Question:

Step 1: Identify the condition for the function to be defined. For $f(x) = \ln(x^2 + ax + 1)$ to be defined for all $x \in \mathbb{R}$, the argument of the logarithm must be strictly positive for every real value of $x$. This means we need: $$x^2 + ax + 1 > 0 \text{ for all } x \in \mathbb{R}$$ Step 2: Apply the discriminant condition for a quadratic to be always positive. For a quadratic expression $x^2 + ax + 1$ to be positive for all real $x$, the parabola (which opens upward since the coefficient of $x^2$ is positive) must not intersect the x-axis. This occurs when the discriminant is negative. The discriminant of $x^2 + ax + 1$ is: $$\Delta = a^2 - 4(1)(1) = a^2 - 4$$ For the quadratic to be always positive, we require: $$\Delta < 0$$ $$a^2 - 4 < 0$$ Step 3: Solve the inequality for $a$. From $a^2 - 4 < 0$, we get: $$a^2 < 4$$ $$|a| < 2$$ $$-2 < a < 2$$ Step 4: Count the integers in the range of $a$. The range of $a$ is the open interval $(-2, 2)$. The integers that lie strictly between $-2$ and $2$ are: $$a \in \{-1, 0, 1\}$$ This gives us **3 integers**. **Final Answer:** The number of integers in the range of $a$ is **3**, which corresponds to **Option 2**.