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. `(sqrt3)/2`
B. `2/(sqrt3)`
C. `sqrt3`
D. `1/(sqrt3)`

Answer

D

Explanation

Since `y=kx^(2)` and `x=ky^(2)` intersect at `(1/k,1/k)`
`int_(0)^(1/k)(sqrt(x/k)-kx^(2))dx=1`
`(x^(2))/(2k)-(kx^(3))/3=1`
`2/(3k^(2))-1/(3k^(2))=1`
`1/(3k^(2))=1`
`k^(2)=1/3`
`k=1/(sqrt3)`