Question

If $\frac{(x^2+ax+3)}{(x^2+x+a)}$ takes all real values for possible real values of $x$, then

• A. $a^3-9a+12\le 0$
• B. $4a^3+39\ge 0$
• C. $a^3-9a\le 0$
• D. $a^3-12\le 0$

Answer 1

Answer: (A)

$a^3-9a+12\le 0$

Let $\frac{x^2+ax+3}{x^2+x+a}=y$
$x^2(y-1)+x(y-a)+ay-3=0$
For any real value of $y$ there must be at least one real $x$ according to the condition in the equation.Therefore, for any value of $y$ at least one real root of $x$ must exist.
For roots to exist, discriminant should be greater than or equal to zero.
$\therefore (y-a{)}^2-4(ay-3\left)\right(y-1)\ge 0$
$y^2(1-4a)+y(2a+12)+a^2-12\ge 0$
For above equation in $y$ to be always greater than equal to zero, discriminant should be less than or equal to zero.
$(2a+12{)}^2-4(1-4a\left)\right(a^2-12)\le 0$
$a^3-9a+12\le 0$