Question

Let a function $f:[0,5]\to R$, be continuous, $f\left(1\right)=3$ and $F$ be defined as: $F\left(x\right)=\begin{array}{l}{\int }_1^x{t}^{2}g\left(t\right)dt\end{array}$ where, $\begin{array}{l}g\left(t\right)={\int }_1^{t}f\left(u\right)du\end{array}$ 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

Explanation for the correct options:

Analyzing the given function $F$ at $x=1$

$F\left(x\right)=\begin{array}{l}{\int }_1^x{t}^{2}g\left(t\right)dt\end{array}$

Differentiating it w.r.t. $x$ we get

$$\begin{gathered} \Rightarrow F\text{'}\left(x\right)=x^{2}g\left(x\right) \\ =x^2{\int }_1^{x}f\left(u\right)du\mathbf{[}\mathbf{\because }\begin{array}{l}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{1}}^{\mathbf{t}}\mathbf{f}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{d}\mathbf{u}\end{array}\mathbf{}\mathbf{}\mathbf{]} \\ \Rightarrow F\text{'}\text{'}\left(x\right)=x^{2}f\left(x\right)+2x{\int }_1^{x}f\left(u\right)du\mathbf{}\mathbf{[}\begin{array}{l}\mathbf{\text{Again differentiatint}}\mathbf{}\mathbf{w}\mathbf{.}\mathbf{r}\mathbf{.}\mathbf{t}\mathbf{.}\mathbf{}\mathbf{x}\mathbf{}\\ \mathbf{\text{Applying}}\mathbf{}\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{(}\mathbf{g}\mathbf{\times }\mathbf{h}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{g}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{h}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{h}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\end{array}\mathbf{}\mathbf{}\mathbf{]} \\ \Rightarrow F\text{'}\text{'}\left(1\right)=1^{2}f\left(1\right)+\left(2\times 1\right)\left[{\left(\int f\left(u\right)du\right)}_{x=1}\right] \\ \Rightarrow F\text{'}\text{'}\left(1\right)=f\left(1\right)+2\times 0 \\ \Rightarrow F\text{'}\text{'}\left(1\right)=f\left(1\right) \\ \Rightarrow F\text{'}\text{'}\left(1\right)=3\left[\because f\left(1\right)=3\text{given}\right] \end{gathered}$$

Since $F\text{'}\left(1\right)=0$ and $F\text{'}\text{'}\left(1\right)=3>0$

Thus $F$ has a local minima at $x=1$

Hence, option C is the correct answer.