Let a function f:[$$0$$,$$5$$]→R be continuous, $$f(1)=3$$ and F be defined as: $$F(x)=$$∫$$1xt2g(t)dt$$, where $$g(t)=$$∫$$1tf(u)du$$ Then for the function F, the point $$x=1$$ is:

Correct option is 2a point of local minima

Questions

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

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detailed solution

Correct option is B

F'(x)=x2g(x)max or min f'(x)=0f'(x)=2xg(x)+x2+g'(x)x2g(x)=0f'(0)=0x=0,g(x)=0f'(1)=2g(1)+g'(1)=3>0g(1)=0f'(1)=0f'(1)>0 Local minima at x=1

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Let a function f:[0,5]→R be continuous, f(1)=3 and F be defined as: F(x)=∫1xt2g(t)dt, where g(t)=∫1tf(u)du Then for the function F, the point x=1 is:

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