In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.(i) `f : R->R ,`defined by `f (x) = 3 4x`(ii) `f : R->R ,`defined by `f(x) =1+x^2`

Explanation

(i) `f:R to R and f(x) = 3 - 4x `
Let `x, y in R and f(x) = f(y)`
`rArr " " 3-4x = 3 - 4y `
`rArr " " - 4x =-4y rArr x =y`
`therefore f` is one-one.
Again, let `f(x) = y` where `y in R`
`rArr " " 3 - 4x =y`
` rArr " " -4 x = y -3`
`rArr " " x = ((y-3))/(-4) in R AA y in R`
`therefore f `is onto.
Therefore, `f` is one-one onto function.
(ii) `f: R to R and f(x) = (1+x^(2))`
Let `x, y in R and f(x) = f(y)`
`rArr " " 1+ x^(2) = 1 + y^(2) rArr x ^(2) = y^(2)`
` therefore f` is not one-one .
Again, let `f(x)= y` where `y in R`
`rArr (1-x^(2)) = y`
`rArr x^(2) = y-1`
`rArr x = pm sqrt(y-1) notin R if y =-2`
`therefore f` is not onto.
Therefore, `f` is neither one-one nor onto.