TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

The solution  of the differential equation (2x-3y+5)dx+(9y+6x-7) dy =0 , is 

Answer:

Explanation:

We have

 (2x-3y+5)dx+(9y+6x-7) dy =0 

$\frac{dy}{dx}=\frac{2x-3y+5}{3(2x-3y)+7}$

Put 2x-3y=z

$\Rightarrow$    $2-\frac{3dy}{dx}=\frac{dz}{dx}$

$\Rightarrow$      $\frac{dy }{dx}=\frac{1}{3}$  $\left(2-\frac{dz}{dx}\right)$

$\therefore$  $\frac{1}{3}\left(2-\frac{dz}{dx}\right)=\frac{z+5}{3z+7}$

$\Rightarrow$    $2-\frac{dz}{dx}=\frac{3z+15}{3z+7}$

$\Rightarrow$    $\frac{dz}{dx}=2-\frac{3z+15}{3z+7}$

$\Rightarrow$     $\frac{dz}{dx}=\frac{3z+1}{3z+7}$

$\Rightarrow$    $\frac{3z+1}{3z+7}dz= dx$

$\Rightarrow$     $\int\left( \frac{3z-1}{3z-1}+\frac{8}{3z-1}\right)dz= dx$

 $=z+\frac{8}{3}log| 3z-1|=x+c$

 $=3z+8log( 3z-1)=3x+c$

 =3x-9y+8 log (6x-9y-1)=c

The differential equation representing the family of circles of constant radius r is 

Answer:

Explanation:

Family of circle of constant radius r is 

 $(x-a)^{2}+(y-b)^{2}=r^{2}$

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

 $\frac{dx}{d\theta}=-r \sin \theta, \frac{dy}{d \theta}= r \cos \theta \Rightarrow \frac{dy}{dx}=-\cot \theta$

 $\frac{dy ^{2}}{dx^{2}}=cosec ^{2} \theta, \frac{d\theta}{d x}= \frac{- cosec ^{2} \theta}{r}$

$\frac{dy ^{2}}{dx^{2}}= \frac{-(1+ \cot ^{2} \theta)^{3/2}}{r}$

 $ry"= -(1+(y')^{2})^{3/2}$

 squaring on both sides , we get

 $r^{2} (y")^{2}=[1+(y')^{2}]^{3}$

Area of the region (in sq units)  bounded by the curve y =$\sqrt{x}$, x= $\sqrt{y}$ and the lines x=1, x=4 , is 

Answer:

Explanation:

We have ,

  y =$\sqrt{x}$, x= $\sqrt{y}$ ,x=1 ,x=4

The graph  of curves are 

 Area of shaded region

$=\int_{1}^{4} (x^{2}-\sqrt{x})dx=\left[\frac{x^{3}}{3}-\frac{2}{3}(x)^{3/2}\right]_{1}^{4}$

 = $\left(\frac{64}{3}-\frac{16}{3}\right)-\left(\frac{1}{3}-\frac{2}{3}\right)$

 = $\frac{64-16-1+2}{3}=\frac{49}{3}$

Consider the function $f(x)=2x^{3}-3x^{2}-x+1$     and the  intervals $I_{1}$=[-1,0],$I_{2}$= [0,1], $I_{3}$  =[1,2], $I_{4}$=[-2,-1]

Then,

Answer:

Explanation:

We have,

 $f(x)=2x^{3}-3x^{2}-x+1$

Let  

  $g(x)=\frac{x^{4}}{2}-x^{3}-\frac{x^{2}}{2}+x$

 $g(-1)=\frac{1}{2}+1-\frac{1}{2}-1=0$

 and g(0)= 0

 $\therefore$    f(x)=0 has roots lie in [-1,0]

Similarly , g(0)=g(1)=0

$\therefore$   f(x)=0 has roots lie in [0,1].

 $g(2)=\frac{16}{2}-8-\frac{4}{2}+2=8-8-2+2=0$

 g(1)= g(2)=0

But  , g(-2)≠ 0

 $\therefore$  f(x)=0 has roots in every interval except $I_{4}$

If $y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+..........\infty}}}}$, then $\frac{dy}{dx}$= 

Answer:

Explanation:

We have,

$y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+..........\infty}}}}$

$y=\sqrt{x+\sqrt{y+y}}\Rightarrow y^{2}=x+\sqrt{2y}$

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

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

 $\Rightarrow$    $y^{4} -2xy^{2}-2y+x^{2}$=0

 On  differentiating  w.r.t to x , we get

  $4 y^{3}\frac{dy}{dx}-2y^{2}-4yx\frac{dy}{dx}-\frac{2dy}{dx}+2x=0$

 $\frac{dy}{dx}(4y^{3}-4xy-2)=-2x+2y^{2}$

$\Rightarrow$      $\frac{dy}{dx}=\frac{y^{2}-x}{2y^{3}-2xy-1}$

• Mathematics - General questions