Question

For $f\left(x\right)=x+\underset{-1}{\overset{1}{\int }}(x-t{x}^2)f\left(t\right)dt$
$\left(I\right)\underset{x\to 0}{lim}f\left(x\right)=0\left(II\right)f^{\prime }(-\frac{1}{3})=1$
$\left(III\right)f$ is continuous and derivable on $R$
$\left(IV\right)$ maximum value of $f\left(x\right)$ does not exist

Which of the following is/are correct?
(correct answer + 2, wrong answer - 0.50)

• A. Only $\left(I\right)$
• B. Only $\left(II\right)$ and $\left(III\right)$
• C. $\left(I\right),\left(II\right)$ and $\left(III\right)$
• D. All of them

Answer 1

The correct option is C $\left(I\right),\left(II\right)$ and $\left(III\right)$

Given : $f\left(x\right)=x+\underset{-1}{\overset{1}{\int }}(x-t{x}^2)f\left(t\right)dt$
$\Rightarrow f\left(x\right)=x+x\underset{-1}{\overset{1}{\int }}f\left(t\right)dt-x^2\underset{-1}{\overset{1}{\int }}tf\left(t\right)dt\Rightarrow f\left(x\right)=ax-b{x}^2$
Where $a=1+\underset{-1}{\overset{1}{\int }}f\left(t\right)dt,b=\underset{-1}{\overset{1}{\int }}tf\left(t\right)dt$

Now,
$a=1+\underset{-1}{\overset{1}{\int }}f\left(t\right)dt\Rightarrow a=1+\underset{-1}{\overset{1}{\int }}at-b{t}^{2}dt\Rightarrow a=1-\frac{2b}{3}\cdots \left(1\right)b=\underset{-1}{\overset{1}{\int }}tf\left(t\right)dt\Rightarrow b=\underset{-1}{\overset{1}{\int }}t(at-b{t}^2)dt\Rightarrow b=\frac{2a}{3}\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get
$a=1-\frac{4a}{9}\Rightarrow a=\frac{9}{13}\Rightarrow b=\frac{6}{13}\Rightarrow f\left(x\right)=\frac{9x-6x^2}{13}\Rightarrow f^{\prime }\left(x\right)=\frac{9-12x}{13}\Rightarrow f^{\prime }(-\frac{1}{3})=\frac{13}{13}=1$
$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)=0$

As $f\left(x\right)$ is a polynomial of degree $2$, so it is continuous and derivable on $R$
As it is a downward parabola, so it has a finite maximum value.