Question 1 |
For the solid S shown below, the value of \int \int_S \int x dxdydz (rounded off to two decimal
places) is ______.
3.52 | |
1.25 | |
2.25 | |
4.25 |
Question 1 Explanation:
x : 0 to 3
y : 0 to 1
z : 0 to 1-y
=\int_{y=0}^{1}\int_{z=0}^{1-y}\int_{x=0}^{3}x\, dx\, dy\, dz=\int_{y=0}^{1}\int_{0}^{1-y}\left ( \frac{x^{2}}{2} \right )^{3}_{0}dz\, dy
=\int_{0}^{1}\frac{9}{2}(z)_{0}^{1-y}dy=\frac{9}{2}\int_{0}^{1}(1-y)dy=\frac{9}{2}\left ( y-\frac{y^{2}}{2} \right )_{0}^{1}
=\frac{9}{2}\left ( 1-\frac{1}{2} \right )=\frac{9}{4}= 2.25
y : 0 to 1
z : 0 to 1-y
=\int_{y=0}^{1}\int_{z=0}^{1-y}\int_{x=0}^{3}x\, dx\, dy\, dz=\int_{y=0}^{1}\int_{0}^{1-y}\left ( \frac{x^{2}}{2} \right )^{3}_{0}dz\, dy
=\int_{0}^{1}\frac{9}{2}(z)_{0}^{1-y}dy=\frac{9}{2}\int_{0}^{1}(1-y)dy=\frac{9}{2}\left ( y-\frac{y^{2}}{2} \right )_{0}^{1}
=\frac{9}{2}\left ( 1-\frac{1}{2} \right )=\frac{9}{4}= 2.25
Question 2 |
The partial derivative of the function
f(x,y,z)=e^{1-x \cos y}+xze^{-1/(1+y^2)}
with respect to x at the point (1,0,e) is
f(x,y,z)=e^{1-x \cos y}+xze^{-1/(1+y^2)}
with respect to x at the point (1,0,e) is
-1 | |
0 | |
1 | |
\frac{1}{e} |
Question 2 Explanation:
\begin{aligned} \text{Given, } f(x,y,z)&=e^{1-x\cos y}+xze^{-1/(1+y^{2})} \\ \frac{\partial f}{\partial x}&=e^{1-x\cos y}(0-\cos y)+ze^{-1/1+y^{2}} \\ \left ( \frac{\partial f }{\partial x} \right )_{(1,0,e)}&=e^{0}(0-1)+e\cdot e^{-1/(1+0)} \\ &=-1+1=0
\end{aligned}
Question 3 |
For a vector field \vec{A}, which one of the following is False?
\vec{A} is solenoidal if \bigtriangledown \cdot \vec{A}=0 | |
\bigtriangledown \times \vec{A} is another vector field. | |
\vec{A} is irrotational if \bigtriangledown ^2 \vec{A}=0. | |
\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A} |
Question 3 Explanation:
Divergence and curl operator is performed on a vector field \vec{A}
Curl operation provides a vector orthogonal to the given vector field \vec{A}
\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A}
If a vector field is irrortational then \bigtriangledown \times \vec{A}=0
If a vector field is solenoidal then \bigtriangledown \cdot \vec{A}=0
If a field is scalar A, then \bigtriangledown ^2 \vec{A}=0, is a laplacian equation.
Hence option (C) is incorrect
Curl operation provides a vector orthogonal to the given vector field \vec{A}
\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A}
If a vector field is irrortational then \bigtriangledown \times \vec{A}=0
If a vector field is solenoidal then \bigtriangledown \cdot \vec{A}=0
If a field is scalar A, then \bigtriangledown ^2 \vec{A}=0, is a laplacian equation.
Hence option (C) is incorrect
Question 4 |
If v_1, v_2,..., v_6 are six vectors in \mathbb{R}^4, which one of the following statements is False?
It is not necessary that these vectors span \mathbb{R}^4. | |
These vectors are not linearly independent. | |
Any four of these vectors form a basis for \mathbb{R}^4. | |
If {v_1, v_3,v_5, v_6} spans \mathbb{R}^4, then it forms a basis for \mathbb{R}^4. |
Question 4 Explanation:
v_1, v_2,..., v_6 are six vectors in \mathbb{R}^4.
For a 4-dimensional vector space,
(i) any four linearly independent vectors form a basis (or)
(ii) Any set of four vectors in \mathbb{R}^4 spans \mathbb{R}^4, then it forms a basis.
Therefore, clearly options (A), (B), (D) are true.
Option (C) is FALSE
For a 4-dimensional vector space,
(i) any four linearly independent vectors form a basis (or)
(ii) Any set of four vectors in \mathbb{R}^4 spans \mathbb{R}^4,
Therefore, clearly options (A), (B), (D) are true.
Option (C) is FALSE
Question 5 |
Consider the line integral
\int _c (xdy-ydx)
the integral being taken in a counterclockwise direction over the closed curve C that forms the boundary of the region R shown in the figure below. The region R is the area enclosed by the union of a 2x3 rectangle and a semi-circle of radius 1. The line integral evaluates to
\int _c (xdy-ydx)
the integral being taken in a counterclockwise direction over the closed curve C that forms the boundary of the region R shown in the figure below. The region R is the area enclosed by the union of a 2x3 rectangle and a semi-circle of radius 1. The line integral evaluates to
6+\pi/2 | |
8+\pi | |
12+\pi | |
16+2 \pi |
Question 5 Explanation:
\begin{array}{l} \text { Given, } \int-y d x+x d y \\ \qquad \begin{aligned} \text { here, } \quad F_{1}&=-y \text { and } \frac{\partial F_{1}}{\partial y}=-1 \\ F_{2} &=x \text { and } \frac{\partial F_{2}}{\partial x}=1 \\ \therefore \int F_{1} d x+F_{2} d y &=\iint\left(\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y}\right) d x d y \\ \int-y d x+x d y &=\iint 1-(-1) d x d y \\ &=2( \text { Area of region R}) \\ &=2\left(6+\frac{\pi}{2}\right)=12+\pi \end{aligned} \end{array}
Question 6 |
Consider a differentiable function f(x) on the set of real numbers such that f(-1)=0 and f'(x)\leq 2. Given these conditions, which one of the following inequalities is necessarily true for all x\in [-2,2]?
f(x)\leq \frac{1}{2}|x+1| | |
f(x)\leq 2|x+1| | |
f(x)\leq \frac{1}{2}|x| | |
f(x)\leq 2|x| |
Question 6 Explanation:
\begin{aligned} \text{Given that, }&\left|f^{\prime}(x)\right| \leq 2 ; &f(-1)=0 \\ &-2 \leq f^{\prime}(x) \leq 2\\ x \in[-2,2] &\rightarrow 2 \leq x \leq 2 \end{aligned}
\therefore By applying mean value theorem in [-1,2]
\begin{array}{l} -2 \leq f^{\prime}(x) \leq 2 \\ -2 \leq \frac{f(2)-f(-1)}{2-(-1)} \leq 2 \\ -2 \leq \frac{f(2)-f(-1)}{3} \leq 2 \\ -6 \leq f(2)-0 \leq 6 \\ -6 \leq f(2) \leq 6 \end{array}
It is satisfied by only option (B).
\therefore By applying mean value theorem in [-1,2]
\begin{array}{l} -2 \leq f^{\prime}(x) \leq 2 \\ -2 \leq \frac{f(2)-f(-1)}{2-(-1)} \leq 2 \\ -2 \leq \frac{f(2)-f(-1)}{3} \leq 2 \\ -6 \leq f(2)-0 \leq 6 \\ -6 \leq f(2) \leq 6 \end{array}
It is satisfied by only option (B).
Question 7 |
The value of the integral \int_{0}^{\pi}\int_{y}^{\pi}\frac{sinx}{x}dxdy, is equal to ______
1 | |
2 | |
3 | |
4 |
Question 7 Explanation:
\begin{array}{l} x=y ; x=\pi\\ y=O; y = \pi \end{array}
\begin{array}{l} =\int_{0}^{\pi} \int_{y}^{\pi} \frac{\sin x}{x} d x d y=\int_{0}^{\pi} \frac{\sin x}{x}(y)_{0}^{x} d x \\ =\int_{0}^{\pi} \frac{\sin x}{x}(x) d x \\ =\int_{0}^{\pi} \sin x d x=(-\cos x)_{0}^{\pi} \\ =-\cos \pi+\cos 0=1+1=2 \end{array}
\begin{array}{l} =\int_{0}^{\pi} \int_{y}^{\pi} \frac{\sin x}{x} d x d y=\int_{0}^{\pi} \frac{\sin x}{x}(y)_{0}^{x} d x \\ =\int_{0}^{\pi} \frac{\sin x}{x}(x) d x \\ =\int_{0}^{\pi} \sin x d x=(-\cos x)_{0}^{\pi} \\ =-\cos \pi+\cos 0=1+1=2 \end{array}
Question 8 |
Let r=x^{2}+y-z and z^{3}-xy+yz+y^{3}=1. Assume that x and y are independent
variables. At (x, y, z) = (2, -1,1), the value (correct to two decimal places) of \frac{\partial r}{\partial x} is _________ .
4.5 | |
5.4 | |
2.4 | |
3.5 |
Question 8 Explanation:
\begin{aligned} r &=x^{2}+y-z &\ldots(i)\\ z^{3}-x y+y z+y^{3} &=1 &\ldots(ii)\\ \frac{\partial r}{\partial x} &=2 x-\frac{\partial z}{\partial x} &\ldots(iii)\\ 3 z^{2} \frac{\partial z}{\partial x}-y+y \frac{\partial z}{\partial x} &=0 \\ \frac{\partial z}{\partial x} &=\frac{y}{3 z^{2}+y} \end{aligned}
By substituting \frac{\partial z}{\partial x} in equation (iii), we get,
\begin{aligned} &=2 x-\frac{y}{3 z^{2}+y} \\ \operatorname{At}(2 .-1.1),& \\ \frac{\partial r}{\partial x}&=2(2)-\frac{(-1)}{3(1)^{2}+(-1)} \\ &=4+\frac{1}{2}=4.50 \end{aligned}
By substituting \frac{\partial z}{\partial x} in equation (iii), we get,
\begin{aligned} &=2 x-\frac{y}{3 z^{2}+y} \\ \operatorname{At}(2 .-1.1),& \\ \frac{\partial r}{\partial x}&=2(2)-\frac{(-1)}{3(1)^{2}+(-1)} \\ &=4+\frac{1}{2}=4.50 \end{aligned}
Question 9 |
A curve passes through the point (x =1, y = 0) and satisfies the differential equation \frac{dy}{dx}=\frac{x^{2}+y^{2}}{2y} + \frac{y}{x}. The equation that describes the curve is
ln(1+\frac{y^{2}}{x^{2}})=x-1 | |
\frac{1}{2}ln(1+\frac{y^{2}}{x^{2}})=x-1 | |
ln(1+\frac{y}{x})=x-1 | |
\frac{1}{2}ln(1+\frac{y}{x})=x-1 |
Question 9 Explanation:
\begin{aligned} \frac{d y}{d x} &=\frac{x^{2}}{2 y}+\frac{y}{2}+\frac{y}{x} \\ \text { Put. } \frac{y}{x}=t \\ \frac{d y}{d x}&=t+x \frac{d t}{d x} \\ t+x \frac{d t}{d x} &=\frac{x}{2 t}+\frac{t x}{2}+t \\ x \frac{d t}{d x} &=x\left(\frac{1}{2 t}+\frac{t}{2}\right) \\ x \frac{d t}{d x} &=x\left(\frac{1+t^{2}}{2 t}\right) \\ \frac{2 t}{1+t^{2}} d t &=\int d x+C \\ \ln \left(1+t^{2}\right) &=x+c \\ t &=\frac{y}{x}\\ \text{So, }\ln \left(1+\frac{y^{2}}{x^{2}}\right)&=x+C \\ \text{At }x=1, y=0 \\ \ln \left(1+\frac{0}{1}\right) &=\ln (1)=0=1+C \\ C &=-1 \\ \text{So,} \ln \left(1+\frac{y^{2}}{x^{2}}\right)&=x-1 \end{aligned}
Question 10 |
Taylor series expansion of f(x)=\int_{0}^{x} e^{-({\frac{t^{2}}{2}})}dt at around x=0 has the form
f(x)=a_{0}+a_{1}x+a_{2}x^{2}+... .
The coefficient a_{2} (correct to two decimal places) is equal to _______.
f(x)=a_{0}+a_{1}x+a_{2}x^{2}+... .
The coefficient a_{2} (correct to two decimal places) is equal to _______.
0 | |
0.05 | |
1 | |
5 |
Question 10 Explanation:
\begin{aligned} f(x) &=\int_{0}^{x} e^{-t^{2} / 2} d t \\ f^{\prime}(x) &=e^{-x^{2} / 2} \text { and } f^{\prime \prime}(x)=e^{-x^{2} / 2}(-x) \\ f^{\prime \prime}(0) &=0 \\ a_{2} &=\frac{f^{\prime \prime}(0)}{2 !}=0 \end{aligned}
There are 10 questions to complete.