TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

The solution of the differential equation  $ydx-xdy+3x^{2}y^{2}e^{x^{3}}dx=0$ satisfying y=1 when x=1, is 

Answer:

Explanation:

 We have,

$ydx-xdy+3x^{2}y^{2}e^{x^{3}}dx=0$

$\frac{ydx-xdy}{y^{2}}+3x^{2}e^{x^{3}}dx=0$

$\Rightarrow$     $d\left(\frac{x}{y}\right)+d(e^{x^{3}})=0$

 On integrating  , we get

 $\frac{x}{y}+e^{x^{3}}$=c

 putting x=y=1, we get

                         c=1+e

$\therefore$       $\frac{x}{y}+e^{x^{3}}$ =1+e

 x+$y\left( e^{x^{3}}-(1+e)\right)+x=0$

 If   $\int\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}dx=\frac{\sqrt{5}}{3} \tan ^{-1}\frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} \tan ^{-1}\frac{x}{\sqrt{2}}+c,$ then $\alpha$ =

Answer:

Explanation:

We have,

$\int\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}dx=\frac{\sqrt{5}}{3} \tan ^{-1}\frac{x}{\sqrt{5}}-\frac{\sqrt{2}}{3} \tan ^{-1}\frac{x}{\sqrt{2}}+c,$ 

 On differentiating  both sides , we get

$\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}$

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

 $\Rightarrow$    $\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}=\frac{1}{3}\left(\frac{5}{5+x^{2}}-\frac{2}{2+x^{2}}\right)$

$\Rightarrow$     $\frac{2x^{2}}{(2x^{2}+\alpha)(x^{2}+5)}=\frac{x^{2}}{(x^{2}+2)(x^{2}+5)}$

 $\Rightarrow$      $\frac{2x^{2}}{(2x^{2}+4)(x^{2}+5)}\Rightarrow\alpha=4$

 The derivative of $\cos h^{-1} x$ with respect to log x at x=5 is 

Answer:

Explanation:

 U= $\cos h^{-1} x$ and v=log x

 $u=\log (x+\sqrt{x^{2}-1}) $  and v=log x

 $\frac{du}{dx}=\left(\frac{1}{x+\sqrt{x^{2}-1}}\right)\left(1+\frac{x}{\sqrt{x^{2}-1}}\right)$  and

$\frac{dv}{dx}=\frac{1}{x}$

$\frac{du}{dx}$= $\frac{1}{\sqrt{x^{2}-1}}$  and $\frac{dv}{dx}=\frac{1}{x} \Rightarrow  \frac{du}{dv}=$

$\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{x}{\sqrt{x^{2}-1}}$

 $\because\left(\frac{du}{dv}\right)_{x=5}=\frac{5}{\sqrt{25-1}}=\frac{5}{\sqrt{24}}=\frac{5}{2\sqrt{6}}$

If f(x)  = $\begin{cases}ax+b, & if x\leq 1\\ax^{2}+c ,& if 1<x\leq2 \\ \frac{dx^{2}+1}{x},&if x\geq 2\end{cases}$

is differentiable on R, then ad-bc=

Answer:

Explanation:

We have ,

f(x)  = $\begin{cases}ax+b, & if x\leq 1\\ax^{2}+c ,& if 1<x\leq2 \\ \frac{dx^{2}+1}{x},&if x\geq 2\end{cases}$

$\because$   f(x)  is differentiable , hence f(x) must be continuous

$\therefore$    $\lim_{x \rightarrow 1-}f(x)=\lim_{x \rightarrow 1+}f(x)$

$\Rightarrow$  a+b=a+c $\Rightarrow$ b=c   ........(i)

and  $\lim_{x \rightarrow 2-}f(x)\lim_{x \rightarrow 2+}f(x)$

 $\Rightarrow$         4a+c = $\frac{4d+1}{2}$

$\Rightarrow$     8a+2c=4d+1    ........(ii)

 $\because$     f(x)  is differentiable on R

 $\therefore$     $f(x)  = \begin{cases}a, & if x< 1\\2ax ,& if 1<x<2 \\d- \frac{1}{x^{2}},&if x> 2\end{cases}$

 f(x) is differentiable at x=1

 $\because$      a=2a $\Rightarrow$  a=0  .....(iii)

 and f(x)  is differentiable at x=2

 $\because$        4a=d-$\frac{1}{4} \Rightarrow$   d= $\frac{1}{4}$  .......(iv)

From Eqs.(ii) and (iv)  ,we get

 c=1=b

$\therefore$  ad-bc=$ 0 \times \frac{1}{4}-1$=-1

The distance between the tangents to the hyperbola  $\frac{x^{2}}{20}- \frac{3y^{2}}{4}$=1 which are parallel to the line x+3y=7 is 

Answer:

Explanation:

 Equation  of tangent to hyperbola  $\frac{x^{2}}{20}- \frac{3y^{2}}{4}$=1

parallel to x+3y=7 is

 $y=-\frac{1}{3}x\pm\sqrt{\frac{20}{9}-\frac{4}{3}}$

$3y=-x\pm2\sqrt{2}$

 $x+3y=2\sqrt{2}$

and  $x+3y=-2\sqrt{2}$

Distance between parallel tangent is 

$d=\frac{2\sqrt{2}+2\sqrt{2}}{\sqrt{1+3^{2}}}=\frac{4\sqrt{2}}{\sqrt{10}}=\frac{4}{\sqrt{5}}$

• Mathematics - General questions