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

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Differential Equations

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) 1 - (lo g_{e} f (t))^{2} d t = e, \forall x \in [0, \frac{π}{2}]

. Then

(6 lo g_{e} f (\frac{π}{6}))^{2}

is equal to

Answer: 27

Setting

x = 0

in the identity gives

f (0) = e

. Differentiating,

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

. With

y = f (x)

this is

\frac{d y}{y 1 - ( ln y ) ^{2}} = - d x

; the substitution

t = ln y

makes the left side

\frac{d t}{1 - t ^{2}}

, so

sin^{- 1} (ln y) = - x + C

. The condition

f (0) = e

gives

sin^{- 1} 1 = C = \frac{π}{2}

, hence

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

. At

x = \frac{π}{6}

this gives

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

, so

(6 ln f (\frac{π}{6}))^{2} = (33)^{2} = 27

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer

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