GATE ME 2014 SET-2

The characteristic equation | A-\lambda D=0
\text{i.e.} \left|\begin{array}{cc}-5-\lambda & 2 \\ -9 & 6-\lambda\end{array}\right|=0
\text{or } (\lambda-6)(\lambda+5)+18=0
\text{or } \quad \lambda^{2}-6 \lambda+5 \lambda-30+18=0
\text{or } \quad \lambda^{2}-\lambda-12=0
\text{or } \lambda=\frac{1 \pm \sqrt{1+48}}{2}=\frac{1 \pm 7}{2}=4,-3
Corresponding to \lambda=4, we have
\; [A-\lambda I] x=\left[\begin{array}{cc}-5-\lambda & 2 \\-9 & 6-\lambda\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]
=0
\text{Or, } \left[\begin{array}{ll}-9 & 2 \\ -9 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=0which gives only one independent equation,
-9 x+2 y=0
\therefore \frac{x}{2}=\frac{y}{9} gives eigen vector (2,9)
Corresponding to \lambda=-3,
=\left[\begin{array}{ll}-2 & 2 \\-9 & 9\end{array}\right]\left[\begin{array}{l}x \\y \end{array}\right]=0
which gives -x+y=0 (only one independent equation)
\therefore \quad \frac{x}{1}=\frac{y}{1} \quad which gives (1,1)
So, the eigen vectors are \left\{\begin{array}{l}2 \\ 9\end{array}\right\} and \left\{\begin{array}{l}1 \\ 1\end{array}\right\}

\lim_{x \rightarrow 0} \frac{\left(e^{2 x}-1\right)}{\sin 4 x}, it is of\left(\frac{0}{0}\right)from
Applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{2 e^{2 x}}{4 \cos 4 x}=\frac{2 \times 1}{4 \times 1}=\frac{1}{2}

\vec{F}=x^{2} z^{2} \hat{i}-2 x y^{2} z \hat{j}+2 y^{2} z^{3} \hat{k}
\nabla \times \vec{F}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{2} z^{2}&-2 x y^{2} z^{2} & 2 y^{2} z^{3} \end{array}\right|
=\hat{i}\left[\frac{\partial}{\partial y}\left(2 y^{2} z^{3}\right)+\frac{\partial}{\partial z}\left(2 x y^{2} z\right)\right]
-\hat{j}\left[\frac{\partial}{\partial x}\left(2 y^{2} z^{3}\right)-\frac{\partial}{\partial z}\left(x^{2} z^{2}\right)\right]
+\hat{k}\left[\frac{\partial}{\partial x}\left(-2 x y^{2} z\right)-\frac{\partial}{\partial y}\left(x^{2} z^{2}\right)\right]
\nabla \times \vec{F}=\hat{i}\left[4 y z^{3}+2 x y^{2}\right]-\hat{j}\left[-2 z x^{2}\right]
=\left(4 y z^{3}+2 x y^{2}\right) \hat{i}+\left(2 x^{2} z\right) \hat{j}-\left(2 y^{2} z\right) \hat{k}

required prob =\frac{^{15} C_{2}}{^{25} C_{2}}=\frac{14 \times 15}{25 \times 24}=\frac{7}{20}

f(x)=e^{-x} \sin (100 x)
P(x)=-e^{-x} \sin (100 x)+e^{-x} \cos (100 x) \times 100
for minima f^{\prime}(x)=0
from here \tan (100 x)=100
\begin{aligned} x &=\frac{1}{100} \tan ^{-1}(100)=0.0156 \\ f(x) &=e^{-0.0156} \sin (100 \times 0.0156)=-0.9844 \end{aligned}