Let f be a positive function. Let I_1 = int_{ | vedclass

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(C) Given $I_1 = \int_{1 - k}^k x f\{x(1 - x)\} dx$ and $I_2 = \int_{1 - k}^k f\{x(1 - x)\} dx$.Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b

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$1/2$

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(C) Given $I_1 = \int_{1 - k}^k x f\{x(1 - x)\} dx$ and $I_2 = \int_{1 - k}^k f\{x(1 - x)\} dx$.Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a + b - x) dx$,we have $a + b = 1 - k + k = 1$.Thus,$I_1 = \int_{1 - k}^k (1 - x) f\{(1 - x)(1 - (1 - x))\} dx$.$I_1 = \int_{1 - k}^k (1 - x) f\{(1 - x)x\} dx$.$I_1 = \int_{1 - k}^k f\{x(1 - x)\} dx - \int_{1 - k}^k x f\{x(1 - x)\} dx$.$I_1 = I_2 - I_1$.$2I_1 = I_2$.Therefore,$I_1/I_2 = 1/2$.

Let $f$ be a positive function. Let $I_1 = \int_{1 - k}^k x f\{x(1 - x)\} dx$ and $I_2 = \int_{1 - k}^k f\{x(1 - x)\} dx$,where $2k - 1 > 0$. Then $I_1/I_2$ is

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