If `(x^2+a x+3)//(x^2+x+a)` takes all real values for possible real values of `x ,` then `4a^2+39<0` b. `4a^5+39 >-0` c. `ageq1/4` d. `a<1/4`

Options

A. -3
B. 2
C. -1
D. -4

Answer

1,4

Explanation

Let `(x^(2) + ax + 3)/(x ^(2) + x + a) = y `
`rArr x^(2) (1 -y) - x (y - a) + 3 - ay = 0 `
`because x in ` R
`(y - a)^(2) - 4 (1 -y) (3 -ay) ge 0 `
`rArr (1 - 4a)y^(2) + (2a + 12) y + a ^(2) - 12 ge 0 ` (1)
Now , (1) is ture for all y `in` R , if A - 4a ` gt 0 and D le 0`. Hence ,
`alt (1)/(4) and 4(a + 6)^(2) - 4 (a^(2) - 12 ) (1 - 4 a) le 0 `
`rArr alt (1)/(4) and 4a^(3) - 36 + 48 le 0 `
`rArr a lt (1)/(4) and 4a^(3) - 36 + 48 le 0 `
`rArr a lt (1)/(4) and 4a^(3) lt 36 ((1)/(4)) - 48 `
`rArr 4a^(3) + 39 lt 0 [ because a lt (1)/(4)]`