TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If a is a unit vector , then $|a \times \hat{i}|^{2} + |a \times \hat{j}|^{2}+ |a \times \hat{k}|^{2}=$

Answer:

Explanation:

We have,

$|a \times \hat{i}|^{2} + |a \times \hat{j}|^{2}+|a \times \hat{k}|^{2}$

 $(|a||\hat{i}| \sin \alpha)^{2}+(|a||\hat{j}| \sin \beta)^{2}+(|a||\hat{k}| \sin \gamma)^{2}$

$= \sin^{2} \alpha+\sin^{2} \beta+\sin^{2} \gamma$   [ $\because |a|=|\hat{i}|=|\hat{j}|=|\hat{k}|=1]$

$=1- \cos^{2} \alpha+1-\cos^{2} \beta+1-\cos^{2} \gamma$

  $=3-(\cos^{2}  \alpha+\cos^{2} \beta+ \cos ^{2} \gamma)$

= 3-1=2       $[ \because   \cos^{2} \alpha+\cos^{2} \beta+ \cos^{2} \gamma=1]$

If $A=\begin{bmatrix}1 & 0&1 \\0 & 2&0\\1&-1&4 \end{bmatrix}$, A=B+C, $B=B^{T}$    and $C=-C^{T}$ , then C= 

Answer:

Explanation:

We have,

$A=\begin{bmatrix}1 & 0&1 \\0 & 2&0\\1&-1&4 \end{bmatrix}$

    $A^{T}=\begin{bmatrix}1 & 0&1 \\0 & 2&-1\\1&0&4 \end{bmatrix}$

 $A+A^{T}=\begin{bmatrix}2 & 0&2\\0 & 4&-1\\2&-1&8\end{bmatrix}$, $A-A^{T}= \begin{bmatrix}0 & 0&0 \\0 & 0&1\\0&-1&0 \end{bmatrix}$

 $A= \frac{1}{2}  (A+A^{T})+\frac{1}{2} (A-A^{T})$

 $A= \begin{bmatrix}1 & 0&1 \\0 & 2&-0.5\\1&-0.5&4 \end{bmatrix}$+$\begin{bmatrix}0 & 0&0 \\0 & 0&0.5\\0&-0.5&0 \end{bmatrix}$

 A= B+C

 $C= \begin{bmatrix}0 & 0&0 \\0 & 0&0.5\\0&-0.5&0\end{bmatrix}$= $-C^{T}$

The probability distribution  of a random variable X is given below

The variance of X is 

Answer:

Explanation:

We have,

 Variance ${\sigma ^{2}}$= $\sum P_{i}X_{i}^{2}-(\sum P_{i}X_{i})^{2}$

$= 3.4 -(1.6)^{2}=3.4-2.56=0.84$

$\int_{0}^{\pi} \frac{xdx}{4\cos^{2}x+9\sin^{2}x}=$

Answer:

Explanation:

We have,

 $I=\int_{0}^{\pi} \frac{xdx}{4 \cos^{2}x+9 \sin^{2}x}$

 $\Rightarrow$       $I=\int_{0}^{\pi} \frac{(\pi-x)dx}{4 \cos^{2}(\pi-x)+9 \sin^{2}(\pi-x)}$  

$\Rightarrow$   $I=\int_{0}^{\pi} \frac{(\pi-x)dx}{4 \cos^{2}x+9 \sin^{2}x}$

$\Rightarrow$     $2I=\int_{0}^{\pi} \frac{\pi dx}{4 \cos^{2}x+9 \sin^{2}x}$

$\Rightarrow$     $2I=\int_{0}^{\pi} \frac{\pi \sec^{2}x dx}{4 +9 \tan^{2}x}$

$\Rightarrow$    $2I=\int_{0}^{\pi/2} \frac{2\pi \sec^{2} x dx}{4 +9 \tan^{2}x}$

            $\left[\because \int_{0}^{2a} f(x) dx-2\int_{0}^{a} f(x) dx\Rightarrow f(2a-x)=f(x)\right]$

$\Rightarrow$      $I= \frac{\pi}{9}\int_{0}^{\pi/2} \frac{\sec^{2} x dx}{\frac{4}{9}+\tan^{2}x }$

 Put   $\tan x=t \Rightarrow  \sec^{2} x dx=dt$

       $x=0, t=0, x=\frac{\pi}{2},t=\infty$

  $\Rightarrow $    $I= \frac{\pi}{9} \int_{0}^{\infty} \frac{dt}{\left(\frac{2}{3}\right)^{2}+t^{2}}$

$\Rightarrow $    $I= \frac{\pi}{9} \times \frac{3}{2}\left[ \tan ^{-1} \frac{3t}{2}\right]_{0}^{\infty}$

$\Rightarrow $     $I= \frac{\pi}{9} \times \frac{3}{2}\times \frac{\pi}{2}= \frac{\pi^{2}}{12}$

If f is differentiable  , $f(x+y)=f(x)f(y)$ for all  $x,y\in IR$, f(3)=3, f'(0)=11, then f'(3)=

Answer:

Explanation:

 We have,

  f(x+y)=f(x)f(y)

 differentiate with respect to x, y as constant , we get

  $f'(x+y)=f'(x) f(y) $

 Put x=0 , y=3 we get

 f'(0+3)=f'(0).f(3)

$\Rightarrow$     f'(3) =f'(0).f(3)

$\Rightarrow$  f'(3)=11 x3 =33    $[\because f'(0)=11, f(3)=3]$

• Mathematics - General questions