Question

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

• A. $-4$
• B. $-2$
• C. $2$
• D. $4$

Answer 1

Answer: (B), (D)

The correct options are
B $-2$
D $4$

$Area={\int }_0^{1-m}(x-x^2-mx)dx$
$={\left[\right(1-m)\cdot \frac{x^2}{2}-\frac{x^3}{3}]}_0^{1-m}$
$=(1-m{)}^3(\frac{1}{2}-\frac{1}{3})=\pm \frac{9}{2}$ (given)
Taking $+ve$ sign, $(1-m{)}^3=27\Rightarrow m=-2$
Taking $-ve$ sign, $(1-m{)}^3=-27\Rightarrow m=4$.