Application of DerivativesmediumFree

Bound on f(4) from f(2) = -4 and f'(x) at least 6 | JEE Advanced LMVT

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Question
Let ff be a function which is continuous and differentiable for all real xx. If f(2)=−4f(2) = -4 and f′(x)≥6f'(x) \ge 6 for all x∈[2,4]x \in [2,4], then
Af(4)<8f(4) < 8
Bf(4)≥8f(4) \ge 8correct
Cf(4)≥12f(4) \ge 12
Dnone of these
Solution
Step 1: Check that Lagrange's mean value theorem applies. ff is continuous on [2,4][2,4] and differentiable on (2,4)(2,4), which is exactly what the theorem needs. Step 2: Write down what it gives. There exists c∈(2,4)c \in (2,4) with
f′(c)=f(4)−f(2)4−2.f'(c) = \frac{f(4)-f(2)}{4-2} .
 Step 3: Substitute the given value f(2)=−4f(2) = -4.
f′(c)=f(4)−(−4)2=f(4)+42.f'(c) = \frac{f(4)-(-4)}{2} = \frac{f(4)+4}{2} .
 Step 4: Apply the bound on the derivative. Since c∈(2,4)⊂[2,4]c \in (2,4) \subset [2,4], we have f′(c)≥6f'(c) \ge 6:
f(4)+42 ≥ 6.\frac{f(4)+4}{2} \ \ge\ 6 .
 Step 5: Solve for f(4)f(4).
f(4)+4≥12  ⟹  f(4)≥8.f(4)+4 \ge 12 \implies f(4) \ge 8 .
 Step 6: Check the options. (2) is exactly this. (1) contradicts it. (3) claims f(4)≥12f(4) \ge 12, which is stronger than what follows - for instance f(x)=6x−16f(x) = 6x - 16 has f(2)=−4f(2) = -4, f′(x)=6≥6f'(x) = 6 \ge 6 and f(4)=8<12f(4) = 8 < 12, so (3) can fail. Answer: (2).
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