Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
The value of \lim _{x \rightarrow \infty} \frac{x \ln (x)}{1+x^{2}} is
0 | |
1 | |
0.5 | |
\infty |
Question 4 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 5 |
The volume determined from \iiint_{V} 8 x y z d V \text { for } V=[2,3] \times[1,2] \times[0,1] will be (in integer) _______
85 | |
35 | |
15 | |
28 |
Question 5 Explanation:
\begin{array}{l} \begin{aligned} \text{Given,}\qquad \qquad \qquad V&=[2,3,] \times[1,2] \times[0,1] \text { i.e. } \\ 2& \lt x \lt 3,1 \lt y \lt 2,0 \lt z \lt 1 \\ I&=\iiint_{V} 8 x y z d V=8 \iiint_{V} x y z d x d y d z\\ So,\qquad \qquad \qquad \qquad &=8 \int_{x=2}^{3} x d x \times \int_{y=1}^{2} y d y \times \int_{z=0}^{1} z d z \\ &=8\left[\frac{x^{2}}{2}\right]_{2}^{3}\left[\frac{y^{2}}{2}\right]_{1}^{2}\left[\frac{z^{2}}{2}\right]_{0}^{1} \\ &=(9-4)(4-1)(1-0) \\ &=5 \times 3 \times 1 \\ &=15 \end{aligned} \end{array}
Question 6 |
Consider the limit:
\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)
The limit (correct up to one decimal place) is _______
\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)
The limit (correct up to one decimal place) is _______
0.25 | |
0.5 | |
0.75 | |
0.95 |
Question 6 Explanation:
\lim _{x \rightarrow 1}\left[\frac{1}{\ln x}-\frac{1}{x-1}\right]=\lim _{x \rightarrow 1}\left[\frac{(x-1)-\ln x}{\ln x(x-1)}\right]\left(\frac{0}{0} \text { form }\right)
So using L-Hospital's rule twices
=\lim _{x \rightarrow 1}\left[\frac{1-\frac{1}{x}}{\log x+(x-1)\left(\frac{1}{x}\right)}\right]=\lim _{x \rightarrow 1}\left(\frac{1}{2+\log x}\right)=0.5
So using L-Hospital's rule twices
=\lim _{x \rightarrow 1}\left[\frac{1-\frac{1}{x}}{\log x+(x-1)\left(\frac{1}{x}\right)}\right]=\lim _{x \rightarrow 1}\left(\frac{1}{2+\log x}\right)=0.5
Question 7 |
The value of \lim_{x \to \infty } \frac{\sqrt{9x^2+2020}}{x+7} is
\frac{7}{9} | |
1 | |
3 | |
Indeterminable |
Question 7 Explanation:
\lim_{x \to \infty }\frac{3x \sqrt{1+\frac{2020}{x^2}}}{x\left ( 1+\frac{7}{x} \right )}=3
Question 8 |
If C represents a line segment between (0, 0, 0) and (1,1, 1) in Cartesian coordinate
system, the value (expressed as integer) of the line integral
\int _C [(y+z)dx+(x+z)dy+(x+y)dz]
is _______.
\int _C [(y+z)dx+(x+z)dy+(x+y)dz]
is _______.
1 | |
2 | |
3 | |
4 |
Question 8 Explanation:
\begin{aligned} I&=\int _c [(ydx+xdy)+(zdx+xdz)+(zdy+ydz)] \\ &=\int _c[d(xy)+d(xz)+d(yz)]\\ &=(xy+yz+zx)_{(0,0,0)}^{(1,1,1)}\\ &=(1+1+1)-(0+0+0)=3 \end{aligned}
Question 9 |
The true value of ln(2) is 0.69. If the value of ln(2) is obtained by linear interpolation
between ln(1) and ln(6), the percentage of absolute error (round off to the nearest integer),
is
35 | |
48 | |
69 | |
84 |
Question 9 Explanation:
True value of \ln 2=0.69=T
\begin{aligned} &x &&y=\ln x \\ &x_0=1 & &0 \\ &x_1=6& &1.79 \end{aligned}
Divided differentiation
\begin{aligned} \frac{1.79-0}{6-1}&=0.358=f[x_0,x_1] \\ \text{Approx:}\;\;\ln 2 &=f[x_0]+(x-x_0)f[x_0,x_1] \\ &= 0+(2-1)0.358\\ &= 0.358=A\\ \% \; error &= \frac{T-A}{T}\times 100=48.11\% \end{aligned}
\begin{aligned} &x &&y=\ln x \\ &x_0=1 & &0 \\ &x_1=6& &1.79 \end{aligned}
Divided differentiation
\begin{aligned} \frac{1.79-0}{6-1}&=0.358=f[x_0,x_1] \\ \text{Approx:}\;\;\ln 2 &=f[x_0]+(x-x_0)f[x_0,x_1] \\ &= 0+(2-1)0.358\\ &= 0.358=A\\ \% \; error &= \frac{T-A}{T}\times 100=48.11\% \end{aligned}
Question 10 |
The value of \lim_{x \to \infty }\frac{x^2-5x+4}{4x^2+2x}
0 | |
\frac{1}{4} | |
\frac{1}{2} | |
1 |
Question 10 Explanation:
It is in \left (\frac{\infty }{\infty } \right )
from so by L-Hospital Rule
\begin{aligned} =&\lim_{x \to \infty }\left ( \frac{2x-5}{8x+2} \right )=\frac{\infty }{\infty }\\ =&\lim_{x \to \infty }\left ( \frac{2}{8} \right )=\frac{1}{4} \end{aligned}
\begin{aligned} =&\lim_{x \to \infty }\left ( \frac{2x-5}{8x+2} \right )=\frac{\infty }{\infty }\\ =&\lim_{x \to \infty }\left ( \frac{2}{8} \right )=\frac{1}{4} \end{aligned}
There are 10 questions to complete.
