Definite Integration Question 333
Question: Let $ f $ be a positive function. Let $ I_1=\int _{1-k}^{k}{xf{ x(1-x) }}dx $ , $ I_2=\int _{1-k}^{k}{f{ x(1-x) }}dx $ when $ 2k-1>0. $ Then $ I_1/I_2 $ is
[IIT 1997 Cancelled]
Options:
A) 2
B) $ k $
C) $ 1/2 $
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
$ I_1=\int _{1-k}^{k}{xf{x(1-x)}dx} $
$ =\int _{1-k}^{k}{(1-k+k-x)f[(1-k+k-x){1-(1-k+k-x)}]dx} $
$ (\because \int_a^{b}{f(x)dx=\int_a^{b}{f(a+b-x)dx)}} $
$ =\int _{1-k}^{k}{(1-x)f{x(1-x)}}dx $
$ =\int _{1-k}^{k}{f{x(1-x)}}dx-\int _{1-k}^{k}{xf{x(1-x)}}dx=I_2-I_1 $
$ \therefore 2I_1=I_2\Rightarrow \frac{I_1}{I_2}=\frac{1}{2} $ .
BETA
