If 'R' is the least value of 'a' such that the function f(x) = x² + ax + 1 is increasing on [1, 2] and 'S' is the greatest value of 'a' such that the function f(x) = x² + ax + 1 is decreasing on [1, 2], then
the value of |R - S| is ___________.

Solution

Maths sir · 4 steps

Here f'(x) = 2x + a is a straight line, so check its sign only at the endpoint where it is most at risk.

1. Derivative. f'(x) = 2x + a. On [1, 2] this grows with x, so its smallest value is at x = 1 and its largest is at x = 2.
2. Increasing, find R. We need 2x + a ≥ 0 for every x in [1, 2]. It is enough that the smallest value is non-negative: 2 + a ≥ 0, so a ≥ -2. The least such a is R = -2.
3. Decreasing, find S. Now we need 2x + a ≤ 0 on all of [1, 2]. It is enough that the largest value is at most zero: 4 + a ≤ 0, so a ≤ -4. The greatest such a is S = -4.

   |R - S| = |-2 - (-4)| = 2

4. So. The required value is 2.

The common slip: 0. Testing f'(1) = 2 + a for both conditions gives R = S = -2 and an answer of 0. For decreasing, the binding endpoint is x = 2.

The idea to keep. For increasing on an interval, the minimum of f' there must be ≥ 0. For decreasing, the maximum of f' must be ≤ 0. Find where f' is smallest or largest first.

Answer from the NTA final answer key. Solution by PrepWiser.