Question

Let $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt,$ where $f$ is continuous function in $[0,3]$ such that $\frac{1}{3}\le f\left(t\right)\le 1$ for all $t\in [0,1]$ and $0\le f\left(t\right)\le \frac{1}{2}$ for all $t\in (1,3]$. The largest possible interval in which $g\left(3\right)$ lies is :

• A. $[1,3]$
• B. $[1,-\frac{1}{2}]$
• C. $[-\frac{3}{2},-1]$
• D. $[\frac{1}{3},2]$

Answer 1

The correct option is D $[\frac{1}{3},2]$

Given : $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
Now,
$g\left(3\right)=\underset{0}{\overset{3}{\int }}f\left(t\right)dt\Rightarrow g\left(3\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{1}{\overset{3}{\int }}f\left(t\right)dt$
We know
$\frac{1}{3}\le f\left(t\right)\le 1$ for all $t\in [0,1]$
$\underset{0}{\overset{1}{\int }}\frac{1}{3}dt\le \underset{0}{\overset{1}{\int }}f\left(t\right)dt\le \underset{0}{\overset{1}{\int }}1dt\Rightarrow \frac{1}{3}\le \underset{0}{\overset{1}{\int }}f\left(t\right)dt\le 1$
Similarly,
$0\le \underset{1}{\overset{3}{\int }}f\left(t\right)dt\le 1$
Therefore,
$g\left(3\right)\in [\frac{1}{3},2]$