Question

Given $f\left(x\right)=\{\begin{array}{cc}x,& 0\le x<\frac{1}{2}\\ \frac{1}{2}& x=\frac{1}{2}\\ 1-x& \frac{1}{2}<x<1\end{array}$
and $g\left(x\right)={(x-\frac{1}{2})}^2,x\in R$. Then the area (in sq. units) of the region bounded by the curves $y=f\left(x\right)$ and $y=g\left(x\right)$ between the lines $2x=1$ to $2x=\sqrt{3}$ is:

• A. $\frac{\sqrt{3}}{4}-\frac{1}{3}$
• B. $\frac{1}{3}+\frac{\sqrt{3}}{4}$
• C. $\frac{1}{2}+\frac{\sqrt{3}}{4}$
• D. $\frac{1}{2}-\frac{\sqrt{3}}{4}$

Answer 1

The correct option is A $\frac{\sqrt{3}}{4}-\frac{1}{3}$

Given $f\left(x\right)=\{\begin{array}{cc}x,& 0\le x<\frac{1}{2}\\ \frac{1}{2},& x=\frac{1}{2}\\ 1-x,& \frac{1}{2}<x<1\end{array}$
$g\left(x\right)={(x-\frac{1}{2})}^2$
The area between $f\left(x\right)$ and $g\left(x\right)$ from $x=\frac{1}{2}$ to $x=\frac{\sqrt{3}}{2}:$

$\therefore$ Required area $=\underset{1/2}{\overset{\sqrt{3}/2}{\int }}\left(f\right(x)-g(x\left)\right)dx$
$=\underset{1/2}{\overset{\sqrt{3}/2}{\int }}((1-x)-{(x-\frac{1}{2})}^2)dx$
$={[x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x}{4}+\frac{x^2}{2}]}_{1/2}^{\sqrt{3}/2}$
$=(\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{24})-(\frac{3}{8}-\frac{1}{24})$
$=\frac{\sqrt{3}}{4}-\frac{1}{3}$ sq.units