Question

Let $f(x)$ be a function defined by
$f(x)=\underset{1}{\overset{x}{\int }}t(t^2-3t+2)dt,1\le x\le 3$, then the range of $f(x)$ is

• A. $[0,2]$
• B. $[-\frac{1}{4},4]$
• C. $[-\frac{1}{4},2]$
• D. None of these

Answer 1

Answer: (C)

Given, $f(x)=\underset{1}{\overset{x}{\int }}t(t^2-3t+2)dt$
$\Rightarrow f^{\prime }(x)=x(x^2-3x+2)=x(x-1)(x-2)$
(using leibnitz rule of derivative)
Now, $f^{\prime }(x)\le 0$ in $1\le x\le 2$
$f^{\prime }(x)\ge 0$ in $2\le x\le 3$

$\therefore f(x)$ decreases in $[1,2]$ and increases in $[2,3]$
$\therefore$ min $(f(x))=f(2)=\underset{1}{\overset{2}{\int }}x(x^2-3x+2)dx$
$=(\frac{x^2}{4}-x^3+x^2{)}_1^2=-\frac{1}{4}$
and max $(f(x))=$ the greatest among $f(1),f(3)$
$\therefore f(1)=\underset{1}{\overset{1}{\int }}x(x^2-3x+2)dx=0$
$f(3)=\underset{1}{\overset{3}{\int }}x(x^2-3x+2)dx=2$
$\therefore$ max $f(x)=2$
$\therefore$ range of $f(x)=[-\frac{1}{4},2]$