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HPAS 2020 Mathematics Optional Paper-1: Complete Solutions

Welcome to the comprehensive solution guide for the Himachal Pradesh Administrative Service (HPAS) 2020 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 2020 Maths Optional Paper-1 Question 1(a)

Obtain the general solution of the following differential equation:
x\frac{dy}{dx}+(2-x)y=e^{3x}, for x > 0.

Solution:

Enter solution here

HPAS 2020 Maths Optional Paper-1 Question 1(b)

Find a linear transformation T:\mathbb{R}^{3}\to\mathbb{R}^{3} such that its image space is the plane x+y+z=0.

Solution:

Standard Result: Image Space and Column Space

Statement: For any linear transformation T:\mathbb{R}^{n}\to\mathbb{R}^{m} represented by standard matrix A such that T(v) = Av, the image space (or range) of T is exactly equal to the column space of A. Therefore, to construct a transformation with a specific image space, the columns of its transformation matrix must form a spanning set for that space.

Step 1: Find a basis for the target image space

The target image space is the plane given by the equation:
x + y + z = 0

We can express one variable in terms of the other two, for instance, z = -x - y.
Any vector v in this plane can be written in the form:
v = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x \\ y \\ -x-y \end{pmatrix}

We can decompose this into a linear combination of two independent vectors by factoring out the free variables x and y:
v = x \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}

Let v_1 = (1, 0, -1)^T and v_2 = (0, 1, -1)^T. These two linearly independent vectors form a basis for the plane x+y+z=0.


Step 2: Construct the Transformation Matrix A

We need a 3 \times 3 matrix A whose column space is spanned by v_1 and v_2. We can achieve this by assigning v_1 and v_2 as the first two columns of A.

To ensure the transformation maps \mathbb{R}^3 exactly to this 2-dimensional plane (meaning the rank of A must be 2), the third column must also lie in the plane. The simplest choice is the zero vector (0,0,0)^T, which trivially lies in every subspace.

Thus, our transformation matrix A is:
A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 0 \end{pmatrix}


Step 3: Define the Linear Transformation

The linear transformation T(x,y,z) is obtained by multiplying the standard matrix A by a generic input vector:

T\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x \\ y \\ -x-y \end{pmatrix}

Step 4: Verification

Let the output of the transformation be (x', y', z') = (x, y, -x-y).
Substitute these into the equation of the plane:
x' + y' + z' = x + y + (-x-y) = 0
Since every output strictly satisfies the equation of the plane, and the rank of the transformation is 2 (matching the dimension of the plane), the image space is exactly the plane x+y+z=0.


Final Answer:

A valid linear transformation satisfying the condition is:
T(x, y, z) = (x, y, -x-y)

HPAS 2020 Maths Optional Paper-1 Question 1(c)

If f:[0,6] \to \mathbb{R} is a continuous function such that f(0)=f(6), then show that f(x)=f(x+3) for some x in [0,3].

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 1(d)

Find the velocity of a particle moving on the surface of a right circular cylinder of radius b.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 2(a)

Define a basis of a vector space. Find a basis of the subspace of the vector space \mathbb{R}^4(\mathbb{R}) generated by the subset:

\{(1,1,0,-1), (2,4,6,0), (-2,-3,-3,1), (-1,-2,-2,2), (4,6,4,-6)\}

Solution:

Part 1: Definition of a Basis

A subset B of a vector space V over a field F is called a basis of V if it satisfies two fundamental conditions:

  1. Linear Independence: No vector in B can be written as a linear combination of the other vectors in B. Mathematically, if c_1 v_1 + c_2 v_2 + \dots + c_k v_k = \mathbf{0} for v_i \in B, then c_1 = c_2 = \dots = c_k = 0.
  2. Spanning: Every vector in the vector space V can be expressed as a finite linear combination of the vectors in B. That is, \text{span}(B) = V.

Part 2: Finding the Basis of the Subspace

Standard Result: Row Space and Basis

Statement: The subspace generated by a set of vectors is identical to the row space of the matrix formed by making these vectors its rows. Elementary row operations do not change the row space of a matrix. Therefore, the non-zero rows of the row echelon form of this matrix form a basis for the subspace generated by the original vectors.


Step 1: Construct the Matrix

Let A be the matrix whose rows are the given vectors:

A = \begin{pmatrix} 1 & 1 & 0 & -1 \\ 2 & 4 & 6 & 0 \\ -2 & -3 & -3 & 1 \\ -1 & -2 & -2 & 2 \\ 4 & 6 & 4 & -6 \end{pmatrix}

Step 2: Apply Elementary Row Operations to reach Row Echelon Form

Use R_1 to eliminate the first entries of the rows below it:
R_2 \to R_2 - 2R_1
R_3 \to R_3 + 2R_1
R_4 \to R_4 + R_1
R_5 \to R_5 - 4R_1

\sim \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 2 & 6 & 2 \\ 0 & -1 & -3 & -1 \\ 0 & -1 & -2 & 1 \\ 0 & 2 & 4 & -2 \end{pmatrix}

Simplify R_2 by dividing by 2 (R_2 \to \frac{1}{2}R_2):

\sim \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -3 & -1 \\ 0 & -1 & -2 & 1 \\ 0 & 2 & 4 & -2 \end{pmatrix}

Use the new R_2 to eliminate the second entries of the rows below it:
R_3 \to R_3 + R_2
R_4 \to R_4 + R_2
R_5 \to R_5 - 2R_2

\sim \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & -2 & -4 \end{pmatrix}

Swap R_3 and R_4 to maintain the echelon staircase structure:

\sim \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -2 & -4 \end{pmatrix}

Use the new R_3 to eliminate the third entry of R_5:
R_5 \to R_5 + 2R_3

\sim \begin{pmatrix} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}

Step 3: Identify the Basis

The matrix is now in row echelon form. The non-zero rows are linearly independent and span the exact same subspace as the original five vectors.

Final Answer:

A basis for the subspace generated by the given subset is the set of vectors:

\{(1, 1, 0, -1), (0, 1, 3, 1), (0, 0, 1, 2)\}

(Note: The dimension of this subspace is 3).

HPAS 2020 Maths Optional Paper-1 Question 2(b)

Let A = \begin{pmatrix} 2 & 2 \\ 1 & 3 \end{pmatrix} . Find an invertible 2 \times 2 matrix P such that PAP^{-1} is a diagonal matrix.

Solution:

Standard Result: Diagonalization Theorem

Statement: An n \times n matrix A is diagonalizable if there exists an invertible matrix P such that PAP^{-1} = D, where D is a diagonal matrix containing the eigenvalues of A. The matrix P is constructed by using the eigenvectors of A as its columns in the same order as their corresponding eigenvalues in D.

Step 1: Find the Eigenvalues

The eigenvalues \lambda are the solutions to the characteristic equation \det(A - \lambda I) = 0:

\det \begin{pmatrix} 2-\lambda & 2 \\ 1 & 3-\lambda \end{pmatrix} = 0 (2-\lambda)(3-\lambda) - 2 = 0 \lambda^2 - 5\lambda + 6 - 2 = 0 \implies \lambda^2 - 5\lambda + 4 = 0 (\lambda - 4)(\lambda - 1) = 0

The eigenvalues are \lambda_1 = 4 and \lambda_2 = 1.


Step 2: Find the Eigenvectors

For \lambda_1 = 4: Solve (A - 4I)v = 0:

\begin{pmatrix} 2-4 & 2 \\ 1 & 3-4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} -2 & 2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0

This implies x_1 = x_2. A corresponding eigenvector is v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.

For \lambda_2 = 1: Solve (A - 1I)v = 0:

\begin{pmatrix} 2-1 & 2 \\ 1 & 3-1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0

This implies x_1 = -2x_2. A corresponding eigenvector is v_2 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}.


Step 3: Construct the Matrix P

The matrix P consists of the eigenvectors as its columns.

P = \begin{pmatrix} v_1 & v_2 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}

Final Answer:

The invertible matrix P is:

P = \begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}

(Note: You can verify that PAP^{-1} results in the diagonal matrix D = \begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}).

HPAS 2020 Maths Optional Paper-1 Question 3(a)

Find the locus of the point of intersection of three mutually perpendicular tangent planes to ax^2+by^2+cz^2=1.

Solution:

Enter solution here

HPAS 2020 Maths Optional Paper-1 Question 3(b)

Show that \lim_{x\to0} \cos\frac{1}{x} does not exist.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 3(c)

Evaluate \lim_{x\to\infty}(\sqrt{x^2+3x}-x) and \lim_{x\to0^{+}}(1-\sin x)^{1/x}.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 4(a)

State a set of sufficient conditions for a local maximum or minimum at a point for a twice continuously differentiable function f(x,y). Test the function f(x,y)=x^3+y^3-9xy+1 for local maximum or minimum.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 4(b)

Let f:\mathbb{R}^2 \to \mathbb{R}^2 be defined by f(x,y)=(x^2+y^2, xy). Compute the total derivative of f at the point (1,2).

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 4(c)

Reduce the following equation to the standard form:

3x^2+5y^2+3z^2+2yz+2zx+2xy-4x-8z+5=0

Find the nature of the conicoid, its center, and the equations of its axes.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 5(a)

Find the volume of the solid region that is interior to both the sphere x^2+y^2+z^2=4 and the cylinder (x-1)^2+y^2=1.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 5(b)

A particle of mass m moves with a central attractive force \mu(r^5-c^4r) towards the origin. It is projected from an apse at distance c with velocity \sqrt{\frac{2\mu}{3}}c^3. Show that the equation of the central orbit is x^4+y^4=c^4.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 6(a)

State the statement of Stokes’ theorem and verify it for the line integral

\oint_C [(x+y)dx + (2x-z)dy + (y+z)dz]

where C is the boundary of the triangle with vertices (2,0,0), (0,3,0), and (0,0,6).

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 6(b)

Define a wrench of a system of forces. Three forces P, Q, and R act along the three straight lines x=0, y-z=a; y=0, z-x=a; and z=0, x-y=a respectively. Find the vector symmetrical equation of the central axis and the pitch of the equivalent wrench.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 7(a)

The tangential acceleration of a particle moving along a circle of radius a is \lambda times the normal acceleration. If its speed at a certain time is u, then prove that it will return to the same point after a time \frac{a}{\lambda u}(1-e^{-2\pi\lambda}).

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 7(b)

Solve the following differential equation:
x\frac{d^2y}{dx^2} - \frac{dy}{dx} = x^2e^x, for x > 0.

Solution:

Enter solution here

HPAS 2020 Maths Optional Paper-1 Question 8(a)

A heavy uniform rod rests with one end against a smooth vertical wall and with a point in its length resting on a smooth peg. Find the position of equilibrium, and show that it is unstable.

Solution:

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HPAS 2020 Maths Optional Paper-1 Question 8(b)

Apply the method of power series to solve the following differential equation:

(1-x^2)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 12y = 0

Solution:

Enter solution here

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