# GATE EC 2006

 Question 1
The rank of the matrix $\begin{bmatrix} 1& 1& 1\\ 1& -1&0 \\ 1& 1& 1 \end{bmatrix}$ is
 A 0 B 1 C 2 D 3
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
$\begin{array}{l} R_{3} \rightarrow R_{1}-R_{3} \\ {\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right]}\\ \therefore \text{Rank}=2 \end{array}$
 Question 2
$\bigtriangledown \times \bigtriangledown \times P$, where P is a vector, is equal to
 A $P \times \bigtriangledown \times P-\bigtriangledown ^{2}P$ B $\bigtriangledown^{2} P+\bigtriangledown (\bigtriangledown \cdot P)$ C $\bigtriangledown^{2} P+\bigtriangledown \times P$ D $\bigtriangledown (\bigtriangledown \cdot P)-\bigtriangledown ^{2}P$
Engineering Mathematics   Calculus
Question 2 Explanation:
From vector triple product.
\begin{aligned} A \times(B \times C) &=B(A . C)-C(A . B) \\ A &=\nabla . B=\nabla, C=P \\ \nabla \times \nabla \times P &=(\nabla P)-P(\nabla \nabla)=\nabla(\nabla P)-\nabla^{2} P \end{aligned}

 Question 3
$\int \int (\bigtriangledown \times P)\cdot ds$, where P is a vector, is equal to
 A $\oint P\cdot dl$ B $\oint \bigtriangledown \times \bigtriangledown \times P\cdot dl$ C $\oint \bigtriangledown \times P\cdot dl$ D $\int \int \int \bigtriangledown \cdot Pdv$
Engineering Mathematics   Calculus
Question 3 Explanation:
$\iint(\Delta x P) d s=\oint P .d l$ (strokes Theorem)
 Question 4
A probability density function is of the form
$p(x)=Ke^{-\alpha |x|},x\in (-\infty ,\infty )$
The value of K is
 A 0.5 B 1 C $0.5\alpha$ D $\alpha$
Engineering Mathematics   Probability and Statistics
Question 4 Explanation:
\begin{aligned} \int_{-\infty}^{\infty} p(x) d x&=1 \\ \int_{-\infty}^{\infty} K e^{-\alpha|x|} d x&=1 \\ \int_{-\infty}^{0} K e^{\mu x} d x+\int_{0}^{\infty} K e^{-u x}&=1\\ \Rightarrow \quad\frac{K}{\alpha}\left[e^{\alpha x}\right]_{-\infty}^{0}+\frac{K}{-\alpha}\left[e^{-\alpha x}\right]_{0}^{\infty}&=1 \\ \Rightarrow \quad \frac{K}{\alpha}+\frac{K}{\alpha}&=1\\ 2 K&=\alpha \\ \Rightarrow \quad K&=0.5 \alpha \end{aligned}
 Question 5
A solution for the differential equation
$\dot{x}(t)+2x(t)=\delta (t)$
with initial condition $x(0^-)=0$ is
 A $e^{-2t}u(t)$ B $e^{2t}u(t)$ C $e^{-t}u(t)$ D $e^{t}u(t)$
Engineering Mathematics   Differential Equations
Question 5 Explanation:
$\dot{x}(t)+2 x(t)=\delta(t)$
Taking L.T. on both sides
\begin{aligned} s X(s)-x(0)+2 X(s)&=1 \\ X(s)[s+2]&=1 \\ X(s)&=\frac{1}{s+2} \\ x(t) &=e^{-2 t} u(t) \end{aligned}

There are 5 questions to complete.