Let f(x) be a function, satisfying f:[0,1]→ℝ, and 5∫01(f(x))2dx+1=10∫01x2f(x)dx then which is/are correct

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$Let f (x) be a function, satisfying f : [0, 1] \to ℝ, and$ $5 \int_{0}^{1} (f (x))^{2} d x + 1 = 10 \int_{0}^{1} x^{2} f (x) d x then which is/are correct$

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$f (x) is concave down everywhere in (0, 1)$

$f (x) = e^{x} has unique solution$

$Maximum value of f (x) is equal to '1$

$f (\frac{1}{2}) = \frac{1}{4}$

answer is A, C.

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Detailed Solution

$\begin{array}{l} Given 5 \int_{0}^{1} {[f (x)]}^{2} d x + 1 = 10 \int_{0}^{1} x^{2} f (x) d x \\ 5 \int_{0}^{1} {[f (x)]}^{2} d x + 5 \int_{0}^{1} x^{4} d x = 10 \int_{0}^{1} x^{2} f (x) d x \\ 5 \int_{0}^{1} {[f (x)]}^{2} - 2 x^{2} f (x) + {(x^{2})}^{2} d x = 0 \\ 5 \int_{0}^{1} {[f (x) - x^{2}]}^{2} d x = 0 \\ f (x) = x^{2} \\ f (\frac{1}{2}) = \frac{1}{4} \\ The function has maximum value 1 in the given interval \end{array}$