TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If   $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, the  $\frac{d^{2}y}{dx^{2}}=$

Answer:

Explanation:

 We have, 

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Let $x=a\cos \theta, y=b \sin \theta$

 $\therefore$      $\frac{dx}{d\theta}=-a \sin \theta, \frac{dy}{d \theta}=b \cos \theta$

 $\frac{dy}{dx}=-\frac{b}{a} \cot\theta$

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

 $\frac{d^{2}y}{dx^{2}}=-\frac{b}{a}(- cosec^{2} \theta) \frac{d \theta}{dx}$

$\Rightarrow$    $\frac{d^{2}y}{dx^{2}}=\frac{b cosec^{2} \theta}{-a^{2} \sin \theta}$

$\Rightarrow$    $\frac{d^{2}y}{dx^{2}}=-\frac{b}{a^{2}\sin^{3} \theta}$

$\Rightarrow$      $\frac{d^{2}y}{dx}=\frac{-b^{4}}{a^{2}y^{3}}$   

                                               $\left[\because \sin \theta=\frac{y}{b}\right]$

The solution  of  $\frac{dy}{dx}=\frac{x+y}{x-y}$  is 

Answer:

Explanation:

We have,

$\frac{dy}{dx}=\frac{x+y}{x-y}$

 Put y=vx

 $\frac{dy}{dx}= v+x \frac{dv}{dx}$

$\therefore$    $v+x \frac{dv}{dx}= \frac{x+vx}{x-vx}=\frac{1+v}{1-v}$

$\Rightarrow$     $x \frac{dv}{dx}=\frac{1+v}{1-v}-v= \frac{1+v-v+v^{2}}{1-v}=\frac{1+v^{2}}{1-v}$

 $\Rightarrow$ $\frac{1-v}{1+v^{2}}dv= \frac{dx}{x}$

 $\Rightarrow$   $\int \frac{1}{1+v^{2}}dv-\frac{1}{2}\int \frac{2v}{1+v^{2}}dv=\int\frac{dx}{x}$

 $\Rightarrow$     $\tan^{-1}v-\frac{1}{2} \log |1+v^{2}|=\log x+C$

 $\Rightarrow$    $\tan^{-1}\frac{y}{x}= \log x+\frac{1}{2} \log \left(1+\frac{y^{2}}{x^{2}}\right)+C$

 $\Rightarrow$     $\tan^{-1}\frac{y}{x}= \log \frac{x(\sqrt{x^{2}+y^{2}})}{x}+C$

 $\Rightarrow$     $\tan^{-1}\frac{y}{x}= \log \sqrt{x^{2}+y^{2}}+C$

If the  perpendicular distance  between  the point (1,1)  to the line 3x+4y+c=0 is 7, then the possible values of C are

Answer:

Explanation:

We lnow that distance from point $(x_{1},y_{1})$ to the line ax+by+c=0 is 

$|\frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}|$

 $\therefore$   Distance from (1,1) to 3x+4y+c=0 is 7

  $\therefore$    $7=|\frac{3(1)+4(1)+c}{\sqrt{3^{2}+4^{2}}}|$

$\Rightarrow$     $7=|\frac{7+c}{5}|$

$\Rightarrow$     $35=|7+c|$

$\Rightarrow$     $7+c=\pm 35$

$\Rightarrow$       $c=-7 \pm 35$

                  c=28,-42

Let $f:(-1,1) \rightarrow IR$ be a differentiable function with f(0)=-1 and f'(0)=1 .If g(x)=(f(2f(x)+2))² , the g'(0)=

Answer:

Explanation:

 We have,

   g(x)=(f(2f(x)+2))²

g'(x)=2(f(2f(x)+2).f'(2f(x)+2)f'(x).2

$\Rightarrow$     g'(0)=2(f(2f(0)+2).f'(2f(0)+2).2f'(0)

$\Rightarrow$    g'(0)=2(f(2(-1)+2).f'(2(-1)+2.2(1)

                                                     $[\because f(0)=-1,f'(0)=1]$

$\Rightarrow$     g'(0)=2[f(-2+2).f'(-2+2)].2

$\Rightarrow$       g'(0)=4f(0).f'(0)

$\Rightarrow$     g'(0)=4 x (-1)(1)

                                =-4

If the imaginary part of $\frac{2z+1}{iz+1}$ is -2  , then the locus of the point representing  z in the complex plane is 

Answer:

Explanation:

Let z=x+iy

 $\therefore$   $\frac{2z+1}{iz+1}=\frac{2(x+iy)+1}{i(x+iy)+1}=\frac{2x+1+iy}{(-y+1)+ix}$

                              =$\frac{[(2x+1)+iy][(-y+1)-ix]}{[(-y+1)+ix][(-y+1)-ix]}$

  =$\frac{(2x+1)(-y+1)-(2x+1)xi+y(-y+1)j+xy}{(-y+1)^{2}+x^{2}}$

  =$\frac{(2x+1)(-y+1)+xy+(-y^{2}+y-2x^{2}-x)i}{(-y+1)^{2}+x^{2}}$

It is given that imaginary part of $\frac{2z+1}{iz+1}=-2$

 $\therefore$    $\frac{-y^{2}+y-2x^{2}-x}{(-y+1)^{2}+x^{2}}=-2$

$\Rightarrow$     $-y^{2}+y-2x^{2}-x=-2(-y+1)^{2}-2x^{2}$

$\Rightarrow$   $-y^{2}+y-x=-2y^{2}-2+4y$

$\Rightarrow$  $y^{2}-3y-x+2=0$

Which represent the equations of parabola

• Mathematics - General questions