Question

For which of the following values of $m$, the area of the region bounded by the curve $y=x-x^2$ and the line $y=mx$ equals $\frac{9}{2}$

• A. $-4$
• B. $-2$
• C. $2$
• D. $4$

Answer 1

The correct option is B $-2$

The equation of curve is $y=x-x^2$
$\Rightarrow x^2-x=y$
$\Rightarrow {(x-\frac{1}{2})}^2=-(y-\frac{1}{4})$
which is a parabola whose vertex is $(\frac{1}{2},\frac{1}{4})$
Hence, finding the point of intersection of the curve and the line,
$x-x^2=mx\Rightarrow x(1-x-m)=0$
i.e., $x=0$ or $x=1-m$
$\therefore \frac{9}{2}={\int }_0^{1-m}(x-x^2-mx)dx$
$={\{\frac{x^2}{2}-\frac{x^3}{3}-m\frac{x^2}{2}\}}_0^{1-m}$
$=(1-m)\frac{(1-m{)}^2}{2}-\frac{(1-m{)}^3}{3}=\frac{(1-m{)}^3}{6}$
$\therefore (1-m{)}^3=\frac{6\times 9}{2}=27$
$\Rightarrow 1-m=(27{)}^{1/3}=3$
Also, $(1-m{)}^3-(3{)}^3=0$
$\therefore (1-m{)}^3=3^3\Rightarrow 1-m=3$
or $m=-2$