Question

Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$ where $\frac{1}{2}\le f\left(t\right)\le 1,tϵ[0,1]$ and $0\le f\left(t\right)\le \frac{1}{2}$ for $tϵ(1,2)$, then [IIT Screening 2000]

• 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 $0\le g\left(2\right)<2$

$g\left(2\right)={\int }_0^{2}f\left(t\right)dt={\int }_0^{1}f\left(t\right)dt+{\int }_1^{2}f\left(t\right)dt$
As $\frac{1}{2}\le f\left(t\right)\le 1$ for $0\le t\le 1$,
${\int }_0^1\frac{1}{2}dt\le {\int }_0^{1}f\left(t\right)dt\le {\int }_0^{1}tdt$ or $\frac{1}{2}\le {\int }_0^{1}f\left(t\right)dt\le 1\dots \dots \left(i\right)$
As $0\le f\left(t\right)\le \frac{1}{2}$ for $1<t\le 2,{\int }_1^{2}0dt\le {\int }_1^{2}f\left(t\right)dt\le {\int }_1^2\frac{1}{2}dt$
$\therefore 0\le {\int }_1^{2}f\left(t\right)dt\le \frac{1}{2}\dots \dots \left(ii\right)$
Adding (i) and (ii), $\frac{1}{2}\le g\left(2\right)\le \frac{3}{2}$
$\therefore$ g(2) satisfies the inequality $0\le g\left(2\right)<2$