Let ‘f’ is a function , continuous on [0,1] such that f(x)≤5∀x∈[0,1] and f(x)≤2x∀x∈[12,1] then the smallest 'a' for which ∫01f(x)dx≤a holds for all 'f' is

2x≤5⇒x∈25,1∴∫01 f(x)dx=∫02/5 f(x)dx+∫2/51 f(x)dx≤525−0+∫2/51 2xdx∫01 f(x)dx≤2+2(ln⁡x)2/151a=2+2ln⁡52

Let ‘f’ is a function , continuous on [0,1] such that f(x)≤5∀x∈[0,1] and f(x)≤2x∀x∈[12,1] then the smallest 'a' for which ∫01f(x)dx≤a holds for all 'f' is

2x≤5⇒x∈25,1∴∫01 f(x)dx=∫02/5 f(x)dx+∫2/51 f(x)dx≤525−0+∫2/51 2xdx∫01 f(x)dx≤2+2(ln⁡x)2/151a=2+2ln⁡52

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Let ‘f’ is a function , continuous on [0,1] such that $f (x) \leq \sqrt{5} \forall x \in [0, 1]$ and $f (x) \leq \frac{2}{x} \forall x \in [\frac{1}{2}, 1]$ then the smallest $' a'$ for which $\int_{0}^{1} f (x) d x \leq a$ holds for all $' f'$ is

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$\sqrt{5}$

$\frac{\sqrt{5}}{2} + 21 n 2$

$2 + 1 n (\frac{\sqrt{5}}{2})$

$2 + 21 n (\frac{\sqrt{5}}{2})$

answer is D.

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$\frac{2}{x} \leq \sqrt{5} \Rightarrow x \in [\frac{2}{\sqrt{5}}, 1)$

$∴ \int_{0}^{1} f (x) d x = \int_{0}^{2 / \sqrt{5}} f (x) d x + \int_{2 / \sqrt{5}}^{1} f (x) d x$

$\leq \sqrt{5} (\frac{2}{\sqrt{5}} - 0) + \int_{2 / \sqrt{5}}^{1} \frac{2}{x} d x$

$\int_{0}^{1} f (x) d x \leq 2 + 2 (\ln x)_{2 / 1 \sqrt{5}}^{1}$

$a = 2 + 2 \ln [\frac{\sqrt{5}}{2}]$

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Let ‘f’ is a function , continuous on [0,1] such that f(x)≤5∀x∈[0,1] and f(x)≤2x∀x∈[12,1] then the smallest 'a' for which ∫01f(x)dx≤a holds for all 'f' is

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Let ‘f’ is a function , continuous on [0,1] such that $f (x) \leq \sqrt{5} \forall x \in [0, 1]$ and $f (x) \leq \frac{2}{x} \forall x \in [\frac{1}{2}, 1]$ then the smallest $' a'$ for which $\int_{0}^{1} f (x) d x \leq a$ holds for all $' f'$ is