Question

If 'R' is the least value of 'a' such that the function \(f(x) = x^2 + ax + 1\) is increasing on \([1, 2]\) and 'S' is the greatest value of 'a' such that the function \(f(x) = x^2 + ax + 1\) is decreasing on \([1, 2]\), then the value of \(|R-S|\) is \dots\dots.

Hint:

For quadratic functions, the vertex is at \(x = -a/2\). For the function to be increasing on \([1, 2]\), the vertex must be at or to the left of the interval (\(-a/2 \leq 1\)). For decreasing, it must be at or to the right (\(-a/2 \geq 2\)).

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
For a differentiable function \(f(x)\), it is increasing on an interval if \(f'(x) \geq 0\) and decreasing if \(f'(x) \leq 0\) throughout that interval.
Step 2: Detailed Explanation:
The derivative of \(f(x)\) is \(f'(x) = 2x + a\).
1. Increasing on \([1, 2]\):
We need \(2x + a \geq 0\) for all \(x \in [1, 2]\).
Since \(2x+a\) is a linear increasing function of \(x\), its minimum value on \([1, 2]\) is at \(x=1\).
\[ 2(1) + a \geq 0 \implies a \geq -2 \]
The least value of \(a\) is \(R = -2\).
2. Decreasing on \([1, 2]\):
We need \(2x + a \leq 0\) for all \(x \in [1, 2]\).
The maximum value of \(2x+a\) on \([1, 2]\) occurs at \(x=2\).
\[ 2(2) + a \leq 0 \implies a \leq -4 \]
The greatest value of \(a\) is \(S = -4\).
3. Calculating \(|R-S|\):
\[ |R - S| = |-2 - (-4)| = |-2 + 4| = 2 \]
Step 3: Final Answer:
The value of \(|R-S|\) is 2.