Question

Let $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$, where $f$ is such that $\frac{1}{2}\le f\left(t\right)\le 1$ for $t\in [0,1]$ and $0\le f\left(t\right)\le \frac{1}{2}$ for $t\in [1,2].$ Then, $g\left(2\right)$ satisfies the inequality

• A. $-\frac{3}{2}\le g\left(2\right)<\frac{1}{2}$
• B. $\frac{1}{2}\le g\left(2\right)<2$
• C. $\frac{3}{2}<g\left(2\right)\le \frac{5}{2}$
• D. $2<g\left(2\right)<4$

Answer 1

The correct option is B $\frac{1}{2}\le g\left(2\right)<2$

Given, $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
$\Rightarrow g\left(2\right)=\underset{0}{\overset{2}{\int }}f\left(t\right)dt\Rightarrow g\left(2\right)=\underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{1}{\overset{2}{\int }}f\left(t\right)dt$
Now,
$\frac{1}{2}\le f\left(t\right)\le 1$ for $t\in [0,1]$
$\Rightarrow \frac{1}{2}\le \underset{0}{\overset{1}{\int }}f\left(t\right)dt\le 1\cdots \left(1\right)$

Also,
$0\le f\left(t\right)\le \frac{1}{2}$ for $t\in [1,2]$
$\Rightarrow 0\le \underset{1}{\overset{2}{\int }}f\left(t\right)dt\le \frac{1}{2}\cdots \left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$, we get
$\frac{1}{2}\le \underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{1}{\overset{2}{\int }}f\left(t\right)dt\le \frac{3}{2}\Rightarrow \frac{1}{2}\le g\left(2\right)\le \frac{3}{2}\therefore \frac{1}{2}\le g\left(2\right)<2$