Let a function $ f:[0,5] \rightarrow R $ be continuous, $ f(1)=3 $ and $ F $ be defined as: $ F(x)=\int_{1}^{x} t^{2} g(t) d t $, where $ g(t)=\int_{1}^{t} f(u) d u . $ Then for the function $ F(x) $, the point $ x=1 $ is:

Question

Let a function $ f:[0,5] \rightarrow R $ be continuous, $ f(1)=3 $ and $ F $ be defined as: $ F(x)=\int_{1}^{x} t^{2} g(t) d t $, where $ g(t)=\int_{1}^{t} f(u) d u . $ Then for the function $ F(x) $, the point $ x=1 $ is:, a point of local minima, not a critical point, a point of local maxima, a point of inflection

MARKS App · Accepted Answer

Given, F x = ∫ 1 x t 2 g t d t By Leibnitz rule we get, F ' x = x 2 g x ⇒ F ' 1 = 1 . g 1 = 0 ∵ g 1 = 0 Now F '' x = 2 x g x + x 2 g ' x ⇒ F '' x = 2 x g x + x 2 f x ∵ g ' x = f x ⇒ F '' 1 = 0 + 1 × 3 ⇒ F '' 1 = 3 F x has a local minimum at x = 1 .

Search any question & find its solution

Looking for more such questions to practice?