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HPAS 2016 Mathematics Optional Paper-1: Complete Solutions

Welcome to the comprehensive solution guide for the Himachal Pradesh Administrative Service (HPAS) 2016 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

HPAS 2016 Maths Optional Paper-1 Question 1(a)

Determine all values of d for which rank of the matrix:

\begin{pmatrix} d & -1 & 0 & 0 \\ 0 & d & -1 & 0 \\ 0 & 0 & d & -1 \\ -6 & 11 & -6 & 1 \end{pmatrix}

is equal to 3.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 1(b)

Show that the vectors v_1=(1,1,2,4), v_2 = (2, -1, -5, 2), v_3=(1,-1,-4,0) and v_4=(2,1,1,6) are linearly dependent in \mathbb{R}^4.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 1(c)

Evaluate:

\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{1/x^2}

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 1(d)

Find the angle between the surfaces:
x \log z = y^2 - 1 and x^2y = 2 - z
at point (1, 1, 1).

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 1(e)

Show that the Legendre polynomial P_n(x) satisfies P_n(-x)=(-1)^n P_n(x).

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 2(a)

Let T be a linear operator on \mathbb{R}^3 defined by
T(x,y,z)=(3x, x-y, 2x+y+z).
Show that T is invertible and determine T^{-1}.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 2(b)

Determine the value of a and b so that the system of equations:

\begin{cases} 2x+3y+5z=9 \\ 7x+3y-2z=8 \\ 2x+3y+az=b \end{cases}

has:
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 3(a)

Show that the function f defined on \mathbb{R} by:

f(x) = \begin{cases} x, & \text{if x is irrational} \\ -x, & \text{if x is rational} \end{cases}

is continuous only at x=0.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 3(b)

Find the asymptotes of the curve:
2x^3 - 5x^2y + 4xy^2 - y^3 + 6x^2 - 7xy + y^2 - x + 5y - 3 = 0.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 4(a)

Find the extreme values of f(x,y,z)=2x+3y+z such that x^2+y^2=5 and x+z=1.

Solution:

Standard Result: Method of Lagrange Multipliers (Multiple Constraints)

Statement: To find the extreme values of a function f(x,y,z) subject to two constraints g(x,y,z) = 0 and h(x,y,z) = 0, we form the Lagrangian function:
L(x,y,z,\lambda,\mu) = f(x,y,z) - \lambda g(x,y,z) - \mu h(x,y,z)
The critical points are found by setting the partial derivatives of L with respect to x, y, z to zero and solving the resulting system of equations alongside the original constraints.

Step 1: Set up the Lagrangian

Let the objective function be f(x,y,z) = 2x + 3y + z.
Let the constraints be g(x,y,z) = x^2 + y^2 - 5 = 0 and h(x,y,z) = x + z - 1 = 0.

The Lagrangian function is:
L = 2x + 3y + z - \lambda(x^2 + y^2 - 5) - \mu(x + z - 1)


Step 2: Find the partial derivatives and set them to zero

Take the partial derivatives of L with respect to x, y, and z:

  • \frac{\partial L}{\partial x} = 2 - 2\lambda x - \mu = 0
  • \frac{\partial L}{\partial y} = 3 - 2\lambda y = 0
  • \frac{\partial L}{\partial z} = 1 - \mu = 0

Step 3: Solve for x and y in terms of \lambda

From the third equation, we immediately get:
\mu = 1

Substitute \mu = 1 into the first equation:
2 - 2\lambda x - 1 = 0 \implies 1 - 2\lambda x = 0 \implies x = \frac{1}{2\lambda}

From the second equation, solve for y:
3 - 2\lambda y = 0 \implies y = \frac{3}{2\lambda}


Step 4: Use the constraints to find \lambda

Substitute the expressions for x and y into the first constraint equation, x^2 + y^2 = 5:
\left(\frac{1}{2\lambda}\right)^2 + \left(\frac{3}{2\lambda}\right)^2 = 5
\frac{1}{4\lambda^2} + \frac{9}{4\lambda^2} = 5
\frac{10}{4\lambda^2} = 5 \implies 20\lambda^2 = 10 \implies \lambda^2 = \frac{1}{2}
Taking the square root gives two possible values for \lambda:
\lambda = \pm \frac{1}{\sqrt{2}}


Step 5: Evaluate the critical points

Case 1: Let \lambda = \frac{1}{\sqrt{2}}

  • x = \frac{1}{2(1/\sqrt{2})} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}
  • y = \frac{3}{2(1/\sqrt{2})} = \frac{3\sqrt{2}}{2} = \frac{3}{\sqrt{2}}
  • From the second constraint (z = 1 - x): z = 1 - \frac{1}{\sqrt{2}}

Substitute these into the objective function f(x,y,z):
f = 2\left(\frac{1}{\sqrt{2}}\right) + 3\left(\frac{3}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right)
f = \frac{2}{\sqrt{2}} + \frac{9}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} + 1 = 5\sqrt{2} + 1

Case 2: Let \lambda = -\frac{1}{\sqrt{2}}

  • x = -\frac{1}{\sqrt{2}}
  • y = -\frac{3}{\sqrt{2}}
  • From the second constraint: z = 1 - \left(-\frac{1}{\sqrt{2}}\right) = 1 + \frac{1}{\sqrt{2}}

Substitute these into the objective function f(x,y,z):
f = 2\left(-\frac{1}{\sqrt{2}}\right) + 3\left(-\frac{3}{\sqrt{2}}\right) + \left(1 + \frac{1}{\sqrt{2}}\right)
f = -\frac{2}{\sqrt{2}} - \frac{9}{\sqrt{2}} + 1 + \frac{1}{\sqrt{2}} = -\frac{10}{\sqrt{2}} + 1 = -5\sqrt{2} + 1


Final Answer:

The extreme values of the function subject to the given constraints are:

  • Maximum Value: 1 + 5\sqrt{2}
  • Minimum Value: 1 - 5\sqrt{2}

HPAS 2016 Maths Optional Paper-1 Question 4(b)

Evaluate the integral:

\int_{0}^{2}\int_{0}^{y^2/2}\frac{y}{\sqrt{x^2+y^2+1}}dxdy

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 5(a)

Determine the general and singular solution of the non-linear differential equation:
y = xy' + (y')^2.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 5(b)

Solve the differential equation:

(D^2 - 2D + 2)y = e^x \tan x

where D = \frac{d}{dx}.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 6(a)

Find the directional derivative of the scalar function \phi = xy^2 + yz^3 at point (2, -1, 1) in the direction of the normal to the surface x \log z - y^2 = -4 at point (-1, 2, 1).

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 6(b)

Show that the field of force given by:
\vec{F} = (y^2 \cos x + z^3)\mathbf{i} + (2y \sin x - 4)\mathbf{j} + (3xz^2 + 2)\mathbf{k}
is conservative and find the work done in moving the particle in the field from a point A (0, 1, 1) to a point B (\pi/2, -1, 2).

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 7(a)

Show that the equation:
2x^2 - 6y^2 - 12z^2 + 18yz + 2zx + xy = 0
represents a pair of planes and find the angle between them.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 7(b)

Find the equations of the tangent planes to the hyperboloid 2x^2 - 6y^2 + 3z^2 = 5 which pass through the lines 3x-3y+6z-5=0 and x+9y-3z=0.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 8(a)

Let V be a finite dimensional inner product space over field F, and let g: V \to F be a linear transformation. Then show that there exists a unique vector y \in V such that g(x) = \langle x,y \rangle for all x \in V.

Solution:

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HPAS 2016 Maths Optional Paper-1 Question 8(b)

A particle moves in a plane in such a manner that its tangential and normal accelerations are always equal and its velocity varies as e^{\tan^{-1}(s/c)}, s being the length of the arc of the curve measured from a fixed point on the curve. Find the path.

Solution:

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