TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

Match the following

Answer:

Explanation:

$I. \int_{-1}^{1} x|x|dx=0$    $[\because x|x| $ is an odd function]

II.Let I= $ \int_{0}^{\pi/2} \left[1+\left( \log \frac{4+3 \sin x}{4+3 \cos x}\right)\right]dx$

  $ I= \int_{0}^{\pi/2}\left(1+ \left[ \log \frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left( \frac{\pi}{2}-x\right)})\right]\right)dx$

$I= \int_{0}^{\pi/2}\left(1+ \left[ \log \frac{4+3 \cos x}{4+3 \sin x}\right]\right)dx$

 $\Rightarrow$     $2 I= \int_{0}^{\pi/2}\left(2+  \log \frac{4+3 \sin x}{4+3 \cos x}+\log \frac{4+3 \cos x}{4+3 \sin x}\right)dx$

    $\Rightarrow$     $2 I= \int_{0}^{\pi/2}\left(2+  \log \frac{(4+3 \sin x)(4+3 \cos x)}{(4+3 \cos x)(4+3 \sin x)}\right)$

 $\Rightarrow$    $2 I= \int_{0}^{\pi/2}(2+ \log 1)dx=\int_{0}^{\pi/2} 2dx= \pi$

 $\Rightarrow$    $I= \frac{\pi}{2}$

 III.    $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$

IV.   $\int_{-a}^{a} f(x) dx= \int_{0}^{a} f(x) dx+\int_{0}^{a} f(-x) dx$

   =  $\int_{0}^{a} [f(x) +f(-x)]dx$

The mean deviation from the mean 10 of the data 6,7,11,12,13, $\alpha$,12,16 is 

Answer:

Explanation:

Given , mean $(\overline{x})$=10

$Mean(\overline{x})= \frac{6+7+10+12+13+\alpha+12+16}{8}$

$\Rightarrow$    $10= \frac{76+ \alpha}{8}$

$\Rightarrow$    $\alpha =4$

$MD(\overline{x})= \frac{ [|6-10|+|7-10|+|10-10|+|12-10|+|13-10|+|4-10|+|12-10|+|16-0|}{8}$

$MD(\overline{x})= \frac{4+3+0+2+3+6+2+6}{8}$

$MD(\overline{x})= \frac{26}{8}=3.25$

Consider the circle $x^{2}+y^{2}-6x+4y=12$. The equations of a tangent of this circle that is parallel to the line x+3y+5=0

Answer:

Explanation:

We have ,

$x^{2}+y^{2}-6x+4y=12$.

$\Rightarrow$  $(x-3)^{2}+(y+2)^{2}=25$

$\Rightarrow$     $(x-3)^{2}+(y+2)^{2}=5^{2}$

 Equation of tangent  whose slope m is 

  $y+2=m(x-3)\pm5\sqrt{m^{2}+1}$   .......(i)

 Now, this tangent is parallel to line 4x+3y+5=0

 $\therefore$  slope of line is -$\frac{4}{3}$

 Put the value of m= -$\frac{4}{3}$ is Eq.(i) , we get

   $y+2=-\frac{4}{3} (x-3)\pm 5\sqrt{\left(\frac{-4}{3}\right)^{2}+1}$

$\Rightarrow$      $y+2=\frac{-4}{3} (x-3)\pm 5\left(\frac{5}{3}\right)$

$\Rightarrow$    $3y+6=-4x+12\pm25$

$\Rightarrow$   $4x+3y=6 \pm 25$

$\Rightarrow$   $4x+3y=31$ or $4x+3y=-19$

Hence , equation  of tangent is 

   $4x+3y-31=0  $ or  $4x+3y+19=0$

If  $\tan 20^{0}= \lambda ,$ then

 $\frac{\tan 160^{0}-\tan 110^{0}}{1+(\tan 160^{0})(\tan 110^{0})}=$

Answer:

Explanation:

Given, $\tan 20^{0}= \lambda$

  $\therefore$     $\frac{\tan 160^{0}-\tan 110^{0}}{1+ (\tan 160^{0})(\tan 110^{0}) }$

  =$\frac{\tan (180^{0}-20^{0})-\tan (90^{0}+20^{0})}{1+ (\tan( 180^{0}-20^{0})(\tan 90^{0}+20^{0}) }$

 $=\frac{-\tan 20^{0}+\cot 20^{0}}{1+ \tan 20^{0} \cot 20^{0}}$

            $[ \because  \tan (180^{0}-\theta)=-\tan \theta; \tan (90^{0}+\theta)=-\cot \theta]$

 = $\frac{ -\lambda+1/\lambda}{1+1}$

   =$\frac{-\lambda^{2}+1}{2 \lambda}$

=$\frac{1- \lambda^{2}}{2 \lambda}$

• Mathematics - General questions