Question

$Letg\left(x\right)={\int }_0^{x}f\left(t\right)dt,fissuchthat\frac{1}{2}\le f\left(t\right)\le 1fortϵ[0,1]and0\le f\left(t\right)\le \frac{1}{2}fortϵ[1,2].Theng\left(2\right)satisfiestheinequality$

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

Answer 1

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

$g\left(x\right)={\int }_0^{x}f\left(t\right)dt$

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

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

As we can form value of $t$ and $f\left(t\right)$, $f$ is a decreasing fraction,

$f\left(2\right)=0,f\left(1\right)=\frac{1}{2},f\left(0\right)=1$.

Range $f\left(t\right)=[0,1]$

Now, $g\left(x\right)=[\frac{f^2\left(x\right)}{2}{]}_0^x$

$g\left(x\right)=\frac{[f\left(1\right){]}^2}{2}-\frac{f\left(0\right){]}^2}{2}$

$\Rightarrow \frac{-1}{2}\le g\left(2\right)\le \frac{3}{2}$ (Range of $f^2\left(x\right)=[0,1]$)

$\Rightarrow \frac{-3}{2}\le g\left(2\right)\le \frac{1}{2}$ (regation)