Question 1 |
Let, f(x,y,z)=4x^2+7xy+3xz^2. The direction in which the function f(x,y,z) increases most rapidly at point P = (1,0,2) is
20\hat{i}+7\hat{j} | |
20\hat{i}+7\hat{j}+12\hat{k} | |
20\hat{i}+12\hat{j} | |
20\hat{i} |
Question 1 Explanation:
Given:
f(x,y,z)=4x^2+7xy+3xz^2
The directional derivative at point P is given by
=\triangledown f|_{point \; P}
\therefore \; \triangledown f=(8x+7y+3z^2)\hat{i}+(0+7x+0)\hat{j}+(0+0+6xz)\hat{k}
at point (1, 0, 2)
\triangledown f|_{(1,0,2)}=20\hat{i}+7\hat{j}+12\hat{k}
The directional derivative at point P is given by
=\triangledown f|_{point \; P}
\therefore \; \triangledown f=(8x+7y+3z^2)\hat{i}+(0+7x+0)\hat{j}+(0+0+6xz)\hat{k}
at point (1, 0, 2)
\triangledown f|_{(1,0,2)}=20\hat{i}+7\hat{j}+12\hat{k}
Question 2 |
Let y^{2}-2y+1=x and \sqrt{x}+y=5. The value of x+\sqrt{y} equals _________. (Give the answer up to three decimal places)
3.732 | |
4.734 | |
5.734 | |
6.732 |
Question 2 Explanation:
\begin{aligned} \sqrt{x}+y&=5\\ \sqrt{x}+1+5x&=5\\ 2\sqrt{x}&=4\\ x&=4\\ y^2-2y+1&=x\\ (y-1)^2&=x\\ y-1&=\sqrt{x}\\ y&=1+\sqrt{x}\\ y&=1+\sqrt{4}=3\\ \therefore \;\;x+\sqrt{y}&=4+\sqrt{3}=5.732 \end{aligned}
Question 3 |
Let x and y be integers satisfying the following equations
2x^{2}+y^{2}=34
x+2y=11
The value of (x+y) is ________.
2x^{2}+y^{2}=34
x+2y=11
The value of (x+y) is ________.
4 | |
3 | |
7 | |
10 |
Question 3 Explanation:
\begin{aligned} x+2y=1\\ x=11-2y\\ 2x^2+y^2=34\\ 2(11-2y)^2+y^2=34\\ 24^2+8y^2-88y+y^2-34=0\\ 9y^2-88y+208=0\\ y=5.77,4\\ x=-0.54\\ x=11-2(4)=3\\ x=3,y=4\\ x+y=7 \end{aligned}
Question 4 |
Only one of the real roots of f(x)=x^{6}-x-1 lies in the interval 1\leq x\leq 2 and bisection method is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is ________.
1000 | |
100 | |
10 | |
1 |
Question 4 Explanation:
\begin{aligned} f(x)&=x^6-x-1\\ a&=1, b=2, \in =0.01=10^{-3} \end{aligned}
The minimum number of iterations by Bisection method is given by
\begin{aligned} \frac{|b-a|}{2^n} &\lt \in\\ \frac{2-1}{2^n} &\lt 10^{-3}\\ \frac{1}{2^n} &\lt \frac{1}{10^{-3}}\\ 2^n &\gt 10^3\\ n &\gt \frac{\ln 1000}{\ln 2}\\ n &\gt 9.96\\ n&=10 \end{aligned}
The minimum number of iterations by Bisection method is given by
\begin{aligned} \frac{|b-a|}{2^n} &\lt \in\\ \frac{2-1}{2^n} &\lt 10^{-3}\\ \frac{1}{2^n} &\lt \frac{1}{10^{-3}}\\ 2^n &\gt 10^3\\ n &\gt \frac{\ln 1000}{\ln 2}\\ n &\gt 9.96\\ n&=10 \end{aligned}
Question 5 |
A differential equation \frac{di}{dt}-0.2i=0 is applicable over -10\lt t \lt 10. If i(4)=10, then i(-5) is _____.
1 | |
0.5 | |
1.6 | |
2.4 |
Question 5 Explanation:
\begin{aligned} \frac{di}{dt} &=0.2i \\ \frac{di}{i}&=0.2 \; dt \\ \int \frac{di}{i} &=\int 0.2 \; dt \\ \log i &=0.2t+ \log C \\ \log i-\log C &=0.2t \\ \log \left ( \frac{i}{C} \right ) &= 0.2t\\ \frac{i}{C} &= e^{0.2t}\\ i &=C e^{0.2t} \\ i(4)=10\;\;i.e.\;\;i &=10 \text{ when } t=4 \\ 10&=Ce^{(0.2)4} \\ 10&= C(2.225)\\ C&=4.493 \\ i&=(4.493)e^{0.2t} \\ \text{ when, } i&=-5 \\ i&=(4.493)e^{(0.2)(-5)}\\ &=1.652 \end{aligned}
There are 5 questions to complete.