If $f$ is a positive function and $I_{1} = \int_{1 - k}^{k}{x\ f\left\lbrack x(1 - x) \right\rbrack\ dx,}$ $I_{2} = \int_{1 - k}^{k}{f\left\lbrack x(1 - x) \right\rbrack\ dx,}$ Where $2k - 1 0,\ then\frac{I_{1}}{I_{2}}$ is

Question

If $f$ is a positive function and $I_{1} = \int_{1 - k}^{k}{x\ f\left\lbrack x(1 - x) \right\rbrack\ dx,}$ $I_{2} = \int_{1 - k}^{k}{f\left\lbrack x(1 - x) \right\rbrack\ dx,}$ Where $2k - 1 > 0,\ then\frac{I_{1}}{I_{2}}$ is

Dr. Abubacker · Accepted Answer

(c)
$I_{1} = \int_{1 - k}^{k}{(1 - x)f\left\lbrack x(1 - x \right)\rbrack\ dx}$
$\Rightarrow \ 2I_{1} = I_{2}\  \Rightarrow \ \frac{I_{1}}{I_{2}} = \frac{1}{2}$

Q. If $f$ is a positive function and $I_{1} = \int_{1 - k}^{k}{x\ f\left\lbrack x(1 - x) \right\rbrack\ dx,}$ $I_{2} = \int_{1 - k}^{k}{f\left\lbrack x(1 - x) \right\rbrack\ dx,}$ Where $2k - 1 > 0,\ then\frac{I_{1}}{I_{2}}$ is

Explanation

Reviewed by Dr. Abubacker

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