Question

Let a function $f:[0,5]\to R$, be continuous, $f\left(1\right)=3$ and $F$ be defined as:
$F\left(x\right)=\underset{1}{\overset{x}{\int }}{t}^{2}g\left(t\right)dt$, where $g\left(t\right)=\underset{1}{\overset{t}{\int }}f\left(u\right)du$. 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

Answer 1

The correct option is C a point of local minima

$F^{\prime }\left(x\right)=x^{2}g\left(x\right)$
Put $x=1$
$\Rightarrow F^{\prime }\left(1\right)=g\left(1\right)=0\cdots \left(1\right)$
Now $F^{\prime \prime }\left(x\right)=2xg\left(x\right)+g^{\prime }\left(x\right)x^2$
$F^{\prime \prime }\left(1\right)=2g\left(1\right)+g^{\prime }\left(1\right)\{\because g^{\prime }(x)=f(x\left)\right\}$
$F^{\prime \prime }\left(1\right)=f\left(1\right)=3\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, $F\left(x\right)$ has local minimum at $x=1$