HPAS 2014 Maths Optional Paper-1 Question 1(a)
Find rank of the matrix :
A=\begin{bmatrix}1&1&1&-1\\ 1&2&3&4\\ 3&4&5&2\end{bmatrix}Solution:
HPAS 2014 Maths Optional Paper-1 Question 1(b)
Show that the set :
W=\{(a,b,c):a-3b+4c=0; a,b,c\in R\} ,
is a subspace of the vector space V_{3}(R).
Solution:
HPAS 2014 Maths Optional Paper-1 Question 1(c)
Examine for continuity of the following function at x=0 :
f(x)=\begin{cases}\frac{x-|x|}{x}&,&x\ne0\\ 1&,&x=0\end{cases}Is it differentiable?
Solution:
HPAS 2014 Maths Optional Paper-1 Question 1(d)
Find the surface represented by the equation :
x^{2}+4y^{2}+z^{2}-4yz+2zx-4xy-2x+4y-2z-3=0Solution:
HPAS 2014 Maths Optional Paper-1 Question 1(e)
Solve :
(D^{4}+2D^{3}-3D^{2})y=x^{3}+2\sin xwhere D\equiv\frac{d}{dx}
Solution:
HPAS 2014 Maths Optional Paper-1 Question 1(f)
Two rods, each of length 2a, have their ends united at an angle \alpha, and are placed in a vertical plane on a sphere of radius r. Prove that the equilibrium is stable or unstable according as \sin \alpha > \text{ or } < \frac{2r}{a}.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 2(a)
Find the characteristic roots and characteristic vectors of the following matrix :
\begin{bmatrix}1&2&3\\ 0&-4&2\\ 0&0&7\end{bmatrix}Solution:
HPAS 2014 Maths Optional Paper-1 Question 2(b)
Solve the following equations using matrix method:
\begin{cases} 2x_{1}+3x_{2}+x_{3}=9 \\ x_{1}+2x_{2}+3x_{3}=6 \\ 3x_{1}+x_{2}+2x_{3}=8 \end{cases}Solution:
HPAS 2014 Maths Optional Paper-1 Question 2(c)
Prove that the mapping :
t:V_{2}(R)\rightarrow V_{3}(R),
which is defined by t(x,y)=(x,y,0) is a linear transformation from vector space V_{2}(R) to the vector space V_{3}(R).
Solution:
HPAS 2014 Maths Optional Paper-1 Question 3(a)
Given the sum of the perimeters of a square and a circle, show that the sum of their areas is least when the side of the square is equal to the diameter of the circle.
Solution:
Standard Result: Second Derivative Test for Single Variable Functions
Statement: For a continuous and twice-differentiable function f(x), the critical points occur where the first derivative is zero, i.e., f'(x) = 0. If the second derivative evaluated at that critical point is strictly positive (f''(x) > 0), then the function has a local minimum at that point.
Step 1: Define Variables and the Constraint
Let the side length of the square be x.
Let the radius of the circle be r.
The perimeter of the square is 4x, and the perimeter (circumference) of the circle is 2\pi r.
We are given that the sum of their perimeters is a constant, say P:
We can express x in terms of r and the constant P to create a single-variable problem:
4x = P - 2\pi r \implies x = \frac{P - 2\pi r}{4} \quad \dots \text{(Equation 1)}
Step 2: Formulate the Objective Function
The total area A is the sum of the area of the square (x^2) and the area of the circle (\pi r^2):
A = x^2 + \pi r^2Substitute Equation 1 into the area formula to make A a function solely of r:
A(r) = \left( \frac{P - 2\pi r}{4} \right)^2 + \pi r^2 A(r) = \frac{1}{16}(P - 2\pi r)^2 + \pi r^2Step 3: Find the Critical Point
To find the minimum area, we differentiate A(r) with respect to r and set it to zero (\frac{dA}{dr} = 0):
\frac{dA}{dr} = \frac{1}{16} \cdot 2(P - 2\pi r) \cdot (-2\pi) + 2\pi r \frac{dA}{dr} = -\frac{\pi}{4}(P - 2\pi r) + 2\pi rSet \frac{dA}{dr} = 0 to find the critical radius:
-\frac{\pi}{4}(P - 2\pi r) + 2\pi r = 0Divide the entire equation by \pi (since \pi \neq 0):
-\frac{1}{4}(P - 2\pi r) + 2r = 0 -\frac{P}{4} + \frac{\pi r}{2} + 2r = 0 r \left( \frac{\pi}{2} + 2 \right) = \frac{P}{4}Multiply by 4 to clear the denominator:
r (2\pi + 8) = PStep 4: Relate x and r at the Critical Point
Now, substitute the original constraint P = 4x + 2\pi r back into the critical point equation:
r(2\pi + 8) = 4x + 2\pi r 2\pi r + 8r = 4x + 2\pi rSubtract 2\pi r from both sides:
8r = 4x \implies x = 2rSince the diameter of a circle is twice its radius (d = 2r), we have found that at the critical point:
x = d
(The side of the square equals the diameter of the circle).
Step 5: Verify it is a Minimum Using the Second Derivative
We must confirm this critical point yields the least (minimum) area, not the maximum. We calculate the second derivative \frac{d^2A}{dr^2}:
\frac{dA}{dr} = -\frac{\pi P}{4} + \frac{\pi^2 r}{2} + 2\pi r \frac{d^2A}{dr^2} = \frac{d}{dr} \left( -\frac{\pi P}{4} + \frac{\pi^2 r}{2} + 2\pi r \right) \frac{d^2A}{dr^2} = 0 + \frac{\pi^2}{2} + 2\piSince \pi is a positive constant, \frac{\pi^2}{2} + 2\pi is strictly positive (> 0).
Because \frac{d^2A}{dr^2} > 0, the function A(r) is concave up, confirming that the critical point is indeed a minimum.
Final Conclusion:
The sum of the areas of the square and the circle is strictly minimized when x = 2r. That is, the total area is least when the side of the square is equal to the diameter of the circle.
Hence Proved.
HPAS 2014 Maths Optional Paper-1 Question 3(b)
Fully examine the nature of the origin on the curve :
x^{2}(x^{2}-4a^{2})=y^{2}(x^{2}-a^{2})and trace the curve.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 3(c)
Find the area of surface of a cone whose semi-vertical angle is \alpha and the base is a circle of radius r.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 4(a)
Evaluate :
\iint_{A}(xy)(x+y)dxdywhere region A is the area between the parabola y=x^{2} and the line y=x.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 4(b)
If u=\tan^{-1}\left(\frac{x^{3}+y^{3}}{x-y}\right), then prove that:
x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2uSolution:
HPAS 2014 Maths Optional Paper-1 Question 4(c)
Prove that the limit of \theta used in Lagrange’s mean value theorem tends to \frac{1}{2} when h\rightarrow0, provided f^{\prime\prime}(x) is continuous and f^{\prime\prime}(x)\ne0.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 5(a)
Find the equation of the sphere which passes through the points :
(1, 0, 0); (0, 1, 0) and (0, 0, 1),
and has its radius as small as possible.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 5(b)
What conic does equation :
13x^{2}-18xy+37y^{2}+2x+14y-2=0represent? Find its centre and the equation to the conic referred to the centre as origin.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 5(c)
Find length and the equations to the shortest distance between the following lines :
\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} \quad \text{and} \quad \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}Solution:
HPAS 2014 Maths Optional Paper-1 Question 6(a)
Discuss the solutions of the following equation :
x^{3}p^{2}+x^{2}yp+a^{3}=0where p\equiv\frac{d}{dx}
Solution:
HPAS 2014 Maths Optional Paper-1 Question 6(b)
Solve by the method of variation of parameters :
x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-y=x^{2}e^{x}Solution:
HPAS 2014 Maths Optional Paper-1 Question 6(c)
For Bessel function, prove that :
x J_{n}^{\prime}(x)=n J_{n}(x)-x J_{n+1}(x)Solution:
HPAS 2014 Maths Optional Paper-1 Question 7(a)
For Beta function, prove that :
B(m,n)=B(m+1,n)+B(m,n+1)Solution:
HPAS 2014 Maths Optional Paper-1 Question 7(b)
If
a=\sin \theta i+\cos \theta j+\theta k, \quad b=\cos \theta i-\sin \theta j-3k \quad \text{and} \quad c=2i+3j-kthen evaluate \frac{d}{d\theta}[a\times(b\times c)] for \theta=0.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 7(c)
Apply Gauss theorem to evaluate :
\int_{s}\{(x^{3}-yz)dydz-2x^{2}y dzdx+zdxdy\}over the surface of a cube bounded by the coordinate planes and the planes x=y=z=a.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 8(a)
The moments of a given system of coplanar forces about three points:
(2, 0), (0, 2) and (2, 2)
in their plane are 3, 4 and 10 units respectively. Find the magnitude of the resultant force and the equation of its line of action.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 8(b)
A particle moves with simple harmonic motion in a straight line. If in the first second after starting from rest it travels a distance a and in the next second it travels a distance b in the same direction, then find the amplitude and period of the motion.
Solution:
HPAS 2014 Maths Optional Paper-1 Question 8(c)
A particle of mass m is falling under gravity through a medium whose resistance is \mu times the velocity. If the particle is released from rest, show that the distance fallen through in time t is :
\frac{gm^{2}}{\mu^{2}}\left(e^{\frac{\mu t}{m}}-1+\frac{\mu t}{m}\right)