Question

Which of the following functions from $A=\left\{x:-1\le x\le 1\right\}$ to itself are bijections?
(a) $f\left(x\right)=\frac{x}{2}$
(b) $g\left(x\right)=\sin \left(\frac{\pi x}{2}\right)$
(c) $h\left(x\right)=\left|x\right|$
(d) $k\left(x\right)=x^2$

Answer 1

$$\begin{gathered} \left(a\right)\text{Range of}f\text{=}\left[\frac{-1}{2},\frac{1}{2}\right]\text{≠}A \\ \mathrm{So},f\text{is not a bijection.} \\ \left(b\right)\text{Range =}\left[\sin \left(\frac{-\mathrm{\pi }}{2}\right),\sin \left(\frac{\mathrm{\pi }}{2}\right)\right]\text{=}\left[-1,1\right]\text{=}A \\ \mathrm{So},g\text{is a bijection.} \\ \left(c\right)h\left(-1\right)=\left|-1\right|=1 \\ \mathrm{and}h\left(1\right)=\left|1\right|=1 \\ \text{⇒-1 and 1 have the same images} \\ \mathrm{So},h\text{is not a bijection.} \\ \left(d\right)k\left(-1\right)={\left(-1\right)}^2=1 \\ \mathrm{and}k\left(1\right)={\left(1\right)}^2=1 \\ \text{⇒-1 and 1 have the same images} \\ \mathrm{So},k\text{is not a bijection.} \end{gathered}$$
So, the answer is (b).