KVPY Sample Paper KVPY Stream-SX Model Paper-27

Let \[f(x)=\int\limits_{0}^{x}{\cos \left( \frac{{{t}^{2}}+2t+1}{5} \right)}\,dt,\] \[0<x<2.\] Then \[f(x)-\]

A) Increases monotonically

B) Decreases monotonically

C) Has one point of local maximum

D) Has one point of local minimum

Correct Answer: C

Solution :

\[f'(x)=\cos \left( \frac{{{x}^{2}}+2x+1}{5} \right)=\cos \left( \frac{{{(x+1)}^{2}}}{5} \right)\]

Since, \[0\le x\le 2,\] \[\frac{1}{5}\le \frac{{{(x+1)}^{2}}}{5}\le \frac{9}{5}\]

\[\cos \frac{{{(x+1)}^{2}}}{5}=0,\]

Only when \[\frac{{{(x+1)}^{2}}}{5}=\frac{\pi }{2}\]

\[f'(x)>0,\] if \[0<x<\sqrt{\frac{5\pi }{2}}-1\]

\[f'(x)<0,\] if \[\sqrt{\frac{5\pi }{2}}-1<x<2\]

\[\therefore x=\sqrt{\frac{5\pi }{2}}-1\] is a point of local maximum.