Question 1 |
Let \vec{E}(x,y,z)=2x^2\hat{i}+5y\hat{j}+3z\hat{k}. The value of \int \int \int _V(\vec{\triangledown }\cdot \vec{E})dV, where V is the
volume enclosed by the unit cube defined by 0\leq x\leq 1,0\leq y\leq 1 \text{ and }0\leq z\leq 1, is
3 | |
8 | |
10 | |
5 |
Question 1 Explanation:
Divergence of \vec{V}, \triangledown \cdot \vec{V}=4x+5+3=4x+8
Now,
\int \int \int( \triangledown \cdot \vec{V})dxdydz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(4x+8)dxdydz=\int_{0}^{1}(4x+8)dx=[2x^2+8x]_0^1=10
Now,
\int \int \int( \triangledown \cdot \vec{V})dxdydz=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}(4x+8)dxdydz=\int_{0}^{1}(4x+8)dx=[2x^2+8x]_0^1=10
Question 2 |
Let f(x)=\int_{0}^{x}e^t(t-1)(t-2)dt . Then f(x) decreases in the interval
x \in (1,2) | |
x \in (2,3) | |
x \in (0,1) | |
x \in (0.5,1) |
Question 2 Explanation:
The function is decreasing, if
f'(x) \lt 0
f'(x)=\frac{d}{dx} \int_{0}^{x}e^t(t-1)(t-2)dt
\Rightarrow \; e^x(x-1)(x-2) \lt 0
It is possible in between 1 & 2. Hence x \in (1,2)
f'(x)=\frac{d}{dx} \int_{0}^{x}e^t(t-1)(t-2)dt
\Rightarrow \; e^x(x-1)(x-2) \lt 0
It is possible in between 1 & 2. Hence x \in (1,2)
Question 3 |
In the open interval \left ( 0,1 \right ), the polynomial p\left ( x \right) =x^{4}-4x^{3}+2 has
two real roots | |
one real root | |
three real roots | |
no real roots |
Question 3 Explanation:
\begin{aligned} x^{4}+2 &=4 x^{3} \\ f_{1}(x) &=x^{4}+2 \\ f_{2}(x) &=4 x^{3} \end{aligned}
It is clear that point of intersection of these graphs is solution (or) root of p(x) = 0

According to intermediate value theorem
P(0) and P(1) are having opposite signs
\therefore a root of p(x) = 0 in (0, 1)
and also from graph, there is only one point of intersection
Hence exactly one real root exists in (0, 1)
It is clear that point of intersection of these graphs is solution (or) root of p(x) = 0

According to intermediate value theorem
P(0) and P(1) are having opposite signs
\therefore a root of p(x) = 0 in (0, 1)
and also from graph, there is only one point of intersection
Hence exactly one real root exists in (0, 1)
Question 4 |
Suppose the circles x^{2}+y^{2}=1 and \left ( x-1\right )^{2}+\left ( y-1 \right )^{2}=r^{2} intersect each other orthogonally at the point \left ( u,v \right ). Then u+v= _______________.
0 | |
1 | |
2 | |
3 |
Question 4 Explanation:
If two curves cut orthogonally then product of slopes = -1
\begin{aligned} x^{2}+y^{2} &=1 \\ 2 x+2 y \frac{d y}{d x} &=0 \\ \left(M_{1}\right)_{u, v} &=\frac{d y}{d x}=\frac{-x}{y}=\frac{-u}{v} \\ (x-1)^{2}+(y-1)^{2} &=r^{2} \\ \left(M_{2}\right)_{u, v} &=\left(\frac{d y}{d x}\right)=\left(\frac{-2(x-1)}{2(y-1)}\right)_{(u . v)}=\frac{1-u}{v-1} \\ \because \qquad \qquad M_{1} M_{2} &=-1\\ \frac{-u}{v} \times \frac{1-u}{v-1}&=-1\\ -u+u^{2} &=-v^{2}+v \\ u^{2}+v^{2} &=v+u \\ \because \qquad \qquad x^{2}+y^{2} &=1 \\ \therefore \qquad \qquad u+v &=1 \end{aligned}
\begin{aligned} x^{2}+y^{2} &=1 \\ 2 x+2 y \frac{d y}{d x} &=0 \\ \left(M_{1}\right)_{u, v} &=\frac{d y}{d x}=\frac{-x}{y}=\frac{-u}{v} \\ (x-1)^{2}+(y-1)^{2} &=r^{2} \\ \left(M_{2}\right)_{u, v} &=\left(\frac{d y}{d x}\right)=\left(\frac{-2(x-1)}{2(y-1)}\right)_{(u . v)}=\frac{1-u}{v-1} \\ \because \qquad \qquad M_{1} M_{2} &=-1\\ \frac{-u}{v} \times \frac{1-u}{v-1}&=-1\\ -u+u^{2} &=-v^{2}+v \\ u^{2}+v^{2} &=v+u \\ \because \qquad \qquad x^{2}+y^{2} &=1 \\ \therefore \qquad \qquad u+v &=1 \end{aligned}
Question 5 |
Let f\left ( x \right )
be a real-valued function such that {f}'\left ( x_{0} \right )=0
for some x _{0} \in\left ( 0,1 \right ), and {f}''\left ( x \right )> 0
for all x \in \left ( 0,1 \right ). Then f\left ( x \right )
has
no local minimum in (0,1) | |
one local maximum in (0,1) | |
exactly one local minimum in (0,1) | |
two distinct local minima in (0,1) |
Question 5 Explanation:
x_{0} \in(0,1), where f(x)=0 is stationary point
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
Question 6 |
Let a_x \; and \; a_y be unit vectors along x and y directions, respectively. A vector function is given by
F = a_xy - a_yx
The line integral of the above function
\int _c F\cdot dl
along the curve C, which follows the parabola y = x^2 as shown below is _______ (rounded off to 2 decimal places).

F = a_xy - a_yx
The line integral of the above function
\int _c F\cdot dl
along the curve C, which follows the parabola y = x^2 as shown below is _______ (rounded off to 2 decimal places).

2 | |
-2 | |
3 | |
-3 |
Question 6 Explanation:
\begin{aligned}
\vec{F}&=y\hat{a}_{x}-x\hat{a}_{y} \\ \vec{r}&=x\hat{i}+y\hat{j} \\ \vec{F}&=y\hat{i}-x\hat{j} \\ d\vec{r}&=dx\hat{i}+dy\hat{j} \\ &=\int _{c}F.d\vec{r}\\ &=\int_{c}F_{1}dx+F_{2}dy \\ &=\int _{c}ydx-xdy \\ &\text{Where C is,}\\ y&=x^{2} \\ dy&=2x dx \\ x &\text{ varies from -1 to 2,}\\ \int _{c}\vec{F}dl&=\int_{-1}^{2}x^{2}dx-x\cdot 2xdx \\ &=\int_{-1}^{2}(x^{2}-2x)dx \\ &=\int_{-1}^{2}-x^{2}dx=\left.\begin{matrix} -\frac{x^{3}}{3} \end{matrix}\right|_{-1}^{2} \\ &=-\frac{8}{3}-\frac{1}{3}=-\frac{9}{3}\\ &=-3
\end{aligned}
Question 7 |
If A=2xi+3yj+4zk and u=x^2+y^2+z^2, then div(uA) at (1,1,1) is____
15 | |
45 | |
30 | |
60 |
Question 7 Explanation:
\begin{aligned} \bigtriangledown \cdot (uA)&=u(\bigtriangledown \cdot A)+(\bigtriangledown A)F\\ &=(x^2+y^2+z^2)[2+3+4]\\ &+(2x\hat{i}+2y\hat{j}+2z\hat{k})(2x\hat{i}+3y\hat{j}+4z\hat{k})\\ &=9(x^2+y^2+z^2)+(4x^2+6y^2+8z^2)\\ At\; (1,1,1)&=9(3)+[4+6+8]\\ &=27+18=45 \end{aligned}
Question 8 |
If f=2x^3+3y^2+4z, the value of line integral \int _c grad \; f\cdot dr evaluated over contour C formed by the segments (-3,-3,2) \rightarrow (2,-3,2) \rightarrow (2,6,2) \rightarrow (2,6,-1) is_______
112 | |
139 | |
156 | |
186 |
Question 8 Explanation:
\begin{aligned} f&=2x^3+3y^2+4z\\ \Delta f&=6x^2\hat{i}+6y\hat{j}+4\hat{k}\\ \text{curl} (\Delta f)&=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ 6x^2& 6y & 4\end{vmatrix}\\ \therefore \;\;\int _c\Delta f\cdot dr&=\int _c d(2x^3+3y^2+4z)=0\\ &=\int_{(-3,-3,2)}^{2,-3,2}d(2x^3+3y^2+4z)\\ &+\int_{(2,-3,2)}^{(2,6,2)}d(2x^3+3y^2+4z)\\ &+\int_{(2,6,2)}^{(2,6,-1)}d(2x^3+3y^2+4z)\\ &=70+81+(-12)=139 \end{aligned}
Question 9 |
Let f(x)=3x^{3}-7x^{2}+5x+6. The maximum value of f(x) over the interval [0, 2] is _______ (up to 1 decimal place).
8.2 | |
12.0 | |
16.2 | |
18.7 |
Question 9 Explanation:
\begin{aligned} f(x)&=3x^3-7x^2+5x+6\\ f'(x)&=9x^2-14x+5\\ f''(x)&=18x-14\\ f'(x)&=0\\ 9x^2-14x+5&=0\\ x&=1,0.55\\ x&=1\\ f''(1)&=18-14=4 \gt 0\; \text{minima}\\ x&=0.55\\ f''(0.55)&=-4.1 \lt 0 \; \text{maxima} \end{aligned}
Maximum [f(0), f'(0.55), f(2)]
Maximum [6,7.13,12]=12
Maximum [f(0), f'(0.55), f(2)]
Maximum [6,7.13,12]=12
Question 10 |
As shown in the figure, C is the arc from the point (3,0) to the point (0,3) on the circle x^{2}+y^{2}=9. The value of the integral \int_{C}(y^{2}+2yx)dx+(2xy+x^{2})dy is _____ (up to 2
decimal places).


0 | |
0.11 | |
0.25 | |
0.66 |
Question 10 Explanation:
\begin{aligned} x^2+y^2&=9\\ x&=3 \cos \theta \\ y&=3 \sin \theta \\ dx&=-3 \sin \theta \; d\theta\\ dy&=3 \cos \theta \;d\theta \end{aligned}
\theta varies from 0 to \frac{\pi}{2}
\int (y^2+2xy)dx+(2xy+x^2)dy =\int_{0}^{\pi/2}(9 \sin ^2 \theta +18 \sin \theta \cos \theta )(-3 \sin \theta d\theta )+(18 \sin \theta \cos \theta +9 \cos ^2 \theta )(3 \cos \theta )d\theta
=\int_{0}^{\pi/2}(-27\sin ^3 \theta -54 \sin ^2 \theta \cos \theta +54 \sin \theta \cos ^2 \theta +27 \cos ^3 \theta )d\theta =0
\theta varies from 0 to \frac{\pi}{2}
\int (y^2+2xy)dx+(2xy+x^2)dy =\int_{0}^{\pi/2}(9 \sin ^2 \theta +18 \sin \theta \cos \theta )(-3 \sin \theta d\theta )+(18 \sin \theta \cos \theta +9 \cos ^2 \theta )(3 \cos \theta )d\theta
=\int_{0}^{\pi/2}(-27\sin ^3 \theta -54 \sin ^2 \theta \cos \theta +54 \sin \theta \cos ^2 \theta +27 \cos ^3 \theta )d\theta =0
There are 10 questions to complete.