Let `A={1,3,5}, B={6,7,8}` and `f={(1,6),(3,7),(5,8)} ` be a function from `A` to `B`. Show that `f` is one to one and onto function.

Explanation

Let `f(x) = (ax^(2) + 3x - 4)/(3x - x^(2) +a)`
`rArr (a + 4y) x^(2) + (3-3y) x - 4 - ay = 0`
Now, x is real. So
D `ge` 0
`rArr 9 (1 -y)^(2) + 4 (a + 4y) (4 + ay )ge`0
`rArr 9 ( + 16 a)y^(2) + (-18 + 4a^(2) + 64)y + (9 + 16a )ge 0, AA y in R (because y` all real values)
`rArr 9 + 16a gt 0 and (4a^(2) + 46)^(2) - 4 (9 + 16 a)^(2) le 0`
`rArr a gt - (9)/(16)`
and `(4a^(2) + 46 - 18 - 32 a) (4a^(2) + 46 + 18 32a )le 0`
`rArr a gt - (9(/(16) and (a^(2) - 8a + 7) (a^(2) + 8a + 16) le 0`
`rArr a gt - (9)/(16) and 1 le a le 7 or a = - 4`
`rArr 1 le a le 7`