If `(y^2-5y+3)(x 62+x+1)<2x` for all `x in R ,` then fin the interval in which `y` m lies.

Explanation

`(y^(2) - 5y+ 3) (x^(2)+x +2) lt 2x, AA x in`R
`rArr y^(2) - 5y + 3 lt (2x)/(x^(2) + x +1) (because x^(2) + x + 1 gt 0 AA x in` R )
L.H.S. must be less then least value of R. H. S. Now lets find the range of R.H.S
Let `(2x)/(x^(2) + x + 1) = p`
or `px^(2)+ (p - 2) x + p = 0`
Since x is real, we have
`(p - 2)^(2) - 4p^(2) ge 0`
or `-2le p le(2)/(3)`
The minimum value of `2x //(x^(2) + x + 1)` is -2. So ,
`y^(2) - 5y + 3 lt - 2`
`rArr y^(2) -5y + 3 lt 0`
`rArr y in ((5-sqrt(5))/(2),(5+ sqrt(5))/(2))`