Question

Let $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$, where $f$ is such that $\frac{1}{2}\le f\left(x\right)\le 1$ 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. $0\le g\left(2\right)<2$
• C. $\frac{1}{2}\le g\left(2\right)\le \frac{3}{2}$
• D. $2<g\left(2\right)<4$

Answer 1

The correct option is C $\frac{1}{2}\le g\left(2\right)\le \frac{3}{2}$

$g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt=\underset{0}{\overset{1/2}{\int }}f\left(t\right)dt+\underset{1/2}{\overset{1}{\int }}f\left(t\right)dt$
As, $t\in [0,1]\Rightarrow \frac{1}{2}\le f\left(t\right)\le 1$
$\Rightarrow \underset{0}{\overset{1}{\int }}\frac{1}{2}dt\le \underset{0}{\overset{1}{\int }}f\left(t\right)dt\le \underset{0}{\overset{1}{\int }}1dt$
$\Rightarrow \frac{1}{2}\le \underset{0}{\overset{1}{\int }}f\left(t\right)dt\le \mathrm{1...}\left(1\right)$
As, $t\in [1,2]\Rightarrow 0\le f\left(t\right)\le \frac{1}{2}$
$\Rightarrow \underset{1}{\overset{2}{\int }}0\cdot dt\le \underset{1}{\overset{2}{\int }}f\left(t\right)dt\le \underset{1}{\overset{2}{\int }}\frac{1}{2}dt$
$\Rightarrow 0\le \underset{1}{\overset{2}{\int }}f\left(t\right)dt\le \frac{1}{2}...\left(2\right)$
Add equation $\left(1\right)$ and $\left(2\right)$
$\Rightarrow \frac{1}{2}\le \underset{0}{\overset{1}{\int }}f\left(t\right)dt+\underset{1}{\overset{2}{\int }}f\left(t\right)dt\le \frac{3}{2}$
$\Rightarrow \frac{1}{2}\le g\left(2\right)\le \frac{3}{2}$