Question 1 |
ax^3+bx^2+cx+d is a polynomial on real x over real coefficients a, b, c, d wherein
a \neq 0. Which of the following statements is true?
d can be chosen to ensure that x = 0 is a root for any given set a, b, c. | |
No choice of coefficients can make all roots identical. | |
a, b, c, d can be chosen to ensure that all roots are complex. | |
c alone cannot ensure that all roots are real. |
Question 1 Explanation:
Given Polynomial ax^{3}+bx^{2}+cx+d=0;\; \; \; a\neq 0
Option (A):
If d=0, then the polynomial equation becomes
\begin{aligned} ax^3+bx^2+cx&=0\\ x(ax^2+bx+c)&=0 \\ x=0 \text{ or } ax^2+bx+c&=0 \end{aligned}
d can be choosen to ensure x=0 is a root of given polynomial.
Hence, Option (A) is correct.
Option B:
A third degree polynomial equation with all root equal is given by
(x+\alpha )^3=0
Thus, by selecting suitable values of a, b, c and d we can have all roots identical.
Hence, option (B) is incorrect.
Option (C): Complex roots always occurs in pairs,
So, the given polynomial will have maximum of 2 complex roots and 1 real root.
Hence, option (C) is incorrect.
Option (D): Nature or roots depends on other coefficients also apart from coefficient 'c'.
Hence, option (D) is correct.
Hence, the correct options are (A) and (D).
Option (A):
If d=0, then the polynomial equation becomes
\begin{aligned} ax^3+bx^2+cx&=0\\ x(ax^2+bx+c)&=0 \\ x=0 \text{ or } ax^2+bx+c&=0 \end{aligned}
d can be choosen to ensure x=0 is a root of given polynomial.
Hence, Option (A) is correct.
Option B:
A third degree polynomial equation with all root equal is given by
(x+\alpha )^3=0
Thus, by selecting suitable values of a, b, c and d we can have all roots identical.
Hence, option (B) is incorrect.
Option (C): Complex roots always occurs in pairs,
So, the given polynomial will have maximum of 2 complex roots and 1 real root.
Hence, option (C) is incorrect.
Option (D): Nature or roots depends on other coefficients also apart from coefficient 'c'.
Hence, option (D) is correct.
Hence, the correct options are (A) and (D).
Question 2 |
Which of the following is true for all possible non-zero choices of integers m, n; m \neq n,
or all possible non-zero choices of real numbers p, q ; p\neq q, as applicable?
\frac{1}{\pi}\int_{0}^{\pi}\sin m\theta \sin n\theta \; d\theta =0 | |
\frac{1}{2\pi}\int_{-\pi/2}^{\pi/2}\sin p\theta \sin q\theta \; d\theta =0 | |
\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin p\theta \cos q\theta \; d\theta =0 | |
\lim_{\alpha \to \infty }\frac{1}{2\alpha }\int_{-\alpha }^{\alpha }\sin p\theta \sin q\theta \; d\theta =0 |
Question 2 Explanation:
\begin{aligned} \because \; p& \neq q\\ &\frac{1}{2\pi}\int_{-\pi}^{\pi} \sin p\theta \cos q\theta d\theta \\ &=\frac{1}{2\pi}\cdot \frac{1}{2}\int_{-\pi}^{\pi} [\sin (p+q)\theta + \sin (p-q)\theta] d\theta \\ &=\frac{1}{4\pi}\left [ \frac{-1}{(p+q)}\cos (p+q)\theta -\frac{1}{(p-q)}\cos (p-q)\theta \right ]_{-\pi}^{\pi}\\ &=\frac{-1}{4\pi} \left \{ \frac{1}{(p+q)}(\cos (p+q) \pi -\cos (p+q)(-\pi)) \right.\\ &+\left. \frac{1}{(p-q)}(\cos (p-q) \pi -\cos (p-q)(-\pi)) \right \}\\ &=0 \end{aligned}
Question 3 |
Which of the following statements is true about the two sided Laplace transform?
It exists for every signal that may or may not have a Fourier transform. | |
It has no poles for any bounded signal that is non-zero only inside a finite time
interval. | |
The number of finite poles and finite zeroes must be equal. | |
If a signal can be expressed as a weighted sum of shifted one sided exponentials,
then its Laplace Transform will have no poles. |
Question 3 Explanation:
It has no poles for any bounded signal that is nonzero in a finite time interval. This is true as we know for finite amplitude finite width signal ROC is entire s plane and ROC never includes any pole.
It implies for such signals there is no poles. Hence the correct answer is option (B).
It implies for such signals there is no poles. Hence the correct answer is option (B).
Question 4 |
Consider a signal x[n]=\left ( \frac{1}{2} \right )^n \; 1[n], where 1[n]=0 if n \lt 0, and 1[n]=1 if n \geq 0. The
z-transform of x[n-k], k \gt 0 is \frac{z^{-k}}{1-\frac{1}{2}z^{-1}}
with region of convergence being
|z| \lt 2 | |
|z| \gt 2 | |
|z| \lt 1/2 | |
|z| \gt 1/2 |
Question 4 Explanation:
\begin{aligned}x(n)&=\left (\frac{1}{2} \right )^{n} u(n)
, \; \; \; \text{ROC of }x(n):\left | z \right | \gt \frac{1}{2} \\ x(n-k)\rightleftharpoons X(z)&=\frac{z^{-k}}{1-\frac{1}{2}z^{-1}}
, \; \; \; \text{ROC of }x(n-k): \left | z \right | \gt \frac{1}{2}\\ \text{For } x(n-k) \; \; \; &\text{ROC will be } \left | z \right |\gt \frac{1}{2}\end{aligned}.
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.