Let f be a differentiable function defined on [ 0 , 2 ] such that f ( x ) 0 and f ( x ) + 0 x f ( t ) 1 - ( log e f ( t ) ) 2 ? d t = e , [ 0 , ] . Then ( 6 log e f ( 6 ) ) 2 is equal to _____ . [2023]

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Solution

Q.
Let $f$ be a differentiable function defined on $[0, \frac{π}{2}]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e, \forall x \in [0, \frac{π}{2}] .$ Then $(6 \log_{e} f {(\frac{π}{6}))}^{2}$ is equal to _____ . [2023]

Ans.
(27)

$Given, f (x) + \int_{0}^{x} f (t) \sqrt{1 - (\log_{e} f {(t))}^{2}} d t = e$ ...(i)

$\Rightarrow f (0) = e$

$Differentiating (i) both sides, we get$

$f' (x) + f (x) \sqrt{1 - (\ln f {(x))}^{2}} = 0$

$Put f (x) = y$

$\Rightarrow \frac{d y}{d x} = - y \sqrt{1 - {(\ln y)}^{2}}$ $\Rightarrow \int \frac{d y}{y \sqrt{1 - {(\ln y)}^{2}}} = - \int d x$

$Put \ln y = t$

$\Rightarrow \int \frac{d t}{\sqrt{1 - t^{2}}} = - x + c \Rightarrow \sin^{- 1} t = - x + c$

$\Rightarrow \sin^{- 1} (\ln y) = - x + c$

$\Rightarrow \sin^{- 1} (\ln f (x)) = - x + c; f (0) = e$

$\Rightarrow \frac{π}{2} = C \Rightarrow \sin^{- 1} (\ln (f (x))) = - x + \frac{π}{2}$

$\Rightarrow \sin^{- 1} (\ln f (\frac{π}{6})) = \frac{- π}{6} + \frac{π}{2} = \frac{π}{3}$

$∴$ $\ln f (\frac{π}{6}) = \sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$

$Thus,$ $(6 \log_{e} f {(\frac{π}{6}))}^{2} = {(6 \times \frac{\sqrt{3}}{2})}^{2} = 27$

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