Q134 / 200 · Manipal (MET) 2012 · Mathematics

Let $f(x)$ be a positive function. Let $\begin{aligned} I_1 & =\int_{1-k}^k x f\{x(1-x)\} d x, \\ I_2 & =\int_{1-k}^k f\{x(1-x)\} d x, \end{aligned}$ where $2 k-1\gt0$, then $\frac{I_1}{I_2}$ is

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

Solution

$\begin{aligned} I_1 & =\int_{1-k}^k x f\{x(1-x)\} d x \\ & =\int_{1-k}^k(1-x) f\{(1-x)(1-(1-x))\} d x \\ & \quad\left(\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right) \\ & =\int_{1-k}^k(1-x) f\{(1-x) x\} d x \\ & =\int_{1-k}^k f\{x(1-x)\} d x -\int_{1-k}^k x f\{x(1-x)\} d x \\ \Rightarrow & \quad I_1=I_2-I_1 \Rightarrow 2 I_1=I_2\end{aligned}$ $\Rightarrow \quad \frac{I_1}{I_2}=\frac{1}{2}$

Hence, the correct answer is option: 'C'.

Relevant Questions