Area Question 14
Question 14
- 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 $y=(g \circ f)(x)$ 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) $\frac{1}{2}(\sqrt{2}-1)$
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Solution:
- 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, $\alpha<\beta$ | $\alpha$ | $=\frac{\pi}{6}$, | |
| $\beta$ | $=\frac{\pi}{3}$ |
322 Area
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
$$ \begin{aligned} A & =\int_{\pi / 6}^{\pi / 3} \cos x d x \ A & =[\sin x]_{\pi / 6}^{\pi / 3} \ A & =\sin \frac{\pi}{3}-\sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2} \ A & =\frac{\sqrt{3}-1}{2} \end{aligned} $$
