Let f be a twice differentiable function on (1, 6). If f(2) = 8, f’(2) = 5, f’(x) ≥ 1 and f''(x) ≥ 4, for all x ∈ (1, 6), then :

Solution

Maths sir · 4 steps

Use the Mean Value Theorem twice, once on f' and once on f, then add the two lower bounds.

1. MVT on f'. Since f is twice differentiable on (1, 6), f' is differentiable on [2, 5]. So there is some c ∈ (2, 5) with f'(5) - f'(2) = 3f''(c).

   f'(5) = f'(2) + 3f''(c) ≥ 5 + 3 × 4 = 17

2. MVT on f. In the same way, there is some d ∈ (2, 5) with f(5) - f(2) = 3f'(d). Since f'(d) ≥ 1, this gives f(5) ≥ 8 + 3 = 11.
3. Add the bounds. Adding the two results gives f(5) + f'(5) ≥ 11 + 17 = 28. The given conditions only ever give lower bounds, so none of the "≤" options can be guaranteed.
4. So. f(5) + f'(5) ≥ 28, which is option (b).

The idea to keep. A lower bound on a derivative over an interval of length h becomes a lower bound on the change in the function: f(b) - f(a) ≥ h × (min of f'). That is MVT at work.

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