MHT CET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If $\int\frac{1}{\sqrt{9-16x^{2}}}dx=\alpha \sin^{-1}(\beta x)+c $  then $\alpha+\frac{1}{\beta}$=

Answer:

Explanation:

Let l= $\int\frac{1}{\sqrt{9-16x^{2}}}dx$

 =$\int\frac{1}{\sqrt{(3)^2-(4x)^{2}}}dx=\frac{1}{4}\frac{1}{\sqrt{(\frac{3}{4})^{2}-(x)^{2}}}dx$

=  $\frac{1}{4}\sin^{-1}\frac{4x}{3}+C$

                          $\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} dx=\sin^{-1}\frac{x}{a}+C\right]$

 But , it is given that

 $\int\frac{1}{\sqrt{9-16x^{2}}}dx=\alpha \sin ^{-1}(\beta x)+C$

$\therefore$   $\alpha \sin ^{-1}(\beta x)+C=\frac{1}{4}\sin^{-1}(\frac{4}{3}x)+C$

On comparing both sides, we get

 $\alpha =\frac{1}{4}$  and  $\beta =\frac{4}{3}$

 $\therefore$      $\alpha+\frac{1}{\beta} =\frac{1}{4}+\frac{1}{4}=\frac{1}{4}+\frac{3}{4}=1$

The inverse of the matrix $\begin{bmatrix}1 & 0&0 \\3 & 3&0\\5&2&-1 \end{bmatrix}$  is 

Answer:

Explanation:

Let A= $\begin{bmatrix}1 & 0&0 \\3 & 3&0\\5&2&-1 \end{bmatrix}$

$\Rightarrow$  $|A|=\begin{bmatrix}1 & 0&0 \\3 & 3&0\\5&2&-1 \end{bmatrix}$

$=1(3\times -1-0)-0(3\times-1-0)+0(3\times2-5\times3)$

$\Rightarrow |A|=1\times-3-0-0=-3$   ...............(i)

Now , adj A

$=\begin{bmatrix}+(3\times-1-0) & -(3\times-1-0)&+(3\times2-5\times3) \\-(0-0) & +(1\times-1-0)&-(2\times1-5\times0)\\+(3\times0-0)&-(1\times0-0)&+(3\times1-0) \end{bmatrix}^T$

$\begin{bmatrix}-3 & 3&6-15 \\0 & -1&-2\\0&0&3\end{bmatrix}^T=\begin{bmatrix}-3 & 3&-9 \\0 & -1&2\\0&0&3\end{bmatrix}^T$

adj A=$\begin{bmatrix}-3 & 0&0 \\3 & -1&0\\9&2&3\end{bmatrix}^T$      ..........(ii)

$\because A^{-1}=\frac{1}{|A|}adj A=\frac{1}{-3}\begin{bmatrix}-3 & 0&0 \\3 & -1&0\\-9&-2&3\end{bmatrix}^T$

                                                                  [from Eqs.(i) and (ii)]

if g(x)  is the inverse function of f(x) and  $f'(x) =\frac{1}{1+x^{4}}$  , then g'(x) is 

Answer:

Explanation:

Given, g(x)=f'(x)

f(g(x))=x

On differentiating both sides w.r.t 'x' , we get

 f'(g(x)).g'(x)=1

$\therefore$    $\frac{1}{1+(g(x))^{4}}g'(x)=1$

      $[\because f'(x)=\frac{1}{1+x^{4}}(given)]$

$\Rightarrow g'(x) =1+[g(x)]^{4}$

$\int_{0}^{1}x \tan ^{-1} x dx=$

Answer:

Explanation:

Ler l=$\int_{0}^{1}x \tan ^{-1} x dx$

 =$[\tan^{-1}x\int x dx]^{1}_{0}-\int_{0}^{1} (\frac{d}{dx}\tan^{-1} x \int x dx) dx$

=$\left[ \tan ^{-1} .x \frac{x^2}{2}\right]^{1}_{0}-\int_{0}^{1}\left(\frac{1}{1+x^{2}}.\frac{x^{2}}{2}\right)dx$

=$\left(\frac{1}{2} \tan^{-1}1-0\right)-\frac{1}{2}\int_{0}^{1}\frac{1+x^{2}-1}{1+x^{2}} dx$

=$\frac{1}{2}\frac{\pi}{4}-\frac{1}{2}\int_{0}^{1} \left(1-\frac{1}{1+x^{2}}\right) dx$

  =$\frac{\pi}{8}-\frac{1}{2}[x-tan^{-1}x]^{1}_{0}$

= $\frac{\pi}{8}-\frac{1}{2}[1-tan^{-1}1-0+0]$

 = $\frac{\pi}{8}-\frac{1}{2}[1-\frac{\pi}{4}]=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}=\frac{\pi}{4}-\frac{1}{2}$

 The number of principal solutions  of $\tan 2\theta$ =1 is 

Answer:

Explanation:

 We have , $\tan 2\theta$ = 1

 Here, the value of $\tan 2\theta$ is positive . So, $\theta$ lies in 1st and 3 rd quadrants

$\therefore$  Number of principal solutions of $\tan 2 \theta$= 1 , is two

• Mathematics - General questions