Question

Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt,$ where $\frac{1}{2}\le f\left(t\right)\le 1$ for $t\in [0,1]$ and $0\le f\left(t\right)\le \frac{1}{2}$ for $t\in (1,2].$ Then

• A. $-\frac{3}{2}\le g\left(2\right)<\frac{1}{2}$
• B. $0\le g\left(2\right)<2$
• 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 $0\le g\left(2\right)<2$

Given $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$
$\Rightarrow g\left(2\right)={\int }_0^{2}f\left(t\right)dt={\int }_0^{1}f\left(t\right)dt+{\int }_1^{2}f\left(t\right)dt$
Now $\frac{1}{2}\le f\left(t\right)\le 1$ for $t\in [0,1]$
We get ${\int }_0^1\frac{1}{2}dt\le {\int }_0^{1}f\left(t\right)dt\le {\int }_0^{1}1dt\Rightarrow \frac{1}{2}\le {\int }_0^{1}f\left(t\right)dt\le 1$ ...(1)
Again $0\le f\left(t\right)\le \frac{1}{2}$ for $t\in [1,2]$
$\Rightarrow {\int }_1^{1}0dt\le {\int }_1^{2}f\left(t\right)dt\le {\int }_1^2\frac{1}{2}dt\Rightarrow 0\le {\int }_1^{2}f\left(t\right)dt\le \frac{1}{2}$ ...(2)
From (1) and (2)
$\frac{1}{2}\le {\int }_0^{1}f\left(t\right)dt+{\int }_1^{2}f\left(t\right)dt\le \frac{3}{2}$$\Rightarrow \frac{1}{2}\le g\left(2\right)\le \frac{3}{2}\Rightarrow 0\le g\left(2\right)<2$