For `x in R` the expression `(x ^(2) +2x +c)/(x ^(2)+ 4x + 3x )` can take all real value if `c in ` :

Options

A. `a in (0, 2)`
B. `a in [0, 1]`
C. `a in [-1, 1]`
D. none of these

Answer

B

Explanation

Let `y = (x^(2)+2x + a)/(x^(2)+4x+3a)`. Then, `x^(2)(y-1)+2(2y-1) x + a(3y-1)=0`
`rArr" "4(2y-1)^(2) - 4(y-1) a (3y-1) ge 0` for all `y in R" "[because x in R]`
`rArr" "(4-3a)y^(2)-4(1-a)y+1 - a ge 0` for all y `in` R
`rArr" "4 - 3a gt 0` and Discriminant `le 0`
`rArr" "a lt (4)/(3) and 16 (1-a)^(2) - 4(1-a)(4-3a) le 0`
`rArr" "a lt (4)/(3) and a (a-1) le 0`
`rArr" "a lt (4)/(3) and 0 le a le 1 rArr 0 le a le 1 rArr a rArr [0, 1]`