Question

Let $f$ be a positive function. Let $I_1={\int }_{1-k}^k\left(x\right)f\left(x\right(1-x\left)\right)dx$; $I_2={\int }_{1-k}^{k}f\left(x\right(1-x\left)\right)dx$, where $2k-1>0$. Then, $\frac{I_2}{I_1}$ is

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

Answer 1

The correct option is D $2$

$I_1={\int }_{1-k}^k\left(x\right)f\left(x\right(1-x\left)\right)dx$

Using the identity:-

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x0dx$

$I_1={\int }_{1-k}^k(1-x)f\left(\right(1-x\left)\right(x\left)\right)dx$

$⟹I_1={\int }_{1-k}^{k}f\left(x\right(1-x\left)\right)dx-{\int }_{1-k}^k\left(x\right)f\left(x\right(1-x\left)\right)dx$

$⟹I_1=I_2-I_1$

$⟹2I_1=I_2$

$⟹\frac{I_2}{I_1}=2$

Hence, answer is option-(D).