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 $x=3y-y^2$ and $x+y=3,$ then the number of values of $a$ is

Answer 1

$x=3y-y^2$ and $x+y=3$
Figure of the above two curves :


Now, shaded area is
$A=\underset{1}{\overset{3}{\int }}\left[\right(3y-y^2)-(3-y\left)\right]dy$
$\Rightarrow A=\underset{1}{\overset{3}{\int }}(4y-y^2-3)dy$
$\Rightarrow A={[2y^2-\frac{y^3}{3}-3y]}_1^3$
$\Rightarrow A=\frac{4}{3}$ sq. units



Now area bounded by $f\left(x\right)$ above the $x-$axis is $\underset{0}{\overset{1}{\int }}(x-x^2)dx$
$={[\frac{x^2}{2}-\frac{x^3}{3}]}_0^1=\frac{1}{6}<\frac{4}{3}$
Hence, $a<0$

Finding point of intersections of the curves $f\left(x\right)$ and $g\left(x\right)$
$x-x^2=ax$
$\Rightarrow x^2+(a-1)x=0$
$x_1+x_2=1-a$
Since $x_1=0,$ therefore $x_2=1-a$
Now, $\underset{0}{\overset{1-a}{\int }}\left(\right(x-x^2)-(ax\left)\right)dx=\frac{4}{3}$
$\Rightarrow {[(1-a)\frac{x^2}{2}-\frac{x^3}{3}]}_0^{1-a}=\frac{4}{3}$
$\Rightarrow \frac{1}{6}(1-a{)}^3=\frac{4}{3}$
$\Rightarrow (1-a{)}^3=8$
$\Rightarrow a=-1$
Hence, number of values of $a$ is $1.$