TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

 The solution of the differential equation  $(x+2y^{3})\frac{dy}{dx}=y$ is 

Answer:

Explanation:

We have   $(x+2y^{3})\frac{dy}{dx}=y$

 $\Rightarrow$              $y \frac{dy}{dX}=x+2y^{3} $

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

 It is linear  differential equation of the firm $\frac{dx}{dy}+Px=Q$

 $\Rightarrow$     $ P= -\frac{1}{y}. Q= 2y^{2}$

   $\therefore$     $IF= e^{\int P dy}=e^{\int \frac{1}{y} dy}=e^{-\log y}=\frac{1}{y}$

 Hence, required  solution is

            x.(IF)= $\int (IF.Q) dy$

     $\Rightarrow$      $\frac{x}{y}=\int\frac{2y^{2}}{y}dy=y^{2}+c$

       $\Rightarrow$     $ x=y^{3}+cy$

The area (in square units) of the region bounded by the curve y=|sin 2x|  and X-axis  in [0, $2 \pi$] is 

Answer:

Explanation:

 Required area   $A=\int_{0}^{2 \pi} | \sin 2x|dx$

 $=4\int_{0}^{ \pi /2}  \sin 2xdx=4\left[ -\frac{\cos 2x}{2}\right]^{\pi/2}_{0}$

  = 2[-(-1-1)]=4

 Hence, option (d) is correct

If the minimum  value of  f(x)= $2x^{2} +\alpha x+8$ is the same as the maximum value of g(x)=$-3x^{2}-4x+\alpha^{2}$ ,  then $\alpha^{2}$=

Answer:

Explanation:

Since , the minimum value of 

f(x)= $2x^{2}+\alpha x+8$   is  $\frac{\alpha^{2}-64}{8}=\frac{64-\alpha^{2}}{8}$

 And the maximum value of g(x)= $-3x^{2}-4x+\alpha^{2}$

 is $-\frac{12\alpha^{2}-16}{-12}=\frac{16+12\alpha^{2}}{12}$

 Now, according to the question

    $\frac{64-\alpha^{2}}{8}=\frac{16+12\alpha^{2}}{12}$

 $\Rightarrow$      $192-3 \alpha^{2}=32+24 \alpha^{2}$

 $\Rightarrow$     $27 \alpha^{2}=160 \Rightarrow  \alpha^{2}= \frac{160}{27}$

 hence3, option(b) is correct

Electric current is measured by tangent galvanometer, the current being proportional to the tangent of the angle $\theta$ of deflection. If the deflection is read as $45^{0}$ and  an error of 1%  is made  in reading it.then  the percentage error in the current  is 

Answer:

Explanation:

According to the given informations   $I\propto \tan\theta$

 So, $\frac{\triangle I}{I}\times 100=sec^{2} \theta \frac{\triangle\theta}{\theta}\times 100$

 at    $\theta=45^{0}.\frac{\triangle \theta}{\theta}\times 100=\frac{\pi}{4}\times\frac{1}{100}\times 100=\frac{\pi}{4}$

 So, percentage error in current is   $\frac{\triangle I}{I}\times 100$

     = $2 \times  \frac{\pi}{4}=\frac{\pi}{2}$

 The coordinates of a a point on the curve  $x=a(\theta +\sin \theta), y =a(1-\cos \theta)$   where the tangent is inclined at an angle $\frac{\pi}{4}$  to the positive X-axis , are

Answer:

Explanation:

 According  to the given information,

    $\frac{dy}{dx}=1  $            ..............(i)

 $\because$   $\frac{dy}{d \theta}= a\sin \theta$ and $\frac{dx}{d \theta}= a(1+\cos \theta)$ 

$\therefore$    $\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{a \sin\theta}{a(1+\cos \theta)}=1$

 $\Rightarrow\frac{ \sin\theta}{1+\cos \theta}=1\Rightarrow\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2}\frac{\theta}{2}}=1\Rightarrow \tan \frac{\theta}{2}=1$

  $\Rightarrow \frac{\theta}{2}=\frac{\pi}{4}\Rightarrow \theta = \frac{\pi}{2}$

 So, the required point on the curve is    $\left(a\left(\frac{\pi}{2}+1\right)a\right)$

 hence  , option (b) is correct

• Mathematics - General questions