A) \[\frac{1}{2}\left( \sqrt{3}-\sqrt{2} \right)\]
B) \[\frac{1}{2}\left( \sqrt{2}-1 \right)\]
C) \[\frac{1}{2}\left( \sqrt{3}-1 \right)\]
D) \[\frac{1}{2}\left( \sqrt{3}+1 \right)\]
Correct Answer: C
Solution :
\[g(x)=\cos {{x}^{2}},f(x)=\sqrt{x}\] \[y=g(f(x))=\cos x\] \[18{{x}^{2}}-9\pi x+{{\pi }^{2}}=0\] \[18{{x}^{2}}-6\pi x-3\pi x+{{\pi }^{2}}=0\] \[(6x-\pi )(3x-\pi )=0\] \[x=\frac{\pi }{6},\frac{\pi }{3}\] \[\alpha =\frac{\pi }{6},\beta =\frac{\pi }{3}\] Req. Area\[=\int_{\pi /6}^{\pi /3}{\cos x.dx=[\sin x]_{\pi /6}^{\pi /3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}}\]You need to login to perform this action.
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