Let f be a differentiable function defined on [ 0,π/2 ] such that f(x) > 0 and f(x) + ∫₀ˣ f(t)√(1 - (log ₑf(t))²) dt = e,∀ x ∈ [ 0,π/2 ]. Then ( 6log ₑf( π/6 ) )² is equal to __________.

Solution

Maths sir · 5 steps

Differentiate to remove the integral, then put y = ln f so the equation separates.

1. Starting value. Put x = 0. The integral from 0 to 0 is zero, so f(0) = e and ln f(0) = 1.
2. Differentiate. Differentiate both sides with respect to x and use the fundamental theorem of calculus. This gives f'(x) + f(x)√(1 - (ln f(x))²) = 0.
3. Substitute y. Let y = ln f(x), so y' = f'(x)/f(x). Dividing the equation by f(x) > 0 gives y' = -√(1 - y²), which separates.

   ∫ dy/√(1 - y²) = -∫ dx ⇒ sin⁻¹ y = -x + C

4. Find C. At x = 0, y = 1, so C = sin⁻¹ 1 = π/2. Then y = sin(π/2 - x) = cos x, so ln f(x) = cos x.
5. So. ln f(π/6) = cos π/6 = √3/2. Then (6 × √3/2)² = (3√3)² = 27.

The common slip: 9. Taking the constant as zero gives ln f(x) = -sin x and the answer 9. The value f(0) = e forces C = π/2.

The idea to keep. When an unknown function sits inside an integral from 0 to x, differentiate once. Then put x = 0 in the original equation to get the starting value that fixes the constant.

Answer from the NTA final answer key. Solution by PrepWiser.