Question 1 |
The value of the integral
\int \int _D 3(x^2+y^2)dxdy
, where D is the shaded triangular region shown in the diagram, is _____ (rounded off to the nearest integer).
\int \int _D 3(x^2+y^2)dxdy
, where D is the shaded triangular region shown in the diagram, is _____ (rounded off to the nearest integer).
128 | |
1024 | |
512 | |
64 |
Question 1 Explanation:
\begin{aligned}
I&=\int_{0}^{4}\int_{-x}^{x}(3x^2+3y^2)dydx\\
&=\int_{0}^{4}\left [ 3x^y+\frac{3y^3}{3} \right ]_{-x}^xdx\\
&=\int_{0}^{4}[3x^2(2x)+2x^3]dx\\
&=\int_{0}^{4}8x^3 dx\\
&=2 \times 4^4\\
&=512
\end{aligned}
Question 2 |
The function f(x)=8 \log _e x-x^2+3 attains its minimum over the interval
[1,e] at x= _________.
(Here \log _e x is the natural logarithm of x=.)
(Here \log _e x is the natural logarithm of x=.)
2 | |
1 | |
e | |
\frac{1+e}{2} |
Question 2 Explanation:
f(x)=8 \log _e x-x^2+3 when
x \in [1,e]
Differentiating both side,
\begin{aligned} f(x)&=\frac{8}{x}-2x=0 \; where\; x \gt 0\\ f'(x)&=0\\ \frac{8}{x}-2x&=0\\ 8-2x^2&=0\\ x^2&=4\\ x&=\pm 2 \end{aligned}
\begin{aligned} f''(x)&=\frac{-8}{x^2}-2 \\ f''(2)&=-6 \lt 0\\ \end{aligned}
f(x) is maximum for x = 2
Minimum of f(x) will be in [1, e] = min [f(1), f(e)]
f(e)=8 \ln e -e^2+3=3.61
Hence, minimum value of f(x) occurs at x=1
Differentiating both side,
\begin{aligned} f(x)&=\frac{8}{x}-2x=0 \; where\; x \gt 0\\ f'(x)&=0\\ \frac{8}{x}-2x&=0\\ 8-2x^2&=0\\ x^2&=4\\ x&=\pm 2 \end{aligned}
\begin{aligned} f''(x)&=\frac{-8}{x^2}-2 \\ f''(2)&=-6 \lt 0\\ \end{aligned}
f(x) is maximum for x = 2
Minimum of f(x) will be in [1, e] = min [f(1), f(e)]
f(e)=8 \ln e -e^2+3=3.61
Hence, minimum value of f(x) occurs at x=1
Question 3 |
Consider the two-dimensional vector field \vec{F}(x,y)=x\vec{i}+y\vec{j}, where \vec{i} and \vec{j} denote
the unit vectors along the x-axis and the y-axis, respectively. A contour C in the x-y plane, as shown in the figure, is composed of two horizontal lines connected at the
two ends by two semicircular arcs of unit radius. The contour is traversed in the
counter-clockwise sense. The value of the closed path integral
\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})
\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})
0 | |
1 | |
8+2 \pi | |
-1 |
Question 3 Explanation:
\oint \vec{F} (x,y)\cdot [dx\vec{i}+dy\vec{j}]
Given \vec{F} (x,y)=x\vec{i}+y\vec{j}
\therefore \int_{c}xdx+ydy=0
Because here vector is conservative.
If the integral function is the total derivative over the closed contoure then it will be zero
Given \vec{F} (x,y)=x\vec{i}+y\vec{j}
\therefore \int_{c}xdx+ydy=0
Because here vector is conservative.
If the integral function is the total derivative over the closed contoure then it will be zero
Question 4 |
Consider the integral
\oint _{c}\frac{sin\left ( x \right )}{x^{2}\left ( x^{2}+4 \right )}dx
where C is a counter-clockwise oriented circle defined as \left | x-i \right |=2. The value of the integral is
\oint _{c}\frac{sin\left ( x \right )}{x^{2}\left ( x^{2}+4 \right )}dx
where C is a counter-clockwise oriented circle defined as \left | x-i \right |=2. The value of the integral is
-\frac{\pi }{8}\sin\left ( 2i \right ) | |
\frac{\pi }{8}\sin\left ( 2i \right ) | |
-\frac{\pi }{4}\sin\left ( 2i \right ) | |
\frac{\pi }{4}\sin\left ( 2i \right ) |
Question 4 Explanation:
MARKS TO ALL AS PER IIT ANSWER KEY
\oint_{c} \frac{\sin x}{x^{2}\left(x^{2}+4\right)} d x, c:|x-i|=2
Poles are given by x^{2}=0 and x^{2}+4=0
\Rightarrow\qquad x=0 is a pole of order '2'
x=2 i are simple nodes
x=0 lies inside 'c'
x=2 i lies inside 'c'
x=-2 i lies outside 'c'
\begin{aligned} \text{Res}_{0} &=\frac{1}{(2-1)} \lim _{x \rightarrow 0} \frac{d}{d z}\left[(x-0)^{2} \frac{\sin x}{x^{2}\left(x^{2}+4\right)}\right] \\ &=\lim _{x \rightarrow 0} \frac{\left(x^{2}+4\right) \cos x-\sin x(2 x)}{\left(x^{2}+4\right)^{2}}=\frac{1}{4} \\ \text{Res}_{2 i} &=\lim _{x \rightarrow 2 i}(x-2 i) \frac{\sin x}{x^{2}(x-2 i)(x+2 i)}=\frac{\sin (2 i)}{(-4)(4 i)} \\ \text{By CRT}\quad \oint_{c} f d x &=2 \pi i\left[\text{Res}_{0}+\text{Res}_{2 i}\right]=2 \pi i\left[\frac{1}{4}+\frac{\sin (2 i)}{-16}\right] \end{aligned}
\oint_{c} \frac{\sin x}{x^{2}\left(x^{2}+4\right)} d x, c:|x-i|=2
Poles are given by x^{2}=0 and x^{2}+4=0
\Rightarrow\qquad x=0 is a pole of order '2'
x=2 i are simple nodes
x=0 lies inside 'c'
x=2 i lies inside 'c'
x=-2 i lies outside 'c'
\begin{aligned} \text{Res}_{0} &=\frac{1}{(2-1)} \lim _{x \rightarrow 0} \frac{d}{d z}\left[(x-0)^{2} \frac{\sin x}{x^{2}\left(x^{2}+4\right)}\right] \\ &=\lim _{x \rightarrow 0} \frac{\left(x^{2}+4\right) \cos x-\sin x(2 x)}{\left(x^{2}+4\right)^{2}}=\frac{1}{4} \\ \text{Res}_{2 i} &=\lim _{x \rightarrow 2 i}(x-2 i) \frac{\sin x}{x^{2}(x-2 i)(x+2 i)}=\frac{\sin (2 i)}{(-4)(4 i)} \\ \text{By CRT}\quad \oint_{c} f d x &=2 \pi i\left[\text{Res}_{0}+\text{Res}_{2 i}\right]=2 \pi i\left[\frac{1}{4}+\frac{\sin (2 i)}{-16}\right] \end{aligned}
Question 5 |
The vector function F\left ( r \right )=-x\hat{i}+y\hat{j}
is defined over a circular arc C shown in the figure.
The line integral of \int _{C} F\left ( r \right ).dr is
The line integral of \int _{C} F\left ( r \right ).dr is
\frac{1}{2} | |
\frac{1}{4} | |
\frac{1}{6} | |
\frac{1}{3} |
Question 5 Explanation:
\begin{aligned} \bar{F} &=-x i+y j \\ \int \vec{F} \cdot \overrightarrow{d r} &=\int_{c}-x d x+y d y \\ &=\int_{\theta=0}^{45^{\circ}}(-\cos \theta(-\sin \theta)+\sin \theta \cos \theta) d \theta \\ \int_{\theta=0}^{\pi / 4} \sin 2 \theta d \theta &\left.=-\frac{\cos 2 \theta}{2}\right]_{0}^{\pi / 4} \\ &=-\frac{1}{2}[0-1]=\frac{1}{2} \end{aligned}
Question 6 |
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 6 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 7 |
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 7 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 8 |
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 8 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 9 |
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 9 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 10 |
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 10 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}
There are 10 questions to complete.