If the area enclosed between the curves `y=kx^2` and `x=ky^2`, where `kgt0`, is 1 square unit. Then k is: (a) `1/sqrt(3)` (b) `sqrt(3)/2` (c) `2/sqrt(3)` (d) `sqrt(3)`

Options

A. `3/4`
B. `1/3`
C. `3`
D. `4/3`

Answer

B

Explanation

`int_(1)^(1/k + 1) ( sqrt((x-1)/(k)) - k (x - 1)^2 ) d x = 1`
` [ ( 2 ( x- 1)^(3//2) )/(3 sqrtk) - (k (x -1)^3)/(3) ]_(1)^(1/k + 1) rArr (2)/(3k^2) - (1)/(3k^2) = 1`
` k^2 = 1/3 , k = (1)/(sqrt3)`