TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If   $A{\frac{\pi}{3}},B{\frac{\pi}{6}}$ are the points on the circle represented in parametric from with centre (0,0)  and radius 12 then the length of the chord AB is 

Answer:

Explanation:

 Parametric  equations  of given circle is $x=12 \cos \theta$, $y=12 \sin \theta$

 $[\because $ Parametric equation $x^{2}+y^{2}=r^{2}$ is $x=r \cos \theta$,$y=r\sin \theta$]

 Now, coordinates  of point A are given by

 $x= 12 \cos \frac{\pi}{3}, y=12 \sin \frac{\pi}{3}$

$\Rightarrow$   $x=12.\frac{1}{2},y=12 .\frac{\sqrt{3}}{2}$

$\Rightarrow$    $x=6; y=6 \sqrt{3}$

ie,        A=$(6,6 \sqrt{3})$

 and coordinates of point B are given by

    $x=12\cos \frac{\pi}{6},y=12 \sin \frac{\pi}{6}$

$\Rightarrow$     $x=12. \frac{\sqrt{3}}{2}.y=12 . \frac{1}{2}$

$\Rightarrow$      $x=6\sqrt{3},y=6$

i.e, $B=(6 \sqrt{3},6)$

 Clearly  , length  of chord

$AB=\sqrt{6(\sqrt{3}-6)^{2}+(6-6\sqrt{3})^{2}}$

    $=\sqrt{2\times6^{2}+(\sqrt{3}-1)^{2}}$

     $=6\sqrt{2}(\sqrt{3}-1)$

=$6(\sqrt{6}-\sqrt{2})$

If $\alpha$ and $\beta$ are the roots of the equation  $ax^{2}+bx+c=0$ and the equation having roots $\frac{1-\alpha}{\alpha}$

and $\frac{1-\beta}{\beta}$ is $px^{2}+qx+r=0$ then r=

Answer:

Explanation:

We have,

$\alpha$,$\beta$ are the roots of the equations $ax^{2}+bx+c=0$

 $\therefore$       $\alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$

  Equations whose roots are $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$ are

  $x^{2}-\left(\frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}\right)x+\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=0$

$\Rightarrow$    $x^{2}-\left(\frac{1}{\alpha}+\frac{1}{\beta}-2\right)x+\left(\frac{1-(\alpha+\beta)+\alpha\beta}{\alpha\beta}\right)=0$

$\Rightarrow$    $x^{2}-\left(\frac{\alpha+\beta-2\alpha\beta}{\alpha\beta}\right)x+\left(\frac{1-(\alpha+\beta)+\alpha\beta}{\alpha\beta}\right)=0$

$\Rightarrow$     $\alpha\beta x^{2}-(\alpha+\beta-2\alpha\beta)x+(1-(\alpha+\beta)+\alpha\beta)=0$

$\Rightarrow$   $\frac{c}{a}x^{2}-\left(-\frac{b}{a}-\frac{2c}{a}\right)x+\left(1+\frac{b}{a}+\frac{c}{a}\right)=0$

$\Rightarrow$     $cx^{2}+(b+2c)x+(a+b+c)=0$

Comparing with $px^{2}+qx+r=0$ we get

                                           r= a+b+c 

If   $\int f(x)\cos x dx= \frac{1}{2} (f(x))^{2}+C$  and f(0)=0 then f'(0)=

Answer:

Explanation:

We have,

$\int f(x) \cos x dx= \frac{1}{2} (f(x))^{2}+C$

On differentiating  w.r.t 'x' , we get

   $f(x) \cos x=\frac{1}{2} \times 2 f(x).f'(x)$

$\Rightarrow$     $f(x) \cos x= f(x).f'(x)$

$\Rightarrow$   $f'(x)=\cos x$

$\Rightarrow$   $f'(0)= \cos 0$

$\Rightarrow$  $f'(0)=1$

If the magnitude  of a,b and a+b  are respectively 3,4 and 5.then the magnitude of (a-b) is

Answer:

Explanation:

We have |a|=3,|b|=4 and |a+b|=5

 Since,        |a+b|=5

$\therefore$    $|a+b|^{2}=25$

$\Rightarrow$      $(a+b).(a+b)=25$

$\Rightarrow$    $a.a+a.b+b+a+b.b=25$

$\Rightarrow$      $|a|^{2}+2a.b+|b|^{2}=25$    [$\therefore$ a.b=b.a]

$\Rightarrow$             $9+2 a.b+16=25$

$\Rightarrow$    $a.b=0$

Now, consider $|a-b|^{2}=(a-b)(a-b)$

= $a.a-a.b-b.a+b.b$

  = $|a|^{2}-2 a.b+|b|^{2}$     [$\because$ a.b=b.a]

 =$9-0+16=25$

$\Rightarrow$    $|a-b|=5$

In $\triangle ABC$ , if a=1,b=2 $\angle C=60^{0}$ , then $4 \triangle^{2}+c^{2}$=

Answer:

Explanation:

Clearly $\triangle =\frac{1}{2}a.b \sin c$

$\Rightarrow$    $\triangle =\frac{1}{2}.2.1. \sin 60^{0}$

$\Rightarrow$   $\triangle =\frac{\sqrt{3}}{2}$

$\Rightarrow$  $\triangle ^{2}=\frac{3}{4}$

$\Rightarrow$     $4\triangle^{2}=3$   .........(i)

By Sine rule, we get

   $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\Rightarrow$    $\frac{\sin A}{1}=\frac{\sin B}{2}=\frac{\sin 60^{0}}{c}$

$\Rightarrow$  $\frac{\sin A}{\sin B}=\frac{1}{2}=\frac{\sin 60^{0}}{c}$

 Considering last two  terms, we get

    $\frac{1}{2}$= $\frac{\frac{\sqrt{3}}{2}}{c}$

$\Rightarrow$        $c=\sqrt{3}$    ..........(ii)

 Thus , $4 \triangle^{2}+c^{2}$=3+3=6   [(using Eqs.(i) and (ii)]

• Mathematics - General questions