A) \[f\left( \frac{1}{2} \right)<\frac{1}{2}and\,\,f\left( \frac{1}{3} \right)>\frac{1}{3}\]
B) \[f\left( \frac{1}{2} \right)>\frac{1}{2}and\,\,f\left( \frac{1}{3} \right)>\frac{1}{3}\]
C) \[f\left( \frac{1}{2} \right)<\frac{1}{2}and\,f\left( \frac{1}{3} \right)<\frac{1}{3}\]
D) \[f\left( \frac{1}{2} \right)>\frac{1}{2}and\,f\left( \frac{1}{3} \right)<\frac{1}{3}\]
Correct Answer: C
Solution :
| We have,\[\int_{0}^{x}{\sqrt{1-{{(f'\left( t \right))}^{2}}}dt=\int_{0}^{x}{f\left( t \right)dt}}\] On differentiating, we get |
| \[\sqrt{1-{{(f'(x))}^{2}}}=f(x)\] \[\Rightarrow \,\,\,\,f'(x)=\pm \sqrt{1-{{(f(x))}^{2}}}\] |
| \[\therefore \] \[f\left( x \right)=\sin x\] |
| \[f\left( x \right)\ne -\sin x\] |
| \[x>\sin x\] |
| \[\therefore \,\,\,\frac{1}{2}>\sin \frac{1}{2}\] \[\Rightarrow \,\,\,\,\,\frac{1}{2}>f\left( \frac{1}{2} \right)and\frac{1}{3}>f\left( \frac{1}{3} \right)\] |
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