# GATE ME 2008

 Question 1
In the Taylor series expansion of $e^{x}$ about x=2, the coefficient of $(x-2)^4$ is
 A $\frac{1}{4!}$ B $\frac{2^4}{4!}$ C $\frac{e^2}{4!}$ D $\frac{e^4}{4!}$
Engineering Mathematics   Calculus
Question 1 Explanation:
$f(x)\text{in the neighbourhood of a}$
$f(x)=\Sigma b_{n}(x-a)^{n} \text{where n from 0 to 0}$
\begin{aligned}\quad\quad\quad\quad b_{n}&=\frac{f^{n}(a)}{n !} \\ f^{4}(x)&=e^{x} ; f^{4}(2)=e^{2}\\ \end{aligned}
$\therefore \text{Coefficient of} (x-2)^{4}=b_{4}=\frac{f^{4}(2)}{4 !}=\frac{e^{2}}{4 !}$
 Question 2
Given that $\ddot{x}+3x=0$, and x(0) = 1, $\dot{x}(0)=0$, what is x(1)?
 A -0.99 B -0.16 C 0.16 D 0.99
Engineering Mathematics   Differential Equations
Question 2 Explanation:
$\ddot{x}+3 x=0$
Auxiliary equation is
$m^{2}+3=0$
$i.e. \qquad m=\pm \sqrt{3}$
$\therefore \qquad x =A \cos \sqrt{3} t+B \sin \sqrt{3} t$
$\dot{x} =\sqrt{3}(B \cos \sqrt{3} t-A \sin \sqrt{3} t)$
$\text{At} \quad t=0$
$\qquad 1=A \qquad ...(i)$
$\qquad 0=B \qquad ...(ii)$
$\therefore \qquad x =\cos \sqrt{3} t$
$x(1) =\cos (\sqrt{3} )=0.99$

 Question 3
The value of $\lim_{x\rightarrow 8}\frac{x^{1/3}-2}{(x-8)}$
 A $1/16$ B $1/12$ C $1/8$ D $1/4$
Engineering Mathematics   Calculus
Question 3 Explanation:
\begin{aligned} \quad \quad(x-8) &=h(\text { say }) \\ \Rightarrow \quad x &=8+h \end{aligned}
$\; \therefore \lim_{h \rightarrow 0} \frac{(8+h)^{1 / 3}-2}{h}$
$\text{Above form in the} \left(\frac{0}{0}\right) \text{by putting the value h=0 }$
$\text{Applying L hospital rule}$
$\lim _{h \rightarrow 0} \frac{\frac{1}{3}(8+h)\left(\frac{1}{3}-1\right)}{1}=\frac{1}{3}(8)^{-\frac{2}{3}}=\frac{1}{12}$
 Question 4
A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?
 A $1/4$ B $1/2$ C $3/8$ D $3/4$
Engineering Mathematics   Probability and Statistics
Question 4 Explanation:
$P(H)=0.5$
Probability of getting head exactly 3 times is
$^{4} C_{3}(0.5)^{3}(0.5)=\frac{1}{4}$
 Question 5
The matrix $\begin{bmatrix} 1 &2 & 4\\ 3 & 0 &6 \\ 1& 1 & p \end{bmatrix}$ has one Eigenvalue equal to 3. The sum of the other two Eigenvalues is
 A p B p-1 C p-2 D p-3
Engineering Mathematics   Linear Algebra
Question 5 Explanation:
Sum of the eigen values of matrix is
= sum of diagonal values present in the matrix
$\therefore 1+0+p=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow p+1=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow \lambda_{2}+\lambda_{3}=p+1-3=p-2$

There are 5 questions to complete.