Area Ques 22
Question
- Let $g(x)=\cos x^{2}, f(x)=\sqrt{x}$ and $\alpha, \beta(\alpha<\beta)$ be the roots of the quadratic equation $18 x^{2}-9 \pi x+\pi^{2}=0$. Then, the area (in sq units) bounded by the curve $y=(g o f)(x)$ and the lines $x=\alpha, x=\beta$ and $y=0$, is
(2018 Main)
(a) $\frac{1}{2}(\sqrt{3}-1)$
(b) $\frac{1}{2}(\sqrt{3}+1)$
(c) $\frac{1}{2}(\sqrt{3}-\sqrt{2})$
(d) $\frac{1}{2}(\sqrt{2}-1)$
Show Answer
Answer:
Correct Answer: 22.(a)
Solution:
Formula:
- We have,
$\Rightarrow$ $18 x^{2}-9 \pi x+\pi^{2}$ $=0$
$\Rightarrow$ $18 x^{2}-6 \pi x-3 \pi x+\pi^{2}$ $=0$
$(6 x-\pi)(3 x-\pi)$ $=0$
$\Rightarrow$ $x$ $=\frac{\pi}{6}, \frac{\pi}{3}$
Now, $\alpha<\beta$
$\alpha$ $=\frac{\pi}{6}$,
$\beta$ $=\frac{\pi}{3}$
Given, $g(x)=\cos x^{2}$ and $f(x)=\sqrt{x}$
$ \begin{aligned} & y =g o f(x) \\ & y =g(f(x))=\cos x \end{aligned} $
Area of region bounded by $x=\alpha, x=\beta, y=0$ and curve $y=g(f(x))$ is
$ \begin{aligned} A & =\int _{\pi / 6}^{\pi / 3} \cos x d x \\ A & =[\sin x] _{\pi / 6}^{\pi / 3} \\ A & =\sin \frac{\pi}{3}-\sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2} \\ A & =(\frac{\sqrt{3}-1}{2}) \end{aligned} $
BETA
