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2022 PEA Q30

f:RRf : \mathbb R \to \mathbb R, f(x)=(x2+1)2022f(x) = (x^2 + 1)^{2022}.

Reveal answer and solution

Answer

D

Solution

  1. 1

    f(x)=((x)2+1)2022=f(x)f(-x) = ((-x)^2 + 1)^{2022} = f(x), so ff is even and hence not one-one (e.g.\ f(1)=f(1)f(1)=f(-1)).

  2. 2

    Also, x2+11x^2 + 1 \geq 1, so f(x)1f(x) \geq 1 for all xx, hence ff is not surjective onto R\mathbb R.

Answer structure / marking notes

Even functions on R\mathbb R cannot be injective unless restricted to [0,)[0,\infty).

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.