PEA 2022 PEA Q24 | ISI MSQE

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2022 PEA Q24

$f : \mathbb R \to \mathbb R$ , $f(x) = \dfrac{x}{1+x^2}$ . Then

Reveal answer and solution

Answer

Solution

1
By AM--GM, $1 + x^2 \geq 2|x|$ , so $|f(x)| \leq 1/2$ . Equality holds at $x = \pm 1$ giving
2
$f(1) = 1/2$ and $f(-1) = -1/2$ . Hence the range is $[-1/2, 1/2]$ .

Answer structure / marking notes

The bound $|x/(1+x^2)| \leq 1/2$ follows from $(|x|-1)^2 \geq 0$ .

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.