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 to 9/2 ?

• A. - 4
• B. - 2
• C. 2
• D. 4

Answer 1

Answer: (B), (D)

The correct options are
B - 2
D 4

The two curves meet at
$mx=x-x^{2}or{x}^2=x(1-m)\therefore x=0,1-m$
${\int }_0^{1-m}(y_1-y_2)dx={\int }_0^{1-m}(x-x^2-mx)dx={[(1-m)\frac{x^2}{2}-\frac{x^3}{3}]}_0^{1-m}$
$\frac{9}{2}$ if m < 1or ${(1-m)}^3[\frac{1}{2}-\frac{1}{3}]=\frac{9}{2}$ or ${(1-m)}^3=27$
$\therefore m=-2$
But if m > 1 then 1 - m is negative then
${[(1-m)\frac{x^2}{2}-\frac{x^{}3}{3}]}_{1-m}^0=\frac{9}{2}-{(1-m)}^3(\frac{1}{2}-\frac{1}{3})=\frac{9}{2}$
$\therefore -{(1-m)}^3=-27$ or $1-m=-3\therefore m=4$