Question 1 |
If any two columns of a determinant P= \begin{vmatrix} 4 & 7 & 8\\ 3 & 1 & 5\\ 9& 6 & 2 \end{vmatrix} are interchanged, which one of the following statements regarding the value of the determinant is CORRECT?
Absolute value remains unchanged but sign will change. | |
Both absolute value and sign will change. | |
Absolute value will change but sign will not change | |
Both absolute value and sign will remain unchanged |
Question 1 Explanation:
Property of determinant: If any two row or column are interchanged, then mangnitude of determinant remains same but sign changes.
Question 2 |
Among the four normal distributions with probability density functions as shown below, which one has the lowest variance?
I | |
II | |
III | |
IV |
Question 2 Explanation:
We know that probability density function of normal distribution is given by
F(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left ( \frac{x-\mu }{\sigma } \right )^2}
For having lowest variance (\sigma ^2), standard deviation (\sigma ) will be lowest. From the density function f(x), we can say that as σ decreases F(x) will increase, so curve having highest peak has lowest standard deviation and variance.
F(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left ( \frac{x-\mu }{\sigma } \right )^2}
For having lowest variance (\sigma ^2), standard deviation (\sigma ) will be lowest. From the density function f(x), we can say that as σ decreases F(x) will increase, so curve having highest peak has lowest standard deviation and variance.
Question 3 |
Simpson's 1/3 rule is used to integrate the function \frac{3}{5}x^{2}+\frac{9}{5} between x = 0 and x = 1 using the least number of equal sub -intervals. The value of the integral is _____________
1 | |
2 | |
3 | |
4 |
Question 3 Explanation:
f(x)=\frac{3}{5} x^{2}+\frac{9}{5}
\begin{array}{|c|c|c|c|}\hline x & 0 & 0.5 & 1 \\\hline f(x) & 1.8 & 1.95 & 2.4 \\\hline\end{array}\\ \Rightarrow\int_{0}^{1} f(x)=\frac{h}{3}\left[y_{0}+4 y_{1}+y_{2}\right]
=\frac{0.5}{3}[1.8+4(1.95)+2.4]=2
\begin{array}{|c|c|c|c|}\hline x & 0 & 0.5 & 1 \\\hline f(x) & 1.8 & 1.95 & 2.4 \\\hline\end{array}\\ \Rightarrow\int_{0}^{1} f(x)=\frac{h}{3}\left[y_{0}+4 y_{1}+y_{2}\right]
=\frac{0.5}{3}[1.8+4(1.95)+2.4]=2
Question 4 |
The value of \lim_{x\rightarrow 0}\frac{1-cos(x^{2})}{2x^{4}} is
0 | |
1/2 | |
1/4 | |
undefined |
Question 4 Explanation:
\lim _{x \rightarrow 0} \frac{1-\cos \left(x^{2}\right)}{2 x^{4}} \quad putting the x \rightarrow 0
we get \frac{0}{0} form
Applying L' Hospital rule
\Rightarrow \lim _{x \rightarrow 0} \frac{2 x \sin \left(x^{2}\right)}{8 x^{3}} \quad
\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{4 x^{2}}
\Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}
\Rightarrow \frac{1}{4} \lim _{x^{2} \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}=\frac{1}{4} \times 1=\frac{1}{4}
we get \frac{0}{0} form
Applying L' Hospital rule
\Rightarrow \lim _{x \rightarrow 0} \frac{2 x \sin \left(x^{2}\right)}{8 x^{3}} \quad
\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{4 x^{2}}
\Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}
\Rightarrow \frac{1}{4} \lim _{x^{2} \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}=\frac{1}{4} \times 1=\frac{1}{4}
Question 5 |
Given two complex numbers z_{1}=5+(5\sqrt{3})i and z_{2}=2/\sqrt{3}+2i, the argument of z_{1}/z_{2} in degrees is
0 | |
30 | |
60 | |
90 |
Question 5 Explanation:
\begin{aligned} z_{1} &=5+(5 \sqrt{3}) i \\ z_{2} &=\frac{2}{\sqrt{3}}+2 i \\ \arg \left(z_{1}\right) &=\theta_{1}=\tan ^{-1}\left(\frac{5 \sqrt{3}}{5}\right) \\ \theta_{1} &=60^{\circ} \\ \arg \left(z_{2}\right) &=\theta_{2}=\tan ^{-1}\left(\frac{2}{2 \sqrt{3}}\right) \\ \theta_{2} &=60^{\circ} \\ \arg \left(\frac{z_{1}}{z_{2}}\right) &=\arg \left(z_{1}\right)-\arg \left(z_{2}\right) \\ &=60-60=0^{\circ} \end{aligned}
There are 5 questions to complete.