Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If ∫₀ˣ √(1 - (f'(t))²) dt = ∫₀ˣ f(t)dt , 0 ≤ x ≤ 1 and f(0) = 0, then lim _(x → 0) 1/x²∫₀ˣ f(t)dt :

Solution

Maths sir · 4 steps

Differentiate the given equation to get f(x) = sin x, then one round of L'Hopital finishes the limit.

1. Remove the integrals. Both sides are integrals from 0 to x, so differentiate them with respect to x. This gives √(1 - (f'(x))²) = f(x). Squaring, (f'(x))² + (f(x))² = 1.
2. Find f. Differentiate again: 2f'(f'' + f) = 0. Near x = 0, f is close to 0, so (f')² is close to 1 and f' ≠ 0. Hence f'' = -f, so f = A sin x + B cos x. The condition f(0) = 0 gives B = 0, then A² = 1, and f ≥ 0 forces A = 1. So f(x) = sin x.
3. Apply L'Hopital. As x → 0, the integral and x² both go to 0, so this is a 0/0 form. Differentiate the top and the bottom. The integral gives back f(x), and x² gives 2x.

   lim(x → 0) (∫₀ˣ f(t) dt)/x² = lim(x → 0) f(x)/2x = lim(x → 0) (sin x)/2x = 1/2

4. So. The limit equals 1/2, which is option (d).

The idea to keep. An integral from 0 to x divided by a power of x is a 0/0 form. Differentiate the top and the bottom: the integral just returns its integrand, so you only need f near 0.

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