Let a function ƒ : [0, 5] → R be continuous, ƒ(1) = 3 and F be defined as :
F(x) = ∫₁ˣ t²g(t)dt , where g(t) = ∫₁ᵗ f(u)du
Then for the function F, the point x = 1 is :

Solution

Maths sir · 4 steps

Differentiate F twice with the Leibniz rule, then check the sign of F'' at x = 1.

1. First derivative. Since f is continuous, g is differentiable with g'(t) = f(t). Differentiating the integral, F'(x) = x² g(x).
2. Critical point. g(1) = ∫₁¹ f(u) du = 0, so F'(1) = 1 × 0 = 0. That makes x = 1 a critical point of F.

   F''(x) = 2x g(x) + x² f(x)

3. Second derivative test. At x = 1: F''(1) = 2 × 1 × 0 + 1 × f(1) = 3. This is positive.
4. So. F'(1) = 0 and F''(1) > 0, so x = 1 is a point of local minima of F, option (c).

The idea to keep. When a function is built from integrals, the lower limit makes the integral vanish there. Use d/dx∫ₐˣ h(t) dt = h(x) to get F', then let the sign of F'' decide.

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