Question

Let $f\left(x\right)=x-x^2$ and $g\left(x\right)=ax$. If the area bounded by $y=f\left(x\right)$ and $y=g\left(x\right)$ is equal to the area bounded by the curves $x=y-y^2$ and $x+y=3$, then the number of possible values of $a$ is

Answer 1

Area between $x=3y-y^2$ and $x+y=3$

$A=\underset{1}{\overset{3}{\int }}\left[\right(3y-y^2)-(3-y\left)\right]dy$
Solving, we get
$A=\frac{4}{3}$

$x-x^2=ax$
$\Rightarrow x^2+(a-1)x=0$
If $x_1,x_2$ are the roots,
then $x_1+x_2=1-a$
$\Rightarrow x_1=0,x_2=1-a$

Given, $\left|\underset{0}{\overset{1-a}{\int }}\left[\right(x-x^2)-(ax\left)\right]dx\right|=\frac{4}{3}$
$\Rightarrow \left|\frac{1}{6}\right(1-a{)}^3|=\frac{4}{3}$
$\Rightarrow |(1-a{)}^3|=8$
$\Rightarrow 1-a=\pm 2$
$\Rightarrow a=-1,3$
$\Rightarrow$ Number of values of $a$ is $2$