Question 1 |
The value of the integral \iint_{R} x y d x d y over the region R, given in the figure, is ___
(rounded off to the nearest integer).
(rounded off to the nearest integer).
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
\begin{aligned} I & =\iint_{R} x y d x d y \\ & =\int_{y=0}^{1} \int_{x=-y}^{y} x y d x d y+\int_{y=1}^{2} \int_{x=y-2}^{2-y} x y d x d y \\ & =\int_{0}^{1} y\left(\frac{x^{2}}{2}\right)_{-y}^{y} d y+\int_{1}^{2} y\left(\frac{x^{2}}{2}\right)_{y-2}^{2-y} d y \\ & =0+0=0 \end{aligned}
Question 2 |
The state equation of a second order system is
\dot{x}(t)=A x(t), x(0) is the initial condition.
Suppose \lambda_{1} and \lambda_{2} are two distinct eigenvalues of A and v_{1} and v_{2} are the corresponding eigenvectors. For constants \alpha_{1} and \alpha_{2}, the solution, x(t), of the state equation is
\dot{x}(t)=A x(t), x(0) is the initial condition.
Suppose \lambda_{1} and \lambda_{2} are two distinct eigenvalues of A and v_{1} and v_{2} are the corresponding eigenvectors. For constants \alpha_{1} and \alpha_{2}, the solution, x(t), of the state equation is
\sum_{i=1}^{2} \alpha_{i} e^{\lambda_{i} t} v_{i} | |
\sum_{i=1}^{2} \alpha_{i} e^{2 \lambda_{i} t} v_{i} | |
\sum_{i=1}^{2} \alpha_{i} e^{3 \lambda_{i} t} v_{i} | |
\sum_{i=1}^{2} \alpha_{i} e^{4 \lambda_{i} t} v_{i} |
Question 3 |
Let x be an n \times 1 real column vector with length l=\sqrt{x^{T} x}. The trace of the matrix P=x x^{T} is
l^{2} | |
\frac{l^{2}}{4} | |
l | |
\frac{l^{2}}{2} |
Question 3 Explanation:
Given,
l=\sqrt{x^{T} x}, P=\left(x x^{T}\right)_{n \times n}
Let
\begin{aligned} (x)_{n \times 1} & =\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{n} \end{array}\right] \\ l & =\sqrt{x^{T} x}=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\ldots x_{n}^{2}} \\ P & =x x^{T} \\ &=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \cdot \\ x_{n} \end{array}\right]\left[x_{1} x_{2} x_{3} \ldots x_{n}\right]\\ P&=\left[ \begin{array}{lllll} x_{1}^{2} & & & & \\ & x_{1}^{2} & & & \\ & & - & & \\ & & & - & \\ & & & & x_{n}^{2} \end{array} \right] \end{aligned}
Trace of P=x_{1}^{2}+x_{2}^{2}+ . . .+ x_{n}^{2}=l^2
l=\sqrt{x^{T} x}, P=\left(x x^{T}\right)_{n \times n}
Let
\begin{aligned} (x)_{n \times 1} & =\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{n} \end{array}\right] \\ l & =\sqrt{x^{T} x}=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\ldots x_{n}^{2}} \\ P & =x x^{T} \\ &=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ \cdot \\ x_{n} \end{array}\right]\left[x_{1} x_{2} x_{3} \ldots x_{n}\right]\\ P&=\left[ \begin{array}{lllll} x_{1}^{2} & & & & \\ & x_{1}^{2} & & & \\ & & - & & \\ & & & - & \\ & & & & x_{n}^{2} \end{array} \right] \end{aligned}
Trace of P=x_{1}^{2}+x_{2}^{2}+ . . .+ x_{n}^{2}=l^2
Question 4 |
The value of the line integral \int_{P}^{Q}\left(z^{2} d x+3 y^{2} d y+2 x z d z\right) along the straight line joining the points P(1,1,2) and Q(2,3,1) is
20 | |
24 | |
29 | |
-5 |
Question 4 Explanation:
\int_{P}^{Q} z^{2} d x+3 y^{2} d y+2 x z d z along the line joining the points P(1,1,2) and Q(2,3,1) is
\begin{aligned} & =\int_{P(1,2)}^{P(2,1)} z^{2} d x+2 x y d z+\int_{y=1}^{3} 3 y^{2} d y \\ & =\left(x z^{2}\right)_{(1,2)}^{(2,1)}+\left(y^{3}\right)_{1}^{3} \\ & =\left(2 \times 1^{2}-1 \times 2^{2}\right)+\left(3^{3}-1^{3}\right) \\ & =-2+26=24 \end{aligned}
\begin{aligned} & =\int_{P(1,2)}^{P(2,1)} z^{2} d x+2 x y d z+\int_{y=1}^{3} 3 y^{2} d y \\ & =\left(x z^{2}\right)_{(1,2)}^{(2,1)}+\left(y^{3}\right)_{1}^{3} \\ & =\left(2 \times 1^{2}-1 \times 2^{2}\right)+\left(3^{3}-1^{3}\right) \\ & =-2+26=24 \end{aligned}
Question 5 |
A random variable X, distributed normally as N(0,1), undergoes the transformation Y=h(X), given in the figure. The form of the probability density function of Y is (In the options given below, a, b, c are non-zero constants and g(y) is piece-wise continuous function)
a \delta(y-1)+b \delta(y+1)+g(y) | |
a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y) | |
a \delta(y+2)+b \delta(y)+c \delta(y-2)+g(y) | |
a \delta(y+1)+b \delta(y-2)+g(y) |
Question 5 Explanation:
X=N(0,1)
f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}
\begin{aligned} Y= & -1 ; x \leq-2 \\ & 0 ;-1 \leq x \leq 1 \\ & 1 ; x \geq 2 \\ & x+1 ;-2 \leq x \leq-1 \\ & x-1 ; 1 \leq x \leq 2 \end{aligned}
Y is taking discrete set of values and a continuous range of values, so it is mixed random variable.
From the given options, density function of 'Y' will be.
f_{Y}(y)=a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y)
f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}
\begin{aligned} Y= & -1 ; x \leq-2 \\ & 0 ;-1 \leq x \leq 1 \\ & 1 ; x \geq 2 \\ & x+1 ;-2 \leq x \leq-1 \\ & x-1 ; 1 \leq x \leq 2 \end{aligned}
Y is taking discrete set of values and a continuous range of values, so it is mixed random variable.
From the given options, density function of 'Y' will be.
f_{Y}(y)=a \delta(y+1)+b \delta(y)+c \delta(y-1)+g(y)
There are 5 questions to complete.