Question 1 |
Numerically integrate, f(x)=10 x-20 x^{2} from lower limit a=0 to upper limit b=0.5. Use Trapezoidal rule with five equal subdivisions. The value (in units,round off to two decimal places) obtained is ____________
0.78 | |
0.65 | |
0.4 | |
0.56 |
Question 1 Explanation:
\begin{aligned} y&=10 x-20 x^{2} \\ a&=0, b=0.5, n=5 \\ \text { So, } \qquad \qquad \qquad h&=\frac{b-a}{n}=0.1 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
Question 2 |
For two n-dimensional real vectors P and Q, the operation s(P,Q) is defined as follows:
s(P,Q) = \displaystyle \sum_{i=1}^n (P[i] \cdot Q[i])
Let \mathcal{L} be a set of 10-dimensional non-zero real vectors such that for every pair of distinct vectors P,Q \in \mathcal{L}, s(p,Q)=0. What is the maximum cardinality possible for the set \mathcal{L}?
s(P,Q) = \displaystyle \sum_{i=1}^n (P[i] \cdot Q[i])
Let \mathcal{L} be a set of 10-dimensional non-zero real vectors such that for every pair of distinct vectors P,Q \in \mathcal{L}, s(p,Q)=0. What is the maximum cardinality possible for the set \mathcal{L}?
9 | |
10 | |
11 | |
100 |
Question 2 Explanation:
Question 3 |
A function is defined in Cartesian coordinate system as f(x, y)=x e^{y}. The value of the directional derivative of the function (in integer) at the point (2,0) along the direction of the straight line segment from point (2,0) to point \left(\frac{1}{2}, 2\right) is _______
0.5 | |
1 | |
1.5 | |
2.2 |
Question 3 Explanation:
\begin{aligned} f(x, y)&=x e^{y}\\ \mathrm{P}(2,0) \text{ and } Q\left(\frac{1}{2}, 2\right)\\ \operatorname{grad} f &=\hat{i}\left(e^{y}\right)+\hat{j}\left(x e^{y}\right)+\hat{k}(0) \\ \Rightarrow \qquad \qquad \qquad (\operatorname{grad} f)_{p} &=\hat{i}+2 \hat{j} ]\\ \overline{P Q}&=\left(\frac{1}{2}-2\right) \hat{i}+(2-0) \hat{j}=-\frac{3}{2} \hat{i}+2 \hat{j}\\ \text{Required directional derivative} &=(\text { grad } f)_{P} \widehat{P Q}\\ &=(\hat{i}+2 \hat{j}) \times \frac{\left(-\frac{3}{2} \hat{i}+2 \hat{j}\right)}{\sqrt{\frac{9}{4}+4}}=\frac{\frac{-3}{2}+4}{\sqrt{\frac{25}{4}}} \\ &=\frac{\frac{-3+8}{2}}{\left(\frac{5}{2}\right)}=1 \end{aligned}
Question 4 |
Find the positive real root of x^3-x-3=0 using Newton-Raphson method. If the starting guess (x_0) is 2, the numerical value of the root after two iterations (x_2) is _______ (round off to two decimal places).
1.67 | |
1.12 | |
2.44 | |
3.25 |
Question 4 Explanation:
\begin{aligned} \text { Given, }\qquad\quad f(x) &=x^{3}-x-3, \quad x_{0}=2 \\ f^{\prime}(x)&=3 x^{2}-1\\ \text { Iteration 1: } \quad x_{1}&=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)}=2-\frac{(8-2-3)}{3(4)-1}=1.72\\ \text { Iteration 2: } \quad x_{2}&=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}=1.72-\frac{\left(1.72^{3}-1.72-3\right)}{3(1.72)^{2}-1}=1.67 \end{aligned}
Question 5 |
Consider the following differential equation
(1+y)\frac{dy}{dx}=y
The solution of the equation that satisfies the condition is y(1)=1 is
(1+y)\frac{dy}{dx}=y
The solution of the equation that satisfies the condition is y(1)=1 is
2ye^y=e^x+e | |
y^2e^y=e^x | |
ye^y=e^x | |
(1+y)e^y=2e^x |
Question 5 Explanation:
\begin{aligned} (1+y) \frac{d y}{d x} &=y \\ \Rightarrow\qquad \left(\frac{1}{y}+1\right) d y &=d x \\ \Rightarrow\qquad \log y+y &=x+c \\ \text { Using, } \qquad y(1) &=1 \\ \quad \log 1+1 &=1+c \quad \Rightarrow c=0 \\ \text { Hence, } \quad \log y+y &=x \\ \Rightarrow\qquad \log y+y \operatorname{loge} &=x \\ \log _{\mathrm{e}}\left(y \cdot e^{y}\right) &=x \\ \Rightarrow\qquad y e^{y} &=e^{x} \end{aligned}
Question 6 |
Let the superscript T represent the transpose operation. Consider the function f(x)=\frac{1}{2}x^T Qx=r^Tx, \; \text{ where } x \text{ and }r \text{ are }n \times 1 vectors and Q is a symmetric n \times n matrix. The stationary point of f(x) is
Q^Tr | |
Q^{-1}r | |
\frac{r}{r^Tr} | |
r |
Question 6 Explanation:
\begin{aligned} \text{Let}\qquad Q=\left[\begin{array}{ll}a & c \\c & b\end{array}\right], x&=\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right], R=\left[\begin{array}{l}r_{1} \\r_{2}\end{array}\right] \\ F(x)&=\frac{1}{2}\left(x_{1}, x_{2}\right)\left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]-\left[r_{1} r_{2}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right] \\ &=\frac{1}{2}\left[a x_{1}^{2}+b x_{2}^{2}+2 c x_{1} x_{2}\right]-\left[r_{1} x_{1}+r_{2} x_{2}\right]\\ \text{i.e.}\qquad \qquad U\left(x_{1}, x_{2}\right)&=\frac{1}{2} a x_{1}^{2}+\frac{1}{2} b x_{2}^{2}+c_{1} x_{1} x_{2}-r_{1} x_{1}-r_{2} x_{2} \end{aligned}
Now, for critical point, \frac{\partial u}{\partial x_{1}}=0 and \frac{\partial u}{\partial x_{2}}=0
\Rightarrow \quad a_{1} x_{1}+c x_{2}-r_{1}=0 \quad \text { and } c x_{2}+c x_{1}-r_{2}=0
In matrix form we can write it as
\begin{aligned} \left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l} x_{1} \\m_{2}\end{array}\right] &=\left[\begin{array}{l}r_{1} \\ r_{2}\end{array}\right] \\\Rightarrow \qquad Q x &=r \end{aligned}
By multiplying both side by Q^{-1}
x=Q^{-1} r
Now, for critical point, \frac{\partial u}{\partial x_{1}}=0 and \frac{\partial u}{\partial x_{2}}=0
\Rightarrow \quad a_{1} x_{1}+c x_{2}-r_{1}=0 \quad \text { and } c x_{2}+c x_{1}-r_{2}=0
In matrix form we can write it as
\begin{aligned} \left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l} x_{1} \\m_{2}\end{array}\right] &=\left[\begin{array}{l}r_{1} \\ r_{2}\end{array}\right] \\\Rightarrow \qquad Q x &=r \end{aligned}
By multiplying both side by Q^{-1}
x=Q^{-1} r
Question 7 |
The smallest eigenvalue and the corresponding eigenvector of the matrix \left[\begin{array}{cc} 2 & -2 \\ -1 & 6 \end{array}\right], respectively, are
1.55 and \left\{\begin{array}{l} 2.00 \\ 0.45 \end{array}\right\} | |
2.00 and \left\{\begin{array}{l} 1.00 \\ 1.00 \end{array}\right\} | |
1.55 and \left\{\begin{array}{l} -2.55 \\ -0.45 \end{array}\right\} | |
1.55 and \left\{\begin{array}{c} 2.00 \\ -0.45 \end{array}\right\} |
Question 7 Explanation:
\begin{aligned} A&=\left[\begin{array}{cc} 2 & -2 \\ -1 & 6 \end{array}\right] \Rightarrow|A-\lambda I|=0 \\ \Rightarrow \qquad \lambda&=(4+\sqrt{6}) \text { and }(4-\sqrt{6})\\ A X&=\lambda X\\ (A-\lambda I) X&=0 \end{aligned}
{\left[\begin{array}{cc} 2-(4-\sqrt{6}) & -2 \\ -1 & 6-(4-\sqrt{6}) \end{array}\right] \left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] =\left[\begin{array}{l} 0 \\ 0 \end{array}\right]}
\begin{aligned} x_{1}&=\left(\frac{2}{-2+\sqrt{6}}\right) x_{2}\\ \text { Let, } \qquad x_{2}&=K \text { then } x_{1}=\left(\frac{2}{-2+\sqrt{6}}\right) \text { K }\\ \Rightarrow \qquad\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]&=\left[\begin{array}{c} \frac{2}{-2+\sqrt{6}} k \\ k \end{array}\right] \approx\left[\begin{array}{c} 2 \\ -2+\sqrt{6} \end{array}\right]=\left[\begin{array}{l} 2.00 \\ 0.45 \end{array}\right] \end{aligned}
{\left[\begin{array}{cc} 2-(4-\sqrt{6}) & -2 \\ -1 & 6-(4-\sqrt{6}) \end{array}\right] \left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right] =\left[\begin{array}{l} 0 \\ 0 \end{array}\right]}
\begin{aligned} x_{1}&=\left(\frac{2}{-2+\sqrt{6}}\right) x_{2}\\ \text { Let, } \qquad x_{2}&=K \text { then } x_{1}=\left(\frac{2}{-2+\sqrt{6}}\right) \text { K }\\ \Rightarrow \qquad\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]&=\left[\begin{array}{c} \frac{2}{-2+\sqrt{6}} k \\ k \end{array}\right] \approx\left[\begin{array}{c} 2 \\ -2+\sqrt{6} \end{array}\right]=\left[\begin{array}{l} 2.00 \\ 0.45 \end{array}\right] \end{aligned}
Question 8 |
The value of \int_{0}^{\pi /2}\int_{0}^{\cos \theta }r \sin \theta dr d\theta is
0 | |
\frac{1}{6} | |
\frac{4}{3} | |
\pi |
Question 8 Explanation:
\begin{aligned} I &=\int_{\theta=0}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=\cos \theta} r \sin \theta d r d \theta \\ &=\int_{\theta=0}^{\frac{\pi}{2}}\left[\frac{r^{2}}{2}\right]_{0}^{\infty \cos \theta} \times \sin \theta d \theta \\ &=\frac{1}{2} \int_{\theta}^{\frac{\pi}{2}} \sin \theta \cdot \cos ^{2} \theta d \theta \\ \text{Let}, \qquad\cos \theta &=t\\ -\sin \theta d \theta &=d t \\ \cos \theta &=t \\ \text{at},\qquad\theta &=\frac{\pi}{2} ; t=0 \\ \theta &=0, t=1 \\ &=\int_{1}^{0} \frac{-t^{2}}{2} d t \\ &=\frac{-1}{2}\left[\frac{t^{3}}{3}\right]_{1}^{0}=\frac{-1}{2} \times\left(\frac{-1}{3}\right) \\ &=\frac{-1}{2}\left[\frac{-1}{3}\right]=\frac{1}{6} \end{aligned}
Question 9 |
If k is a constant, the general solution of \frac{d y}{d x}-\frac{y}{x}=1 will be in the form of
y=x ln(kx) | |
y=k ln(kx) | |
y=x ln(x) | |
y=xk ln(k) |
Question 9 Explanation:
\begin{aligned} \frac{d y}{d x}-\frac{y}{x} &=1 \\ \frac{d y}{d x}+P y &=Q \\ P &=-\frac{1}{x}, Q=1 \\ I F &=e^{\int P d x}=e^{\int \frac{-1}{x} d x}=\frac{1}{x} \\ y(I F) &=\int Q(I F) d x+c \\ y\left(\frac{1}{x}\right) &=\int 1 \cdot \frac{1}{x} d x+\ln k \\ y &=x \ln (x k) \end{aligned}
Question 10 |
Suppose that P is a 4x5 matrix such that every solution of the equation Px=0 is a scalar multiple of \begin{bmatrix} 2 & 5 & 4 &3 & 1 \end{bmatrix}^T. The rank of P is _______
1 | |
2 | |
3 | |
4 |
Question 10 Explanation:
There are 10 questions to complete.
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