Let `f: RtoR ,f(x)=(x-a)/((x-b)(x-c)),b > c `
If `f` is onto, then prove that `a in (c , b)`

Explanation

`f(x)=(x-a)/((x-b)(x-c))`
Since f(x) is onto, range must be R.
Now, `yx^(2)-[(b+c)y+1]x+bcy +a=0`
Now, x is real. Therefore,
`D ge 0`
`or [(b+c)y+1]^ (2)-4y(bcy+a)ge 0 AA y in R`
`" "`[as given that `f(x)` is an onto function]
`implies (b-c)^(2)y^(2)+2(b+c-2a)y+1 ge 0 AA y in R`
` :. D le 0`
` or 4(b+c-2a)^(2)-4(b-c)^(2) le 0`
`or (b+c-2a-b+c)(b+c-2a+b-c) le0`
`or (c-a) (b-a) le 0`
i.e., ` c le a and b ge a or c ge a and b le a`
` or c le a le b " "("as " b gt c)`
`or a in (c,b)`