Definite Integration Question 427
Question: If $ f(x)=A\sin ( \frac{\pi x}{2} )+B, $
$ {f}’( \frac{1}{2} )=\sqrt{2} $ and $ \int_0^{1}{f(x)dx=\frac{2A}{\pi },} $ then the constants $ A $ and $ B $ are respectively
[IIT 1995]
Options:
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 } $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=A\sin ( \frac{\pi x}{2} )+B,f’( \frac{1}{2} )=\sqrt{2} $ ,
$ \int_0^{1}{f(x)dx=\frac{2A}{\pi }} $ ,
(given)
therefore $ \int_0^{1}{{ A\sin ( \frac{\pi x}{2} )+B }dx=\frac{2A}{\pi }} $
therefore $ | -\frac{2A}{\pi }\cos \frac{\pi x}{2}+Bx |_0^{1}=\frac{2A}{\pi } $
therefore $ B-( \frac{-2A}{\pi } )=\frac{2A}{\pi }\Rightarrow B=0 $
$ \therefore f(x)=A\sin \frac{\pi x}{2}\Rightarrow f’(x)=\frac{\pi A}{2}\cos \frac{\pi x}{2} $
$ \therefore f’( \frac{1}{2} )=\frac{\pi A}{2}( \frac{1}{\sqrt{2}} )=\sqrt{2}\Rightarrow \pi A=4\Rightarrow A=\frac{4}{\pi } $
Hence $ A=\frac{4}{\pi },B=0 $ .
BETA
