Discussion Forum

Let $g(x)=\cos x ^{2}, f(x)=\sqrt{x}$ and $\alpha ,\beta (\alpha <\beta)$ be the roots of the quadratic equation  

$18x^{2}-9\pi x +\pi^{2}=0$ .Then , the area (in sq units) bounded by the curve  $y= (gof) (x)$ and the lines x=α, x=β

 and y=0, is 

Answer:

Explanation:

We have 

$\Rightarrow$    $18x^{2}-9\pi x +\pi^{2}=0$

$\Rightarrow$   $18x^{2}-6\pi x-3\pi x +\pi^{2}=0$

 $(6x-\pi) (3x-\pi)=0$

  $\Rightarrow$     $x=\frac{\pi}{6},\frac{\pi}{3}$

 Now ,  $\alpha<\beta =\frac{\pi}{6},\beta=\frac{\pi}{3}$

  Given  $g(x)=\cos x^{2}$ and  $f(x)=\sqrt{x}$

       $y=g Of(x)$

       $\therefore$    $y=g (f(x) )=\cos x$

    Area of region bounded by x=α,

  x=β , y=0 and curve  $y=g (f(x) )$ is

    $A=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x dx$

    $A=[\sin x]_\frac{\pi}{6}^\frac{\pi}{3}$

    $A=\sin \frac{\pi}{3}-\sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2}$

      $A=(\frac{\sqrt{3}-1}{2})$