Definite Integration Question 336
Question: If $ \int_0^{x}{f(t)dt}=x+\int_x^{1}{tf(t)dt,} $ then the value of $ f(1) $ is
[IIT 1998; AMU 2005]
Options:
A) 1/2
B) 0
C) 1
D) -1/2
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_0^{x}{f(t)dt=x+\int_x^{1}{tf(t)dt\Rightarrow \int_0^{x}{f(t)dt=x-\int_1^{x}{tf(t)dt}}}} $
Differentiating w.r.t x, we get $ f(x)=1+{0-xf(x)} $
therefore $ f(x)=1-xf(x)\Rightarrow (1+x)f(x)=1\Rightarrow f(x)=\frac{1}{1+x} $
$ f(1)=\frac{1}{1+1}=\frac{1}{2} $ .
BETA
