Correct Answer - d
In `f : R to R, f(x) = x^(4)`
Let `x, y in R and f(x) = f(y)`
`rArr " " x^(4) = y^(4) rArr x = pm y`
`therefore f` is not one-one.
`rArr f` is many one.
Again let `f(x) =y ` where `y in R`
`rArr " " x^(4) = y `
`rArr " " x = (y)^(1//4) notin R if y = -1 `
`therefore f ` is not onto.
Therefore, `f` is neither one-one nor onto.