Definite Integration Question 33
Question 33
- Let $f$ be a positive function.
If $I_{1}=\int_{1-k}^{k} x f[x(1-x)] d x$ and
$$ I_{2}=\int_{1-k}^{k} f[x(1-x)] d x \text {, where } 2 k-1>0 $$
Then, $\frac{I_{1}}{I_{2}}$ is
(1997C, 2M) (a) 2 (b) $k$ (c) $1 / 2$ (d) 1
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Solution:
- Given, $I_{1}=\int_{1-k}^{k} x f[x(1-x)] d x$
$$ \begin{aligned} \Rightarrow \quad I_{1} & =\int_{1-k}^{k}(1-x) f[(1-x) x] d x \ & \left.\left.=\int_{1-k}^{k} f[(1-x)] d x\right]-\int_{1-k}^{k} x f(1-x)\right] d x \ \Rightarrow \quad I_{1} & =I_{2}-I_{1} \Rightarrow \frac{I_{1}}{I_{2}}=\frac{1}{2} \end{aligned} $$
