MHT CET 2016 :: Mathematics :: General questions

• Mathematics - General questions

Derivative of log (sec $\theta$+ tan $\theta$)  with respect to $\sec \theta$  at $\theta$ =$\frac{\pi}{4}$

Answer:

Explanation:

Let u=log (sec $\theta$+ tan $\theta$)   and v= $\sec \theta$

 On differentiating both sides w.r.t  $\theta$   , we get

$\frac{du}{d\theta}=\frac{1}{(\sec \theta+\tan \theta)}(\sec \theta\tan \theta+\sec^{2} \theta)$  and 

$\frac{dv}{d\theta}=\sec \theta\tan \theta$

 $\therefore$  $\frac{dv}{d\theta}=\frac{\frac{du}{d\theta}}{\frac{dv}{d\theta}}$

 =  $\frac{\sec \theta(\tan \theta+\sec \theta)}{( \sec\theta+\tan \theta)\times \sec \theta \tan \theta}= \cot \theta$

$\Rightarrow$    $\frac{du}{dv(\theta=\pi/4)}=\cot \frac{\pi}{4}=1$

In  $\triangle ABC.(a-b)^{2}\cos^{2}\frac{C}{2}+(a+b)^{2}\sin^{2}\frac{C}{2}$   is equal to

Answer:

Explanation:

$(a-b)^{2}\cos^{2}\frac{C}{2}+(a+b)^{2}\sin^{2}\frac{C}{2}$

=  $(a^{2}+b^{2}-2ab)\cos^{2}\frac{C}{2}+(a^{2}+b^{2}+2ab)\sin^{2}\frac{C}{2}$

 =  $a^{2}\left(\cos^{2}\frac{C}{2}+\sin^{2}\frac{C}{2}\right)+b^{2}\left(\cos^{2}\frac{C}{2}+\sin^{2}\frac{C}{2}\right)-2ab\left(\cos^{2}\frac{C}{2}-\sin^{2}\frac{C}{2}\right)$

 = $a^{2}+b^{2}-2ab \cos C$

 $\left[\because\cos^{2}\frac{C}{2}-\sin^{2}\frac{C}{2}=\cos C\right]$

     =   $a^{2}+b^{2}-2ab \left(\frac{a^{2}+b^{2}-c^{2}}{2ab}\right)=c^{2}$

                                                                             $\left[\because \cos C =\frac{a^{2}+b^{2}}{2ab}\right]$

 The  approximate value of $f(x) = x^{3}+5x^{2}-7x+9$  at x=1. 1 is 

Answer:

Explanation:

Given,  $f(x) = x^{3}+5x^{2}-7x+9$

 On differentiating both sides w.r.t x , we get

   f'(x) = $3x^{2}+10x-7$

Let x=1  and $\triangle x$=0.1 , so that

 $f(x+\triangle x)=f(1+0.1)=f(1.1)$

We know that  ,

 $f(x+\triangle x)=f(x)+\triangle x f'(x)$

 = $x^{3}+5x^{2}-7x+9+\triangle x\times (3x^{2}+10x-7)$

  Put x=1  and $\triangle x$=0.1 , we get

 f(1+0.1)

  = $1^{3}+5(1)^{2}-7(1)+9+0.1\times (3\times 1^{2}+10\times1-7)$

$\Rightarrow$      $f(1.1)=1+5-7+9+0.1(3+10-7)=8+0.1(6)=8+0.6=8.6$

  Direction  cosines of the line  $\frac{x+2}{2}=\frac{2y-5}{3}=z$  =-1 are 

Answer:

Explanation:

 Given equation of line is,

 $\frac{x+2}{2}=\frac{2y-5}{3}, z+1=0$

or   $\frac{x+2}{2}=\frac{y-\frac{5}{2}}{\frac{3}{2}},z+1=0$

 Hence, DR of a line are $<2 , \frac{3}{2} ,0 >$

Now,   $\sqrt{2^{2}+\left(\frac{3}{2}\right)^{2}+0}=\sqrt{4+\frac {9}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{2}$

 $\therefore$    DC of a line are   $< \frac{2}{5/2}, \frac{3/2}{5/2},0 > or < \frac{4}{5},\frac{3}{5},0>$

 If  $2 \tan ^{-}(\cos x)=\tan^{-1}(2 cosec x)$ then sin x+cosx is equal to

Answer:

Explanation:

 Given ,  $2 \tan ^{-}(\cos x)=\tan^{-1}(2 cosec x)$ 

$\Rightarrow$     $tan^{-1}\frac{2\cos x}{1- \cos^{2} x}=\tan^{-1}\left(\frac{2}{\sin x}\right)$

 $\Rightarrow$   $\frac{2\cos x}{1- \cos^{2} x}=\frac{2}{\ sin x}\Rightarrow \frac {\cos x}{sin ^{2} x}\Rightarrow \frac {1}{\sin x}$

   $\Rightarrow  \frac {\cos x}{sin x}=1$    $[\because \sin x\neq0]$

  $\Rightarrow$      $\tan x=1\Rightarrow x= \frac{\pi}{4}$

Now , $\sin x+\cos x= \sin \frac{\pi}{4}+\cos \frac {\pi}{4}$

  $=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$

• Mathematics - General questions