PEA 2022 PEA Q22 | ISI MSQE

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2022 PEA Q22

$f : [0,\infty) \to \mathbb R$ , $f(0) = 0$ , $f''(x) > 0$ for $x > 0$ . Then $f(x)/x$ is

Reveal answer and solution

Answer

Solution

1
Let $h(x) = f(x)/x$ . Then
2
$h'(x) = \frac{x f'(x) - f(x)}{x^2}.$
3
Define $\phi(x) = x f'(x) - f(x)$ , $\phi(0) = 0$ , and $\phi'(x) = f'(x) + x f''(x) - f'(x) = x f''(x) > 0$ .
4
Hence $\phi(x) > 0$ for $x > 0$ , so $h'(x) > 0$ and $f(x)/x$ is strictly increasing on $(0,\infty)$ .

Answer structure / marking notes

The constant $1$ in (C)/(D) is irrelevant; it is the convexity and $f(0)=0$ that drive monotonicity globally.

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.