Question

Consider the function $f\left(x\right)$ such that $f\left(x\right)=x^2+{\int }_0^1(x+t)f\left(t\right)dt$, then the minimum value of $f\left(x\right)$ is,

• A. $\frac{-109}{3}$
• B. $\frac{-109}{6}$
• C. $\frac{-109}{12}$
• D. $\frac{-15}{2}$

Answer 1

Answer: (C)

$\frac{-109}{12}$

$y\left(x\right)=x^2+ax+b,$ where $a={\int }_0^{1}y\left(t\right)dt,b={\int }_0^{1}t.y\left(t\right)dt$

$\therefore a={\int }_0^1(t^2+at+b)dt,b={\int }_0^{1}t.(t^2+at+b)dt={\int }_0^1(t^3+a{t}^2+bt)dt$

$\Rightarrow a={[\frac{t^3}{3}+\frac{a{t}^2}{2}+bt]}_0^1,b={[\frac{t^4}{4}+\frac{a{t}^3}{3}+\frac{b{t}^2}{2}]}_0^1$

$a=\frac{1}{3}+\frac{a}{2}+b$ and $b=\frac{1}{4}+\frac{a}{3}+\frac{b}{2}$

$\Rightarrow \frac{a}{2}-b=\frac{1}{3}$ and $\frac{b}{2}-\frac{a}{3}=\frac{1}{4}$

By solving we get

substituting $b=\frac{a}{2}-\frac{1}{3}$ in $\frac{b}{2}-\frac{a}{3}=\frac{1}{4}$

we get

$\frac{b}{2}-\frac{a}{3}=\frac{1}{4}$

$\Rightarrow \frac{(\frac{a}{2}-\frac{1}{3})}{2}-\frac{a}{3}=\frac{1}{4}$

$\Rightarrow \frac{a}{4}-\frac{1}{6}-\frac{a}{3}=\frac{1}{4}$

$\Rightarrow \frac{-a}{12}=\frac{10}{24}$

$\Rightarrow a=-5$

$\therefore b=\frac{a}{2}-\frac{1}{3}=\frac{-5}{2}-\frac{1}{3}=\frac{-17}{6}$

$\therefore f\left(x\right)=x^2-5x-\frac{17}{6}$

$={(x-\frac{5}{2})}^2-\frac{25}{4}-\frac{17}{6}$

$={(x-\frac{5}{2})}^2-\frac{109}{12}$

$\therefore f_{min}=-\frac{109}{12}$