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JEE Main 2023 · 24 Jan · Shift 2 · numerical answer

JEE Main 2023 Mathematics question on Differential Equations

Let ff be a differentiable function defined on [0,π2]\left[0, \dfrac{\pi}{2}\right] such that f(x)>0f(x) > 0 and

f(x)+∫0xf(t)1−(log⁡ef(t))2 dt=e,∀x∈[0,π2]f(x) + \int_0^x f(t)\sqrt{1 - \left(\log_e f(t)\right)^2}\, dt = e, \quad \forall x \in \left[0, \dfrac{\pi}{2}\right]

Then (6log⁡ef ⁣(π6))2\left(6\log_e f\!\left(\dfrac{\pi}{6}\right)\right)^2 is equal to __________.
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Answer: 27

Solution

Differentiate both sides with respect to x:

f'(x) + f(x)√(1 - (ln f(x))²) = 0

Let u = ln f(x), so f'(x)/f(x) = u', giving:

u' + √(1 - u²) = 0 → du/√(1 - u²) = -dx

Integrating: arcsin(u) = -x + C

At x = 0: f(0) + 0 = e, so f(0) = e, meaning u(0) = ln e = 1, so arcsin(1) = π/2 = C.

Thus arcsin(ln f(x)) = π/2 - x → ln f(x) = sin(π/2 - x) = cos x.

At x = π/6: ln f(π/6) = cos(π/6) = √3/2.

Therefore (6 ln f(π/6))² = (6 · √3/2)² = (3√3)² = 27.

[JEE Main 2023 · 24 Jan · Shift 2]

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