Question:

**Step 1:** For $f(x)$ to be surjective onto $\mathbb{R}$, for every $y \in \mathbb{R}$, there must exist at least one real number $x$ in the domain $D$ such that $f(x)=y$. The domain $D$ is defined as $D = \{x \in \mathbb{R} \mid x^2 + x + 2c \neq 0\}$. **Step 2:** Rearrange the function equation to form a quadratic in $x$: $$y = \frac{x^2 - x + c}{x^2 + x + 2c}$$ $$y(x^2 + x + 2c) = x^2 - x + c$$ $$yx^2 + yx + 2cy = x^2 - x + c$$ $$(y-1)x^2 + (y+1)x + (2cy-c) = 0$$ **Step 3:** Analyze the case where $y=1$. If $y=1$, the equation from Step 2 simplifies to: $$(1-1)x^2 + (1+1)x + (2c(1)-c) = 0$$ $$2x + c = 0$$ $$x = -\frac{c}{2}$$ For $y=1$ to be in the range of $f(x)$, this value of $x$ must be in the domain $D$. This means the denominator $x^2+x+2c$ must not be zero for $x=-c/2$. Substitute $x=-c/2$ into the denominator: $$\left(-\frac{c}{2}\right)^2 + \left(-\frac{c}{2}\right) + 2c = \frac{c^2}{4} - \frac{c}{2} + 2c = \frac{c^2}{4} + \frac{3c}{2} = \frac{c(c+6)}{4}$$ For $y=1$ to be in the range, we must have $\frac{c(c+6)}{4} \neq 0$, which implies $c \neq 0$ and $c \neq -6$. **Step 4:** Analyze the case where $y \neq 1$. For $y \neq 1$, the equation $(y-1)x^2 + (y+1)x + (2cy-c) = 0$ is a quadratic in $x$. For it to have real solutions for $x$, its discriminant $\Delta_x$ must be non-negative: $$\Delta_x = (y+1)^2 - 4(y-1)(2cy-c) \ge 0$$ Expanding this expression: $$\Delta_x = (y^2 + 2y + 1) - 4(2cy^2 - cy - 2cy + c)$$ $$\Delta_x = y^2 + 2y + 1 - 4(2cy^2 - 3cy + c)$$ $$\Delta_x = y^2 + 2y + 1 - 8cy^2 + 12cy - 4c$$ $$\Delta_x = (1-8c)y^2 + (2+12c)y + (1-4c)$$ For $f(x)$ to be surjective, this inequality $\Delta_x \ge 0$ must hold for all $y \in \mathbb{R}$ (excluding $y=1$, which is handled in Step 3). For a quadratic $Ay^2 + By + C \ge 0$ for all $y$, two conditions must be met: 1. The leading coefficient must be positive: $1-8c > 0 \implies c < \frac{1}{8}$. 2. The discriminant of this quadratic in $y$ must be non-positive: $\Delta_y \le 0$. $$\Delta_y = (2+12c)^2 - 4(1-8c)(1-4c)$$ $$\Delta_y = (4 + 48c + 144c^2) - 4(1 - 4c - 8c + 32c^2)$$ $$\Delta_y = 4 + 48c + 144c^2 - 4(1 - 12c + 32c^2)$$ $$\Delta_y = 4 + 48c + 144c^2 - 4 + 48c - 128c^2$$ $$\Delta_y = 16c^2 + 96c$$ So, we require $16c^2 + 96c \le 0$. $$16c(c+6) \le 0$$ This inequality holds for $-6 \le c \le 0$. **Step 5:** Combine all conditions. From Step 4, we have $c < 1/8$ and $-6 \le c \le 0$. Combining these two conditions yields $-6 \le c \le 0$. From Step 3, we established that $c \neq 0$ and $c \neq -6$ are necessary for $y=1$ to be in the range of $f(x)$. Therefore, combining all conditions, the possible values for $c$ are in the interval $-6 < c < 0$. The integral values of $c$ in this interval are $\{-5, -4, -3, -2, -1\}$.