A) \[\frac{\pi }{2}\]and\[\frac{\pi }{2}\]
B) \[\frac{2}{\pi }\]and\[\frac{3}{\pi }\]
C) \[\frac{4}{\pi }\]and\[0\]
D) \[0\]and\[\frac{-4}{\pi }\]
Correct Answer: C
Solution :
We have,\[f(x)=A\sin \left( \frac{\pi x}{2} \right)+B,\,\,f'\left( \frac{1}{2} \right)=\sqrt{2}\] and\[\int_{0}^{1}{f(x)}=\frac{2A}{\pi }\] \[\Rightarrow \] \[\int_{0}^{1}{A\sin \left( \frac{\pi x}{2} \right)+B=\frac{2A}{\pi }}\] \[\Rightarrow \] \[\left[ \frac{-2A\cos \left( \frac{\pi x}{2} \right)}{\pi }+Bx \right]=\frac{2A}{\pi }\] \[\Rightarrow \] \[-\frac{2A}{\pi }\cos \left( \frac{\pi }{2} \right)+B+\frac{2A}{\pi }\cos {{0}^{o}}=\frac{2A}{\pi }\] \[\Rightarrow \] \[B+\frac{2A}{\pi }=\frac{2A}{\pi }\] \[\Rightarrow \] \[B=0\] Now, \[f(x)=A\sin \left( \frac{\pi x}{2} \right)\] \[\Rightarrow \] \[f'(x)=A\sin \left( \frac{\pi x}{2} \right)\] \[\Rightarrow \] \[f'(x)=\frac{\pi A}{2}\cos \left( \frac{\pi x}{2} \right)\] \[\Rightarrow \] \[f'\left( \frac{1}{2} \right)=\frac{\pi A}{2}\cos \left( \frac{\pi }{4} \right)\] \[\Rightarrow \] \[A=\frac{\pi }{4}\] \[\therefore \] \[A=\frac{4}{\pi },\,\,B=0\]
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