# Calculus

 Question 1
Find the positive real root of $x^3-x-3=0$ using Newton-Raphson method. If the starting guess $(x_0)$ is 2, the numerical value of the root after two iterations $(x_2)$ is _______ (round off to two decimal places).
 A 1.67 B 1.12 C 2.44 D 3.25
GATE ME 2021 SET-2   Engineering Mathematics
Question 1 Explanation:
\begin{aligned} \text { Given, }\qquad\quad f(x) &=x^{3}-x-3, \quad x_{0}=2 \\ f^{\prime}(x)&=3 x^{2}-1\\ \text { Iteration 1: } \quad x_{1}&=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)}=2-\frac{(8-2-3)}{3(4)-1}=1.72\\ \text { Iteration 2: } \quad x_{2}&=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}=1.72-\frac{\left(1.72^{3}-1.72-3\right)}{3(1.72)^{2}-1}=1.67 \end{aligned}
 Question 2
Let the superscript T represent the transpose operation. Consider the function $f(x)=\frac{1}{2}x^T Qx=r^Tx, \; \text{ where } x \text{ and }r \text{ are }n \times 1$ vectors and $Q$ is a symmetric $n \times n$ matrix. The stationary point of $f(x)$ is
 A $Q^Tr$ B $Q^{-1}r$ C $\frac{r}{r^Tr}$ D $r$
GATE ME 2021 SET-2   Engineering Mathematics
Question 2 Explanation:
\begin{aligned} \text{Let}\qquad Q=\left[\begin{array}{ll}a & c \\c & b\end{array}\right], x&=\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right], R=\left[\begin{array}{l}r_{1} \\r_{2}\end{array}\right] \\ F(x)&=\frac{1}{2}\left(x_{1}, x_{2}\right)\left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]-\left[r_{1} r_{2}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right] \\ &=\frac{1}{2}\left[a x_{1}^{2}+b x_{2}^{2}+2 c x_{1} x_{2}\right]-\left[r_{1} x_{1}+r_{2} x_{2}\right]\\ \text{i.e.}\qquad \qquad U\left(x_{1}, x_{2}\right)&=\frac{1}{2} a x_{1}^{2}+\frac{1}{2} b x_{2}^{2}+c_{1} x_{1} x_{2}-r_{1} x_{1}-r_{2} x_{2} \end{aligned}
Now, for critical point, $\frac{\partial u}{\partial x_{1}}=0$ and $\frac{\partial u}{\partial x_{2}}=0$
$\Rightarrow \quad a_{1} x_{1}+c x_{2}-r_{1}=0 \quad \text { and } c x_{2}+c x_{1}-r_{2}=0$
In matrix form we can write it as
\begin{aligned} \left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l} x_{1} \\m_{2}\end{array}\right] &=\left[\begin{array}{l}r_{1} \\ r_{2}\end{array}\right] \\\Rightarrow \qquad Q x &=r \end{aligned}
By multiplying both side by $Q^{-1}$
$x=Q^{-1} r$
 Question 3
The value of $\int_{0}^{\pi /2}\int_{0}^{\cos \theta }r \sin \theta dr d\theta$ is
 A 0 B $\frac{1}{6}$ C $\frac{4}{3}$ D $\pi$
GATE ME 2021 SET-2   Engineering Mathematics
Question 3 Explanation:
\begin{aligned} I &=\int_{\theta=0}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=\cos \theta} r \sin \theta d r d \theta \\ &=\int_{\theta=0}^{\frac{\pi}{2}}\left[\frac{r^{2}}{2}\right]_{0}^{\infty \cos \theta} \times \sin \theta d \theta \\ &=\frac{1}{2} \int_{\theta}^{\frac{\pi}{2}} \sin \theta \cdot \cos ^{2} \theta d \theta \\ \text{Let}, \qquad\cos \theta &=t\\ -\sin \theta d \theta &=d t \\ \cos \theta &=t \\ \text{at},\qquad\theta &=\frac{\pi}{2} ; t=0 \\ \theta &=0, t=1 \\ &=\int_{1}^{0} \frac{-t^{2}}{2} d t \\ &=\frac{-1}{2}\left[\frac{t^{3}}{3}\right]_{1}^{0}=\frac{-1}{2} \times\left(\frac{-1}{3}\right) \\ &=\frac{-1}{2}\left[\frac{-1}{3}\right]=\frac{1}{6} \end{aligned}
 Question 4
Let $f(x)=x^2-2x+2$ be a continuous function defined on $x \in [1,3]$. The point $x$ at which the tangent of $f(x)$ becomes parallel to the straight line joining $f(1)$ and $f(3)$ is
 A 0 B 1 C 2 D 3
GATE ME 2021 SET-1   Engineering Mathematics
Question 4 Explanation:
By Lagrangian mean value theorem,
\begin{aligned} \frac{f(3)-f(1)}{3-1} &=f^{\prime}(c) \\ \Rightarrow\quad \frac{5-1}{3-1} &=2 x-2\\ \Rightarrow\quad 2 x-2&=4\\ \Rightarrow\quad x&=2 \in(1,3) \end{aligned}
 Question 5
The value of $\lim_{x \to 0}\left ( \frac{1- \cos x}{x^2} \right )$ is
 A $\frac{1}{4}$ B $\frac{1}{3}$ C $\frac{1}{2}$ D 1
GATE ME 2021 SET-1   Engineering Mathematics
Question 5 Explanation:
\begin{aligned} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)&=? \;\;\;\;\;\;\left(\frac{0}{0} \text { form }\right) \\ \text { Applying } L \cdot H \text { rule } & =\lim _{x \rightarrow 0} \frac{\sin x}{2 x}\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \end{aligned}
 Question 6
The directional derivative $f(x,y,z)=xyz$ at point (-1,1,3) in the direction of vector $\hat{i}-2\hat{j}+2\hat{k}$ is
 A $3\hat{i}-3\hat{j}-\hat{k}$ B $-\frac{7}{3}$ C $\frac{7}{3}$ D 7
GATE ME 2020 SET-2   Engineering Mathematics
Question 6 Explanation:
\begin{aligned} f &=x y z \\ \nabla f &=f_{x} \bar{i}+f_{y} \bar{j}+f_{z} \bar{k}=y z \bar{i}+x \bar{z} \bar{j}+x y \bar{k} \\ (\nabla f)_{(-1,1,3)} &=3 \bar{i}-3 \bar{j}-\bar{k} \\ \vec{a} &=\bar{i}-2 \bar{j}+2 \bar{k} \end{aligned}
Directional derivative of f in direction of $\vec{a}$
$\underset{\vec{a}}{D f}=\nabla f \times \frac{\vec{a}}{|\vec{a}|}=\frac{3(1)+(-3)(-2)+(-1)(2)}{\sqrt{(1)^{2}+(-2)^{2}+(2)^{2}}}=\frac{7}{3}$
 Question 7
Let $I=\int_{x=0}^{1}\int_{y=0}^{x^2}xy^2dydx$ then, $I$ may also be expressed as
 A $I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}xy^2dxdy$ B $I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}yx^2dxdy$ C $I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}xy^2dxdy$ D $I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}yx^2dxdy$
GATE ME 2020 SET-2   Engineering Mathematics
Question 7 Explanation:
$I=\int_{0}^{1} \int_{0}^{x^{2}} x y^{2} d y d x$

Change on rules, $I=\int_{y=0}^{1} \int_{x=\sqrt{y}}^{1} x y^{2} d x d y$
 Question 8
A vector field is defined as
$\vec{f}(x,y,z)=\frac{x}{[x^2+y^2+z^2]^{\frac{3}{2}}}\hat{i}$ $+\frac{y}{[x^2+y^2+z^2]^{\frac{3}{2}}}\hat{j}$ $+\frac{z}{[x^2+y^2+z^2]^{\frac{3}{2}}}\hat{k}$
where, $\hat{i},\hat{j},\hat{k}$ are unit vectors along the axes of a right-handed rectangular/Cartesian coordinate system. The surface integral $\int \int \vec{f}\cdot d\vec{S}$ (where $d\vec{S}$ is an elemental surface area vector) evaluated over the inner and outer surfaces of a spherical shell formed by two concentric spheres with origin as the center, and internal and external radii of 1 and 2, respectively, is
 A 0 B $2 \pi$ C $4 \pi$ D $8 \pi$
GATE ME 2020 SET-1   Engineering Mathematics
Question 8 Explanation:
\begin{aligned} \vec{F} &=\frac{\vec{r}}{r^{3}} \\ \nabla \cdot \vec{F} &=\nabla \cdot \frac{\vec{r}}{r^{3}}=0\\ \text{By divergence theorem}\\ \iint_{S} \bar{f} \cdot \vec{d} S&=\iiint_{R} \nabla \vec{F} d x d y d z=0 \end{aligned}
 Question 9
For three vectors $\vec{A}=2\hat{j} - 3 \hat{k}$, $\vec{B}=-2\hat{i} + \hat{k}$, $\vec{C}=3\hat{i} - \hat{j}$, where $\hat{i}, \hat{j} \; and \; \hat{k}$ are unit vectors along the axes of a right-handed rectangular/Cartesian coordinate system, the value of $(\vec{A}\cdot(\vec{B}\times \vec{C})+6)$ is __________.
 A 2 B 4 C 6 D 8
GATE ME 2020 SET-1   Engineering Mathematics
Question 9 Explanation:
$\vec{A}\cdot (\vec{B}\times \vec{C})=\begin{vmatrix} 0 &2 &-3 \\ -2 &0 &1 \\ 3 &-1 &0 \end{vmatrix}=0-2[0-3]-3[2-0]=0$
$\vec{A}\cdot (\vec{B}\times \vec{C})+6=0+6=6$
 Question 10
The value of $\lim_{x \to 1} \left ( \frac{1-e^{-c(1-x)}}{1-xe^{-c(1-x)}} \right )$ is
 A c B c+1 C $\frac{c}{c+1}$ D $\frac{c+1}{c}$
GATE ME 2020 SET-1   Engineering Mathematics
Question 10 Explanation:
Applying L Hospital rule
$\lim_{x \to 1}\left ( \frac{1-e^{-c(1-x)}}{1-xe^{-c(1-x)}} \right )=\lim_{x \to 1}\left ( \frac{1-e^{-c+cx}}{-x(ce^{-c+x})-(e^{-c+cx})} \right ) =\frac{-c}{-c-1}=\frac{c}{c+1}$

There are 10 questions to complete.

### 12 thoughts on “Calculus”

1. The question posted have minor mistakes in it, in answers it show up something else other than what mentioned in question. Please look into it ,otherwise its a good platform to practise

• Thank you for your suggestions.
Can you please share the questions numbers for the above suggestions.

• question number 10

• question no 9 mein b vector, wrong type ho gaya hai

2. in que 10 x->1 not zero

• Thank You Jashwanth Reddy Earla,
We have updated the question.

3. que 35 f(x)= 2x^3-3x^2

• Thank You Jashwanth Reddy Earla,
We have updated the question.