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

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

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

## UPSC Maths Optional Question Papers

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खण्ड A
SECTION A
Question:-01 (a) सिद्ध कीजिए कि $\mathrm{n}$$\mathrm{n}$n\mathrm{n}$\mathrm{n}$ विमीय सदिश समष्टि $\mathrm{V}$$\mathrm{V}$V\mathrm{V}$\mathrm{V}$ के लिए $\mathrm{n}$$\mathrm{n}$n\mathrm{n}$\mathrm{n}$ रैखिकत: स्वतंत्र सदिशों का कोई भी समुच्चय $\mathrm{V}$$\mathrm{V}$V\mathrm{V}$\mathrm{V}$ के लिए एक आधार बनाता है ।
Question:-01 (a) Prove that any set of $\mathrm{n}$$\mathrm{n}$n\mathrm{n}$\mathrm{n}$ linearly independent vectors in a vector space $\mathrm{V}$$\mathrm{V}$V\mathrm{V}$\mathrm{V}$ of dimension $\mathrm{n}$$\mathrm{n}$n\mathrm{n}$\mathrm{n}$ constitutes a basis for $\mathrm{V}$$\mathrm{V}$V\mathrm{V}$\mathrm{V}$.
Question:-01 (b) माना $\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{3}$$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{3}$T:R^(2)rarrR^(3)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^3$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{3}$ एक रैखिक रूपांतरण, ऐसा है कि $\mathrm{T}\left(\begin{array}{l}1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 2\\ 3\end{array}\right)$$\mathrm{T}\left(\begin{array}{l}1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 2\\ 3\end{array}\right)$T([1],[0])=([1],[2],[3])\mathrm{T}\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$\mathrm{T}\left(\begin{array}{l}1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 2\\ 3\end{array}\right)$ तथा $\mathrm{T}\left(\begin{array}{l}1\\ 1\end{array}\right)=\left(\begin{array}{r}-3\\ 2\\ 8\end{array}\right)$$\mathrm{T}\left(\begin{array}{l}1\\ 1\end{array}\right)=\left(\begin{array}{r}-3\\ 2\\ 8\end{array}\right)$T([1],[1])=([-3],[2],[8])\mathrm{T}\left(\begin{array}{l}1 \\ 1\end{array}\right)=\left(\begin{array}{r}-3 \\ 2 \\ 8\end{array}\right)$\mathrm{T}\left(\begin{array}{l}1\\ 1\end{array}\right)=\left(\begin{array}{r}-3\\ 2\\ 8\end{array}\right)$ है । $\mathrm{T}\left(\begin{array}{l}2\\ 4\end{array}\right)$$\mathrm{T}\left(\begin{array}{l}2\\ 4\end{array}\right)$T([2],[4])\mathrm{T}\left(\begin{array}{l}2 \\ 4\end{array}\right)$\mathrm{T}\left(\begin{array}{l}2\\ 4\end{array}\right)$ को ज्ञात कीजिए ।
Question:-01(b) Let $\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{3}$$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{3}$T:R^(2)rarrR^(3)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^3$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{3}$ be a linear transformation such that $\mathrm{T}\left(\begin{array}{l}1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 2\\ 3\end{array}\right)$$\mathrm{T}\left(\begin{array}{l}1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 2\\ 3\end{array}\right)$T([1],[0])=([1],[2],[3])\mathrm{T}\left(\begin{array}{l}1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)$\mathrm{T}\left(\begin{array}{l}1\\ 0\end{array}\right)=\left(\begin{array}{l}1\\ 2\\ 3\end{array}\right)$ and $\mathrm{T}\left(\begin{array}{l}1\\ 1\end{array}\right)=\left(\begin{array}{r}-3\\ 2\\ 8\end{array}\right)$$\mathrm{T}\left(\begin{array}{l}1\\ 1\end{array}\right)=\left(\begin{array}{r}-3\\ 2\\ 8\end{array}\right)$T([1],[1])=([-3],[2],[8])\mathrm{T}\left(\begin{array}{l}1 \\ 1\end{array}\right)=\left(\begin{array}{r}-3 \\ 2 \\ 8\end{array}\right)$\mathrm{T}\left(\begin{array}{l}1\\ 1\end{array}\right)=\left(\begin{array}{r}-3\\ 2\\ 8\end{array}\right)$. Find $\mathrm{T}\left(\begin{array}{l}2\\ 4\end{array}\right)$$\mathrm{T}\left(\begin{array}{l}2\\ 4\end{array}\right)$T([2],[4])\mathrm{T}\left(\begin{array}{l}2 \\ 4\end{array}\right)$\mathrm{T}\left(\begin{array}{l}2\\ 4\end{array}\right)$
Question:-01(c) $\underset{x\to \mathrm{\infty }}{lim}{\left({e}^{x}+x\right)}^{\frac{1}{x}}$$\underset{x\to \mathrm{\infty }}{lim} {\left({e}^{x}+x\right)}^{\frac{1}{x}}$lim_(x rarr oo)(e^(x)+x)^((1)/(x))\lim _{x \rightarrow \infty}\left(e^x+x\right)^{\frac{1}{x}}$\underset{x\to \mathrm{\infty }}{lim}{\left({e}^{x}+x\right)}^{\frac{1}{x}}$ का मान निकालिए ।
Question:-01(c) Evaluate $\underset{x\to \mathrm{\infty }}{lim}{\left({e}^{x}+x\right)}^{\frac{1}{x}}$$\underset{x\to \mathrm{\infty }}{lim} {\left({e}^{x}+x\right)}^{\frac{1}{x}}$lim_(x rarr oo)(e^(x)+x)^((1)/(x))\lim _{x \rightarrow \infty}\left(e^x+x\right)^{\frac{1}{x}}$\underset{x\to \mathrm{\infty }}{lim}{\left({e}^{x}+x\right)}^{\frac{1}{x}}$
Question:-01(d) ${\int }_{0}^{2}\frac{dx}{\left(2x-{x}^{2}\right)}$${\int }_{0}^{2} \frac{dx}{\left(2x-{x}^{2}\right)}$int_(0)^(2)(dx)/((2x-x^(2)))\int_0^2 \frac{d x}{\left(2 x-x^2\right)}${\int }_{0}^{2}\frac{dx}{\left(2x-{x}^{2}\right)}$ की अभिसारिता का परीक्षण कीजिए ।
Question:-01(d) Examine the convergence of ${\int }_{0}^{2}\frac{dx}{\left(2x-{x}^{2}\right)}$${\int }_{0}^{2} \frac{dx}{\left(2x-{x}^{2}\right)}$int_(0)^(2)(dx)/((2x-x^(2)))\int_0^2 \frac{d x}{\left(2 x-x^2\right)}${\int }_{0}^{2}\frac{dx}{\left(2x-{x}^{2}\right)}$.
Question:-01(e) एक चर समतल एक स्थिर बिन्दु $\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)$$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)$(a,b,c)(\mathrm{a}, \mathrm{b}, \mathrm{c})$\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)$ से गुज़रता है तथा अक्षों को क्रमशः $\mathrm{A},\mathrm{B}$$\mathrm{A},\mathrm{B}$A,B\mathrm{A}, \mathrm{B}$\mathrm{A},\mathrm{B}$$\mathrm{C}$$\mathrm{C}$C\mathrm{C}$\mathrm{C}$ बिन्दुओं पर मिलता है । बिन्दुओं $\mathrm{O},\mathrm{A},\mathrm{B}$$\mathrm{O},\mathrm{A},\mathrm{B}$O,A,B\mathrm{O}, \mathrm{A}, \mathrm{B}$\mathrm{O},\mathrm{A},\mathrm{B}$ तथा $\mathrm{C}$$\mathrm{C}$C\mathrm{C}$\mathrm{C}$ से गुज़रते हुए गोले के केन्द्र का बिन्दुपथ ज्ञात कीजिए, जहाँ $\mathrm{O}$$\mathrm{O}$O\mathrm{O}$\mathrm{O}$ मूल-बिन्दु है ।
Question:-01(e) A variable plane passes through a fixed point (a, b, c) and meets the axes at points A, B and C respectively. Find the locus of the centre of the sphere passing through the points $\mathrm{O},\mathrm{A},\mathrm{B}$$\mathrm{O},\mathrm{A},\mathrm{B}$O,A,B\mathrm{O}, \mathrm{A}, \mathrm{B}$\mathrm{O},\mathrm{A},\mathrm{B}$ and $\mathrm{C},\mathrm{O}$$\mathrm{C},\mathrm{O}$C,O\mathrm{C}, \mathrm{O}$\mathrm{C},\mathrm{O}$ being the origin.

Question:-02(a) निम्नलिखित समीकरण निकाय के सभी हलों को पंक्ति-समानीत विधि से ज्ञात कीजिए :
$\begin{array}{rl}& {\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{array}$$\begin{array}{r}{\mathrm{x}}_{1}+2{\mathrm{x}}_{2}-{\mathrm{x}}_{3}=2\\ 2{\mathrm{x}}_{1}+3{\mathrm{x}}_{2}+5{\mathrm{x}}_{3}=5\\ -{\mathrm{x}}_{1}-3{\mathrm{x}}_{2}+8{\mathrm{x}}_{3}=-1\end{array}${:[x_(1)+2x_(2)-x_(3)=2],[2x_(1)+3x_(2)+5x_(3)=5],[-x_(1)-3x_(2)+8x_(3)=-1]:}\begin{aligned} &\mathrm{x}_1+2 \mathrm{x}_2-\mathrm{x}_3=2 \\ &2 \mathrm{x}_1+3 \mathrm{x}_2+5 \mathrm{x}_3=5 \\ &-\mathrm{x}_1-3 \mathrm{x}_2+8 \mathrm{x}_3=-1 \end{aligned}$\begin{array}{rl}& {\mathrm{x}}_{1}+2{\mathrm{x}}_{2}-{\mathrm{x}}_{3}=2\\ & 2{\mathrm{x}}_{1}+3{\mathrm{x}}_{2}+5{\mathrm{x}}_{3}=5\\ & -{\mathrm{x}}_{1}-3{\mathrm{x}}_{2}+8{\mathrm{x}}_{3}=-1\end{array}$
Question:-02(a) Find all solutions to the following system of equations by row-reduced method :
$\begin{array}{rl}& {\mathrm{x}}_{1}+2{\mathrm{x}}_{2}-{\mathrm{x}}_{3}=2\\ & 2{\mathrm{x}}_{1}+3{\mathrm{x}}_{2}+5{\mathrm{x}}_{3}=5\\ & -{\mathrm{x}}_{1}-3{\mathrm{x}}_{2}+8{\mathrm{x}}_{3}=-1\end{array}$$\begin{array}{r}{\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{array}${:[x_(1)+2x_(2)-x_(3)=2],[2x_(1)+3x_(2)+5x_(3)=5],[-x_(1)-3x_(2)+8x_(3)=-1]:}\begin{aligned} &\mathrm{x}_1+2 \mathrm{x}_2-\mathrm{x}_3=2 \\ &2 \mathrm{x}_1+3 \mathrm{x}_2+5 \mathrm{x}_3=5 \\ &-\mathrm{x}_1-3 \mathrm{x}_2+8 \mathrm{x}_3=-1 \end{aligned}$\begin{array}{rl}& {\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{array}$
Question:-02(b) एक $l$$l$ll$l$ लम्बाई के तार को दो भागों में काटकर क्रमशः एक वर्ग तथा एक वृत्त के रूप में मोड़ा गया है । लग्रांज की अनिर्धारित गुणक विधि का प्रयोग करके, इस तरह से प्राप्त किए गए क्षेत्रफलों के योगफल का न्यूनतम मान ज्ञात कीजिए ।
Question:-02(b) A wire of length $l$$l$ll$l$ 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 $\mathrm{P},\mathrm{Q},\mathrm{R};{\mathrm{P}}^{\mathrm{\prime }},{\mathrm{Q}}^{\mathrm{\prime }},{\mathrm{R}}^{\mathrm{\prime }}$$\mathrm{P},\mathrm{Q},\mathrm{R};{\mathrm{P}}^{\mathrm{\prime }},{\mathrm{Q}}^{\mathrm{\prime }},{\mathrm{R}}^{\mathrm{\prime }}$P,Q,R;P^(‘),Q^(‘),R^(‘)\mathrm{P}, \mathrm{Q}, \mathrm{R} ; \mathrm{P}^{\prime}, \mathrm{Q}^{\prime}, \mathrm{R}^{\prime}$\mathrm{P},\mathrm{Q},\mathrm{R};{\mathrm{P}}^{\mathrm{\prime }},{\mathrm{Q}}^{\mathrm{\prime }},{\mathrm{R}}^{\mathrm{\prime }}$, एक बिन्दु से दीर्घवृत्तज $\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\text{}\mathrm{b}}^{2}}+\frac{{\mathrm{z}}^{2}}{{\mathrm{c}}^{2}}=1$$\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\text{}\mathrm{b}}^{2}}+\frac{{\mathrm{z}}^{2}}{{\mathrm{c}}^{2}}=1$(x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}+\frac{\mathrm{z}^2}{\mathrm{c}^2}=1 पर छः (सिक्स) अभिलम्ब पाद हैं तथा $l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n}\mathrm{z}=\mathrm{p}$$l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n}\mathrm{z}=\mathrm{p}$lx+my+nz=pl \mathrm{x}+\mathrm{my}+\mathrm{nz}=\mathrm{p}$l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n}\mathrm{z}=\mathrm{p}$ से समतल $\mathrm{P}\mathrm{Q}\mathrm{R}$$\mathrm{P}\mathrm{Q}\mathrm{R}$PQR\mathrm{PQR}$\mathrm{P}\mathrm{Q}\mathrm{R}$ निरूपित है, दर्शाइए कि $\frac{\mathrm{x}}{{\mathrm{a}}^{2}l}+\frac{\mathrm{y}}{{\mathrm{b}}^{2}\text{}\mathrm{m}}+\frac{\mathrm{z}}{{\mathrm{c}}^{2}\mathrm{n}}+\frac{1}{\mathrm{p}}=0$$\frac{\mathrm{x}}{{\mathrm{a}}^{2}l}+\frac{\mathrm{y}}{{\mathrm{b}}^{2}\text{}\mathrm{m}}+\frac{\mathrm{z}}{{\mathrm{c}}^{2}\mathrm{n}}+\frac{1}{\mathrm{p}}=0$(x)/(a^(2)l)+(y)/(b^(2)(m))+(z)/(c^(2)n)+(1)/(p)=0\frac{\mathrm{x}}{\mathrm{a}^2 l}+\frac{\mathrm{y}}{\mathrm{b}^2 \mathrm{~m}}+\frac{\mathrm{z}}{\mathrm{c}^2 \mathrm{n}}+\frac{1}{\mathrm{p}}=0, समतल ${\mathrm{P}}^{\mathrm{\prime }}{\mathrm{Q}}^{\mathrm{\prime }}{\mathrm{R}}^{\mathrm{\prime }}$${\mathrm{P}}^{\mathrm{\prime }}{\mathrm{Q}}^{\mathrm{\prime }}{\mathrm{R}}^{\mathrm{\prime }}$P^(‘)Q^(‘)R^(‘)\mathrm{P}^{\prime} \mathrm{Q}^{\prime} \mathrm{R}^{\prime}${\mathrm{P}}^{\mathrm{\prime }}{\mathrm{Q}}^{\mathrm{\prime }}{\mathrm{R}}^{\mathrm{\prime }}$ को निरूपित करता है ।
Question:-02(c) If $\mathrm{P},\mathrm{Q},\mathrm{R};{\mathrm{P}}^{\mathrm{\prime }},{\mathrm{Q}}^{\mathrm{\prime }},{\mathrm{R}}^{\mathrm{\prime }}$$\mathrm{P},\mathrm{Q},\mathrm{R};{\mathrm{P}}^{\mathrm{\prime }},{\mathrm{Q}}^{\mathrm{\prime }},{\mathrm{R}}^{\mathrm{\prime }}$P,Q,R;P^(‘),Q^(‘),R^(‘)\mathrm{P}, \mathrm{Q}, \mathrm{R} ; \mathrm{P}^{\prime}, \mathrm{Q}^{\prime}, \mathrm{R}^{\prime}$\mathrm{P},\mathrm{Q},\mathrm{R};{\mathrm{P}}^{\mathrm{\prime }},{\mathrm{Q}}^{\mathrm{\prime }},{\mathrm{R}}^{\mathrm{\prime }}$ are feet of the six normals drawn from a point to the ellipsoid $\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\text{}\mathrm{b}}^{2}}+\frac{{\mathrm{z}}^{2}}{{\mathrm{c}}^{2}}=1$$\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\text{}\mathrm{b}}^{2}}+\frac{{\mathrm{z}}^{2}}{{\mathrm{c}}^{2}}=1$(x^(2))/(a^(2))+(y^(2))/(b^(2))+(z^(2))/(c^(2))=1\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}+\frac{\mathrm{z}^2}{\mathrm{c}^2}=1, and the plane $\mathrm{P}\mathrm{Q}\mathrm{R}$$\mathrm{P}\mathrm{Q}\mathrm{R}$PQR\mathrm{PQR}$\mathrm{P}\mathrm{Q}\mathrm{R}$ is represented by $lx+my+nz=p$$lx+my+nz=p$lx+my+nz=pl x+m y+n z=p$lx+my+nz=p$, show that the plane ${P}^{\mathrm{\prime }}{Q}^{\mathrm{\prime }}{R}^{\mathrm{\prime }}$${P}^{\mathrm{\prime }}{Q}^{\mathrm{\prime }}{R}^{\mathrm{\prime }}$P^(‘)Q^(‘)R^(‘)P^{\prime} Q^{\prime} R^{\prime}${P}^{\mathrm{\prime }}{Q}^{\mathrm{\prime }}{R}^{\mathrm{\prime }}$ is given by $\frac{\mathrm{x}}{{\mathrm{a}}^{2}l}+\frac{\mathrm{y}}{{\mathrm{b}}^{2}\text{}\mathrm{m}}+\frac{\mathrm{z}}{{\mathrm{c}}^{2}\mathrm{n}}+\frac{1}{\mathrm{p}}=0$$\frac{\mathrm{x}}{{\mathrm{a}}^{2}l}+\frac{\mathrm{y}}{{\mathrm{b}}^{2}\text{}\mathrm{m}}+\frac{\mathrm{z}}{{\mathrm{c}}^{2}\mathrm{n}}+\frac{1}{\mathrm{p}}=0$(x)/(a^(2)l)+(y)/(b^(2)(m))+(z)/(c^(2)n)+(1)/(p)=0\frac{\mathrm{x}}{\mathrm{a}^2 l}+\frac{\mathrm{y}}{\mathrm{b}^2 \mathrm{~m}}+\frac{\mathrm{z}}{\mathrm{c}^2 \mathrm{n}}+\frac{1}{\mathrm{p}}=0.

Question:-03(a) माना समुच्चय $\mathrm{P}=\left\{\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)\mid \begin{array}{c}\mathrm{x}-\mathrm{y}-\mathrm{z}=0\text{तथा}\\ 2\mathrm{x}-\mathrm{y}+\mathrm{z}=0\end{array}\right\}${:P={[x],[y],[z])∣[x-y-z=0″ तथा “],[2x-y+z=0]}\left.\mathrm{P}=\left\{\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right) \mid \begin{array}{c}\mathrm{x}-\mathrm{y}-\mathrm{z}=0 \text { तथा } \\ 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=0\end{array}\right\} सदिश समष्टि ${\mathbb{R}}^{3}\left(\mathbb{R}\right)$${\mathbb{R}}^{3}\left(\mathbb{R}\right)$R^(3)(R)\mathbb{R}^3(\mathbb{R})${\mathbb{R}}^{3}\left(\mathbb{R}\right)$ के सदिशों का एक समूह है । तब
(i) सिद्ध कीजिए कि $\mathrm{P},{\mathbb{R}}^{3}$$\mathrm{P},{\mathbb{R}}^{3}$P,R^(3)\mathrm{P}, \mathbb{R}^3$\mathrm{P},{\mathbb{R}}^{3}$ की एक उपसमष्टि है ।
(ii) $\mathrm{P}$$\mathrm{P}$P\mathrm{P}$\mathrm{P}$ का एक आधार तथा विमा ज्ञात कीजिए ।
Question:-03(a) Let the set $P=\left\{\begin{array}{c}x\\ y\\ z\end{array}\right)\mid \begin{array}{c}x-y-z=0\text{and}\\ 2x-y+z=0\end{array}\right\}${:P={[x],[y],[z])∣[x-y-z=0″ and “],[2x-y+z=0]}\left.P=\left\{\begin{array}{c}x \\ y \\ z\end{array}\right) \mid \begin{array}{c}x-y-z=0 \text { and } \\ 2 x-y+z=0\end{array}\right\} be the collection of vectors of a vector space ${\mathbb{R}}^{3}\left(\mathbb{R}\right)$${\mathbb{R}}^{3}\left(\mathbb{R}\right)$R^(3)(R)\mathbb{R}^3(\mathbb{R})${\mathbb{R}}^{3}\left(\mathbb{R}\right)$. Then
(i) prove that $\mathrm{P}$$\mathrm{P}$P\mathrm{P}$\mathrm{P}$ is a subspace of ${\mathbb{R}}^{3}$${\mathbb{R}}^{3}$R^(3)\mathbb{R}^3${\mathbb{R}}^{3}$.
(ii) find a basis and dimension of $P$$P$PP$P$.
Question:-03(b) द्विशः समाकलन का उपयोग करके, वृत्त ${\mathrm{x}}^{2}+{\mathrm{y}}^{2}=4$${\mathrm{x}}^{2}+{\mathrm{y}}^{2}=4$x^(2)+y^(2)=4\mathrm{x}^2+\mathrm{y}^2=4${\mathrm{x}}^{2}+{\mathrm{y}}^{2}=4$ तथा परवलय ${\mathrm{y}}^{2}=3\mathrm{x}$${\mathrm{y}}^{2}=3\mathrm{x}$y^(2)=3x\mathrm{y}^2=3 \mathrm{x}${\mathrm{y}}^{2}=3\mathrm{x}$ के उभयनिष्ठ क्षेत्रफल का परिकलन कीजिए ।
Question:-03(b)Use double integration to calculate the area common to the circle ${x}^{2}+{y}^{2}=4$${x}^{2}+{y}^{2}=4$x^(2)+y^(2)=4x^2+y^2=4${x}^{2}+{y}^{2}=4$ and the parabola ${y}^{2}=3x$${y}^{2}=3x$y^(2)=3xy^2=3 x${y}^{2}=3x$.
Question:-03(c) लघुतम संभाव्य त्रिज्या के गोले का समीकरण ज्ञात कीजिए जो सरल रेखाओं : $\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}-8}{-1}=\frac{\mathrm{z}-3}{1}$$\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}-8}{-1}=\frac{\mathrm{z}-3}{1}$(x-3)/(3)=(y-8)/(-1)=(z-3)/(1)\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}-8}{-1}=\frac{\mathrm{z}-3}{1}$\frac{\mathrm{x}-3}{3}=\frac{\mathrm{y}-8}{-1}=\frac{\mathrm{z}-3}{1}$ तथा $\frac{\mathrm{x}+3}{-3}=\frac{\mathrm{y}+7}{2}=\frac{\mathrm{z}-6}{4}$$\frac{\mathrm{x}+3}{-3}=\frac{\mathrm{y}+7}{2}=\frac{\mathrm{z}-6}{4}$(x+3)/(-3)=(y+7)/(2)=(z-6)/(4)\frac{\mathrm{x}+3}{-3}=\frac{\mathrm{y}+7}{2}=\frac{\mathrm{z}-6}{4}$\frac{\mathrm{x}+3}{-3}=\frac{\mathrm{y}+7}{2}=\frac{\mathrm{z}-6}{4}$ को स्पर्श करता है ।
Question:-03(c)Find the equation of the sphere of smallest possible radius which touches the straight lines : $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$(x-3)/(3)=(y-8)/(-1)=(z-3)/(1)\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$(x+3)/(-3)=(y+7)/(2)=(z-6)/(4)\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$

Question:-04(a) एक रैखिक प्रतिचित्र $\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$T:R^(2)rarrR^(2)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ ज्ञात कीजिए जो कि ${\mathbb{R}}^{2}$${\mathbb{R}}^{2}$R^(2)\mathbb{R}^2${\mathbb{R}}^{2}$ के प्रत्येक सदिश को $\theta$$\theta$theta\theta$\theta$ कोण से घुमा देता है । यह भी सिद्ध कीजिए कि $\theta =\frac{\pi }{2}$$\theta =\frac{\pi }{2}$theta=(pi)/(2)\theta=\frac{\pi}{2}$\theta =\frac{\pi }{2}$ के लिए, $\mathrm{T}$$\mathrm{T}$T\mathrm{T}$\mathrm{T}$ का कोई भी अभिलक्षणिक मान (आइगेनमान) $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ में नहीं है ।
Question:-04(a) Find a linear map $\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$T:R^(2)rarrR^(2)\mathrm{T}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$\mathrm{T}:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ which rotates each vector of ${\mathbb{R}}^{2}$${\mathbb{R}}^{2}$R^(2)\mathbb{R}^2${\mathbb{R}}^{2}$ by an angle $\theta$$\theta$theta\theta$\theta$. Also, prove that for $\theta =\frac{\pi }{2},\text{}\mathrm{T}$$\theta =\frac{\pi }{2},\text{}\mathrm{T}$theta=(pi)/(2),T\theta=\frac{\pi}{2}, \mathrm{~T} has no eigenvalue in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$.
Question:-04(b) वक्र ${\mathrm{y}}^{2}{\mathrm{x}}^{2}={\mathrm{x}}^{2}-{\mathrm{a}}^{2}$${\mathrm{y}}^{2}{\mathrm{x}}^{2}={\mathrm{x}}^{2}-{\mathrm{a}}^{2}$y^(2)x^(2)=x^(2)-a^(2)\mathrm{y}^2 \mathrm{x}^2=\mathrm{x}^2-\mathrm{a}^2${\mathrm{y}}^{2}{\mathrm{x}}^{2}={\mathrm{x}}^{2}-{\mathrm{a}}^{2}$ का अनुरेख (ट्रेस) कीजिए, जहाँ $\mathrm{a}$$\mathrm{a}$a\mathrm{a}$\mathrm{a}$ एक वास्तविक अचर है ।
Question:-04(b) Trace the curve ${y}^{2}{x}^{2}={x}^{2}-{a}^{2}$${y}^{2}{x}^{2}={x}^{2}-{a}^{2}$y^(2)x^(2)=x^(2)-a^(2)y^2 x^2=x^2-a^2${y}^{2}{x}^{2}={x}^{2}-{a}^{2}$, where $a$$a$aa$a$ is a real constant.
Question:-04(c) यदि समतल $ux+vy+wz=0$$ux+vy+wz=0$ux+vy+wz=0u x+v y+w z=0$ux+vy+wz=0$, शंकु $a{x}^{2}+b{y}^{2}+{\mathrm{c}\mathrm{z}}^{2}=0$$a{x}^{2}+b{y}^{2}+{\mathrm{c}\mathrm{z}}^{2}=0$ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+\mathrm{cz}^2=0$a{x}^{2}+b{y}^{2}+{\mathrm{c}\mathrm{z}}^{2}=0$ को लंब जनकों में काटता है, तो सिद्ध कीजिए कि $\left(b+c\right){u}^{2}+\left(c+a\right){v}^{2}+\left(a+b\right){w}^{2}=0$$\left(b+c\right){u}^{2}+\left(c+a\right){v}^{2}+\left(a+b\right){w}^{2}=0$(b+c)u^(2)+(c+a)v^(2)+(a+b)w^(2)=0(b+c) u^2+(c+a) v^2+(a+b) w^2=0$\left(b+c\right){u}^{2}+\left(c+a\right){v}^{2}+\left(a+b\right){w}^{2}=0$.
Question:-04(c) If the plane $ux+vy+wz=0$$ux+vy+wz=0$ux+vy+wz=0u x+v y+w z=0$ux+vy+wz=0$ cuts the cone $a{x}^{2}+b{y}^{2}+c{z}^{2}=0$$a{x}^{2}+b{y}^{2}+c{z}^{2}=0$ax^(2)+by^(2)+cz^(2)=0a x^2+b y^2+c z^2=0$a{x}^{2}+b{y}^{2}+c{z}^{2}=0$ in perpendicular generators, then prove that $\left(b+c\right){u}^{2}+\left(c+a\right){v}^{2}+\left(a+b\right){w}^{2}=0$$\left(b+c\right){u}^{2}+\left(c+a\right){v}^{2}+\left(a+b\right){w}^{2}=0$(b+c)u^(2)+(c+a)v^(2)+(a+b)w^(2)=0(b+c) u^2+(c+a) v^2+(a+b) w^2=0$\left(b+c\right){u}^{2}+\left(c+a\right){v}^{2}+\left(a+b\right){w}^{2}=0$.
खण्ड B
SECTION B

Question:-05(a) दर्शाइए कि अवकल समीकरण $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}\mathrm{y}=\mathrm{Q}$$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}\mathrm{y}=\mathrm{Q}$(dy)/(dx)+Py=Q\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}\mathrm{y}=\mathrm{Q}$ का व्यापक हल
$\mathrm{y}=\frac{\mathrm{Q}}{\mathrm{P}}-{\mathrm{e}}^{-\int \mathrm{P}dx}\left\{\mathrm{C}+\int {\mathrm{e}}^{\int \mathrm{P}dx}\text{}\mathrm{d}\left(\frac{\mathrm{Q}}{\mathrm{P}}\right)\right\}$$\mathrm{y}=\frac{\mathrm{Q}}{\mathrm{P}}-{\mathrm{e}}^{-\int \mathrm{P}dx}\left\{\mathrm{C}+\int {\mathrm{e}}^{\int \mathrm{P}dx}\text{}\mathrm{d}\left(\frac{\mathrm{Q}}{\mathrm{P}}\right)\right\}$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 $\mathrm{P},\mathrm{Q},\mathrm{x}$$\mathrm{P},\mathrm{Q},\mathrm{x}$P,Q,x\mathrm{P}, \mathrm{Q}, \mathrm{x}$\mathrm{P},\mathrm{Q},\mathrm{x}$ के शून्येतर फलन हैं तथा $\mathrm{C}$$\mathrm{C}$C\mathrm{C}$\mathrm{C}$ एक स्वेच्छ अचर है ।
Question:-05(a)Show that the general solution of the differential equation $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}\mathrm{y}=\mathrm{Q}$$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}\mathrm{y}=\mathrm{Q}$(dy)/(dx)+Py=Q\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{P}\mathrm{y}=\mathrm{Q}$ can be written in the form $y=\frac{Q}{P}-{e}^{-\int Pdx}\left\{C+\int {e}^{\int Pdx}d\left(\frac{Q}{P}\right)\right\}$$y=\frac{Q}{P}-{e}^{-\int Pdx}\left\{C+\int {e}^{\int Pdx}d\left(\frac{Q}{P}\right)\right\}$y=(Q)/(P)-e^(-int Pdx){C+inte^(int Pdx)d((Q)/(P))}y=\frac{Q}{P}-e^{-\int P d x}\left\{C+\int e^{\int P d x} d\left(\frac{Q}{P}\right)\right\}$y=\frac{Q}{P}-{e}^{-\int Pdx}\left\{C+\int {e}^{\int Pdx}d\left(\frac{Q}{P}\right)\right\}$, where $\mathrm{P},\mathrm{Q}$$\mathrm{P},\mathrm{Q}$P,Q\mathrm{P}, \mathrm{Q}$\mathrm{P},\mathrm{Q}$ are non-zero functions of $\mathrm{x}$$\mathrm{x}$x\mathrm{x}$\mathrm{x}$ and $\mathrm{C}$$\mathrm{C}$C\mathrm{C}$\mathrm{C}$, an arbitrary constant.
Question:-05(b) दर्शाइए कि परवलयों के निकाय : ${\mathrm{x}}^{2}=4\mathrm{a}\left(\mathrm{y}+\mathrm{a}\right)$${\mathrm{x}}^{2}=4\mathrm{a}\left(\mathrm{y}+\mathrm{a}\right)$x^(2)=4a(y+a)\mathrm{x}^2=4 \mathrm{a}(\mathrm{y}+\mathrm{a})${\mathrm{x}}^{2}=4\mathrm{a}\left(\mathrm{y}+\mathrm{a}\right)$ के लंबकोणीय संछेदी, उसी निकाय में स्थित होते हैं ।
Question:-05(b) Show that the orthogonal trajectories of the system of parabolas : ${\mathrm{x}}^{2}=4\mathrm{a}\left(\mathrm{y}+\mathrm{a}\right)$${\mathrm{x}}^{2}=4\mathrm{a}\left(\mathrm{y}+\mathrm{a}\right)$x^(2)=4a(y+a)\mathrm{x}^2=4 \mathrm{a}(\mathrm{y}+\mathrm{a})${\mathrm{x}}^{2}=4\mathrm{a}\left(\mathrm{y}+\mathrm{a}\right)$ belong to the same system.
Question:-05(c) $\mathrm{w}$$\mathrm{w}$w\mathrm{w}$\mathrm{w}$ भार का एक पिंड, $\theta$$\theta$theta\theta$\theta$ कोण से झुके हुए एक रूक्ष समतल पर स्थित है, घर्षण गुणांक $\mu$$\mu$mu\mu$\mu$, $\mathrm{tan}\theta$$\mathrm{tan}\theta$tan theta\tan \theta$\mathrm{tan}\theta$ से अधिक है । पिंड को समतल पर ऊपर की तरफ ‘ $b$$b$bb$b$ ‘ दूरी तक धीरे-धीरे खींचने तथा वापस आरम्भिक बिन्दु तक खींचने में किए गए कार्य को ज्ञात कीजिए, जहाँ लगाया गया बल प्रत्येक दशा में समतल के समान्तर है ।
Question:-05(c)A body of weight $w$$w$ww$w$ rests on a rough inclined plane of inclination $\theta$$\theta$theta\theta$\theta$, the coefficient of friction, $\mu$$\mu$mu\mu$\mu$, being greater than $\mathrm{tan}\theta$$\mathrm{tan}\theta$tan theta\tan \theta$\mathrm{tan}\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{2\mathrm{g}\mathrm{h}}$$\sqrt{2\mathrm{g}\mathrm{h}}$sqrt(2gh)\sqrt{2 \mathrm{gh}}$\sqrt{2\mathrm{g}\mathrm{h}}$ वेग के साथ बिन्दु $\mathrm{O}$$\mathrm{O}$O\mathrm{O}$\mathrm{O}$ से प्रक्षेपित किया गया तथा समतल के बिन्दु $\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)$$\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)$P(x,y)\mathrm{P}(\mathrm{x}, \mathrm{y})$\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)$ पर स्पर्श-रेखा से टकराता है जहाँ अक्ष $\mathrm{O}\mathrm{X}$$\mathrm{O}\mathrm{X}$OX\mathrm{OX}$\mathrm{O}\mathrm{X}$ तथा $\mathrm{O}\mathrm{Y}$$\mathrm{O}\mathrm{Y}$OY\mathrm{OY}$\mathrm{O}\mathrm{Y}$ क्रमशः बिन्दु $\mathrm{O}$$\mathrm{O}$O\mathrm{O}$\mathrm{O}$ से क्षैतिज तथा अधोमुखी ऊर्ध्वाधर रेखाएँ हैं । यदि प्रक्षेपण की दो संभव दिशाएँ समकोण पर हों, तो दर्शाइए कि ${\mathrm{x}}^{2}=2\mathrm{h}\mathrm{y}$${\mathrm{x}}^{2}=2\mathrm{h}\mathrm{y}$x^(2)=2hy\mathrm{x}^2=2 \mathrm{hy}${\mathrm{x}}^{2}=2\mathrm{h}\mathrm{y}$ तथा प्रक्षेपण की संभव दिशाओं में से एक, कोण POX को द्विभाजित करती है ।
Question:-05(d)A projectile is fired from a point $\mathrm{O}$$\mathrm{O}$O\mathrm{O}$\mathrm{O}$ with velocity $\sqrt{2\mathrm{g}\mathrm{h}}$$\sqrt{2\mathrm{g}\mathrm{h}}$sqrt(2gh)\sqrt{2 \mathrm{gh}}$\sqrt{2\mathrm{g}\mathrm{h}}$ and hits a tangent at the point $\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)$$\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)$P(x,y)\mathrm{P}(\mathrm{x}, \mathrm{y})$\mathrm{P}\left(\mathrm{x},\mathrm{y}\right)$ in the plane, the axes $\mathrm{O}\mathrm{X}$$\mathrm{O}\mathrm{X}$OX\mathrm{OX}$\mathrm{O}\mathrm{X}$ and $\mathrm{O}\mathrm{Y}$$\mathrm{O}\mathrm{Y}$OY\mathrm{OY}$\mathrm{O}\mathrm{Y}$ being horizontal and vertically downward lines through the point $\mathrm{O}$$\mathrm{O}$O\mathrm{O}$\mathrm{O}$, respectively. Show that if the two possible directions of projection be at right angles, then ${\mathrm{x}}^{2}=2\mathrm{h}\mathrm{y}$${\mathrm{x}}^{2}=2\mathrm{h}\mathrm{y}$x^(2)=2hy\mathrm{x}^2=2 \mathrm{hy}${\mathrm{x}}^{2}=2\mathrm{h}\mathrm{y}$ and then one of the possible directions of projection bisects the angle POX.
Question:-05(e) दर्शाइए कि $\stackrel{\to }{\mathrm{A}}=\left(6\mathrm{x}\mathrm{y}+{\mathrm{z}}^{3}\right)\stackrel{^}{\mathrm{i}}+\left(3{\mathrm{x}}^{2}-\mathrm{z}\right)\stackrel{^}{\mathrm{j}}+\left(3x{\mathrm{z}}^{2}-\mathrm{y}\right)\stackrel{^}{\mathrm{k}}$$\stackrel{\to }{\mathrm{A}}=\left(6\mathrm{x}\mathrm{y}+{\mathrm{z}}^{3}\right)\stackrel{^}{\mathrm{i}}+\left(3{\mathrm{x}}^{2}-\mathrm{z}\right)\stackrel{^}{\mathrm{j}}+\left(3x{\mathrm{z}}^{2}-\mathrm{y}\right)\stackrel{^}{\mathrm{k}}$vec(A)=(6xy+z^(3)) hat(i)+(3x^(2)-z) hat(j)+(3xz^(2)-y) hat(k)\overrightarrow{\mathrm{A}}=\left(6 \mathrm{xy}+\mathrm{z}^3\right) \hat{\mathrm{i}}+\left(3 \mathrm{x}^2-\mathrm{z}\right) \hat{\mathrm{j}}+\left(3 x \mathrm{z}^2-\mathrm{y}\right) \hat{\mathrm{k}}$\stackrel{\to }{\mathrm{A}}=\left(6\mathrm{x}\mathrm{y}+{\mathrm{z}}^{3}\right)\stackrel{^}{\mathrm{i}}+\left(3{\mathrm{x}}^{2}-\mathrm{z}\right)\stackrel{^}{\mathrm{j}}+\left(3x{\mathrm{z}}^{2}-\mathrm{y}\right)\stackrel{^}{\mathrm{k}}$ अघूर्णी है । $\varphi$$\varphi$phi\phi$\varphi$ को भी ज्ञात कीजिए जबकि $\stackrel{\to }{\mathrm{A}}=\mathrm{\nabla }\varphi$$\stackrel{\to }{\mathrm{A}}=\mathrm{\nabla }\varphi$vec(A)=grad phi\overrightarrow{\mathrm{A}}=\nabla \phi$\stackrel{\to }{\mathrm{A}}=\mathrm{\nabla }\varphi$.
Question:-05(e)Show that $\stackrel{\to }{\mathrm{A}}=\left(6xy+{z}^{3}\right)\stackrel{^}{i}+\left(3{x}^{2}-z\right)\stackrel{^}{j}+\left(3x{z}^{2}-y\right)\stackrel{^}{k}$$\stackrel{\to }{\mathrm{A}}=\left(6xy+{z}^{3}\right)\stackrel{^}{i}+\left(3{x}^{2}-z\right)\stackrel{^}{j}+\left(3x{z}^{2}-y\right)\stackrel{^}{k}$vec(A)=(6xy+z^(3)) hat(i)+(3x^(2)-z) hat(j)+(3xz^(2)-y) hat(k)\overrightarrow{\mathrm{A}}=\left(6 x y+z^3\right) \hat{i}+\left(3 x^2-z\right) \hat{j}+\left(3 x z^2-y\right) \hat{k}$\stackrel{\to }{\mathrm{A}}=\left(6xy+{z}^{3}\right)\stackrel{^}{i}+\left(3{x}^{2}-z\right)\stackrel{^}{j}+\left(3x{z}^{2}-y\right)\stackrel{^}{k}$ is irrotational. Also find $\varphi$$\varphi$phi\phi$\varphi$ such that $\stackrel{\to }{\mathrm{A}}=\mathrm{\nabla }\varphi$$\stackrel{\to }{\mathrm{A}}=\mathrm{\nabla }\varphi$vec(A)=grad phi\overrightarrow{\mathrm{A}}=\nabla \phi$\stackrel{\to }{\mathrm{A}}=\mathrm{\nabla }\varphi$.

Question:-06(a) $2l$$2l$2l2 l$2l$ लम्बाई का एक तार (केबिल) जिसका भार $\mathrm{w}$$\mathrm{w}$w\mathrm{w}$\mathrm{w}$ प्रति इकाई (यूनिट) लम्बाई है, एक क्षैतिज रेखा के दो बिन्दुओं $\mathrm{P}$$\mathrm{P}$P\mathrm{P}$\mathrm{P}$ तथा $\mathrm{Q}$$\mathrm{Q}$Q\mathrm{Q}$\mathrm{Q}$ से लटकी हुई है । दर्शाइए कि तार की विस्तृति (स्पैन) $2l\left(1-\frac{2{\text{}\mathrm{h}}^{2}}{3{l}^{2}}\right)$$2l\left(1-\frac{2{\text{}\mathrm{h}}^{2}}{3{l}^{2}}\right)$2l(1-(2h^(2))/(3l^(2)))2 l\left(1-\frac{2 \mathrm{~h}^2}{3 l^2}\right) है, जहाँ $\mathrm{h}$$\mathrm{h}$h\mathrm{h}$\mathrm{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 $2l$$2l$2l2 l$2l$ hangs from two points $\mathrm{P}$$\mathrm{P}$P\mathrm{P}$\mathrm{P}$ and $\mathrm{Q}$$\mathrm{Q}$Q\mathrm{Q}$\mathrm{Q}$ in the same horizontal line. Show that the span of the cable is $2l\left(1-\frac{2{h}^{2}}{3{l}^{2}}\right)$$2l\left(1-\frac{2{h}^{2}}{3{l}^{2}}\right)$2l(1-(2h^(2))/(3l^(2)))2 l\left(1-\frac{2 h^2}{3 l^2}\right)$2l\left(1-\frac{2{h}^{2}}{3{l}^{2}}\right)$, where $h$$h$hh$h$ is the sag in the middle of the tightly stretched position.
Question:-06(b) प्राचल-विचरण विधि का उपयोग करके, निम्नलिखित अवकल समीकरण :
$\left({x}^{2}-1\right)\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}+2y={\left({x}^{2}-1\right)}^{2}$$\left({x}^{2}-1\right)\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}+2y={\left({x}^{2}-1\right)}^{2}$(x^(2)-1)(d^(2)y)/(dx^(2))-2x(dy)/(dx)+2y=(x^(2)-1)^(2)\left(x^2-1\right) \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}+2 y=\left(x^2-1\right)^2$\left({x}^{2}-1\right)\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}+2y={\left({x}^{2}-1\right)}^{2}$
को हल कीजिए, जहाँ समानीत समीकरण का एक हल $\mathrm{y}=\mathrm{x}$$\mathrm{y}=\mathrm{x}$y=x\mathrm{y}=\mathrm{x}$\mathrm{y}=\mathrm{x}$ दिया गया है ।
Question:-06(b) Solve the following differential equation by using the method of variation of parameters : $\left({x}^{2}-1\right)\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}+2y={\left({x}^{2}-1\right)}^{2}$$\left({x}^{2}-1\right)\frac{{d}^{2}y}{d{x}^{2}}-2x\frac{dy}{dx}+2y={\left({x}^{2}-1\right)}^{2}$(x^(2)-1)(d^(2)y)/(dx^(2))-2x(dy)/(dx)+2y=(x^(2)-1)^(2)\left(x^2-1\right) \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}+2 y=\left(x^2-1\right)^2$\left({x}^{2}-1\right)\frac{{d}^{2}y}{d{x}^{2}