Question

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

• A. $-4$
• B. $-2$
• C. $2$
• D. $4$

Answer 1

Answer: (D)

$4$

The two curves meet at $mx=x-x^2$ or $x^2=x(1-m)$
$\Rightarrow x^2=x-mx$
$\therefore x=0,1-m$
${\int }_0^{1-m}(y_1-y_2)dx$
$={\int }_0^{1-m}(x-x^2-mx)dx$
$={[(1-m)\frac{x^2}{2}-\frac{x^2}{3}]}_0^{1-m}=\frac{9}{2}$(given)
If $m<1$
or $={(1-m)}^3[\frac{1}{2}-\frac{1}{3}]=\frac{9}{2}$
or $-{(1-m)}^3=-27$
or $1-m=-3$
$\Rightarrow m=4$