# Calculus

 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 _______
 A 0.5 B 1 C 1.5 D 2.2
GATE CE 2021 SET-2   Engineering Mathematics
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 ________
 A 1.2 B 3.1 C 2.4 D 0
GATE CE 2021 SET-2   Engineering Mathematics
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
 A $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$ B $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ C $\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}$ D $\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}$
GATE CE 2021 SET-2   Engineering Mathematics
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
 A 0 B 1 C 0.5 D $\infty$
GATE CE 2021 SET-2   Engineering Mathematics
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) _______
 A 85 B 35 C 15 D 28
GATE CE 2021 SET-1   Engineering Mathematics
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 _______
 A 0.25 B 0.5 C 0.75 D 0.95
GATE CE 2021 SET-1   Engineering Mathematics
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
$=\lim _{x \rightarrow 1}\left[\frac{1-\frac{1}{x}}{\log x+(x-1)\left(\frac{1}{x}\right)}\right]=\lim _{x \rightarrow 1}\left(\frac{1}{2+\log x}\right)=0.5$
 Question 7
The value of $\lim_{x \to \infty } \frac{\sqrt{9x^2+2020}}{x+7}$ is
 A $\frac{7}{9}$ B 1 C 3 D Indeterminable
GATE CE 2020 SET-2   Engineering Mathematics
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 _______.
 A 1 B 2 C 3 D 4
GATE CE 2020 SET-1   Engineering Mathematics
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
 A 35 B 48 C 69 D 84
GATE CE 2020 SET-1   Engineering Mathematics
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}
 Question 10
The value of $\lim_{x \to \infty }\frac{x^2-5x+4}{4x^2+2x}$
 A 0 B $\frac{1}{4}$ C $\frac{1}{2}$ D 1
GATE CE 2020 SET-1   Engineering Mathematics
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}
There are 10 questions to complete.