Question 1 |
F(t) is a periodic square wave function as shown. It
takes only two values, 4 and 0, and stays at each of
these values for 1 second before changing. What is
the constant term in the Fourier series expansion of
F(t)?
1 | |
2 | |
3 | |
4 |
Question 1 Explanation:
The constant term in the Fourier series expansion of
F(t) is the average value of F(t) in one fundamental
period i.e.,
\frac{\int_{0}^{1}4dt+\int_{1}^{2}0dt}{2}=\frac{4}{2}=2
\frac{\int_{0}^{1}4dt+\int_{1}^{2}0dt}{2}=\frac{4}{2}=2
Question 2 |
Consider a cube of unit edge length and sides
parallel to co-ordinate axes, with its centroid at the
point (1, 2, 3). The surface integral \int_{A}^{}\vec{F}.d\vec{A} of a
vector field \vec{F}=3x\hat{i}+5y\hat{j}+6z\hat{k} over the entire
surface A of the cube is ______.
14 | |
27 | |
28 | |
31 |
Question 2 Explanation:
Given,
\begin{aligned} \bar{F} &=3x\bar{i}+5y\bar{j}+6z\bar{k} \\ \triangledown \cdot \bar{F}&= \frac{\partial }{\partial x}(3x)+\frac{\partial }{\partial x}(5y)+\frac{\partial }{\partial x}(6z)\\ &= 3+5+6=14 \end{aligned}
By gauss divers once Theorem
\begin{aligned} \int_{A}^{}\bar{F}\cdot dA &=\int \int \int (\triangledown \cdot F)dV =\int \int \int 14\; dv\\ &=14 \times \text{volume of a cube of side 1 unit } \\ &=14 \times (1)^3=14\end{aligned}
\begin{aligned} \bar{F} &=3x\bar{i}+5y\bar{j}+6z\bar{k} \\ \triangledown \cdot \bar{F}&= \frac{\partial }{\partial x}(3x)+\frac{\partial }{\partial x}(5y)+\frac{\partial }{\partial x}(6z)\\ &= 3+5+6=14 \end{aligned}
By gauss divers once Theorem
\begin{aligned} \int_{A}^{}\bar{F}\cdot dA &=\int \int \int (\triangledown \cdot F)dV =\int \int \int 14\; dv\\ &=14 \times \text{volume of a cube of side 1 unit } \\ &=14 \times (1)^3=14\end{aligned}
Question 3 |
Consider the definite integral
\int_{1}^{2}(4x^2+2x+6)dx
Let I_e be the exact value of the integral. If the same integral is estimated using Simpson's rule with 10 equal subintervals, the value is I_s . The percentage error is defined as e=100 \times (I_e-I_s)/I_e . The value of e is
\int_{1}^{2}(4x^2+2x+6)dx
Let I_e be the exact value of the integral. If the same integral is estimated using Simpson's rule with 10 equal subintervals, the value is I_s . The percentage error is defined as e=100 \times (I_e-I_s)/I_e . The value of e is
2.5 | |
3.5 | |
1.2 | |
0 |
Question 3 Explanation:
Exact value
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}
Question 4 |
Given \int_{-\infty }^{\infty }e^{-x^2}dx=\sqrt{\pi}
If a and b are positive integers, the value of \int_{-\infty }^{\infty }e^{-a(x+b)^2}dx is ___.
If a and b are positive integers, the value of \int_{-\infty }^{\infty }e^{-a(x+b)^2}dx is ___.
\sqrt{\pi a} | |
\sqrt{\frac{\pi}{a}} | |
b \sqrt{\pi a} | |
b \sqrt{\frac{\pi}{a}} |
Question 4 Explanation:
\begin{aligned}
&\text{ Let }(x+b)=t\\
&\Rightarrow \; dx=dt\\
&\text{When ,} x=-\infty ;t=-\infty \\
&\int_{-\infty }^{-\infty }e^{-n(x+b)^2}dx=\int_{-\infty }^{-\infty }e^{-at^2}dt\\
&\text{Let, }at^2=y^2\Rightarrow t=\frac{y}{\sqrt{a}}\\
&2at\;dt=3y\;dy\\
&dt=\frac{ydy}{at}=\frac{ydy}{a\frac{y}{\sqrt{a}}}=\frac{y}{\sqrt{a}}\\
&\int_{-\infty }^{-\infty }e^{-at^2}dt=\int_{-\infty }^{-\infty }e^{-y^2}\cdot \frac{dy}{\sqrt{a}}=\sqrt{\frac{\pi}{a}}
\end{aligned}
Question 5 |
A polynomial \phi (s)=a_{n}s^{n}+a_{n-1}s^{n-1}+...+a_{1}s+a_0 of
degree n \gt 3 with constant real coefficients a_n, a_{n-1},...a_0
has triple roots at s=-\sigma . Which one of the
following conditions must be satisfied?
\phi (s)=0 at all the three values of s satisfying s^3+\sigma ^3=0 | |
\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^2 \phi (s)}{ds^2}=0 \text{ at }s=-\sigma | |
\phi (s)=0, \frac{d \phi (s)}{ds}=0, \text{ and }\frac{d^4 \phi (s)}{ds^4}=0 \text{ at }s=-\sigma | |
\phi (s)=0, \text{ and }\frac{d^3 \phi (s)}{ds^3}=0 \text{ at }s=-\sigma |
Question 5 Explanation:
Since \varphi (s) has a triple roots at s=-\sigma
Therefore, \varphi (s)=(s+\sigma )^3\psi (s)
It satisfies all the conditions in option (B) is correct.
Therefore, \varphi (s)=(s+\sigma )^3\psi (s)
It satisfies all the conditions in option (B) is correct.
There are 5 questions to complete.