(3sin x-cot x)(dy)/(dx)+2ysin^(2)x=e^(-cos x)sin^(2)xfrac{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 x
को हल कीजिए।
(ii) t^(-1//2)t^{-1 / 2} तथा t^(1//2)t^{1 / 2} का लाप्लास रूपांतर ज्ञात कीजिए। सिद्ध कीजिए कि t^(n+(1)/(2))t^{n+frac{1}{2}} का लाप्लास रूपांतर (Gamma(n+1+(1)/(2)))/(s^(n+1+(1)/(2)))frac{Gammaleft(n+1+frac{1}{2}right)}{s^{n+1+frac{1}{2}}}
होता है, जहाँ n inNn in mathbf{N}.
(i) Solve the differential equation (d^(2)y)/(dx^(2))+(3sin x-cot x)(dy)/(dx)+2ysin^(2)x=e^(-cos x)sin^(2)xfrac{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 x
(ii) Find the Laplace transforms of t^(-1//2)t^{-1 / 2} and t^(1//2)t^{1 / 2}. Prove that the Laplace transform of t^(n+(1)/(2))t^{n+frac{1}{2}}, where n inNn in mathbb{N}, is (Gamma(n+1+(1)/(2)))/(s^(n+1+(1)/(2)))frac{Gammaleft(n+1+frac{1}{2}right)}{s^{n+1+frac{1}{2}}}
7(a) समीकरण x^(2)y^(”)-2xy^(‘)+2y=x^(3)sin xx^2 y^{prime prime}-2 x y^{prime}+2 y=x^3 sin 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^(”)-2xy^(‘)+2y=x^(3)sin xx^2 y^{prime prime}-2 x y^{prime}+2 y=x^3 sin 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=au tan alphax=a cos u, y=a sin u, z=a u tan alpha 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 ux=a cos u, y=a sin u,z=au tan alphay=a sin u, z=a u tan alpha.
(c) yy-अक्ष की दिशा में गतिमान एक कण का मूलबिंदु की ओर त्वरण FyF y है, जहाँ F,yF, y का एक धनात्मक एवं सम फलन है। जब कण y=-ay=-a तथा y=ay=a के बीच में कंपन करता है, तब उसका आवर्तकाल TT है। दर्शाइए कि (2pi)/(sqrt(F_(1))) < T < (2pi)/(sqrt(F_(2)))frac{2 pi}{sqrt{F_1}}<T<frac{2 pi}{sqrt{F_2}}
जहाँ F_(1)F_1 एवं F_(2)F_2 परास [-a,a][-a, a] में FF के अधिकतम एवं न्यूनतम मान हैं। आगे दर्शाइए कि जब लंबाई ll का एक सरल लोलक ऊर्ध्वाधर रेखा के किसी भी ओर 30^(@)30^{circ} तक दोलन करता है, तब T,2pisqrt(l//g)T, 2 pi sqrt{l / g} तथा 2pisqrt(l//g)sqrt(pi//3)2 pi sqrt{l / g} sqrt{pi / 3} के बीच में रहता है।
A particle moving along the yy-axis has an acceleration FyF y towards the origin, where FF is a positive and even function of yy. The periodic time, when the particle vibrates between y=-ay=-a and y=ay=a, is TT. Show that (2pi)/(sqrt(F_(1))) < T < (2pi)/(sqrt(F_(2)))frac{2 pi}{sqrt{F_1}}<T<frac{2 pi}{sqrt{F_2}}
where F_(1)F_1 and F_(2)F_2 are the greatest and the least values of FF within the range [-a,a][-a, a]. Further, show that when a simple pendulum of length ll oscillates through 30^(@)30^{circ} on either side of the vertical line, TT lies between 2pisqrt(l//g)2 pi sqrt{l / g} and 2pisqrt(l//g)sqrt(pi//3)2 pi sqrt{l / g} sqrt{pi / 3}
8(a) अवकल समीकरण ((dy)/(dx))^(2)((y)/(x))^(2)cot^(2)alpha-2((dy)/(dx))((y)/(x))+((y)/(x))^(2)cosec^(2)alpha=1left(frac{d y}{d x}right)^2left(frac{y}{x}right)^2 cot ^2 alpha-2left(frac{d y}{d x}right)left(frac{y}{x}right)+left(frac{y}{x}right)^2 operatorname{cosec}^2 alpha=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 ((dy)/(dx))^(2)((y)/(x))^(2)cot^(2)alpha-2((dy)/(dx))((y)/(x))+((y)/(x))^(2)cosec^(2)alpha=1left(frac{d y}{d x}right)^2left(frac{y}{x}right)^2 cot ^2 alpha-2left(frac{d y}{d x}right)left(frac{y}{x}right)+left(frac{y}{x}right)^2 operatorname{cosec}^2 alpha=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 (mu)/(” (दूरी “^(2))frac{mu}{text { (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 (mu)/(” (distance) “^(2))frac{mu}{text { (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 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} के लिए x^(2)+y^(2)=4,z=0x^2+y^2=4, z=0 और z=3z=3 से घिरे हुए क्षेत्र में सत्यापित कीजिए।
(ii) स्टोक्स प्रमेय के द्वारा oint_(C)e^(x)dx+2ydy-dzoint_C e^x d x+2 y d y-d z का मान ज्ञात कीजिए, जहाँ CC, वक्र x^(2)+y^(2)=4x^2+y^2=4, z=2z=2 है।
(i) State Gauss divergence theorem. Verify this theorem for 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}, taken over the region bounded by x^(2)+y^(2)=4,z=0x^2+y^2=4, z=0 and z=3z=3.
(ii) Evaluate by Stokes’ theorem oint_(C)e^(x)dx+2ydy-dzoint_C e^x d x+2 y d y-d z, where CC is the curve x^(2)+y^(2)=4,z=2x^2+y^2=4, z=2.
UPSC Maths Optional Paper – 01 2018
upsc-m2018-1-08bbcd08-2ec0-4a0d-8659-5c6f186e3a0d
खण्ड-A / SECTION-A
1(a) मान लीजिये कि AA एक 3xx23 times 2 आव्यूह है और BB एक 2xx32 times 3 आव्यूह है। दर्शाइये कि C=A*BC=A cdot B एक अव्युत्क्रमणणीय आव्यूह है।
Let AA be a 3xx23 times 2 matrix and BB a 2xx32 times 3 matrix. Show that C=A*BC=A cdot B is a singular matrix.
(b) आधार सदिशों e_(1)=(1,0)e_1=(1,0) और e_(2)=(0,1)e_2=(0,1) को alpha_(1)=(2,-1)alpha_1=(2,-1) एवं alpha_(2)=(1,3)alpha_2=(1,3) के रैखिक संयोग के रूप में ब्यक्त कीजिये।
Express basis vectors e_(1)=(1,0)e_1=(1,0) and e_(2)=(0,1)e_2=(0,1) as linear combinations of alpha_(1)=(2,-1)alpha_1=(2,-1) and alpha_(2)=(1,3)alpha_2=(1,3).
(c) निर्धारित कीजिये कि lim_(z rarr1)(1-z)tan (pi z)/(2)lim _{z rightarrow 1}(1-z) tan frac{pi z}{2} का अस्तित्व है या कि नहीं। अगर यह सीमा विघ्यमान है, तो इसका मान ज्ञात कीजिये।
Determine if lim_(z rarr1)(1-z)tan (pi z)/(2)lim _{z rightarrow 1}(1-z) tan frac{pi z}{2} exists or not. If the limit exists, then find its value.
(d) सीमा 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} का मान ज्ञात कीजिये।
Find the limit 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}.
(e) सरल रेखा (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)frac{x-1}{2}=frac{y-1}{3}=frac{z+1}{-1} का समतल x+y+2z=6x+y+2 z=6 पर प्रक्षेपण ज्ञात कीजिये।
Find the projection of the straight line (x-1)/(2)=(y-1)/(3)=(z+1)/(-1)frac{x-1}{2}=frac{y-1}{3}=frac{z+1}{-1} on the plane x+y+2z=6x+y+2 z=6.
2(a) अगर AA और BB समरूप n xx nn times n आव्यूह हैं, तो दर्शाइये कि उनके आइगेन मान एक ही हैं।
Show that if AA and BB are similar n xx nn times n matrices, then they have the same eigenvalues.
(b) बिन्दु (1,0)(1,0) से परबलय y^(2)=4xy^2=4 x तक की न्यूनतम दूरी ज्ञात कीजिये।
Find the shortest distance from the point (1,0)(1,0) to the parabola y^(2)=4xy^2=4 x.
(c) दीर्घवृत्त (x^(2))/(a^(2))+(y^(2))/(b^(2))=1xfrac{x^2}{a^2}+frac{y^2}{b^2}=1 x-अक्ष के चारों तरफ परिभ्रमण कर रहा है। परिक्रमित घन का आयतन ज्ञात कीजिये।
The ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1frac{x^2}{a^2}+frac{y^2}{b^2}=1 revolves about the xx-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]:}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}
और zz-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]:}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}
and the z-axis.
3(a) रेखिक समीकरण निकाय {:[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}
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+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}
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)=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}
निर्धारित कीजिये कि (del f)/(del x)(0,1)frac{partial f}{partial x}(0,1) और (del f)/(del y)(0,1)frac{partial f}{partial y}(0,1) में से किसका अस्तित्व है और किसका अस्तित्व नहीं है।
Let {:[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}
Determine which of (del f)/(del x)(0,1)frac{partial f}{partial x}(0,1) and (del f)/(del y)(0,1)frac{partial f}{partial y}(0,1) exists and which does not exist.
(c) परबलयज (x+y+z)(2x+y-z)=6z(x+y+z)(2 x+y-z)=6 z की उन जनक रेखाओं के समीकरणों को ज्ञात कीजिये, जो बिन्दु (1,1,1)(1,1,1) में से गुज्जरती है।
Find the equations to the generating lines of the paraboloid (x+y+z)(2x+y-z)=6z(x+y+z)(2 x+y-z)=6 z which pass through the point (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) और (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) and (1,2,3)(1,2,3).
4.(a) अन्तराल [2,3][2,3] पर x^(4)-5x^(2)+4x^4-5 x^2+4 के अधिकतम और न्यूनतम मान ज्ञात कीजिये।
Find the maximum and the minimum values of x^(4)-5x^(2)+4x^4-5 x^2+4 on the interval [2,3][2,3].
(b) समाकल 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} का मान निकालिये।
Evaluate the integral 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}.
(c) उस शंकु, जिसका शीर्ष (0,0,1)(0,0,1) है और जिसका निर्देशक वक्र 2x^(2)-y^(2)=4,z=02 x^2-y^2=4, z=0 है, का समीकरण ज्ञात कीजिये।
Find the equation of the cone with (0,0,1)(0,0,1) as the vertex and 2x^(2)-y^(2)=4,z=02 x^2-y^2=4, z=0 as the guiding curve.
(d) 3x-y+3z=83 x-y+3 z=8 के समांतर और बिन्दु (1,1,1)(1,1,1) में से गुजरते हुये समतल का समीकरण ज्ञात कीजिये।
Find the equation of the plane parallel to 3x-y+3z=83 x-y+3 z=8 and passing through the point (1,1,1)(1,1,1).
खण्ड-B / SECTION-B
5. (a) हल कीजिये/Solve : y^(”)-y=x^(2)e^(2x)y^{prime prime}-y=x^2 e^{2 x}
(b) x=3t,y=3t^(2),z=3t^(3)x=3 t, y=3 t^2, z=3 t^3 समीकरणों वाले बक्र के एक आय बिन्दु पर स्पर्शं-रेखा और रेखा y=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=3t,y=3t^(2),z=3t^(3)x=3 t, y=3 t^2, z=3 t^3 and the line y=z-x=0y=z-x=0.
(c) हल कीजिये/Solve: 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}
(d) (i) f(t)=(1)/(sqrtt)f(t)=frac{1}{sqrt{t}} का लाप्लास रूपान्तर ज्ञात कीजिये। Find the Laplace transform of f(t)=(1)/(sqrtt)f(t)=frac{1}{sqrt{t}}.
(ii) (5s^(2)+3s-16)/((s-1)(s-2)(s+3))frac{5 s^2+3 s-16}{(s-1)(s-2)(s+3)} का विलोम लाप्लास रुपान्तर ज्ञात कीजिये। Find the inverse Laplace transform of (5s^(2)+3s-16)/((s-1)(s-2)(s+3))frac{5 s^2+3 s-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 dd दूरी पर है और जिसकी ऊँचाई hh है, को छूते हुये पार करता है। अगर यह कण ऊर्ध्वाधर तल पर गतिमान है और इसकी क्षैतिज पहुँच RR है, तो प्रक्षेपण की उच्चता ज्ञात कीजिये।
A particle projected from a given point on the ground just clears a wall of height hh at a distance dd from the point of projection. If the particle moves in a vertical plane and if the horizontal range is RR, find the elevation of the projection.
6(a) हल कीजिये /Solve : ((dy)/(dx))^(2)y+2(dy)/(dx)x-y=0left(frac{d y}{d x}right)^2 y+2 frac{d y}{d x} x-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 और x_(2)x_2 की दूरी पर वेग क्रमशः v_(1)v_1 और v_(2)v_2 है। उसकी गति का आवर्तकाल ज्ञात कीजिये।
A particle moving with simple harmonic motion in a straight line has velocities v_(1)v_1 and v_(2)v_2 at distances x_(1)x_1 and x_(2)x_2 respectively from the centre of its path. Find the period of its motion.
(c) हल कीजिये/Solve : y^(”)+16 y=32 sec 2xy^{prime prime}+16 y=32 sec 2 x
(d) अगर गोलक x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2 का पृष्ठ SS है, तो गाउस के अपसरण प्रमेय का इस्तेमाल करते हुये ∬_(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]
का मान निकालिये।
If SS is the surface of the sphere x^(2)+y^(2)+z^(2)=a^(2)x^2+y^2+z^2=a^2, then evaluate ∬_(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]
using Gauss’ divergence theorem.
7. (a) हल कीजिये/Solve : (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))
(b) बक्र 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}
की वक्रता और विमोटन ज्ञात कीजिये।
Find the curvature and torsion of the curve 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}
(c) प्रारंभिक मान समस्या {:[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}
को हल कीजिये।
Solve the initial value problem {:[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}
(d) alphaalpha और betabeta को, जिसके लिये x^( alpha)y^( beta)x^alpha y^beta समीकरण (4y^(2)+3xy)dx-(3xy+2x^(2))dy=0left(4 y^2+3 x yright) d x-left(3 x y+2 x^2right) d y=0 का एक समाकलन गुण्पक है, ज्ञात कीजिये और समीकरण हल कीजिये।
Find alphaalpha and betabeta such that x^( alpha)y^( beta)x^alpha y^beta is an integrating factor of (4y^(2)+3xy)dx-(3xy+2x^(2))dy=0left(4 y^2+3 x yright) d x-left(3 x y+2 x^2right) d y=0 and solve the equation.
8. (a) मान लीजिये कि 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} है। दर्शाइये कि 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}.
Let 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}. Show that curl(curl {:( vec(v)))=grad(div vec(v))-grad^(2) vec(v)left.vec{v}right)=operatorname{grad}(operatorname{div} vec{v})-nabla^2 vec{v}.
(b) स्टोकस प्रमेय का इस्तेमाल करते हुये रेखा समाकल int _(C)-y^(3)dx+x^(3)dy+z^(3)dzint_C-y^3 d x+x^3 d y+z^3 d z का मान निकालिये। यहाँ सिलिन्डर x^(2)+y^(2)=1x^2+y^2=1 और समतल x+y+z=1x+y+z=1 का प्रतिच्छेद CC है। CC पर अभिकिन्यास xyx y-समतल में वामावर्त गति के संगत है।
Evaluate the line integral int _(C)-y^(3)dx+x^(3)dy+z^(3)dzint_C-y^3 d x+x^3 d y+z^3 d z using Stokes’ theorem. Here CC is the intersection of the cylinder x^(2)+y^(2)=1x^2+y^2=1 and the plane x+y+z=1x+y+z=1. The orientation on CC corresponds to counterclockwise motion in the xyx y-plane.
(c) मान लीजिये कि vec(F)=xy^(2) vec(i)+(y+x) vec(j)vec{F}=x y^2 vec{i}+(y+x) vec{j} है। ग्रीन के प्रमेय का इस्तेमाल करते हुये प्रथम चतुर्थाश में वक्रों y=x^(2)y=x^2 और y=xy=x द्वारा परिबद्ध क्षेत्र पर (grad xx vec(F))* vec(k)(nabla times vec{F}) cdot vec{k} का समाकलन कीजिये।
Let vec(F)=xy^(2) vec(i)+(y+x) vec(j)vec{F}=x y^2 vec{i}+(y+x) vec{j}. Integrate (grad xx vec(F))* vec(k)(nabla times vec{F}) cdot vec{k} over the region in the first quadrant bounded by the curves y=x^(2)y=x^2 and y=xy=x using Green’s theorem.
(d) f(y)f(y), जिसके लिये समीकरण (2xe^(y)+3y^(2))dy+(3x^(2)+f(y))dx=0left(2 x e^y+3 y^2right) d y+left(3 x^2+f(y)right) d x=0 सथातथ्य है, ज्ञात कीजिये और हाल निकालिये।
Find f(y)f(y) such that (2xe^(y)+3y^(2))dy+(3x^(2)+f(y))dx=0left(2 x e^y+3 y^2right) d y+left(3 x^2+f(y)right) d x=0 is exact and hence solve.
खण्ड A
SECTION A
Question:-01 (a) सिद्ध कीजिए कि nmathrm{n} विमीय सदिश समष्टि Vmathrm{V} के लिए nmathrm{n} रैखिकत: स्वतंत्र सदिशों का कोई भी समुच्चय Vmathrm{V} के लिए एक आधार बनाता है ।
Question:-01 (a) Prove that any set of nmathrm{n} linearly independent vectors in a vector space Vmathrm{V} of dimension nmathrm{n} constitutes a basis for Vmathrm{V}.
Answer:
To prove that any set of nn linearly independent vectors in a vector space VV of dimension nn constitutes a basis for VV, we need to show two things:
The set spans VV.
The set is linearly independent.
Given:
A vector space VV of dimension nn.
A set S={v_(1),v_(2),dots,v_(n)}S = { mathbf{v}_1, mathbf{v}_2, ldots, mathbf{v}_n } of nn linearly independent vectors in VV.
Proof:
Step 1: SS is Linearly Independent (Given)
We are given that SS is a set of nn linearly independent vectors. Therefore, no vector in SS can be written as a linear combination of the other vectors in SS.
Step 2: SS Spans VV
To show that SS spans VV, we need to show that any vector wmathbf{w} in VV can be written as a linear combination of vectors in SS.
Let’s consider another basis BB for VV. Since VV has dimension nn, BB contains exactly nn vectors. We can write wmathbf{w} as a linear combination of vectors in BB: 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}_n
Now, each b_(i)mathbf{b}_i can also be written as a linear combination of vectors in SS because SS and BB are both bases for VV: 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}_n
Substituting this into the equation for wmathbf{w}, we get: 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)
Simplifying, we find that wmathbf{w} can indeed be written as a linear combination of vectors in SS, proving that SS spans VV.
Conclusion
Since SS is both linearly independent and spans VV, it constitutes a basis for VV.
Thus, we have proven that any set of nn linearly independent vectors in a vector space VV of dimension nn constitutes a basis for VV.
Question:-01 (b) माना T:R^(2)rarrR^(3)mathrm{T}: mathbb{R}^2 rightarrow mathbb{R}^3 एक रैखिक रूपांतरण, ऐसा है कि T([1],[0])=([1],[2],[3])mathrm{T}left(begin{array}{l}1 \ 0end{array}right)=left(begin{array}{l}1 \ 2 \ 3end{array}right) तथा T([1],[1])=([-3],[2],[8])mathrm{T}left(begin{array}{l}1 \ 1end{array}right)=left(begin{array}{r}-3 \ 2 \ 8end{array}right) है । T([2],[4])mathrm{T}left(begin{array}{l}2 \ 4end{array}right) को ज्ञात कीजिए ।
Question:-01(b) Let T:R^(2)rarrR^(3)mathrm{T}: mathbb{R}^2 rightarrow mathbb{R}^3 be a linear transformation such that T([1],[0])=([1],[2],[3])mathrm{T}left(begin{array}{l}1 \ 0end{array}right)=left(begin{array}{l}1 \ 2 \ 3end{array}right) and T([1],[1])=([-3],[2],[8])mathrm{T}left(begin{array}{l}1 \ 1end{array}right)=left(begin{array}{r}-3 \ 2 \ 8end{array}right). Find T([2],[4])mathrm{T}left(begin{array}{l}2 \ 4end{array}right)
Answer:
Introduction: In this problem, we are given a linear transformation T:R^(2)rarrR^(3)T: mathbb{R}^2 rightarrow mathbb{R}^3 and are asked to find the image of a vector under this transformation. We are given the images of two vectors under Tmathrm{T} and will use this information along with the linearity property of the transformation to find the desired image.
Assumptions: We assume that Tmathrm{T} is a linear transformation.
Definition:
Linear Transformation: A function T:V rarr WT: V rightarrow W between two vector spaces VV and WW is called a linear transformation if for all vectors u,v in Vu, v in V and scalars cc, the following conditions hold:
a. T(u+v)=T(u)+T(v)T(u+v)=T(u)+T(v)
b. T(cu)=cT(u)T(c u)=c T(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 Tmathrm{T}.
Work/Calculations: The calculations have already been provided in the original answer. I’ll restate them here for clarity:
Represent the vector ([2],[4])left(begin{array}{c}2 \ 4end{array}right) as a linear combination of the given vectors:
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]).]:}begin{aligned}
& Tleft(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}
Conclusion: In this problem, we found the image of the vector ([2],[4])left(begin{array}{c}2 \ 4end{array}right) under the linear transformation T:R^(2)rarrR^(3)T: mathbb{R}^2 rightarrow mathbb{R}^3. 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 TT is ([-14],[4],[26])left(begin{array}{c}-14 \ 4 \ 26end{array}right).
1.(a) यदि A=[[1,-1,1],[2,-1,0],[1,0,0]]A=left[begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0end{array}right] है, तो A^(-1)A^{-1} को ज्ञात किए बिना दर्शाइए कि A^(2)=A^(-1)A^2=A^{-1}
If A=[[1,-1,1],[2,-1,0],[1,0,0]]A=left[begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0end{array}right], then show that A^(2)=A^(-1)A^2=A^{-1} (without finding A^(-1)A^{-1} ).
Answer: Introduction
We are given a 3×3 matrix AA and asked to prove that the square of this matrix, A^(2)A^2, is equal to its inverse, A^(-1)A^{-1}, without explicitly finding the inverse of AA. Assumptions
The problem assumes that the matrix AA is invertible, i.e., it has an inverse A^(-1)A^{-1}. Definition
The inverse of a matrix AA, denoted A^(-1)A^{-1}, is a unique matrix such that when it is multiplied by AA, the result is the identity matrix II. 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 is indeed the inverse of AA, then the product AA^(2)AA^2 should be the identity matrix II. We will calculate A^(2)A^2 and AA^(2)AA^2 to verify this. Work/Calculations
First, let’s calculate A^(2)A^2: 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 & 0end{array}right] cdot left[begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0end{array}right]
The calculation yields: 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 & 1end{array}right]
Next, we calculate AA^(2)AA^2: 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 & 0end{array}right] cdot left[begin{array}{ccc}0 & 0 & 1 \ 0 & -1 & 2 \ 1 & -1 & 1end{array}right]
The calculation yields: 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 & 1end{array}right] Conclusion
The product of AA and A^(2)A^2 is the identity matrix II. Therefore, we have shown that A^(2)A^2 is indeed the inverse of AA, i.e., A^(2)=A^(-1)A^2 = A^{-1}, without explicitly finding A^(-1)A^{-1}.Page Break
1.(b) क्रमित आधारक 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) पर परिभाषित रैखिक संकारक : T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c) से संबन्धित आव्यूह ज्ञात कीजिए ।
Find the matrix associated with the linear operator on V_(3)(R)V_3(R) defined by T(a,b,c)=(a+b,a-b,2c)T(a, b, c)=(a+b, a-b, 2 c) 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)}.
Answer:
Introduction
We are tasked with finding the matrix representation of the linear operator T:V_(3)(R)rarrV_(3)(R)T: V_3(mathbb{R}) to V_3(mathbb{R}) defined by 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)}.
Assumptions
TT is a linear operator.
V_(3)(R)V_3(mathbb{R}) is a vector space over the real numbers Rmathbb{R}.
The ordered basis BB consists of vectors in V_(3)(R)V_3(mathbb{R}).
Method/Approach
To find the matrix representation of TT with respect to the basis BB, we will:
Apply TT to each vector in the basis BB.
Express the resulting vectors in terms of the basis BB.
Use the coefficients as columns of the matrix representation of TT.
Work/Calculations
Step 1: Apply TT to each vector in BB
Let’s apply TT to the vectors (0,1,1)(0,1,1), (1,0,1)(1,0,1), and (1,1,0)(1,1,0).
T(0,1,1)=(0+1,0-1,2xx1)T(0,1,1) = (0+1, 0-1, 2 times 1)
T(1,0,1)=(1+0,1-0,2xx1)T(1,0,1) = (1+0, 1-0, 2 times 1)
T(1,1,0)=(1+1,1-1,2xx0)T(1,1,0) = (1+1, 1-1, 2 times 0)
After calculating, we get the following transformed vectors:
T(0,1,1)=(1,-1,2)T(0,1,1) = (1, -1, 2)
T(1,0,1)=(1,1,2)T(1,0,1) = (1, 1, 2)
T(1,1,0)=(2,0,0)T(1,1,0) = (2, 0, 0)
Step 2: Express the resulting vectors in terms of BB
We need to express each of these vectors as a linear combination of the basis vectors in BB.
The matrix representation [T]_(B)[T]_B of TT with respect to the basis BB is formed by using these coefficients as columns: [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}
Conclusion
The matrix representation of the linear operator TT 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)} is: [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}
(a) माना समुच्चय VV में सभी n xx nn times n के वास्तविक मैजिक वर्ग हैं। दिखाइए कि समुच्चय V,RV, R पर एक सदिश समष्टि है। दो भिन्न-भिन्न 2xx22 times 2 मैजिक वर्ग के उदाहरण दीजिए।
Consider the set VV of all n xx nn times n real magic squares. Show that VV is a vector space over RR. Give examples of two distinct 2xx22 times 2 magic squares.
Answer:
Introduction
The problem asks us to prove that the set VV of all n xx nn times n real magic squares is a vector space over Rmathbb{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, and ZZ are n xx nn times n magic squares with general entries as follows: 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] 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] 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]
And let aa and bb be real numbers.
Work/Calculations
Property 1: Closure under Addition and Scalar Multiplication
X+Y in MS(n)X + Y in operatorname{MS}(n)
Let’s substitute the values: (X+Y)=[x_(ij)+y_(ij)](X + Y) = [x_{ij} + y_{ij}]
The sum of each row, column, and diagonal in XX and YY is the same constant kk. Therefore, the sum of each corresponding row, column, and diagonal in X+YX + Y will be 2k2k, which means X+YX + Y is also a magic square.
aX in MS(n)aX in operatorname{MS}(n)
Let’s substitute the values: aX=[ax_(ij)]aX = [ax_{ij}]
If we multiply XX by a scalar aa, each row, column, and diagonal sum becomes akak, which means aXaX is also a magic square.
After calculating, we find that both X+YX + Y and aXaX are in MS(n)operatorname{MS}(n), proving closure under addition and scalar multiplication.
Property 2: Commutativity of Addition
X+Y=Y+XX + Y = Y + X
This is straightforward because matrix addition is commutative.
Property 3: Associativity of Addition
X+(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 0mathbf{0} be the n xx nn times n magic square where every entry is zero. Then, X+0=0+X=XX + mathbf{0} = mathbf{0} + X = X
Property 5: Existence of Additive Inverse
Let X^(‘)=-XX’ = -X. Then, X+X^(‘)=X^(‘)+X=0X + X’ = X’ + X = mathbf{0}
Property 6: Distributive Law 1
a(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=aX+bX(a + b)X = aX + bX
This is also a property of matrices.
Property 8: Associativity of Scalar Multiplication
(ab)X=a(bX)(ab)X = a(bX)
This is true for matrices.
Property 9: Multiplication by Identity
1X=X1X = X
This is true for any matrix XX.
We have shown that the set VV of all n xx nn times n real magic squares satisfies all the properties required for it to be a vector space over Rmathbb{R}. Therefore, VV is indeed a vector space over Rmathbb{R}.
For a 2xx22 times 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 2xx22 times 2 magic square MM with entries a,b,c,a, b, c, and dd: M=[[a,b],[c,d]]M = left[begin{array}{cc}
a & b \
c & d
end{array}right]
For MM to be a magic square, the following conditions must be met:
The sum of each row must be the same: a+b=c+da + b = c + d
The sum of each column must be the same: a+c=b+da + c = b + d
The sum of the diagonals must be the same: a+d=b+ca + d = b + c
Example 1
Let’s choose a=1,b=1,c=1,a = 1, b = 1, c = 1, and d=1d = 1. All the sums are 1+1=21 + 1 = 2, so it’s a magic square. M_(1)=[[1,1],[1,1]]M_1 = left[begin{array}{cc}
1 & 1 \
1 & 1
end{array}right]
Example 2
Let’s choose a=2,b=2,c=2,a = 2, b = 2, c = 2, and d=2d = 2. All the sums are 2+2=42 + 2 = 4, so it’s a magic square. M_(2)=[[2,2],[2,2]]M_2 = left[begin{array}{cc}
2 & 2 \
2 & 2
end{array}right]
Conclusion
Therefore, we conclude that the set VV of all n xx nn times n real magic squares is a valid vector space over Rmathbb{R}. This conclusion is underpinned by the rigorous application of vector space properties and the concrete examples provided for 2xx22 times 2 magic squares, affirming the validity of this mathematical concept.
Page Break
(b) माना M_(2)(R)M_2(R) सभी 2xx22 times 2 वास्तविक आव्यूहों का सदिश समष्टि है। माना B=[[1,-1],[-4,4]]B=left[begin{array}{cc}1 & -1 \ -4 & 4end{array}right]. माना T:M_(2)(R)rarrM_(2)(R)T: M_2(R) rightarrow M_2(R) एक रैखिक रूपांतरण है, जो T(A)=BAT(A)=B A द्वारा परिभाषित है। TT की कोटि (रिक) व शून्यता (नलिटि) ज्ञात कीजिए। आव्यूह AA ज्ञात कीजिए, जो शून्य आव्यूह को प्रतिचित्रित करता है।
Let M_(2)(R)M_2(R) be the vector space of all 2xx22 times 2 real matrices. Let B=[[1,-1],[-4,4]]B=left[begin{array}{cc}1 & -1 \ -4 & 4end{array}right]. Suppose T:M_(2)(R)rarrM_(2)(R)T: M_2(R) rightarrow M_2(R) is a linear transformation defined by T(A)=BAT(A)=B A. Find the rank and nullity of TT. Find a matrix AA which maps to the null matrix.
Answer:
Introduction
We are given a vector space M_(2)(R)M_2(mathbb{R}) of all 2xx22 times 2 real matrices and a specific matrix BB. A linear transformation T:M_(2)(R)rarrM_(2)(R)T: M_2(mathbb{R}) rightarrow M_2(mathbb{R}) is defined as T(A)=BAT(A) = BA. We are asked to find the rank and nullity of TT and to find a matrix AA that maps to the null matrix under TT.
Work/Calculations
Finding a Matrix AA that Maps to the Null Matrix
To find a matrix AA that maps to the null matrix under TT, we need to find AA such that BA=0BA = 0.
Let A=[[a,b],[c,d]]A = left[begin{array}{cc} a & b \ c & d end{array}right].
Then BABA is: 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]
For BABA to be the null matrix, we need a-c=0a – c = 0, b-d=0b – d = 0, -4a+4c=0-4a + 4c = 0, and -4b+4d=0-4b + 4d = 0.
Solving these equations, we find a=ca = c and b=db = d.
Therefore, any matrix AA of the form [[a,b],[a,b]]left[begin{array}{cc} a & b \ a & b end{array}right] will map to the null matrix under TT.
Nullity of TT
The nullity of TT is the dimension of the null space of TT, denoted as N(T)N(T). The null space consists of all matrices AA such that BA=0BA = 0.
Any matrix AA of the form [[a,b],[a,b]]left[begin{array}{cc} a & b \ a & b end{array}right] will map to the null matrix under TT.
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]]left[begin{array}{ll}x_1 & x_2 \ x_3 & x_4end{array}right]=left[begin{array}{ll}p & q \ p & qend{array}right]=pleft[begin{array}{ll}1 & 0 \ 1 & 0end{array}right]+qleft[begin{array}{ll}0 & 1 \ 0 & 1end{array}right]
Thus, the null space is spanned by the matrices [[1,0],[1,0]]left[begin{array}{ll}1 & 0 \ 1 & 0end{array}right] and [[0,1],[0,1]]left[begin{array}{ll}0 & 1 \ 0 & 1end{array}right], and its dimension is 2. Hence, “Nullity”(T)=2text{Nullity}(T) = 2.
Rank of TT
By the rank-nullity theorem, we have: rho(T)+N(T)=4Longrightarrowrho(T)=4-2=2rho(T) + N(T) = 4 implies rho(T) = 4 – 2 = 2
So, the rank of TT is 2.
Finding Matrix AA that Maps to the Null Matrix
Any matrix AA of the form [[a,b],[a,b]]left[begin{array}{cc} a & b \ a & b end{array}right] will map to the null matrix under TT.
Lets take an example :
To find out if AA maps to the null matrix under TT, we need to compute BABA.
Given B=[[1,-1],[-4,4]]B = left[begin{array}{cc}1 & -1 \ -4 & 4end{array}right] and A=[[2,3],[2,3]]A = left[begin{array}{ll}2 & 3 \ 2 & 3end{array}right], BA=[[1,-1],[-4,4]][[2,3],[2,3]]=[[0,0],[0,0]]BA = left[begin{array}{cc}1 & -1 \ -4 & 4end{array}right] left[begin{array}{ll}2 & 3 \ 2 & 3end{array}right] = left[begin{array}{ll}0 & 0 \ 0 & 0end{array}right]
After calculating, we find that BABA is indeed the null matrix.
Conclusion
The rank of TT is 1.
The nullity of TT is 1.
A matrix AA that maps to the null matrix under TT is of the form [[a,b],[a,b]]left[begin{array}{cc} a & b \ a & b end{array}right] Example is A=[[2,3],[2,3]]A = left[begin{array}{ll}2 & 3 \ 2 & 3end{array}right] .
(a) माना कि f:[0,(pi)/(2)]rarrRf:left[0, frac{pi}{2}right] rightarrow mathbb{R} एक संतत फलन है, जैसा कि
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((pi)/(2))fleft(frac{pi}{2}right) का मान ज्ञात कीजिए।
Let f:[0,(pi)/(2)]rarrRf:left[0, frac{pi}{2}right] rightarrow mathbb{R} be a continuous function such that 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}
Find the value of f((pi)/(2))fleft(frac{pi}{2}right).
Answer:
Introduction
The problem asks us to find the value of f((pi)/(2))fleft(frac{pi}{2}right) for a given function f(x)=(cos^(2)x)/(4x^(2)-pi^(2))f(x) = frac{cos^2 x}{4x^2 – pi^2} defined on the interval [0,(pi)/(2)]left[0, frac{pi}{2}right]. The function is continuous except at x=(pi)/(2)x = frac{pi}{2} because the denominator becomes zero at that point. To find f((pi)/(2))fleft(frac{pi}{2}right), we’ll need to evaluate the limit of f(x)f(x) as xx approaches (pi)/(2)frac{pi}{2}.
Work/Calculations
Step 1: Define the Function and the Limit
The function f(x)f(x) is given as: f(x)=(cos^(2)x)/(4x^(2)-pi^(2))f(x) = frac{cos^2 x}{4x^2 – pi^2}
We need to find: f((pi)/(2))=lim_(x rarr(pi)/(2))f(x)fleft(frac{pi}{2}right) = lim_{{x to frac{pi}{2}}} f(x)
Step 2: Simplify the Function
To find the limit, we can use L’Hôpital’s Rule, which states that if lim_(x rarr a)(f(x))/(g(x))lim_{{x to a}} frac{f(x)}{g(x)} is an indeterminate form (0)/(0)frac{0}{0} or (oo )/(oo)frac{infty}{infty}, then: 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)}
provided the limits on the right-hand side exist.
First, let’s check if the function f(x)f(x) is in indeterminate form at x=(pi)/(2)x = frac{pi}{2}: “Numerator at “x=(pi)/(2):cos^(2)((pi)/(2))=0text{Numerator at } x = frac{pi}{2}: cos^2 left(frac{pi}{2}right) = 0 “Denominator at “x=(pi)/(2):4((pi)/(2))^(2)-pi^(2)=0text{Denominator at } x = frac{pi}{2}: 4 left(frac{pi}{2}right)^2 – pi^2 = 0
Both the numerator and the denominator are zero, so we have an indeterminate form (0)/(0)frac{0}{0}.
Step 3: Apply L’Hôpital’s Rule
Let’s differentiate the numerator and the denominator with respect to xx.
f^(‘)(x)f'(x) (Derivative of cos^(2)xcos^2 x) = -2cos(x)sin(x)-2 cos(x) sin(x)
g^(‘)(x)g'(x) (Derivative of 4x^(2)-pi^(2)4x^2 – pi^2) = 8x8x
Now, we can find the limit: 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}
Let’s substitute the values into the formula and calculate the limit.
After substituting the values, we get: 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 cosleft(frac{pi}{2}right) sinleft(frac{pi}{2}right)}{8 times frac{pi}{2}}
After calculating, we get: lim_(x rarr(pi)/(2))(-2cos(x)sin(x))/(8x)=0lim_{{x to frac{pi}{2}}} frac{-2 cos(x) sin(x)}{8x} = 0
Conclusion
The value of f((pi)/(2))fleft(frac{pi}{2}right) is 00. We used L’Hôpital’s Rule to evaluate the limit of the function f(x)f(x) as xx approaches (pi)/(2)frac{pi}{2}, and found that the limit is 00. Therefore, f((pi)/(2))=0fleft(frac{pi}{2}right) = 0.
Page Break
(b) माना कि f:D(subeR^(2))rarrRf: Dleft(subseteq mathbb{R}^2right) rightarrow mathbb{R} एक फलन है और (a,b)in D(a, b) in D. अगर f(x,y)f(x, y) बिंदु (a,b)(a, b) पर संतत है, तो दर्शाइए कि फलन f(x,b)f(x, b) और f(a,y)f(a, y) क्रमशः x=ax=a और y=by=b पर संतत हैं।
Let f:D(subeR^(2))rarrRf: Dleft(subseteq mathbb{R}^2right) rightarrow mathbb{R} be a function and (a,b)in D(a, b) in D. If f(x,y)f(x, y) is continuous at (a,b)(a, b), then show that the functions f(x,b)f(x, b) and f(a,y)f(a, y) are continuous at x=ax=a and at y=by=b respectively.
Answer:
Introduction
The problem asks us to prove that if a function f(x,y)f(x, y) is continuous at a point (a,b)(a, b), then the functions f(x,b)f(x, b) and f(a,y)f(a, y) are continuous at x=ax = a and y=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) is said to be continuous at (a,b)(a, b) if for every epsilon > 0epsilon > 0, there exists delta > 0delta > 0 such that: sqrt((x-a)^(2)+(y-b)^(2)) < deltaLongrightarrow|f(x,y)-f(a,b)| < epsilonsqrt{(x-a)^2 + (y-b)^2} < delta implies |f(x, y) – f(a, b)| < epsilon
Step 2: Prove Continuity for f(x,b)f(x, b) at x=ax = a
We need to show that for every epsilon > 0epsilon > 0, there exists delta > 0delta > 0 such that: |x-a| < deltaLongrightarrow|f(x,b)-f(a,b)| < epsilon|x – a| < delta implies |f(x, b) – f(a, b)| < epsilon
To prove this, let’s consider epsilon > 0epsilon > 0. Since f(x,y)f(x, y) is continuous at (a,b)(a, b), there exists delta > 0delta > 0 such that: sqrt((x-a)^(2)+(y-b)^(2)) < deltaLongrightarrow|f(x,y)-f(a,b)| < epsilonsqrt{(x-a)^2 + (y-b)^2} < delta implies |f(x, y) – f(a, b)| < epsilon
Now, consider |x-a| < delta|x – a| < delta. We can rewrite this as sqrt((x-a)^(2)+(b-b)^(2)) < deltasqrt{(x-a)^2 + (b-b)^2} < delta.
By the continuity of f(x,y)f(x, y) at (a,b)(a, b), this implies: |f(x,b)-f(a,b)| < epsilon|f(x, b) – f(a, b)| < epsilon
Thus, we have shown that f(x,b)f(x, b) is continuous at x=ax = a.
Step 3: Prove Continuity for f(a,y)f(a, y) at y=by = b
Similarly, we need to show that for every epsilon > 0epsilon > 0, there exists delta > 0delta > 0 such that: |y-b| < deltaLongrightarrow|f(a,y)-f(a,b)| < epsilon|y – b| < delta implies |f(a, y) – f(a, b)| < epsilon
Again, consider epsilon > 0epsilon > 0. By the continuity of f(x,y)f(x, y) at (a,b)(a, b), there exists delta > 0delta > 0 such that: sqrt((x-a)^(2)+(y-b)^(2)) < deltaLongrightarrow|f(x,y)-f(a,b)| < epsilonsqrt{(x-a)^2 + (y-b)^2} < delta implies |f(x, y) – f(a, b)| < epsilon
Now, consider |y-b| < delta|y – b| < delta. We can rewrite this as sqrt((a-a)^(2)+(y-b)^(2)) < deltasqrt{(a-a)^2 + (y-b)^2} < delta.
By the continuity of f(x,y)f(x, y) at (a,b)(a, b), this implies: |f(a,y)-f(a,b)| < epsilon|f(a, y) – f(a, b)| < epsilon
Thus, we have shown that f(a,y)f(a, y) is continuous at y=by = b.
Conclusion
We have successfully proven that if f(x,y)f(x, y) is continuous at (a,b)(a, b), then f(x,b)f(x, b) is continuous at x=ax = a and f(a,y)f(a, y) is continuous at y=by = b. We used the definition of continuity to establish these results.
(a) मान लीजिये कि AA एक 3xx23 times 2 आव्यूह है और BB एक 2xx32 times 3 आव्यूह है। दर्शाइये कि C=A*BC=A cdot B एक अव्युत्क्रमणणीय आव्यूह है।
Let AA be a 3xx23 times 2 matrix and BB a 2xx32 times 3 matrix. Show that C=A*BC=A cdot B is a singular matrix.
Answer:
Introduction
The problem asks us to show that the product C=A*BC = A cdot B is a singular matrix, given that AA is a 3xx23 times 2 matrix and BB is a 2xx32 times 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 CC
First, let’s find the dimensions of the resulting matrix CC when AA and BB are multiplied.
The dimensions of CC can be determined by the outer dimensions of AA and BB. In this case, AA is 3xx23 times 2 and BB is 2xx32 times 3, so CC will be 3xx33 times 3.
Step 2: Rank of CC
The rank of CC is limited by the smaller of the two ranks of AA and BB. Since AA is 3xx23 times 2, its rank can be at most 2. Similarly, BB is 2xx32 times 3, so its rank can also be at most 2.
Therefore, the rank of CC can be at most 2. “Rank”(C) <= min(“Rank”(A),”Rank”(B)) <= 2text{Rank}(C) leq min(text{Rank}(A), text{Rank}(B)) leq 2
Step 3: Determinant of CC
For a 3xx33 times 3 matrix to be invertible (non-singular), its rank must be 3. However, we’ve established that the rank of CC can be at most 2. Therefore, CC must be singular.
To confirm, the determinant of a singular matrix is zero: “Det”(C)=0text{Det}(C) = 0
Conclusion
We have shown that the matrix C=A*BC = A cdot B will be a 3xx33 times 3 matrix with a rank of at most 2. Since the rank is less than 3, CC is a singular matrix, and its determinant is zero. Therefore, C=A*BC = A cdot B is indeed a singular matrix.
Page Break
(b) आधार सदिशों e_(1)=(1,0)e_1=(1,0) और e_(2)=(0,1)e_2=(0,1) को alpha_(1)=(2,-1)alpha_1=(2,-1) एवं alpha_(2)=(1,3)alpha_2=(1,3) के रैखिक संयोग के रूप में ब्यक्त कीजिये।
Express basis vectors e_(1)=(1,0)e_1=(1,0) and e_(2)=(0,1)e_2=(0,1) as linear combinations of alpha_(1)=(2,-1)alpha_1=(2,-1) and alpha_(2)=(1,3)alpha_2=(1,3).
Answer:
Introduction
The problem asks us to express the basis vectors e_(1)=(1,0)e_1 = (1, 0) and e_(2)=(0,1)e_2 = (0, 1) as linear combinations of the vectors alpha_(1)=(2,-1)alpha_1 = (2, -1) and alpha_(2)=(1,3)alpha_2 = (1, 3). In other words, we want to find constants c_(1)c_1 and c_(2)c_2 such that: e_(1)=c_(1)alpha_(1)+c_(2)alpha_(2)e_1 = c_1 alpha_1 + c_2 alpha_2 e_(2)=d_(1)alpha_(1)+d_(2)alpha_(2)e_2 = d_1 alpha_1 + d_2 alpha_2
Work/Calculations
Step 1: Setting up the Equations for e_(1)e_1
The equation for e_(1)e_1 can be written as: (1,0)=c_(1)(2,-1)+c_(2)(1,3)(1, 0) = c_1 (2, -1) + c_2 (1, 3)
Breaking it down into components, we get: 1=2c_(1)+c_(2)1 = 2c_1 + c_2 0=-c_(1)+3c_(2)0 = -c_1 + 3c_2
Step 2: Solving for c_(1)c_1 and c_(2)c_2
After Calculating, we get: c_(1)=(3)/(7),quadc_(2)=(1)/(7)c_1 = frac{3}{7}, quad c_2 = frac{1}{7}
Step 3: Setting up the Equations for e_(2)e_2
The equation for e_(2)e_2 can be written as: (0,1)=d_(1)(2,-1)+d_(2)(1,3)(0, 1) = d_1 (2, -1) + d_2 (1, 3)
Breaking it down into components, we get: 0=2d_(1)+d_(2)0 = 2d_1 + d_2 1=-d_(1)+3d_(2)1 = -d_1 + 3d_2
Step 4: Solving for d_(1)d_1 and d_(2)d_2
After Calculating, we get: d_(1)=-(1)/(7),quadd_(2)=(2)/(7)d_1 = -frac{1}{7}, quad d_2 = frac{2}{7}
Conclusion
The basis vector e_(1)=(1,0)e_1 = (1, 0) can be expressed as a linear combination of alpha_(1)alpha_1 and alpha_(2)alpha_2 as follows: e_(1)=(3)/(7)alpha_(1)+(1)/(7)alpha_(2)e_1 = frac{3}{7} alpha_1 + frac{1}{7} alpha_2
Similarly, the basis vector e_(2)=(0,1)e_2 = (0, 1) can be expressed as: e_(2)=-(1)/(7)alpha_(1)+(2)/(7)alpha_(2)e_2 = -frac{1}{7} alpha_1 + frac{2}{7} alpha_2
Thus, both e_(1)e_1 and e_(2)e_2 can be represented as linear combinations of alpha_(1)alpha_1 and alpha_(2)alpha_2.
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