TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

 The angle between the curves $x^{2}=8y$  and xy=8 is 

Answer:

Explanation:

Given equation of curves are

            $x^{2}=8y$  .......(i)

    and xy=8      ..........(ii)

 On substituting the value of y from Eq.(i) in Eq.(ii)  ,

we get

                 $x\left(\frac{x^{2}}{8}\right)=8$

$\Rightarrow$        $x^{3}=64$

   $\Rightarrow$      x=4

 On substituting  x=4 in Eq.(ii)  , we get  y=2

 Thus , the point of intersection of given curves is (4,2)

 Now, let $m_{1}$  be the slope of tangent to the curve (i) at point (4,2)  and $m_{2}$  be the slope of tangent to the curve (ii)

at point (4,2)

 Then, $m_{1}=\frac{4}{4}=1$

                          $\left[ \because x^{2}=8y \Rightarrow 2x=8\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{x}{4}\right]$

 and $m_{2}= \frac{-2}{4}=-\frac{1}{2}$

                    $\left[ \because xy=8 \Rightarrow x\frac{dy}{dx}+y=0\Rightarrow \frac{dy}{dx}=\frac{-y}{4}\right]$

 Now, the angle $\theta$  between the curves is given by

             $\tan \theta= \left(\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right)$

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

    $\Rightarrow$      $\tan \theta=-3$

   $\Rightarrow$=   $\theta= tan^{-1}(-3)$

Using the letters of the word TRICK, a five-letter word with distinct letters is formed such that C is the middle . In how many ways this is possible?

Answer:

Explanation:

 If we fix C in the middle, then the rest 4 letters can be arranged in $^{4}P_{4}=4!$ ways =24 ways 

If the slope of the tangent of the circle $S=x^{2}+y^{2}-13=0$ at (2,3)  is m, then the point  $\left(m,\frac{-1}{m}\right)$   is 

Answer:

Explanation:

Given  the equation of circle is 

  $S=x^{2}+y^{2}-13=0$  ...........(i)

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

   $2x+2y \frac{dy}{dx}=0$

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

 Now , slope of tangent ,   $m= \frac{dy}{dx}|_{at(2,3)}= \frac{-2}{3}$

  $\Rightarrow$               $m=\frac{-2}{3}$

$\therefore$          $\left(m, -\frac{1}{m}\right)=\left(-\frac{2}{3},\frac{3}{2}\right)$

 On substituting this point in LHS  eq.(i), we get

 LHS= $\frac{4}{9}+\frac{9}{4}-13$

 $=\frac{16+81-468}{36}<0$

$\therefore$      $\left(m, -\frac{1}{m}\right)$   is an internal point with respect to the circle S=0

$\alpha$  and $\beta$ are the roots of $x^{2}+2x+c=0$. If $\alpha^{3}+\beta^{3}=4$, then the value of c is 

Answer:

Explanation:

Given  , $\alpha$ and $\beta$ are roots of $x^{2}+2x+c=0$

 $\therefore$    Sum of roots , $\alpha+\beta$= -coefficient of x / coefficient of x²

 $\Rightarrow$     $\alpha+\beta=\frac{-2}{1}=-2$    .......(i)

  and products of roots , $\alpha \beta$= costant term/ coefficient of x²

 $\Rightarrow$     $\alpha \beta =\frac{c}{1}=c$         .......(ii)

 Since , $\alpha^{2}+\beta^{2}=4$         [given]

 $\Rightarrow$  $(\alpha+\beta)(\alpha^{2}+\beta^{2}-\alpha \beta)=4$

                                $\left[\because a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)\right]$

  $\Rightarrow$      $(\alpha+\beta)[(\alpha^{2}+\beta^{2}-3\alpha\beta)]=4$

                                             $\left[\because a^{2}+b^{2}=(a+b)^{2}-2\alpha\beta)\right]$

    $\Rightarrow$     $ (-2)[(-2)^{2}-3 \times c]=4$   [ From Eqs,(i) and (ii)]

 $\Rightarrow$             (-2)[4-3c]=4

 $\Rightarrow$            $4-3c=\frac{-4}{2}$

 $\Rightarrow$                 $4-3c=-2$

 $\Rightarrow$          $-3c=-2-4$

 $\Rightarrow$        -3c=-6

                         c=2

$\sin^{-1}\frac{\sqrt{3}}{2}+\sin^{-1}\sqrt{\frac{2}{3}}$= 

Answer:

Explanation:

We have, 

$\sin^{-1}\frac{\sqrt{3}}{2}+\sin^{-1}\sqrt{\frac{2}{3}}$

         $=\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}\sqrt{1-\left(\sqrt{\frac{2}{3}}\right)^{2}}+\frac{\sqrt{3}}{2}\sqrt{1-\left(\sqrt{\frac{3}{2}}\right)^{2}}\right]$

 As    0< $\frac{\sqrt{3}}{2}=0.866, \sqrt{\frac{2}{3}}=0.816 \leq$ 1   and

  $\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\sqrt{\frac{2}{3}}\right)^{2}=1.4167>1$

                   $[ \because   \sin^{-1} x+\sin^{-1}y=\pi-\sin^{-1}$

                 $x-\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}, if 0<x,y\leq1$    and $x^{2}+y^{2} >1$]

    = $\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}\sqrt{1-\frac{2}{3}^{}}+\frac{\sqrt{2}}{3}\sqrt{1-{\frac{3}{4}}}\right]$

   = $\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}\sqrt{\frac{3-2}{3}^{}}+\frac{\sqrt{2}}{3}\sqrt{{\frac{4-3}{4}}}\right]$

      =$\pi-\sin^{-1}\left[\frac{\sqrt{3}}{2}{\frac{1}{\sqrt{3}}^{}}+\frac{\sqrt{2}}{\sqrt{3}}{\frac{1}{2}}\right]$

  $=\pi-\sin^{-1}\left[\frac{\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\right]$

• Mathematics - General questions