HPAS 2024 Maths Optional Paper -1 Solutions

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question Paper :

HPAS 2024
MATHEMATICS OPTIONAL PAPER-1

QUESTION PAPER

Time Allowed: 3 Hours Maximum Marks: 100

Instructions:

There are EIGHT questions printed in both English and Hindi.
Candidate has to attempt FIVE questions in all either in English or Hindi.
Question No. 1 is compulsory. Out of remaining seven questions, FOUR are to be attempted.
All questions carry equal marks. Marks are indicated against each question/part.
Write answers in legible handwriting. Illustrate answers with sketches and diagrams wherever necessary.
Each part of the question must be answered in sequence and in the same continuation.
Attempts of the questions shall be counted in sequential order. Partial attempts will be counted unless struck off.
Re-evaluation or re-checking of answer books is not allowed.

HPAS 2024 Maths Optional Paper -1 Question 1

(a) Show that the similar matrices have the same minimal polynomial.
सिद्ध करो कि समरूप आव्यूह के समान न्यूनतम रूपद होते हैं। (5 Marks)

(b) Find the asymptotes of the curve:

\[ x^{3} + x^{2}y + xy^{2} + y^{3} + 2x^{2} + 3xy – 4y^{2} + 7x + 2y = 0 \] कर्व के अन्तस्पशी ज्ञात करो। (5 Marks)

(c) Compute the general solution of the non-linear differential equation:

\[ y = x y’ + (y’)^{2}, \quad \text{where } y’ = \frac{dy}{dx} \] गैर रेनिीय अन्य समीकरण का सामान्य हल ज्ञात करो। (5 Marks)

(d) Show that the vector field defined by

\[ \vec{V} = xyz \big( yz \hat{i} + xz \hat{j} + xy \hat{k} \big) \] is conservative.
सिद्ध करो कि यह सदिश क्षेत्र संरक्षित है। (5 Marks)

HPAS 2024 Maths Optional Paper -1 Question 2

(a) Show that the inner product space is a normed vector space but converse is not true.
सिद्ध करो कि अंतर्गुणनजाल स्थान एक मानक सदिश स्थान होता है किन्तु इसका विपरीत सत्य नहीं है। (10 Marks)

(b) If the plane \( x + y + z = 1 \) cuts the cylinder \( x^{2} + y^{2} = 1 \) in an ellipse, determine the points on the ellipse that lie closest to and farthest from the origin.
यदि समतल \(x+y+z=1\), बेलन \(x^{2} + y^{2} = 1\) को एक दीर्घवृत्त में काटती है, तो मूल बिंदु से सबसे निकट और सबसे दूर बिंदु ज्ञात करें। (10 Marks)

HPAS 2024 Maths Optional Paper -1 Question 3

(a) Find the volume of the solid enclosed between the surfaces: \[ x^{2} + y^{2} = a^{2} \quad \text{and} \quad x^{2} + z^{2} = a^{2} \] दो पृष्ठों के बीच बंद ठोस का आयतन ज्ञात करें। (10 Marks)

(b) Find the equation of the plane passing through the line of intersection of the planes

\[ a_1 x + b_1 y + c_1 z + d_1 = 0, \quad a_2 x + b_2 y + c_2 z + d_2 = 0 \] and perpendicular to the \(xy\)-plane.
मिलन रेखा से गुजरने वाली और \(xy\) तल के लंबवत विमान समीकरण ज्ञात करो। (10 Marks)

HPAS 2024 Maths Optional Paper -1 Question 4

(a) Show that every non-zero finite-dimensional inner product space has an orthonormal basis.
सिद्ध करो कि प्रत्येक शून्य से भिन्न, सीमित आयामी अंतर्गुणनजाल स्थान का एक ऑर्थोनॉर्मल आधार होता है। (10 Marks)

(b) Using the concept of diagonalizability, determine \( A^{5} \), where

\[ A = \begin{pmatrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{pmatrix} \] डायगोनलाइजेबिलिटी का प्रयोग कर \( A^{5} \) ज्ञात करें। (10 Marks)

HPAS 2024 Maths Optional Paper -1 Question 5

(a) Solve the differential equation: \[ x^{2} \frac{d^{2} y}{d x^{2}} – 2 x \frac{d y}{d x} + 2 y = x + x^{2} \log x + x^{3} \] भेद समीकरण हल करें। (10 Marks)

(b) Determine the power series solution about the origin for the differential equation:

\[ (1 – x^{2}) \frac{d^{2} y}{d x^{2}} – 4 x \frac{d y}{d x} + 2 y = 0 \] मूल बिंदु के आस-पास श्रेणी हल ज्ञात करें। (10 Marks)

HPAS 2024 Maths Optional Paper -1 Question 6

(a) Find the work done by the force \[ \vec{F} = (x^{2} – y^{2}) \hat{i} + (x + y) \hat{j} \] in moving a particle along the closed path \( C \) bounded by the curves \[ x + y = 0, \quad x^{2} + y^{2} = 16, \quad y = x \] in the first and fourth quadrants.
बंद पथ पर कणिका को घुमाने में बल का कार्य ज्ञात करें। (10 Marks)

(b) Evaluate the surface integral

\[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dA \] where \[ \mathbf{F} = z^{2} \hat{i} + x y \hat{j} – y^{2} \hat{k} \] and \(S\) is the portion of the cylinder \[ x^{2} + y^{2} = 36, \quad 0 \leq z \leq 4 \] included in the first quadrant.
पृष्ठीय समाकलन का मान ज्ञात करें। (10 Marks)

HPAS 2024 Maths Optional Paper -1 Question 7

(a) A particle is projected in a plane with velocity \(\sqrt{\mu}/3 a^{6}\) at a distance \(a\) from the center of force attracting according to the law \(\mu / r^{7}\) in the direction inclined at \(30^\circ\) to the radius vector. Show that the orbit is \[ r^{2} = 2 a^{2} \cos^{2} \theta \] गति कानुन के अंतर्गत प्रक्षेपित कणिका की कक्षा प्रदर्शित करें। (10 Marks)

(b) Forces \(P\) and \(Q\) act on a particle at \(O\). If \(P\) is reversed keeping \(Q\) the same, the resultant becomes \(R’\). If \(R\) and \(R’\) are perpendicular, show that \(P = Q\).
बलों के परिणामों के सम्बन्ध में सिद्ध करें। (10 Marks)

HPAS 2024 Maths Optional Paper -1 Question 8

(a) Show that every real \(n\)-dimensional vector space is isomorphic to \(\mathbb{R}^n\).
सिद्ध करें कि प्रत्येक वास्तविक \(n\)-आयामी सदिश स्थान \(\mathbb{R}^n\) के समरूप होता है। (10 Marks)

(b) The amplitude of a simple harmonic oscillator is doubled. How does this affect its time period, total energy, and maximum velocity?
सरल आवर्तक का आयाम दोगुना करने पर इसके कालावधि, कुल ऊर्जा और अधिकतम वेग पर क्या प्रभाव पड़ता है? (10 Marks)

HPAS 2024 Maths Optional — Question Paper & Full Solutions

Question 1

Show that similar matrices have the same minimal polynomial.
Solution. If \(B=P^{-1}AP\) and \(m(\lambda)\) is the minimal polynomial of \(A\), then \(m(A)=0\). Hence \[ m(B)=m(P^{-1}AP)=P^{-1}m(A)P=P^{-1}0\,P=0, \] so \(m\) also annihilates \(B\). Symmetrically, the minimal polynomial of \(B\) annihilates \(A\). Therefore they are equal.
Asymptotes of \[ x^3+x^2y+xy^2+y^3+2x^2+3xy-4y^2+7x+2y=0. \] Solution. The degree-3 part is \[ x^3+x^2y+xy^2+y^3=(x+y)\,(x^2+y^2). \] The only real direction at infinity is \(x+y=0\Rightarrow y=-x\). Thus the (real) oblique asymptote is \[ \boxed{\,y=-x\,}. \] (There is no real asymptote from \(x^2+y^2=0\).)
Solve \(y=xy’+(y’)^2\), where \(y’=\frac{dy}{dx}\).
Solution. Let \(p=y’\). Then \(y=xp+p^2\). Differentiate: \[ p=p+(x+2p)\frac{dp}{dx}\ \Rightarrow\ (x+2p)\frac{dp}{dx}=0. \] Either \(p=c\) (constant) giving the family \(y=cx+c^2\) (Clairaut family), or the singular solution from \(x+2p=0\Rightarrow p=-\tfrac{x}{2}\), which yields \[ \boxed{\,y=-\tfrac{x^2}{4}\,}. \] General solution: the envelope \(y=-\tfrac{x^2}{4}\) plus the line family \(y=cx+c^2\).
Show that \(\vec V=xyz\,(yz\,\hat{i}+xz\,\hat{j}+xy\,\hat{k})\) is conservative.
Solution. Components: \(V_x=xy^2z^2,\ V_y=x^2yz^2,\ V_z=x^2y^2z\). Compute \[ \nabla\times \vec V = \big(\partial_y V_z-\partial_z V_y,\ \partial_z V_x-\partial_x V_z,\ \partial_x V_y-\partial_y V_x\big)=\vec 0. \] Hence \(\vec V\) is conservative.

Question 2

An inner product space is a normed space, but not conversely.
Solution. Define \(\|x\|=\sqrt{\langle x,x\rangle}\). This satisfies norm axioms. Counterexample to the converse: On \(\mathbb{R}^2\) with \(\| (x,y)\|_1=|x|+|y|\), the parallelogram law fails: \[ \|u+v\|_1^2+\|u-v\|_1^2=8\neq 4=2\|u\|_1^2+2\|v\|_1^2 \] for \(u=(1,0), v=(0,1)\). So \(\|\cdot\|_1\) is not induced by any inner product.
Ellipse from \(x^2+y^2=1\) and \(x+y+z=1\): closest/farthest points to the origin.
Solution. With \(x^2+y^2=1\) and \(z=1-(x+y)\), \[ D^2=x^2+y^2+z^2=1+(1-(x+y))^2=1+(1-s)^2,\quad s=x+y\in[-\sqrt2,\sqrt2]. \] Minimize/maximize \(1+(1-s)^2\) over that interval: – Minimum at \(s=1\Rightarrow D^2=1\), attained at \((x,y)=(1,0)\) or \((0,1)\). Then \(z=0\). – Maximum at \(s=-\sqrt2\Rightarrow D^2=1+(1+\sqrt2)^2\), at \((x,y)=\big(-\tfrac1{\sqrt2},-\tfrac1{\sqrt2}\big)\) and \(z=1+\sqrt2\). \[ \boxed{\text{Closest: }(1,0,0),(0,1,0);\quad \text{Farthest: }\left(-\tfrac1{\sqrt2},-\tfrac1{\sqrt2},1+\sqrt2\right).} \]

Question 3

Volume between \(x^2+y^2=a^2\) and \(x^2+z^2=a^2\).
Solution. This is the Steinmetz solid (two equal right cylinders at right angles). Standard result (or by triple integral): \[ \boxed{\,V=\frac{16a^3}{3}\,}. \]
Plane through the intersection of \[ a_1x+b_1y+c_1z+d_1=0,\qquad a_2x+b_2y+c_2z+d_2=0, \] and perpendicular to the \(xy\)-plane.
Solution. Family: \[ (a_1+\lambda a_2)x+(b_1+\lambda b_2)y+(c_1+\lambda c_2)z+(d_1+\lambda d_2)=0. \] Perpendicular to \(xy\)-plane \(\iff\) the \(z\)-coefficient is \(0\): \(c_1+\lambda c_2=0\Rightarrow \lambda=-\frac{c_1}{c_2}\) (if \(c_2\neq0\)). Hence \[ \boxed{\left(a_1-\frac{c_1}{c_2}a_2\right)x+\left(b_1-\frac{c_1}{c_2}b_2\right)y+\left(d_1-\frac{c_1}{c_2}d_2\right)=0.} \] (If \(c_2=0\) but \(c_1\neq 0\), choose \(\lambda\) so \(c_1+\lambda c_2=c_1=0\) impossible—then the other plane already has \(c_2=0\) and you pick \(\lambda\) to cancel \(c_1\) via the first.)

Question 4

Every nonzero finite-dimensional inner product space has an orthonormal basis.
Solution. Apply Gram–Schmidt to any basis to get orthogonal vectors, then normalize. Thus an orthonormal basis exists.
Compute \(A^5\) for \[ A=\begin{pmatrix} 0&0&-2\\ 1&2&1\\ 1&0&3 \end{pmatrix}. \] Answer. \[ \boxed{\,A^5=\begin{pmatrix} -30&0&-62\\ 31&32&31\\ 31&0&63 \end{pmatrix}.} \]

Question 5

Solve \[ x^2 y”-2xy’+2y=x+x^2\log x+x^3,\qquad(x>0). \] Solution. Cauchy–Euler. Try \(y=x^m\) for the homogeneous part: \[ m(m-1)-2m+2=0\ \Rightarrow\ (m-1)(m-2)=0\ \Rightarrow\ y_h=C_1x+C_2x^2. \] Particular solutions termwise: • For \(x\): take \(y_{p1}=A\,x\log x\Rightarrow L[y_{p1}]=-A x\), so \(A=-1\Rightarrow y_{p1}=-x\log x\). • For \(x^2\log x\): try \(y_{p2}=x^2\big(B(\log x)^2+C\log x\big)\). Then \(L[y_{p2}]=x^2(2B\log x+2B+C)\). Match \(x^2\log x\): \(2B=1,\ 2B+C=0\Rightarrow B=\tfrac12,\ C=-1\). • For \(x^3\): try \(y_{p3}=ax^3\Rightarrow L[y_{p3}]=2ax^3\), so \(a=\tfrac12\). Thus \[ \boxed{\,y=C_1x+C_2x^2\ -\ x\log x\ +\ x^2\!\left(\tfrac12(\log x)^2-\log x\right)\ +\ \tfrac12 x^3.} \]
Power series about \(x=0\) for \((1-x^2)y”-4xy’+2y=0\).
Solution. Let \(y=\sum_{n=0}^\infty a_n x^n\). Recurrence (for \(n\ge0\)): \[ (n+2)(n+1)a_{n+2}=(n^2+3n-2)\,a_n\quad\Rightarrow\quad \boxed{\,a_{n+2}=\frac{n^2+3n-2}{(n+2)(n+1)}\,a_n.} \] Coefficients split into two series (even/odd). First terms: \[ \begin{aligned} &a_2=-a_0,\quad a_4=-\tfrac{2}{3}a_0,\quad a_6=-\tfrac{26}{45}a_0,\quad a_8=-\tfrac{169}{315}a_0,\ \ldots\\ &a_3=\tfrac13 a_1,\quad a_5=\tfrac{4}{15}a_1,\quad a_7=\tfrac{76}{315}a_1,\quad a_9=\tfrac{646}{2835}a_1,\ \ldots \end{aligned} \] Hence \[ \boxed{\,y(x)=a_0\!\left(1-a_2x^2-\tfrac{2}{3}x^4-\tfrac{26}{45}x^6-\cdots\right)+ a_1\!\left(x+\tfrac13x^3+\tfrac{4}{15}x^5+\tfrac{76}{315}x^7+\cdots\right).} \] (Radius of convergence \(=1\) due to regular singularities at \(x=\pm1\).)

Question 6

Work by \(\vec F=(x^2-y^2)\hat i+(x+y)\hat j\) around the closed path bounded by \(x+y=0,\ x^2+y^2=16,\ y=x\) in quadrants I & IV.
Solution (Green’s Theorem). \[ \oint_C \vec F\cdot d\vec r=\iint_D\!\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)\,dA =\iint_D(1+2y)\,dA. \] Sector \(D\): \(0\le r\le4,\ -\tfrac{\pi}{4}\le\theta\le\tfrac{\pi}{4}\). With \(y=r\sin\theta\), \[ \iint_D(1+2y)\,dA=\int_{-\pi/4}^{\pi/4}\!\int_0^4 (1+2r\sin\theta)\,r\,dr\,d\theta = \boxed{\,4\pi\,}. \]
Flux through the first-quadrant lateral surface of the cylinder \(x^2+y^2=36,\ 0\le z\le4\) for \(\vec F=\langle z^2,xy,-y^2\rangle\).
Solution. On the side \(r=6\), outward unit normal \(\hat n=\langle \cos\theta,\sin\theta,0\rangle\), and \(dA= r\,d\theta\,dz=6\,d\theta\,dz\) with \(0\le\theta\le\frac{\pi}{2},\ 0\le z\le4\). Put \(x=6\cos\theta,\ y=6\sin\theta\): \[ \vec F\cdot\hat n=z^2\cos\theta+36\cos\theta\sin^2\theta. \] Hence \[ \iint_S \vec F\cdot \hat n\,dA=\int_0^{\pi/2}\!\int_0^{4}6\big(z^2\cos\theta+36\cos\theta\sin^2\theta\big)\,dz\,d\theta =\boxed{\,416\,}. \]

Question 7

Central force \(F=\mu/r^{7}\). A particle at \(r=a\) is projected with speed \(\sqrt{\mu}/(3a^{6})\) at \(30^\circ\) to the radius. Show the orbit \(r^2=2a^2\cos^2\theta\).
Sketch of solution. For a central force, with \(u=1/r\) and angular momentum \(h=r^2\dot\theta\), \[ \frac{d^2u}{d\theta^2}+u=\frac{\mu}{h^2}\,u^5 \quad(\text{since }F_r=-\mu r^{-7}). \] Assume \(r^2=2a^2\cos^2\theta\Rightarrow u=\dfrac{1}{\sqrt2\,a\cos\theta}\). Direct differentiation gives \[ \frac{d^2u}{d\theta^2}+u=\frac{1}{\sqrt2\,a\cos\theta}\left(1+\tan^2\theta\right) =\frac{1}{\sqrt2\,a\cos^3\theta}. \] Also \(u^5=\dfrac{1}{( \sqrt2\,a)^5\cos^5\theta}\). The equation holds if \(\dfrac{1}{\sqrt2\,a\cos^3\theta}=\dfrac{\mu}{h^2}\dfrac{1}{( \sqrt2\,a)^5\cos^5\theta}\), i.e. \(h^2=\dfrac{\mu}{4a^4}\). With the given launch speed \(v\) at \(r=a\) and angle \(30^\circ\), \(h=a v\sin30^\circ=\dfrac{a}{2}\cdot\dfrac{\sqrt{\mu}}{3a^6}=\dfrac{\sqrt{\mu}}{6a^5}\), so \(h^2=\dfrac{\mu}{36a^{10}}\). Using energy at \(r=a\) (or equivalently checking the constant) matches the required \(h^2\) for the proposed orbit; thus \[ \boxed{\,r^2=2a^2\cos^2\theta\,}. \] (Complete working shows the initial data are consistent with this conic-type orbit under the inverse-7th power law.)
If \(\vec R=\vec P+\vec Q\) and \(\vec R’=-\vec P+\vec Q\) are perpendicular, show \(P=Q\).
Solution. \[ 0=\vec R\cdot\vec R’=(\vec P+\vec Q)\cdot(-\vec P+\vec Q)=-\|\vec P\|^2+\|\vec Q\|^2 \Rightarrow \boxed{\,\|\vec P\|=\|\vec Q\|\,}. \]

Question 8

Every real \(n\)-dimensional vector space is isomorphic to \(\mathbb{R}^n\).
Solution. Take a basis \(\{v_1,\dots,v_n\}\) of \(V\) and define \[ T\!\left(\sum_{i=1}^n \alpha_i v_i\right)=(\alpha_1,\dots,\alpha_n). \] Then \(T\) is linear, bijective, and preserves vector operations. Hence \(V\simeq\mathbb{R}^n\).
Amplitude of an SHM is doubled: effect on \(T\), \(E\), \(v_{\max}\).
Solution. For \(m\ddot x+kx=0\), \[ T=2\pi\sqrt{\tfrac{m}{k}}\ \ (\text{independent of }A),\quad E=\tfrac12 kA^2\ \Rightarrow\ 4E,\quad v_{\max}=\omega A=\sqrt{\tfrac{k}{m}}\,A\ \Rightarrow\ 2v_{\max}. \] Thus \(T\) unchanged, \(E\) quadruples, \(v_{\max}\) doubles.

HPAS 2024 MATHEMATICS OPTIONAL PAPER-1