Question

Let $f\left(x\right)$ be a function defined by $f\left(x\right)={\int }_1^{x}t(t^2-3t+2)dt,1\le x\le 3$.

Then, range of $f\left(x\right)$ is?

• A. $[0,2]$
• B. $[-1/4,4]$
• C. $[-1/4,2]$
• D. none of these

Answer 1

Answer: (C)

$[-1/4,2]$

We have,
$f\left(x\right)={\int }_1^x(t^3-3t^2+2t)dt$
$\Rightarrow f^{\prime }\left(x\right)=x^3-3x^2+2x=x(x-2)(x-1)$
The changes in signs of $f^{\prime }\left(x\right)$ for different values of $x$ are given in figure.
Clearly, $f\left(x\right)$ is decreasing in $[1,2]$ and increasing in $[2,3]$.
$\therefore Min.f\left(x\right)=f\left(2\right)={\int }_1^{2}t(t^2-3t+2)dt={[\frac{t^4}{4}-t^3+t^2]}_1^2=-\frac{1}{4}$
and $Max.f\left(x\right)=f\left(3\right)={\int }_1^{3}t(t^2-3t+2)dt={[\frac{t^4}{4}-t^3+t^2]}_1^3=2$
Since $f\left(x\right)$ is continuous on $[1,3]$. Therefore, range of $f\left(x\right)=f\left[f\right(2),f(3\left)\right]=[-1/4,2]$.