HPAS 2023 Mathematics Optional Paper-1: Complete Solutions
Welcome to the comprehensive solution guide for the Himachal Pradesh Administrative Service (HPAS) 2023 Mathematics Optional Paper-1. This resource provides detailed, step-by-step solutions designed specifically for civil service aspirants to master the core mathematical concepts and methodologies required for the exam.
Whether you are revising key theorems, practicing previous year questions, or mastering advanced analytical geometry and calculus, these carefully structured solutions will help streamline your preparation. Use the index below to jump directly to specific questions and topics.
Table of Contents
- Question 1(a): Roots of Legendre polynomial
- Question 1(b): Closed set of continuous function
- Question 1(c): Conservative vector field
- Question 1(d): Diagonalizable matrix without distinct eigenvalues
- Question 1(e): Max velocity in SHM
- Question 2(a): Cauchy-Schwarz inequality for cosine
- Question 2(b): Annihilator dimension theorem
- Question 2(c): Intersection of plane and cone
- Question 3(a): Radius of curvature of lemniscate
- Question 3(b): Limit evaluation
- Question 3(c): Rank-Nullity Theorem
- Question 4(a): Partial derivatives relation
- Question 4(b): Shortest distance between lines
- Question 5(a): General and singular solution of DE
- Question 5(b): DE in terms of Bessel functions
- Question 5(c): Variation of parameters
- Question 6(a): Stokes’s theorem evaluation
- Question 6(b): Directional derivative
- Question 7(a): 3-dimensional vector space over R
- Question 7(b): Three forces in equilibrium
- Question 8(a): Total energy of particle in SHM
- Question 8(b): Central force expression
HPAS 2023 Maths Optional Paper-1 Question 1(a)
Show that the roots of the equation P_n(x)=0 are real and lie between -1 and 1, where P_n(x) is the Legendre’s polynomial of degree n, and n is a positive integer.
Solution:
To prove that the roots of the Legendre polynomial P_n(x) are real, distinct, and lie in the interval (-1, 1), we utilize the properties of the Legendre differential equation and the orthogonality of Legendre polynomials.
1. The Legendre Differential Equation
The Legendre polynomial P_n(x) is a solution to the following second-order linear differential equation:
\frac{d}{dx} \left[ (1-x^2) \frac{d}{dx} P_n(x) \right] + n(n+1)P_n(x) = 02. Orthogonality Property
Legendre polynomials satisfy the orthogonality condition on the interval [-1, 1]:
\int_{-1}^{1} P_n(x) P_m(x) \, dx = 0 \quad \text{for } n \neq mSince P_0(x) = 1, we have \int_{-1}^{1} P_n(x) \, dx = 0 for all n \geq 1. This implies that P_n(x) must change sign at least once in the interval (-1, 1).
3. Proof by Contradiction
Suppose P_n(x) changes sign at m points x_1, x_2, \dots, x_m within the interval (-1, 1), where m \lt n.
- Let Q(x) = (x-x_1)(x-x_2)\dots(x-x_m) be a polynomial of degree m.
- The product P_n(x)Q(x) does not change sign on (-1, 1), implying \int_{-1}^{1} P_n(x) Q(x) \, dx \neq 0.
- By the definition of orthogonal polynomials, P_n(x) must be orthogonal to any polynomial of degree less than n. Since m \lt n, we must have \int_{-1}^{1} P_n(x) Q(x) \, dx = 0.
Conclusion
The contradiction implies that m must be at least n. Since P_n(x) is a polynomial of degree n, it has exactly n real, distinct roots in the interval (-1, 1).
HPAS 2023 Maths Optional Paper-1 Question 1(b)
Let the function f be continuous on the real line \mathbb{R}. Then show that the set A = \{x : f(x) = 0\} is closed.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 1(c)
For what values of a and b is the vector field \vec{F} = (x+z)\mathbf{i} + a(y+z)\mathbf{j} + b(x+y)\mathbf{k} a conservative field?
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 1(d)
Give an example of a diagonalizable matrix that does not have distinct eigen values.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 1(e)
A particle executes Simple Harmonic Motion with a period of 10 seconds and an amplitude of 5 cm. Calculate the maximum velocity.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 2(a)
Using the Cauchy-Schwarz inequality, show that the cosine of an angle has an absolute value of at most 1.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 2(b)
Let V be a finite-dimensional vector space and W be a subspace of V. Show that \dim A(W) = \dim V - \dim W, where A(W) is the annihilator of W.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 2(c)
Find the equations of the lines in which the plane 2x+y-z=0 cuts the cone 4x^2 - y^2 + 3z^2 = 0.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 3(a)
Show that the radius of curvature of the lemniscate (x^2+y^2)^2 = a^2(x^2-y^2) at any point where the tangent is parallel to the x-axis, is \frac{\sqrt{2}a}{3}.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 3(b)
Evaluate
\lim_{x\to0} x^m (\log_e x)^nwhere m and n are positive integers.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 3(c)
Let T: V \to W be a linear transformation. Then show that \text{Rank}(T) + \text{Nullity}(T) = \dim(V).
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 4(a)
If
u(x,y) = \sin^{-1}\left(\left(\frac{x^{1/3}+y^{1/3}}{x^{1/2}+y^{1/2}}\right)^{1/2}\right)then show that
x^2\frac{\partial^2 u}{\partial x^2} + 2xy\frac{\partial^2 u}{\partial x \partial y} + y^2\frac{\partial^2 u}{\partial y^2} = \frac{1}{144}\tan u(13 + \tan^2 u)Solution:
Step 1: Identify the Homogeneous Function
Rewrite the expression to isolate u:
\sin u = \left( \frac{x^{1/3} + y^{1/3}}{x^{1/2} + y^{1/2}} \right)^{1/2}Let V(x, y) = \sin u. Check the degree of homogeneity n by substituting x \to tx and y \to ty:
V(tx, ty) = \left( \frac{(tx)^{1/3} + (ty)^{1/3}}{(tx)^{1/2} + (ty)^{1/2}} \right)^{1/2} = \left( \frac{t^{1/3}(x^{1/3} + y^{1/3})}{t^{1/2}(x^{1/2} + y^{1/2})} \right)^{1/2} V(tx, ty) = \left( t^{1/3 - 1/2} \right)^{1/2} V(x, y) = \left( t^{-1/6} \right)^{1/2} V(x, y) = t^{-1/12} V(x, y)Thus, V is a homogeneous function of degree n = -1/12.
Step 2: Apply Euler’s Theorem (First Order)
By Euler’s Theorem, x \frac{\partial V}{\partial x} + y \frac{\partial V}{\partial y} = nV. Since V = \sin u:
x (\cos u \frac{\partial u}{\partial x}) + y (\cos u \frac{\partial u}{\partial y}) = -\frac{1}{12} \sin uDividing by \cos u:
x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = -\frac{1}{12} \tan uLet G(u) = -\frac{1}{12} \tan u.
Step 3: Apply the Second-Order Extension
For any function satisfying x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = G(u), the following relation holds:
x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = G(u) [G'(u) - 1]First, calculate G'(u):
G'(u) = \frac{d}{du} \left( -\frac{1}{12} \tan u \right) = -\frac{1}{12} \sec^2 uNow substitute G(u) and G'(u) into the second-order relation:
\text{RHS} = \left( -\frac{1}{12} \tan u \right) \left[ -\frac{1}{12} \sec^2 u - 1 \right] = \frac{1}{12} \tan u \left[ \frac{\sec^2 u + 12}{12} \right]Using the identity \sec^2 u = 1 + \tan^2 u:
\text{RHS} = \frac{1}{144} \tan u (1 + \tan^2 u + 12) = \frac{1}{144} \tan u (13 + \tan^2 u)Final Conclusion:
This matches the target expression. Hence proved:
x^2\frac{\partial^2 u}{\partial x^2} + 2xy\frac{\partial^2 u}{\partial x \partial y} + y^2\frac{\partial^2 u}{\partial y^2} = \frac{1}{144}\tan u(13 + \tan^2 u)HPAS 2023 Maths Optional Paper-1 Question 4(b)
Find the magnitude and the equations of the shortest distance between the lines:
\frac{x}{2} = \frac{-y}{3} = \frac{z}{1} \quad \text{and} \quad \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}Solution:
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HPAS 2023 Maths Optional Paper-1 Question 5(a)
Determine the general and singular solutions of the differential equation 9p^2(2-y)^2 = 4(3-y), where p = \frac{dy}{dx}.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 5(b)
Solve the differential equation
x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \frac{xy}{2} = 0in terms of Bessel functions.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 5(c)
Using the method of variation of parameters, solve the differential equation (D^2 - 2D + 2)y = e^x \tan x, where D = \frac{d}{dx}.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 6(a)
Given \vec{F} = y\mathbf{i} - z^3\mathbf{j} + x^2\mathbf{k}, use Stokes’s theorem to evaluate \int_C \vec{F} \cdot d\vec{r}, where C is the boundary of the area S formed by the part of the plane x+4y+z=4 that lies in the first octant. The integration around the boundary C is in the clockwise direction.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 6(b)
Find the directional derivative of f(x,y,z) = x^2 + 3y^2 + 2z^2 in the direction of the vector 2\mathbf{i} - \mathbf{j} - 2\mathbf{k} and determine its value at the point (1, -3, 2).
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 7(a)
Let V be the vector space of real-valued functions y=f(x) satisfying
\frac{d^3y}{dx^3} - 6\frac{d^2y}{dx^2} + 11\frac{dy}{dx} - 6y = 0Then show that V is a 3-dimensional vector space over \mathbb{R}.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 7(b)
Three forces P, Q, and R act on a particle and keep it in equilibrium. If the angle between P and Q, and between Q and R, is 120^\circ each, then show that P=Q=R.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 8(a)
A particle performing Simple Harmonic Motion has a mass of 2.5 gm and a frequency of vibration of 10 Hz. It is oscillating with an amplitude of 2 cm. Calculate the total energy of the particle.
Solution:
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HPAS 2023 Maths Optional Paper-1 Question 8(b)
The motion of a particle under the influence of a central force is described by r = a \sin\theta. Find an expression for the force.
Solution:
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