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2022 PEA Q19

f:R→Rf : \mathbb R \to \mathbb R is C2C^2 with f′′(x)>0f''(x) > 0 for all xx, f(1)=1f(1)=1, f(2)=2f(2)=2. Then,

Reveal answer and solution

Answer

B

Solution

  1. 1

    By the Mean Value Theorem, there exists c∈(1,2)c \in (1,2) with f′(c)=f(2)−f(1)2−1=1f'(c) = \frac{f(2)-f(1)}{2-1} = 1.

  2. 2

    Since f′′>0f'' > 0, f′f' is strictly increasing. Hence for any x>cx > c, in particular at x=2x = 2,

  3. 3

    f′(2)>f′(c)=1f'(2) > f'(c) = 1.

Answer structure / marking notes

Convexity makes f′f' monotonically increasing, not bounded above by 11.

Content note

Imported from public/resources/isi/msqe/solutions/pea/2022/ISI_MSQE_PEA_2022_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.