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

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