If f is continuous in [0, 1] and \underset{x \rightarrow 0}{Lim} \frac{\int_{0}^{x} f \left(\right. \sqrt{1 - x^{2}} \left.\right) \textrm{ } \left(cost\right)^{2} \textrm{ } dt}{\left(tan\right)^{- 1} \left(\right. 2 x \left.\right)} = 4 then f (1) is equal to

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

Question

If f is continuous in [0, 1] and $\underset{x \to 0}{Lim} \frac{\int_{0}^{x} f (\sqrt{1 - x^{2}}) {cost}^{2} dt}{\tan^{- 1} (2 x)}$ = 4 then f (1) is equal to

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$\underset{x \to 0}{Lim} \frac{f (\sqrt{1 - x^{2}}) \int_{0}^{x} {cost}^{2} dt}{\tan^{- 1} (2 x)}$

$= 4 \underset{x \to 0}{Lim} \frac{f (\sqrt{1 - x^{2}}) {cosx}^{2} + \int_{0}^{x} {cost}^{2} dt \cdot f' (\sqrt{1 - x^{2}}) (\frac{- 2 x}{2 \sqrt{1 - x^{2}}})}{\frac{2}{1 + 4 x^{2}}} = 4$

= 4 ⇒ f (1) = 8