A) \[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) 2
D) 1
Correct Answer: C
Solution :
\[\int\limits_{\sin t}^{1}{{{x}^{2}}g(x)\,dx=(1-\sin t)}\]c Differentiate w.r.t. t \[0-{{\sin }^{2}}\,t.g(\sin t)\,\cos t=-\cos t\] \[\therefore \,\,g(\sin t)\,=\frac{1}{{{\sin }^{2}}t}\] Put \[t=\frac{\pi }{4}\] \[\therefore \,\,g\left( \frac{1}{\sqrt{2}} \right)=2\]You need to login to perform this action.
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