Definite Integration Question 325
Question: Let $ f(x) $ be a non-negative continous function such that the area bounded by the curve $ y=f(x) $ , x-axis and the ordinates $ x=\frac{\pi }{4} $ , $ x=\beta >\frac{\pi }{4} $ is $ ( \beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta ) $ . Then $ f\ ( \frac{\pi }{2} ) $ is
[AIEEE 2005]
Options:
A) $ ( 1-\frac{\pi }{4}-\sqrt{2} ) $
B) $ ( 1-\frac{\pi }{4}+\sqrt{2} ) $
C) $ ( \frac{\pi }{4}+\sqrt{2}-1 ) $
D) $ ( \frac{\pi }{4}-\sqrt{2}+1 ) $
Show Answer
Answer:
Correct Answer: B
Solution:
Given that, $ \int _{\pi /4}^{\beta }{f\ (x)dx} $
$ =\beta \sin \beta +\frac{\pi }{4}\cos \beta +\sqrt{2}\beta $
Differentiating w.r.t. b, we get
$ \therefore $ $ f(\beta )=\sin \beta +\beta \cos \beta -\frac{\pi }{4}\sin \beta +\sqrt{2} $ ,
Hence, $ f\ ( \frac{\pi }{2} )=( 1-\frac{\pi }{4}+\sqrt{2} ) $ .
BETA
