Question

Let $f$ be a positive function.
If $ I_{1}=\int\limits_{1-k}^{k} x f[x(1-x)] d x$ and $I_{2}=\int\limits_{1-k}^{k} f[x(1-x)] d x$, where $2 k-1>0 .$
Then, $\frac{I_{1}}{I_{2}}$ is

• A. 2
• B. k
• C. $ \frac{1}{2} $
• D. 1

Answer 1

Answer: (C)

Given, $ I_{1}=\int\limits_{1-k}^{k} x f[x(1-x)] d x$
$\Rightarrow I_{1}=\int\limits_{1-k}^{k}(1-x) f[(1-x) x] d x$
$\left.\left.=\int\limits_{1-k}^{k} f[(1-x)] d x\right]-\int\limits_{1-k}^{k} x f(1-x)\right] d x$
$\Rightarrow I_{1}=I_{2}-I_{1} \Rightarrow \frac{I_{1}}{I_{2}}=\frac{1}{2}$