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).
1.67 | |
1.12 | |
2.44 | |
3.25 |
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
Q^Tr | |
Q^{-1}r | |
\frac{r}{r^Tr} | |
r |
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
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
0 | |
\frac{1}{6} | |
\frac{4}{3} | |
\pi |
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
0 | |
1 | |
2 | |
3 |
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}
\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
\frac{1}{4} | |
\frac{1}{3} | |
\frac{1}{2} | |
1 |
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
3\hat{i}-3\hat{j}-\hat{k} | |
-\frac{7}{3} | |
\frac{7}{3} | |
7 |
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}
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
I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}xy^2dxdy | |
I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}yx^2dxdy | |
I=\int_{y=0}^{1}\int_{x=\sqrt{y}}^{1}xy^2dxdy | |
I=\int_{y=0}^{1}\int_{x=0}^{\sqrt{y}}yx^2dxdy |
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
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
\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
0 | |
2 \pi | |
4 \pi | |
8 \pi |
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 __________.
2 | |
4 | |
6 | |
8 |
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
\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
c | |
c+1 | |
\frac{c}{c+1} | |
\frac{c+1}{c} |
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}
\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.
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
in que 10 x->1 not zero
Thank You Jashwanth Reddy Earla,
We have updated the question.
que 35 f(x)= 2x^3-3x^2
Thank You Jashwanth Reddy Earla,
We have updated the question.
Question 101- Answer is maximum(x,-x)
Thank You Ajay,
We have updated the answer.
Error in statement of Q.40
error in que 40