For which of the following values of `m` is the area of the regions bounded by the curve `y=x-x^2` and the line `y=m x` equal `9/2?` `-4` (b) `-2` (c) 2 (d) 4

Options

A. -4
B. -2
C. 2
D. 4

Answer

B, D

Explanation

Case I When m = 0
In this case, ` y=x-x^(2) " "…(i), `
and `y=0 " "…(ii)`
are two given curves, `y gt 0` is total region above X-axis.
Therefore, area between `y=x-x^(2)` and ` y=0`
is area between `y=x-x^(2)` and above the X-axis

` therefore A=int_(0)^(1)(x-x^(2))dx=[(x^(2))/(2)-(x^(3))/(3)]_(0)^(1)=(1)/(2)-(1)/(3)=(1)/(6) ne (9)/(2) `
Hence, no solutation exists.
Case II When `m lt 0 `
In this case, area between ` y=x-x^(2)` and ` y=mx` is OABCO and points of intersection are (0,0) and
` {1-m,m(1-m)}`.
` therefore " Area of curve OABCO " = int_(0)^(1-m)[x-x^(2)-mx]dx `

`=[(1-m)(x^(2))/(2)-(x^(3))/(3)]_(0)^(1-m) `
` = (1)/(2)(1-m)^(3)-(1)/(3)(1-m)^(3)=(1)/(6)(1-m)^(3) `
` therefore (1)/(6)(1-m)^(3)=(9)/(2) " "["given"]`
` rArr (1-m)^(3)=27`
` rArr 1-m=3 `
` rArr m=-2 `
Case III When ` m gt 0 `
In this case, ` y=mx ` and ` y=x-x^(2)` intersect in (0,0) and
`{(1-m), m(1-m)}` as shown in figure

` therefore " Area of shaded region " =int_(1-m)^(0)(x-x^(2)-mx)dx `
`=[(1-m)(x^(2))/(2)-(x^(3))/(3)]_(1-m)^(0) `
`=-(1)/(2)(1-m)(1-m)^(2)+(1)/(3)(1-m)^(3)`
`=-(1)/(6)(1-m)^(3)`
`rArr (9)/(2)=-(1)/(6)(1-m)^(3) " "["given"]`
` rArr (1-m)^(3)=-27`
`rArr (1-m)=-3 `
`rArr m=3+1=4 `
Therefore, (b) and (d) are the answers.