Let f:[-π/2, π/2] → R be a differentiable function such that f(0)=1/2. If the lim x → 0 (x ∫₀ˣ f(t) dt)/(e^(x²)-1)=α, then 8 α² is equal to :

Solution

Maths sir · 4 steps

Split off (e^(x²)-1)/x² → 1, and what is left is the average of f near 0, which is f(0).

1. Regroup the fraction. Both top and bottom go to 0. Multiply and divide by x² so the standard limit appears on its own.

   (x∫₀ˣ f(t) dt)/(e^(x²)-1) = (∫₀ˣ f(t) dt)/x × x²/(e^(x²)-1)

2. Second factor. Put u = x². Then u/(eᵘ-1) → 1 as u → 0.
3. First factor. It is of the form 0/0, so apply L'Hopital's rule. By the fundamental theorem of calculus the derivative of ∫₀ˣ f(t) dt is f(x), so the limit is lim(x → 0) f(x) = f(0) = 1/2. This uses only that f is continuous, which it is because it is differentiable.
4. So. α = 1/2 × 1 = 1/2, and 8α² = 8 × 1/4 = 2, option (b).

The idea to keep. Pull standard limits like (eᵘ-1)/u → 1 out first. Then 1/x∫₀ˣ f(t) dt → f(0), because the derivative of an integral with variable upper limit is the integrand.

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