A) \[\frac{2}{\pi }\]
B) \[\frac{4}{\pi }\]
C) 0
D) 1
Correct Answer: C
Solution :
| [c] Given function \[f(x)=A\,\,\sin \left( \frac{\pi x}{2} \right)+B\] |
| Differentiating w.r.t. x |
| \[f'(x)=A\,\,\cos \left( \frac{\pi x}{2} \right).\frac{\pi }{2}\] |
| \[f'\left( \frac{1}{2} \right)=\sqrt{2}=A\left( \cos \frac{\pi }{4} \right)\frac{\pi }{2}=A.\frac{1}{\sqrt{2}}.\frac{\pi }{2}\] |
| \[\Rightarrow A=\frac{(\sqrt{2}\times \sqrt{2})\times 2}{\pi }=\frac{4}{\pi }\] |
| Now, \[\int_{0}^{1}{f(x)dx=\frac{2A}{\pi }}\] |
| \[\Rightarrow \int_{0}^{1}{\left\{ A\sin \left( \frac{\pi x}{2} \right)+B \right\}dx=\frac{2\times 4}{{{\pi }^{2}}}}\] |
| \[\Rightarrow \left[ -A\cos \frac{\pi x}{2}.\frac{2}{\pi }+Bx \right]_{0}^{1}=\frac{8}{{{\pi }^{2}}}\] |
| \[\Rightarrow -\frac{4}{\pi }.\frac{2}{\pi }\cos \frac{\pi }{2}+B+\frac{4}{\pi }.\frac{2}{\pi }\cos 0=\frac{8}{{{\pi }^{2}}}\] |
| \[\Rightarrow B+\frac{8}{{{\pi }^{2}}}=\frac{8}{{{\pi }^{2}}}\Rightarrow B=0\] |
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