Question

Consider the functions $f,g:R\to R$ defined by
$f(x)=x^2+\frac{5}{12}\text{ and }g(x)=\{\begin{array}{ll}2\left(1-\frac{4|x|}{3}\right),& |x|\le \frac{3}{4},\\ 0,& |x|>\frac{3}{4}.\end{array}$
If $\alpha$ is the area of the region
$\left\{(x,y)\in R\times R:|x|\le \frac{3}{4},0\le y\le \min \{f(x),g(x)\}\right\},$
then the value of $9\alpha$ is _____

Answer 1

Answer: 6

$x^2+\frac{5}{12}=\frac{2-8x}{3}$
$x^2+\frac{8x}{3}+\frac{5}{12}-2=0$

$12x^2+32x-19=0$
$12x^2+38x-6x-19=0$
$2x(6x+19)-1(6x+19)=0$
$(6x+19)(2x-1)=0$
$x=\frac{1}{2}$
$\alpha =2A_1+A_2$
$\alpha =2\left(\underset{0}{\overset{1/2}{\int }}{x}^2+\frac{5}{12}dx+\frac{1}{2}\times \frac{1}{4}\times \frac{2}{3}\right)$
$\Rightarrow \alpha =2\left[{\left(\frac{x^3}{3}+\frac{5x}{12}\right)}_0^{1/2}+\frac{1}{12}\right]$
$\Rightarrow \alpha =2\left[\frac{1}{24}+\frac{5}{24}+\frac{1}{12}\right]$
$\Rightarrow \alpha =2\left[\frac{1+5+2}{24}\right]\Rightarrow \alpha =2\times \frac{8}{24}\Rightarrow 9\alpha =9\times \frac{8}{12}$
$\Rightarrow 9\alpha =6$