Which one of the following functions is a bij | vedclass

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(D) function is a bijection if it is both one-to-one (injective) and onto (surjective). $(a)$ $f(x) = \sqrt{\{x\}}$. Since $\{x\}$ is periodic with pe

Vedclass · Accepted Answer

$f: [0,4] \rightarrow [0,4]$ defined by $f(x) = \sqrt{16-x^2}$

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(D) function is a bijection if it is both one-to-one (injective) and onto (surjective). $(a)$ $f(x) = \sqrt{\{x\}}$. Since $\{x\}$ is periodic with period $1$, $f(0.1) = f(1.1)$, so it is many-to-one. $(b)$ $f(x) = -(x^2-4x+4)+1 = 1-(x-2)^2$. This is a parabola opening downwards, which is many-to-one. $(c)$ $f(x) = \frac{1}{\sqrt{x-5}}$. The range is $(0, \infty)$, which is not equal to the codomain $R \setminus \{0\}$, so it is into. $(d)$ $f(x) = \sqrt{16-x^2}$. For $x \in [0,4]$, $f(x)$ is strictly decreasing from $4$ to $0$. Thus, it is one-to-one and onto. Hence, it is a bijection.

Which one of the following functions is a bijection?

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