Question

Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$, 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(2) satisfies the inequality,

• A. $-\frac{3}{2}\le g\left(2\right)<\frac{1}{2}$
• B. $0\le g\left(2\right)<2$
• C. $\frac{3}{2}<\left(2\right)\le \frac{5}{2}$
• D. $2<g\left(2\right)<4$

Answer 1

Answer: (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, $\frac{1}{2}\le f\left(t\right)\le$ for $t\in [0,1]$

We get ${\int }_0^1\frac{1}{2}dt\le {\int }_0^{1}f\left(t\right)dt\le {\int }_0^{1}1dt\Rightarrow \frac{1}{2}\le {\int }_0^{1}f\left(t\right)dt\le 1$ ...(i)

Again, $0\le f\left(t\right)\le \frac{1}{2}$ for $t\in [1,2]$ ...(ii)

$\Rightarrow {\int }_1^{2}0dt\le {\int }_1^{2}f\left(t\right)dt\le {\int }_1^2\frac{1}{2}dt$ $\Rightarrow 0\le {\int }_1^{2}f\left(t\right)dt\le \frac{1}{2}$

From Eqs.(i) and (ii), we get

$\frac{1}{2}\le {\int }_0^{1}f\left(t\right)dt+{\int }_1^{2}f\left(t\right)dt\le \frac{3}{2}$ $\Rightarrow \frac{1}{2}\le g\left(2\right)\le \frac{3}{2}$ $\Rightarrow 0\le g\left(2\right)<2.$