Question

If $\frac{x-1}{x^2+2x+a}$ takes all real values for possible values of x, then find $a$.

Answer 1

-3

Answer 2

Let $y = \frac{x-1}{x^2+2x+a}$. For $y$ to take all real values, the equation $y(x^2+2x+a) = x-1$ must have real solutions for $x$ for every real value of $y$, provided the denominator is non-zero.

Rearranging, we get $yx^2 + (2y-1)x + (ay+1) = 0$.

For $y \ne 0$, this is a quadratic in $x$. It has real roots if the discriminant $\Delta = (2y-1)^2 - 4y(ay+1) \ge 0$.

$\Delta = (4-4a)y^2 - 8y + 1$.

For $\Delta \ge 0$ for all $y \ne 0$, the quadratic in $y$ must satisfy $4-4a \ge 0$ and $(-8)^2 - 4(4-4a)(1) \le 0$.

$a \le 1$ and $64 - 16 + 16a \le 0 \implies 48 + 16a \le 0 \implies a \le -3$.

Thus, for $y \ne 0$, real roots exist if $a \le -3$.

For $y=0$, the equation is $-x+1=0$, so $x=1$. For $y=0$ to be in the range, the denominator at $x=1$ must be non-zero: $1^2+2(1)+a = 3+a \ne 0$, so $a \ne -3$.

Combining the conditions for all real $y$ to be possible, we need $a \le -3$ and $a \ne -3$, which means $a < -3$.

If $a < -3$, the range is $\mathbb{R}$.

If $a = -3$, the range is $\mathbb{R} \setminus \{0, 1/4\}$.

The question asks for "find $a$", implying a specific value. The value $a=-3$ is the boundary between the region where the range is $\mathbb{R}$ ($a<-3$) and where it is not. Assuming the question expects this critical value, the answer is -3.