Question 1 |
Let the probability density function of a random variable x be given as
f(x)=ae^{-2|x|}
The value of 'a' is _________
f(x)=ae^{-2|x|}
The value of 'a' is _________
0.5 | |
1 | |
1.5 | |
2 |
Question 1 Explanation:
f(x)=\left\{\begin{matrix}
ae^{2x} &x \lt 0 \\
ae^{-2x} &x \gt 0
\end{matrix}\right.
Therefore,
\begin{aligned} \int_{-\infty }^{\infty }f(x)dx&=1\\ \int_{-\infty }^{0}ae^{2x}dx+\int_{0 }^{\infty }ae^{-2x}dx&=1\\ a\left [ \left [ \frac{e^{2x}}{2} \right ]_{-\infty }^0 +[ \left [ \frac{e^{-2x}}{-2} \right ]^{\infty }_0\right ]&=1\\ a\left [ \frac{1}{2}+\frac{1}{2} \right ]&=1\\ a&=1 \end{aligned}
Therefore,
\begin{aligned} \int_{-\infty }^{\infty }f(x)dx&=1\\ \int_{-\infty }^{0}ae^{2x}dx+\int_{0 }^{\infty }ae^{-2x}dx&=1\\ a\left [ \left [ \frac{e^{2x}}{2} \right ]_{-\infty }^0 +[ \left [ \frac{e^{-2x}}{-2} \right ]^{\infty }_0\right ]&=1\\ a\left [ \frac{1}{2}+\frac{1}{2} \right ]&=1\\ a&=1 \end{aligned}
Question 2 |
Let \vec{E}(x,y,z)=2x^2\hat{i}+5y\hat{j}+3z\hat{k}. The value of \int \int \int _V(\vec{\triangledown }\cdot \vec{E})dV, where V is the
volume enclosed by the unit cube defined by 0\leq x\leq 1,0\leq y\leq 1 \text{ and }0\leq z\leq 1, is
3 | |
8 | |
10 | |
5 |
Question 2 Explanation:
Divergence of \vec{V}, \triangledown \cdot \vec{V}=4x+5+3=4x+8
Now,
\int \int \int( \triangledown \cdot \vec{V})dxdydz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(4x+8)dxdydz=\int_{0}^{1}(4x+8)dx=[2x^2+8x]_0^1=10
Now,
\int \int \int( \triangledown \cdot \vec{V})dxdydz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(4x+8)dxdydz=\int_{0}^{1}(4x+8)dx=[2x^2+8x]_0^1=10
Question 3 |
Let R be a region in the first quadrant of the xy plane enclosed by a closed curve C
considered in counter-clockwise direction. Which of the following expressions does
not represent the area of the region R?
\int \int _R dxdy | |
\oint _c xdy | |
\oint _c ydx | |
\frac{1}{2}\oint _c( xdy-ydx) |
Question 3 Explanation:
Using green theorem?s
\oint _cF_1dx+F_2dy=\int \int _R\left ( \frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\right )dxdy
Check all the options:
\oint xdy=\int \int _R\left ( \frac{\partial x}{\partial x} -0\right )dxdy=\int \int _Rdxdy
\frac{1}{2}\oint xdy-ydx=\frac{1}{2}\int \int _R(1+1)dxdy=\int \int _Rdxdy
\oint ydx=\int \int _R(-1)dxdy=-\int \int _Rdxdy
Hence, \oint ydx is not represent the area of the region.
\oint _cF_1dx+F_2dy=\int \int _R\left ( \frac{\partial F_2}{\partial x} -\frac{\partial F_1}{\partial y}\right )dxdy
Check all the options:
\oint xdy=\int \int _R\left ( \frac{\partial x}{\partial x} -0\right )dxdy=\int \int _Rdxdy
\frac{1}{2}\oint xdy-ydx=\frac{1}{2}\int \int _R(1+1)dxdy=\int \int _Rdxdy
\oint ydx=\int \int _R(-1)dxdy=-\int \int _Rdxdy
Hence, \oint ydx is not represent the area of the region.
Question 4 |
Let, f(x,y,z)=4x^2+7xy+3xz^2. The direction in which the function f(x,y,z) increases most rapidly at point P = (1,0,2) is
20\hat{i}+7\hat{j} | |
20\hat{i}+7\hat{j}+12\hat{k} | |
20\hat{i}+12\hat{j} | |
20\hat{i} |
Question 4 Explanation:
Given:
f(x,y,z)=4x^2+7xy+3xz^2
The directional derivative at point P is given by
=\triangledown f|_{point \; P}
\therefore \; \triangledown f=(8x+7y+3z^2)\hat{i}+(0+7x+0)\hat{j}+(0+0+6xz)\hat{k}
at point (1, 0, 2)
\triangledown f|_{(1,0,2)}=20\hat{i}+7\hat{j}+12\hat{k}
The directional derivative at point P is given by
=\triangledown f|_{point \; P}
\therefore \; \triangledown f=(8x+7y+3z^2)\hat{i}+(0+7x+0)\hat{j}+(0+0+6xz)\hat{k}
at point (1, 0, 2)
\triangledown f|_{(1,0,2)}=20\hat{i}+7\hat{j}+12\hat{k}
Question 5 |
Consider a matrix A=\begin{bmatrix}
1 & 0 & 0\\
0 & 4 & -2\\
0&1 &1
\end{bmatrix}
The matrix A satisfies the equation 6A^{-1}=A^2+cA+dI where c and d are scalars and I is the identity matrix.
Then (c+d) is equal to
The matrix A satisfies the equation 6A^{-1}=A^2+cA+dI where c and d are scalars and I is the identity matrix.
Then (c+d) is equal to
5 | |
17 | |
-6 | |
11 |
Question 5 Explanation:
Characteristic equation:
\begin{aligned} |A-\lambda I|&=0 \\ \begin{vmatrix} 1-\lambda & 0 &0 \\ 0 & 4-\lambda & 2\\ 0 & -1 & 1-\lambda \end{vmatrix}&=0 \\ (1-\lambda )[(4-\lambda )(1-\lambda )+2] &=0\\ \lambda ^3-6\lambda ^2+11\lambda -6&=0 \end{aligned}
By cayley hamilton theorem
\begin{aligned} A^3-6A^2+11A-6 &=0 \\ A^2-6A+11I&=6A^{-1} \end{aligned}
On comparison : c= -6 and d = 11
Therefore, c + d = -6 + 11 = 5
\begin{aligned} |A-\lambda I|&=0 \\ \begin{vmatrix} 1-\lambda & 0 &0 \\ 0 & 4-\lambda & 2\\ 0 & -1 & 1-\lambda \end{vmatrix}&=0 \\ (1-\lambda )[(4-\lambda )(1-\lambda )+2] &=0\\ \lambda ^3-6\lambda ^2+11\lambda -6&=0 \end{aligned}
By cayley hamilton theorem
\begin{aligned} A^3-6A^2+11A-6 &=0 \\ A^2-6A+11I&=6A^{-1} \end{aligned}
On comparison : c= -6 and d = 11
Therefore, c + d = -6 + 11 = 5
Question 6 |
Let f(x)=\int_{0}^{x}e^t(t-1)(t-2)dt . Then f(x) decreases in the interval
x \in (1,2) | |
x \in (2,3) | |
x \in (0,1) | |
x \in (0.5,1) |
Question 6 Explanation:
The function is decreasing, if
f'(x) \lt 0
f'(x)=\frac{d}{dx} \int_{0}^{x}e^t(t-1)(t-2)dt
\Rightarrow \; e^x(x-1)(x-2) \lt 0
It is possible in between 1 & 2. Hence x \in (1,2)
f'(x)=\frac{d}{dx} \int_{0}^{x}e^t(t-1)(t-2)dt
\Rightarrow \; e^x(x-1)(x-2) \lt 0
It is possible in between 1 & 2. Hence x \in (1,2)
Question 7 |
e^A denotes the exponential of a square matrix A. Suppose \lambda is an eigenvalue and v is the corresponding eigen-vector of matrix A.
Consider the following two statements:
Statement 1: e^\lambda is an eigenvalue of e^A.
Statement 2: v is an eigen-vector of e^A.
Which one of the following options is correct?
Consider the following two statements:
Statement 1: e^\lambda is an eigenvalue of e^A.
Statement 2: v is an eigen-vector of e^A.
Which one of the following options is correct?
Statement 1 is true and statement 2 is false. | |
Statement 1 is false and statement 2 is true | |
Both the statements are correct. | |
Both the statements are false. |
Question 7 Explanation:
Eigen value will change but eigen vector not
change.
Question 8 |
Consider a 3 x 3 matrix A whose (i,j)-th element, a_{i,j}=(i-j)^3. Then the matrix A will be
symmetric. | |
skew-symmetric. | |
unitary | |
null. |
Question 8 Explanation:
for \; i=j\Rightarrow a_{ij}=(i-i)^3=0\forall i
for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}
\therefore \; A_{3 \times 3 } is skew symmetric matrix.
for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}
\therefore \; A_{3 \times 3 } is skew symmetric matrix.
Question 9 |
Let A be a 10\times10 matrix such that A^{5} is a null matrix, and let I be the 10\times10 identity matrix. The determinant of \text{A+I} is ___________________.
1 | |
2 | |
4 | |
8 |
Question 9 Explanation:
\begin{aligned} \text{Given}:\qquad A^{5}&=0\\ A x &=\lambda x \\ \Rightarrow \qquad \qquad A^{5} x &=\lambda^{5} x \quad(\because x \neq 0) \\ \Rightarrow \qquad \qquad \lambda^{5} &=0 \\ \Rightarrow \qquad \qquad \lambda &=0 \end{aligned}
Eigen values of A+I given \lambda+1
\because Eigen values of I_{A}=1
Hence |A+I|= Product of eigen values =1 \times 1 \times 1 \times \ldots 10 times
=1
Eigen values of A+I given \lambda+1
\because Eigen values of I_{A}=1
Hence |A+I|= Product of eigen values =1 \times 1 \times 1 \times \ldots 10 times
=1
Question 10 |
Let \left ( -1 -j \right ), \left ( 3 -j \right ), \left ( 3 + j \right )
and \left ( -1+ j \right ) be the vertices of a rectangle C in the complex plane. Assuming that C is traversed in counter-clockwise direction, the value of the contour integral \oint _{C}\dfrac{dz}{z^{2}\left ( z-4 \right )} is
j\pi /2 | |
0 | |
-j\pi /8 | |
j\pi /16 |
Question 10 Explanation:
\oint_{C} \frac{d z}{z^{2}(z-4)}
Singularities are given by z^{2}(z-4)=0
\Rightarrow \qquad\qquad z=0,4
z=0 is pole of order m=2 lies inside contour 'c'
z=4 is pole of order m=1 lies outside 'c'
\begin{aligned} \text{Res}_{0} &=\frac{1}{(2-1) !} \text{lt}_{\rightarrow 0} \frac{d^{2-1}}{d z^{2-1}}\left[(z-0)^{2} \frac{1}{z^{2}(z-4)}\right] \\ &=\frac{-1}{(0.4)^{2}}=\frac{-1}{16}\\ \text{By CRT}\\ \oint_{C} f(z) d z &=2 \pi j \text{Res}_{0}=2 \pi j\left[\frac{-1}{16}\right] \\ &=\frac{-j \pi }{8} \end{aligned}
Singularities are given by z^{2}(z-4)=0
\Rightarrow \qquad\qquad z=0,4
z=0 is pole of order m=2 lies inside contour 'c'
z=4 is pole of order m=1 lies outside 'c'
\begin{aligned} \text{Res}_{0} &=\frac{1}{(2-1) !} \text{lt}_{\rightarrow 0} \frac{d^{2-1}}{d z^{2-1}}\left[(z-0)^{2} \frac{1}{z^{2}(z-4)}\right] \\ &=\frac{-1}{(0.4)^{2}}=\frac{-1}{16}\\ \text{By CRT}\\ \oint_{C} f(z) d z &=2 \pi j \text{Res}_{0}=2 \pi j\left[\frac{-1}{16}\right] \\ &=\frac{-j \pi }{8} \end{aligned}
There are 10 questions to complete.