Key Concepts and Formulas

• Monotonicity and Derivatives: A differentiable function $f(x)$ is increasing on an interval $[c, d]$ if $f'(x) \ge 0$ for all $x \in [c, d]$, and decreasing if $f'(x) \le 0$ for all $x \in [c, d]$.
• Vertex of a Parabola: The x-coordinate of the vertex of the parabola $f(x) = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$.

Step-by-Step Solution

Step 1: Find the derivative of f(x)

We are given the function $f(x) = x^2 + ax + 1$. To determine where the function is increasing or decreasing, we need to find its derivative. $f'(x) = \frac{d}{dx}(x^2 + ax + 1) = 2x + a$

Step 2: Find the least value of 'a' (R) for f(x) to be increasing on [1, 2]

For $f(x)$ to be increasing on $[1, 2]$, we require $f'(x) \ge 0$ for all $x \in [1, 2]$. This means $2x + a \ge 0$ for $x \in [1, 2]$. Since $2x + a$ is a linear function with a positive slope, its minimum value on the interval $[1, 2]$ occurs at $x = 1$. Therefore, we need $2(1) + a \ge 0$, which gives us $2 + a \ge 0$, so $a \ge -2$. The least value of 'a' that satisfies this condition is $a = -2$. Thus, $R = -2$.

Step 3: Find the greatest value of 'a' (S) for f(x) to be decreasing on [1, 2]

For $f(x)$ to be decreasing on $[1, 2]$, we require $f'(x) \le 0$ for all $x \in [1, 2]$. This means $2x + a \le 0$ for $x \in [1, 2]$. Since $2x + a$ is a linear function with a positive slope, its maximum value on the interval $[1, 2]$ occurs at $x = 2$. Therefore, we need $2(2) + a \le 0$, which gives us $4 + a \le 0$, so $a \le -4$. The greatest value of 'a' that satisfies this condition is $a = -4$. Thus, $S = -4$.

Step 4: Calculate |R - S|

We have $R = -2$ and $S = -4$. We need to find $|R - S| = |-2 - (-4)| = |-2 + 4| = |2| = 2$.

Common Mistakes & Tips

• Confusing increasing/decreasing conditions: Remember that for a function to be increasing, its derivative must be greater than or equal to zero, and for a function to be decreasing, its derivative must be less than or equal to zero.
• Endpoint analysis: When dealing with linear functions on closed intervals, the maximum and minimum values occur at the endpoints of the interval. This simplifies the analysis.
• Sign errors: Be careful with signs when solving inequalities.

Summary

We found the derivative of the given quadratic function and then used the conditions for increasing and decreasing functions to determine the least value of 'a' (R) and the greatest value of 'a' (S). We then calculated the absolute difference between R and S to arrive at the final answer.

Final Answer

The final answer is \boxed{2}.