Integral Calculus Question 227
Question: If $ f(x)=A\sin ( \frac{\pi x}{2} )+B $ and $ f’( \frac{1}{2} )=\sqrt{2} $ and $ \int_0^{1}{f(x)dx=\frac{2A}{\pi }} $ , then what is the value of B?
Options:
A) $ \frac{2}{\pi } $
B) $ \frac{4}{\pi } $
C) 0
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given function $ f(x)=A\sin ( \frac{\pi x}{2} )+B $ Differentiating w.r.t. x $ f’(x)=A\cos ( \frac{\pi x}{2} ).\frac{\pi }{2} $ $ f’( \frac{1}{2} )=\sqrt{2}=A( \cos \frac{\pi }{4} )\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}{{ A\sin ( \frac{\pi x}{2} )+B }dx=\frac{2\times 4}{{{\pi }^{2}}}} $
$ \Rightarrow [ -A\cos \frac{\pi x}{2}.\frac{2}{\pi }+Bx ]_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 $
BETA
