TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

The solution  of the differential equation $\frac{dy}{dx}=1- \cos (y-x) \cot (y-x) $ is 

Answer:

Explanation:

We have   $\frac{dy}{dx}=1- \cos (y-x) \cot (y-x) $

Put  $y-x = v \Rightarrow  \frac{dy}{dx}=1+\frac{dv}{dx}$

 $\therefore$    $1+ \frac{dv}{dx}=1- \cos v \cot v $  $\Rightarrow$ $\frac{dv}{dx}= -\frac{\cos ^{2} v}{\sin v}$

 $\Rightarrow$     $-\int\frac{\sin v}{\cos^{2} v}dv=\int dx\Rightarrow -\int \sec v \tan v dv=dx$

 $\Rightarrow -\sec v=x+c\Rightarrow x+\sec(y-x)=c$

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}  \cot^{9} x dx=$

Answer:

Explanation:

I= $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}  \cot^{9} x dx$

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

 put sin x=t, cos x dx=dt

 $x=\frac{\pi}{4}, t= \frac{1}{ \sqrt{2}}, x= \frac{\pi}{2},t=1  $

 $\therefore$    $I= \int_{\frac{1}{\sqrt{2}}}^{1}  \frac{(1-t^{2}) ^{4}}{t^{9}}  dt$

 $I= \int_{\frac{1}{\sqrt{2}}}^{1}\left(  \frac{1-4t^{2}+6t^{4}-4t^{6}+t^{8}}{t^{9}}\right)  dt$

$I= \int_{\frac{1}{\sqrt{2}}}^{1}\left( t^{-9}-4t^{-7}+6t^{-5}-4t^{-3}+ \frac{1}{t}\right)  dt$

  $I=\left[\frac{1}{8t^{8}}-\frac{4}{6t^{6}}+\frac{6}{4t^{4}}-\frac{4}{2t^{2}}+\log t\right]^{1}_{1/\sqrt{2}}$

 I= $\left[\left(\frac{1}{8}-\frac{4}{6}+\frac{6}{4}-\frac{4}{2}+0\right)-\left(2-\frac{32}{6}-6-4-\frac{1}{2}\log 2\right)\right]$

I= $\frac{-7}{24}+\frac{1}{2} log 2$

$\int\frac{dx}{(1+x)\sqrt{8+7x-x^{2}}}$=

Answer:

Explanation:

We have

 I= $\int\frac{dx}{(1+x)\sqrt{8+7x-x^{2}}}$

$I=\int\frac{dx}{(1+x)\sqrt{(8-x)(1+x)}}$

    $I=\int\frac{dx}{\sqrt{\frac{8-x}{1+x}(1+x)^{2}}}$

Put   $\frac{8-x}{1+x}=t^{2}\Rightarrow\left(\frac{(1+x)(-1)-(8-x)}{(1+x)^{2}}\right)dx=2t dt$

    $\Rightarrow$     $\frac{-9}{(1+x)^{2}}dx=2t dt$

$\therefore$      $I=\frac{-2}{9}\int\frac{tdt}{t}$

$I =\frac{-2}{9}\int dt=-\frac{2}{9}t+c$

$\therefore$ I=$-\frac{2}{9}\sqrt{\frac{8-x}{1+x}}+c$

Let $f:D\rightarrow R,D\subseteq R, c \in D$  and r be a non zero real number . Consider the following statements:

I: c is  an extreme  point of f $\Rightarrow$ c is  an extreme point of rf

II. c is an extreme point of f $\Rightarrow$ c  is an extreme  point of r+f

Which of the following is correct?

Answer:

Explanation:

I. Let the function 

            y= f(x)

c is extreme point of f 

 $\therefore$    f'(c)=0

 Now, y=r f'(x)

                               $\frac{dy}{dx}= r f'(x)$

 $\left(\frac{dy}{dx}\right)_{x=c}=rf'(c)$

   f'(c)=0

 $\therefore$ C is also extreme point  of rf.

II. Let               y=r+f(x)

            $\frac{dy}{dx}=f'(x)$

c is extreme  point of f(x).

 $\because$       $\left(\frac{dy}{dx}\right)_{x=c}=0$

$\Rightarrow$     c is also extreme point of r+f . Hence both statement are true.

If $\frac{dy}{dx}=4$  and $\frac{d^{2}y}{d^{2}x}=-3$ at a point P on the curve y=f(x), then 

$\left(\frac{d^{2}x}{dy^{2}}\right)=$

Answer:

Explanation:

$\frac{dx}{dy}=\left(\frac{dy}{dx}\right)^{-1}$

  $\frac{d^{2}x}{dy^{2}}=\frac{-1}{\left(\frac{dy}{dx}\right)^{2}}\times\frac{d}{dx}\left(\frac{dy}{dx}\right)\times\left(\frac{dx}{dy}\right)$

 $\Rightarrow$     $\frac{d^{2}x}{dy^{2}}=\frac{-d^{2}y}{dx^{2}}/\left(\frac{dy}{dx}\right)^{3}$

$\left(\frac{d^{2}x}{dy^{2}}\right)_{P}=-\left(\frac{-3}{(4)^{3}}\right)=\frac{3}{64}$

                                                                                    $\left[\because \frac{dy}{dx}=4,\frac{d^{2}x}{dx^{2}}=-3\right]$

• Mathematics - General questions