If the straight line x=b divide the area enclosed by `y=(1-x)^(2),y=0 " and " x=0 " into two parts " R_(1)(0le x le b) " and " R_(2)(b le x le 1) " such that " R_(1)-R_(2)=(1)/(4). ` Then, b equals

Options

A. `(3)/(4)`
B. `(1)/(2)`
C. `(1)/(3)`
D. `(1)/(4)`

Answer

B

Explanation

Here, area between 0 to b is `R_(1)` and b to 1 is ` R_(2).`
`therefore int_(0)^(b)(1-x)^(2)dx-int_(b)^(1)(1-x)^(2)dx = (1)/(4) `
`rArr [((1-x)^(3))/(-3)]_(0)^(b)-[((1-x)^(3))/(-3)]_(b)^(1)=(1)/(4) `
`rArr -(1)/(3)[(1-b)^(3)-1]+(1)/(3)[0-(1-b)^(3)]=(1)/(4) `
` rArr -(2)/(3)(1-b)^(3)=-(1)/(3)+(1)/(4)=-(1)/(12) rArr (1-b)^(3)=(1)/(8) `
` rArr (1-b)=(1)/(2) rArr b=(1)/(2) `