JEE Main On 08 April 2018 Question 19
Question: 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=(g\circ f)(x) $ and the lines $ x=\alpha ,x=\beta $ and $ y=0 $ , is: [JEE Main Online 08-04-2018]
Options:
A) $ \frac{1}{2}( \sqrt{3}-\sqrt{2} ) $
B) $ \frac{1}{2}( \sqrt{2}-1 ) $
C) $ \frac{1}{2}( \sqrt{3}-1 ) $
D) $ \frac{1}{2}( \sqrt{3}+1 ) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ g(x)=\cos x^{2},f(x)=\sqrt{x} $
$ y=g(f(x))=\cos x $ $ 18x^{2}-9\pi x+{{\pi }^{2}}=0 $
$ 18x^{2}-6\pi x-3\pi x+{{\pi }^{2}}=0 $
$ (6x-\pi )(3x-\pi )=0 $ $ x=\frac{\pi }{6},\frac{\pi }{3} $
$ \alpha =\frac{\pi }{6},\beta =\frac{\pi }{3} $
Req. Area $ =\int _{\pi /6}^{\pi /3}{\cos x.dx=[\sin x] _{\pi /6}^{\pi /3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}} $
