# GATE ME 2015 SET-1

 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?
 A Absolute value remains unchanged but sign will change. B Both absolute value and sign will change. C Absolute value will change but sign will not change D Both absolute value and sign will remain unchanged
Engineering Mathematics   Linear Algebra
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? A I B II C III D IV
Engineering Mathematics   Probability and Statistics
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.

 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 _____________
 A 1 B 2 C 3 D 4
Engineering Mathematics   Numerical Methods
Question 3 Explanation:
$f(x)=\frac{3}{5} x^{2}+\frac{9}{5}$
$\begin{array}{|c|c|c|c|}\hline x & 0 & 0.5 & 1 \\\hline f(x) & 1.8 & 1.95 & 2.4 \\\hline\end{array}\\ \Rightarrow\int_{0}^{1} f(x)=\frac{h}{3}\left[y_{0}+4 y_{1}+y_{2}\right]$
$=\frac{0.5}{3}[1.8+4(1.95)+2.4]=2$
 Question 4
The value of $\lim_{x\rightarrow 0}\frac{1-cos(x^{2})}{2x^{4}}$ is
 A 0 B $1/2$ C $1/4$ D undefined
Engineering Mathematics   Calculus
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
$\Rightarrow \lim _{x \rightarrow 0} \frac{2 x \sin \left(x^{2}\right)}{8 x^{3}} \quad$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{4 x^{2}}$
$\Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}$
$\Rightarrow \frac{1}{4} \lim _{x^{2} \rightarrow 0} \frac{\sin \left(x^{2}\right)}{x^{2}}=\frac{1}{4} \times 1=\frac{1}{4}$
 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
 A 0 B 30 C 60 D 90
Engineering Mathematics   Complex Variables
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.