Question 1 |
Numerically integrate, f(x)=10 x-20 x^{2} from lower limit a=0 to upper limit b=0.5. Use Trapezoidal rule with five equal subdivisions. The value (in units,round off to two decimal places) obtained is ____________
0.78 | |
0.65 | |
0.4 | |
0.56 |
Question 1 Explanation:
\begin{aligned} y&=10 x-20 x^{2} \\ a&=0, b=0.5, n=5 \\ \text { So, } \qquad \qquad \qquad h&=\frac{b-a}{n}=0.1 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
Question 2 |
A function is defined in Cartesian coordinate system as f(x, y)=x e^{y}. The value of the directional derivative of the function (in integer) at the point (2,0) along the direction of the straight line segment from point (2,0) to point \left(\frac{1}{2}, 2\right) is _______
0.5 | |
1 | |
1.5 | |
2.2 |
Question 2 Explanation:
\begin{aligned} f(x, y)&=x e^{y}\\ \mathrm{P}(2,0) \text{ and } Q\left(\frac{1}{2}, 2\right)\\ \operatorname{grad} f &=\hat{i}\left(e^{y}\right)+\hat{j}\left(x e^{y}\right)+\hat{k}(0) \\ \Rightarrow \qquad \qquad \qquad (\operatorname{grad} f)_{p} &=\hat{i}+2 \hat{j} ]\\ \overline{P Q}&=\left(\frac{1}{2}-2\right) \hat{i}+(2-0) \hat{j}=-\frac{3}{2} \hat{i}+2 \hat{j}\\ \text{Required directional derivative} &=(\text { grad } f)_{P} \widehat{P Q}\\ &=(\hat{i}+2 \hat{j}) \times \frac{\left(-\frac{3}{2} \hat{i}+2 \hat{j}\right)}{\sqrt{\frac{9}{4}+4}}=\frac{\frac{-3}{2}+4}{\sqrt{\frac{25}{4}}} \\ &=\frac{\frac{-3+8}{2}}{\left(\frac{5}{2}\right)}=1 \end{aligned}
Question 3 |
The smallest eigenvalue and the corresponding eigenvector of the matrix \left[\begin{array}{cc} 2 & -2 \\ -1 & 6 \end{array}\right], respectively, are
1.55 and \left\{\begin{array}{l} 2.00 \\ 0.45 \end{array}\right\} | |
2.00 and \left\{\begin{array}{l} 1.00 \\ 1.00 \end{array}\right\} | |
1.55 and \left\{\begin{array}{l} -2.55 \\ -0.45 \end{array}\right\} | |
1.55 and \left\{\begin{array}{c} 2.00 \\ -0.45 \end{array}\right\} |
Question 3 Explanation:
\begin{aligned} A&=\left[\begin{array}{cc} 2 & -2 \\ -1 & 6 \end{array}\right] \Rightarrow|A-\lambda I|=0 \\ \Rightarrow \qquad \lambda&=(4+\sqrt{6}) \text { and }(4-\sqrt{6})\\ A X&=\lambda X\\ (A-\lambda I) X&=0 \end{aligned}
{\left[\begin{array}{cc} 2-(4-\sqrt{6}) & -2 \\ -1 & 6-(4-\sqrt{6}) \end{array}\right] \left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] =\left[\begin{array}{l} 0 \\ 0 \end{array}\right]}
\begin{aligned} x_{1}&=\left(\frac{2}{-2+\sqrt{6}}\right) x_{2}\\ \text { Let, } \qquad x_{2}&=K \text { then } x_{1}=\left(\frac{2}{-2+\sqrt{6}}\right) \text { K }\\ \Rightarrow \qquad\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]&=\left[\begin{array}{c} \frac{2}{-2+\sqrt{6}} k \\ k \end{array}\right] \approx\left[\begin{array}{c} 2 \\ -2+\sqrt{6} \end{array}\right]=\left[\begin{array}{l} 2.00 \\ 0.45 \end{array}\right] \end{aligned}
{\left[\begin{array}{cc} 2-(4-\sqrt{6}) & -2 \\ -1 & 6-(4-\sqrt{6}) \end{array}\right] \left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] =\left[\begin{array}{l} 0 \\ 0 \end{array}\right]}
\begin{aligned} x_{1}&=\left(\frac{2}{-2+\sqrt{6}}\right) x_{2}\\ \text { Let, } \qquad x_{2}&=K \text { then } x_{1}=\left(\frac{2}{-2+\sqrt{6}}\right) \text { K }\\ \Rightarrow \qquad\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]&=\left[\begin{array}{c} \frac{2}{-2+\sqrt{6}} k \\ k \end{array}\right] \approx\left[\begin{array}{c} 2 \\ -2+\sqrt{6} \end{array}\right]=\left[\begin{array}{l} 2.00 \\ 0.45 \end{array}\right] \end{aligned}
Question 4 |
If k is a constant, the general solution of \frac{d y}{d x}-\frac{y}{x}=1 will be in the form of
y=x ln(kx) | |
y=k ln(kx) | |
y=x ln(x) | |
y=xk ln(k) |
Question 4 Explanation:
\begin{aligned} \frac{d y}{d x}-\frac{y}{x} &=1 \\ \frac{d y}{d x}+P y &=Q \\ P &=-\frac{1}{x}, Q=1 \\ I F &=e^{\int P d x}=e^{\int \frac{-1}{x} d x}=\frac{1}{x} \\ y(I F) &=\int Q(I F) d x+c \\ y\left(\frac{1}{x}\right) &=\int 1 \cdot \frac{1}{x} d x+\ln k \\ y &=x \ln (x k) \end{aligned}
Question 5 |
The value (round off to one decimal place) of \int_{-1}^{1} x e^{|x|} d x is ________
1.2 | |
3.1 | |
2.4 | |
0 |
Question 5 Explanation:
\begin{aligned} \int_{-1}^{1} x e^{-|x|} d x&=0\\ \text { As } \qquad \qquad \qquad f(-x)&=x e^{x} \text { is an odd function. } \end{aligned}
Question 6 |
If A is a square matrix then orthogonality property mandates
A A^{T}=I | |
A A^{T}=0 | |
A A^{T}=A^{-1} | |
A A^{T}=A^{2} |
Question 6 Explanation:
\text { If, } \qquad \qquad A A^{\top}=I \quad \text { or } A^{-1}=A^{T}
The matrix is orthogonal.
The matrix is orthogonal.
Question 7 |
The unit normal vector to the surface X^{2}+Y^{2}+Z^{2}-48=0 at the point (4,4,4) is
\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} | |
\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} | |
\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}} | |
\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}} |
Question 7 Explanation:
\begin{aligned} \phi &=x^{2}+y^{2}+z^{2}-48, P(4,4,4) \\ \operatorname{grad} \phi &=\vec{\nabla} \phi=\hat{i} \frac{\partial \phi}{\partial x}+\hat{j} \frac{\partial \phi}{\partial y}+\hat{k} \frac{\partial \phi}{\partial z} \\ &=(2 x) \hat{i}+(2 y) \hat{j}+(2 z) \hat{k} \\ \vec{n} &=(\operatorname{grad} \phi)_{P}=8 \hat{i}+8 \hat{j}+8 \hat{k} \\ \hat{n} &=\frac{\vec{n}}{|\vec{n}|}=\frac{8 \hat{i}+8 \hat{j}+8 \hat{k}}{\sqrt{64+64+64}}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \\ & \simeq\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\right) \end{aligned}
Question 8 |
The rank of the matrix \left[\begin{array}{cccc} 5 & 0 & -5 & 0 \\ 0 & 2 & 0 & 1 \\ -5 & 0 & 5 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] is
1 | |
2 | |
3 | |
4 |
Question 8 Explanation:
\begin{aligned} \left[\begin{array}{cccc} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ -5 & 0 & -1 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] & \stackrel{R_{1} \longleftrightarrow R_{1}+R_{3}}{\longrightarrow}\left[\begin{array}{llll} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] \\ & \stackrel{R_{4} \longleftrightarrow R_{4}-\frac{1}{2} R_{2}}{\longrightarrow}\left[\begin{array}{llll} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{2} \end{array}\right]\\ &R_{3} \longleftrightarrow R_{4}\left[\begin{array}{llll}5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{2} \\ 0 & 0 & 0 & 0\end{array}\right] \end{aligned}
Rank(A) = 3
Rank(A) = 3
Question 9 |
The value of \lim _{x \rightarrow \infty} \frac{x \ln (x)}{1+x^{2}} is
0 | |
1 | |
0.5 | |
\infty |
Question 9 Explanation:
\begin{aligned} &\lim _{x \rightarrow \infty}\left(\frac{x \ln x}{x^{2}+1}\right) \qquad \qquad \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ &=\lim _{x \rightarrow \infty}\left(\frac{x\left(\frac{1}{x}\right)+\ln x}{2 x}\right) \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ \lim _{x \rightarrow \infty}\left(\frac{0+\frac{1}{x}}{2}\right)&=\lim _{x \rightarrow \infty}\left(\frac{1}{2 x}\right)=\frac{1}{2 \times \infty}=0 \end{aligned}
Question 10 |
The value of \int_{0}^{1} e^{x} d x using the trapezoidal rule with four equal subintervals is
1.718 | |
1.727 | |
2.192 | |
2.718 |
Question 10 Explanation:
Let \int_{a}^{b}f(x)dx=\int_{0}^{1}e^x \; dx \text{ and }n=4
Then a=0,b=1,f(x)=e^x \text{ and }h=\frac{b-a}{n}=\frac{1-0}{4}=0.25
\begin{array}{|c|c|c|c|c|c|} \hline x&0&0.25&0.50&0.75&1\\ \hline y=f(x=e^x)&1&1.284&1.649&2.117&2.718\\ \hline \end{array}
The formula of trapezoidal rule to the given data is given by
\int_{a}^{b}f(x)dx\simeq \int_{a}^{b}p(x)dx=\frac{h}{2}\left \{ (y_0+y_n)+2(y_1+y_2+y_3) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (1+2.718)+2(1.284+1.640+2.117) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (3.718)+2(5.03) \right \}=1.727
Then a=0,b=1,f(x)=e^x \text{ and }h=\frac{b-a}{n}=\frac{1-0}{4}=0.25
\begin{array}{|c|c|c|c|c|c|} \hline x&0&0.25&0.50&0.75&1\\ \hline y=f(x=e^x)&1&1.284&1.649&2.117&2.718\\ \hline \end{array}
The formula of trapezoidal rule to the given data is given by
\int_{a}^{b}f(x)dx\simeq \int_{a}^{b}p(x)dx=\frac{h}{2}\left \{ (y_0+y_n)+2(y_1+y_2+y_3) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (1+2.718)+2(1.284+1.640+2.117) \right \}
\int_{0}^{1}e^x \; dx\simeq \int_{0}^{1}p(x)dx=\frac{0.25}{2}\left \{ (3.718)+2(5.03) \right \}=1.727
There are 10 questions to complete.