Question

Let a function \(f : [0, 5] \to R\) be continuous, \(f(1) = 3\) and F be defined as :
\(F(x) = \int_1^x t^2 g(t)\,dt\), where \(g(t) = \int_1^t f(u)\,du\).
Then for the function F, the point x = 1 is :

• A. a point of inflection.
• B. a point of local maxima.
• C. a point of local minima.
• D. not a critical point.

Hint:

Differentiate F twice using the fundamental theorem of calculus and evaluate at x = 1, where g(1) = 0.

Answer 1

The Correct Option is C

\(F'(x) = x^2 g(x)\) and \(g(1) = \int_1^1 f(u)\,du = 0\), so \(F'(1) = 0\): x = 1 is a critical point.
\(g'(x) = f(x)\), so \(F''(x) = 2x\,g(x) + x^2 f(x)\).
\(F''(1) = 2 \cdot 1 \cdot 0 + 1 \cdot f(1) = 3 \gt 0\).
Since \(F'(1) = 0\) and \(F''(1) \gt 0\), x = 1 is a point of local minima.
Answer: (3) a point of local minima.