Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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
Question 6 |
Let f be a real-valued function of a real variable defined as f (x)=x-[x], where [x]
denotes the largest integer less than or equal to x. The value of \int_{0.25}^{1.25}f(x)dx is _______ (up to 2 decimal places).
0.25 | |
0.5 | |
0.75 | |
0.85 |
Question 6 Explanation:
\begin{aligned} \int_{0.25}^{1.25} f(x)dx&=\int_{0.25}^{1}(x-[x])dx+\int_{1}^{1.25}(x-[x])dx \\ &= \int_{0.25}^{1.25}(xdx)-\left [ \int_{0.25}^{1}[x]dx+\int_{1}^{1.25}[x]dx \right ]\\ &=\left.\begin{matrix} \left ( \frac{x^2}{2} \right ) \end{matrix}\right|_{0.25}^{1.25}-\left ( \int_{0.25}^{1}0dx+\int_{1}^{1.25}1dx \right ) \\ &=\frac{(1.25)^2}{2}-\left ( \frac{(0.25)^2}{2} \right )-(0+0.25) \\ &=\frac{1}{2}[(1.5625-0.0625)]-0.25 \\ &= 0.5 \end{aligned}
Question 7 |
The value of the directional derivative of the function \phi (x,y,z)=xy^{2}+yz^{2}+zx^{2} at the
point (2,-1,1) in the direction of the vector p=i+ 2j + 2k is
1 | |
0.95 | |
0.93 | |
0.9 |
Question 7 Explanation:
\begin{aligned} \phi &= xy^2+yz^2+zx^2\\ \bigtriangledown \phi &=\bar{i}\frac{\partial \phi }{\partial x}+\bar{j}\frac{\partial \phi }{\partial y} +\bar{k}\frac{\partial \phi }{\partial z} \\ &=\bar{i}(y^2+2xz)+\bar{j}(2xy+z^2)\\&+\bar{k}(2yz+x^2) \\ \bigtriangledown \phi _{(2,-1,1)} &= \bar{i}(1+4)+\bar{j}(-4+1)+\bar{k}(-2+4)\\ &= 5\bar{i}-3\bar{j}+2\bar{k}\\ \bar{P} &= \bar{i}+2\bar{j}+2\bar{k}\\ |\bar{P}|&=\sqrt{1+4+4}=3 \end{aligned}
THe directional derivative of f(x,y,z) at (2,-1,1) in the direction of \bar{P} is \bigtriangledown \phi _{at \; P} \cdot \frac{\bar{P}}{|\bar{P}|}= (5\bar{i}-3\bar{j}+2\bar{k})\left ( \frac{\bar{i}+2\bar{j}+2\bar{k}}{3} \right )=\frac{5-6+4}{3}=1
THe directional derivative of f(x,y,z) at (2,-1,1) in the direction of \bar{P} is \bigtriangledown \phi _{at \; P} \cdot \frac{\bar{P}}{|\bar{P}|}= (5\bar{i}-3\bar{j}+2\bar{k})\left ( \frac{\bar{i}+2\bar{j}+2\bar{k}}{3} \right )=\frac{5-6+4}{3}=1
Question 8 |
Let f be a real-valued function of a real variable defined as f(x)=x^{2} for x \gt 0, and f(x)=-x^{2} for x \lt 0. Which one of the following statements is true?
f(x) is discontinuous at x=0 | |
f(x) is continuous but not differentiable at x=0 | |
f(x) is differentiable but its first derivative is not continuous at x=0 | |
f(x) is differentiable but its first derivative is not differentiable at x=0 |
Question 8 Explanation:
\begin{aligned} f(x)&=\left\{\begin{matrix} x^2 & x \geq 0\\ -x^2 & x \lt 0 \end{matrix}\right.\\ f'(x)&=\left\{\begin{matrix} 2x & x \geq 0\\ -2x & x \lt 0 \end{matrix}\right.\\ f''(x)&=\left\{\begin{matrix} 2 & x \geq 0\\ -2 & x \lt 0 \end{matrix}\right. \end{aligned}
The first derivation of f (i.e) f'(x) is not derivable at x=0.
The first derivation of f (i.e) f'(x) is not derivable at x=0.
Question 9 |
Let
g(x)=\left\{\begin{matrix} -x & x\leq 1\\ x+1, &x\geq 1 \end{matrix}\right. and
f(x)=\left\{\begin{matrix} 1-x & x\leq 0\\ x^{2}, & x \gt 0 \end{matrix}\right.
Consider the composition of f and g, i.e., (f\circ g)(x)=f(g(x)). The number of discontinuities in (f\circ g)(x) present in the interval (-\infty ,0) is:
g(x)=\left\{\begin{matrix} -x & x\leq 1\\ x+1, &x\geq 1 \end{matrix}\right. and
f(x)=\left\{\begin{matrix} 1-x & x\leq 0\\ x^{2}, & x \gt 0 \end{matrix}\right.
Consider the composition of f and g, i.e., (f\circ g)(x)=f(g(x)). The number of discontinuities in (f\circ g)(x) present in the interval (-\infty ,0) is:
0 | |
1 | |
2 | |
4 |
Question 9 Explanation:
\begin{aligned} f(x)=1-x; &x \lt 0\\ g(x)=-x;&x \lt 0 \end{aligned}
(Noth are continous for x \lt 0)
Therefore, fog(x) is continous for x \lt 0
The composite function of two continous function is always continous . Therefore the number of discontinuities are zero.
(Noth are continous for x \lt 0)
Therefore, fog(x) is continous for x \lt 0
The composite function of two continous function is always continous . Therefore the number of discontinuities are zero.
Question 10 |
Consider a function f (x, y,z) given by
f(x,y,z)=(x^{2}+y^{2}-2z^{2})(y^{2}+z^{2})
The partial derivative of this function with respect to x at the point, x = 2, y = 1 and z = 3 is ________
f(x,y,z)=(x^{2}+y^{2}-2z^{2})(y^{2}+z^{2})
The partial derivative of this function with respect to x at the point, x = 2, y = 1 and z = 3 is ________
13 | |
40 | |
36 | |
4 |
Question 10 Explanation:
\begin{aligned} f(x,y,z)&=(x^2+y^2-2z^2)(y^2+z^2)\\ \frac{\partial f}{\partial x}&=(x^2+y^2-2z^2)(0)+(y^2+z^2)(2x)\\ &=0+(y^2+z^2)(2x)\\ \left.\begin{matrix} \frac{\partial f}{\partial x} \end{matrix}\right|_{x=2,y=1,z=3}&=(1+9)(2)(2)=40 \end{aligned}
There are 10 questions to complete.