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Let $f$ be a positive function and $I_{1} = \int_{1 - k}^{k} x f (x (1 - x)) d x, I_{2} = \int_{1 - k}^{k} f (x (1 - x)) d x$ where $2 k - 1 > 0$ then $I_{1} / I_{2}$ is

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Correct option is C

Since ∫ab f(x)dx=∫ab f(a+b−x)dx we have I1=∫1−kk (k+1−k−x)f((k+1−k−x)1-k+1-k-x=∫1−kk (1−x)f((1−x)x)dx=∫1−kk f((1−x)x)dx−∫1−kk xf((1−x)x)dx=I2−I1.So 2I1=I2 and II1/I2=1/2.

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Let f be a positive function and I1=∫1−kk xf(x(1−x))dx,I2=∫1−kk f(x(1−x))dx where 2k−1>0 then I1/I2 is

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Ready to Test Your Skills?

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Let $f$ be a positive function and $I_{1} = \int_{1 - k}^{k} x f (x (1 - x)) d x, I_{2} = \int_{1 - k}^{k} f (x (1 - x)) d x$ where $2 k - 1 > 0$ then $I_{1} / I_{2}$ is