Question

Let f be a positive function. Let
$I_1={\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dx,I_2={\int }_{1-k}^{k}f\left\{x\right(1-x\left)\right\}dx$
$when2k-1>0.Then\frac{I_1}{I_2}is$ [IIT 1997 Cancelled]

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

Answer 1

The correct option is C

$\frac{1}{2}$

$I_1={\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dx$
$={\int }_{1-k}^k(1-k+k-x)f\left[\right(1-k+k-x\left)\right\{1-(1-k+k-x)\left\}\right]dx$
$={\int }_{1-k}^k(1-x)f\left\{x\right(1-x\left)\right\}dx$
$={\int }_{1-k}^{k}f\left\{x\right(1-x\left)\right\}dx-{\int }_{1-k}^{k}xf\left\{x\right(1-x\left)\right\}dx=I_2-I_1$
$\therefore 2I_1=I_2\Rightarrow \frac{I_1}{I_2}=\frac{1}{2}$