Question 1 |
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 1 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 2 |
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 2 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}
Question 3 |
Let p\left ( z\right )=z^{3}+\left ( 1+j \right )z^{2}+\left ( 2+j \right )z+3, where z is a complex number.
Which one of the following is true?
Which one of the following is true?
\text{conjugate}\:\left \{ p\left ( z \right ) \right \}=p\left ( \text{conjugate} \left \{ z \right \} \right ) for all z | |
The sum of the roots of p\left ( z \right )=0 is a real number | |
The complex roots of the equation p\left ( z \right )=0
come in conjugate pairs | |
All the roots cannot be real |
Question 3 Explanation:
Since sum of the roots is a complex number
\Rightarrow absent one root is complex
So all the roots cannot be real.
\Rightarrow absent one root is complex
So all the roots cannot be real.
Question 4 |
The real numbers, x and y with y = 3x^2 + 3x + 1, the maximum and minimum value
of y for x \in [-2, 0] are respectively ________
7 and 1/4 | |
7 and 1 | |
-2 and -1/2 | |
1 and 1/4 |
Question 4 Explanation:
\begin{aligned}
y&=3x^{2}+3x+1 \; \; in \; [-2,0] \\ \frac{\partial y}{\partial x}&=6x+3,\; \; \frac{\partial^2 y}{\partial x^2}=6 \\ \frac{\mathrm{d} y}{\mathrm{d} x}&=0\\ 6x+3&=0 \\ x&=\frac{-1}{2} \\ \frac{d^{2} y}{dx^{2}}&=6 \gt 0\text{ minimum} \end{aligned}
Maximum value of y in [-2, 0] is maximum {f(-2), f(0)}
max{7, 1} = 7
Minimum value of y in [-2, 0]
min\begin{Bmatrix} f(-2) & f(0) &f(-\frac{1}{2}) \\ \downarrow, &\downarrow, &\downarrow \\ 7& 1 & \frac{1}{4} \end{Bmatrix}+=\frac{1}{4}
Maximum value 7, minimum value \frac{1}{4}
Maximum value of y in [-2, 0] is maximum {f(-2), f(0)}
max{7, 1} = 7
Minimum value of y in [-2, 0]
min\begin{Bmatrix} f(-2) & f(0) &f(-\frac{1}{2}) \\ \downarrow, &\downarrow, &\downarrow \\ 7& 1 & \frac{1}{4} \end{Bmatrix}+=\frac{1}{4}
Maximum value 7, minimum value \frac{1}{4}
Question 5 |
The value of the following complex integral, with C representing the unit circle centered
at origin in the counterclockwise sense, is:
\int_{c}\frac{z^2+1}{z^2-2z}dz
\int_{c}\frac{z^2+1}{z^2-2z}dz
8 \pi i | |
-8 \pi i | |
- \pi i | |
\pi i |
Question 5 Explanation:
\begin{aligned} I&=\int _C \frac{z^2+1}{z^2-2z}dz\;\;\;|z|=1 \\ \text{Using } & \text{Cauchy's integral theorem}\\ \int _C\frac{F(z)}{z-a}dz&=2 \pi i (Re_{(z=a)})\;\;\;...(i)\\ I&=\int _C \frac{z^2+1}{z(z-2)}dz \end{aligned}
Poles are at z=0 and 2 but only z=0 lies inside the unit circle.
Residue at (z=0)=\lim_{z \to 0}\frac{z^2+1}{z(z-2)}
Re_{(z=0)}=-\frac{1}{2}
Using equation (i)
\int _C \frac{z^2+1}{z^2-2z}dz=2 \pi i \times \left ( \frac{-1}{2} \right )=-\pi i
Poles are at z=0 and 2 but only z=0 lies inside the unit circle.
Residue at (z=0)=\lim_{z \to 0}\frac{z^2+1}{z(z-2)}
Re_{(z=0)}=-\frac{1}{2}
Using equation (i)
\int _C \frac{z^2+1}{z^2-2z}dz=2 \pi i \times \left ( \frac{-1}{2} \right )=-\pi i
There are 5 questions to complete.