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GATE CSE 2021 Set 2 Q35: Upper bound on f(3) via the mean value theorem

Question

Suppose that f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} is a continuous function on the interval [−3,3][-3, 3] and a differentiable function in the interval (−3,3)(-3, 3) such that for every xx in the interval, f′(x)≤2f'(x) \leq 2. If f(−3)=7f(-3) = 7, then f(3)f(3) is at most ________.

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Numerical answer. No negative marking.
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Answer: 19 (official GATE 2021 final answer key)

Concept used: Mean value theorem: f(3)−f(−3)=6f′(c)f(3) - f(-3) = 6 f'(c) for some cc in (−3,3)(-3, 3).

First step: Bound f′(c)≤2f'(c) \leq 2, so f(3)≤7+6×2f(3) \leq 7 + 6 \times 2.

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Source: official GATE 2021 Set 2 CS question paper and final answer key, organised by IIT Bombay. Question paper · Answer key