Question : The state variable formulation of a system is given as \[\begin{bmatrix}\dot{x_1} \\ \dot{x_2} \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u, \quad x_1(0)=0,\,x_2(0)=0, \quad y=\begin{bmatrix}1 & 0\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}.\] The system is
controllable but not observable
not controllable but observable
both controllable and observable
both not controllable and not observable
Solution:
The system has matrices \(A=\begin{pmatrix}-2&0\\0&-1\end{pmatrix}\), \(B=\begin{pmatrix}1\\1\end{pmatrix}\), \(C=\begin{pmatrix}1&0\end{pmatrix}\).
Controllability matrix: \[Q_c=[B\; AB]=\begin{bmatrix}1 & -2 \\ 1 & -1\end{bmatrix},\qquad \det(Q_c)=1\neq 0.\] Hence controllable.
Observability matrix: \[Q_o=\begin{bmatrix} C \\ CA \end{bmatrix} =\begin{bmatrix} 1&0 \\ -2&0 \end{bmatrix},\qquad \det(Q_o)=0.\] Hence not observable.
Correct option: A.
Question : A function \(y=5x^2+10x\) is defined over \((1,2)\). At least at one point in the interval, \(\frac{dy}{dx}\) is
\(20\)
\(25\)
\(30\)
\(35\)
Solution:
By Mean Value Theorem: \[f'(c)=\frac{f(2)-f(1)}{2-1}.\]
\(f(1)=15,\ f(2)=40\) Thus, \[f'(c)=40-15=25.\]
Correct option: B.
Question : \(\displaystyle \int \frac{z^2-4}{z^2+4}\,dz\) evaluated anticlockwise around \(|z-i|=2\) is
\(-4\pi\)
0
\(2+\pi\)
\(2+2i\)
Solution:
Poles at \(z=\pm 2i\). Inside \(|z-i|=2\) only \(z=2i\) lies.
Residue at \(z=2i\): \[\text{Res}= \lim_{z\to 2i}\frac{z^2-4}{z+2i} =\frac{(2i)^2-4}{4i}=\frac{-8}{4i}=2i.\]
Integral: \[I=2\pi i(2i)=4\pi i^2=-4\pi.\]
Correct option: A.
Question : A matrix has eigenvalues \(-1\) and \(-2\) with eigenvectors \(\begin{bmatrix}1\\-1\end{bmatrix}\) and \(\begin{bmatrix}1\\-2\end{bmatrix}\). The matrix is
\(\begin{bmatrix}1 & 1 \\ -1 & -2\end{bmatrix}\)
\(\begin{bmatrix}1 & 2 \\ -2 & 4\end{bmatrix}\)
\(\begin{bmatrix}-1 & 0 \\ 0 & -2\end{bmatrix}\)
\(\begin{bmatrix}0 & 1 \\ -2 & -3\end{bmatrix}\)
Solution:
Solving \(A x=\lambda x\) gives \(a=0,\ b=1,\ c=-2,\ d=-3\).
Thus \[A=\begin{pmatrix}0&1\\-2&-3\end{pmatrix}.\]
Correct option: D.
Question : Given \(\mathbf{F}=y^2x\,\mathbf{a_x}-yz\,\mathbf{a_y}-x^2\,\mathbf{a_z}\), evaluate \(\int \mathbf{F}\cdot d\mathbf{l}\) along the \(x\)-axis from \(x=1\) to \(x=2\).
\(-2.33\)
0
2.33
7
Solution:
Along \(x\)-axis: \(y=0,\ z=0,\ dy=0,\ dz=0\).
Thus \[\mathbf{F}\cdot d\mathbf{l} = (0)dx=0.\]
Integral: \[I=\int_1^2 0\,dx=0.\]
Correct option: B.
Question : \(\begin{bmatrix}2 & -2\\1 & -1\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}=0\) has
no solution
only the zero solution
non-zero unique solution
multiple solutions
Solution:
Determinant: \[\det(A)=2(-1)-(-2)(1)=0.\]
Rank \(=1<2\) → infinite solutions.
Solutions: \((x_1,x_2)=(k,k)\).
Correct option: D.
Question : Square roots of \(-i\) are
\(i,-i\)
\(\cos(-\pi/4)+i\sin(-\pi/4)\) and \(\cos(3\pi/4)+i\sin(3\pi/4)\)
\(\cos(\pi/4)+i\sin(3\pi/4)\) and \(\cos(3\pi/4)+i\sin(\pi/4)\)
\(\cos(3\pi/4)+i\sin(-\pi/4)\) and \(\cos(-\pi/4)+i\sin(3\pi/4)\)
Solution:
\(-i=e^{-i\pi/2}\). Square roots:
\[z=e^{-i\pi/4},\qquad z=e^{i3\pi/4}.\]
Correct option: B.
Question : A continuous RV \(X\) has pdf \(f(x)=e^{-x}\) for \(x>0\). Find \(P(X>1)\).
\(1/e\)
\(e\)
1
Data insufficient
Solution:
\[P(X>1)=\int_1^\infty e^{-x}\,dx=e^{-1}.\]
Correct option: A.