Question

Which of the following function is one-one?

• A. $f:R\to R$ given by$f\left(x\right)=|x-1|$ for all $x\in R$
• B. $g:[-\frac{\pi }{2},\frac{\pi }{2}]\to R$ given by $g\left(x\right)=|sinx|$ for all $x\in [\frac{-\pi }{2},\frac{\pi }{2}]$
• C. $h:[\frac{-\pi }{2},\frac{\pi }{2}]\in R$ given by $h\left(x\right)=sinx$ for all $x\in [\frac{-\pi }{2},\frac{\pi }{2}]$
• D. $\varphi :R\to R$ given by $f\left(x\right)=x^2-4$ for all $x\in R$

Answer 1

The correct option is C $h:[\frac{-\pi }{2},\frac{\pi }{2}]\in R$ given by $h\left(x\right)=sinx$ for all $x\in [\frac{-\pi }{2},\frac{\pi }{2}]$

Option A - Consider $x=2$ and $x=0$ , the value of $f\left(x\right)$ is same. Hence it is not one-one
Option B - If we replace $x$ by $-x$, then the value of $g\left(x\right)$ remains the same. Hence it is not one-one
Option C - $h\left(x\right)$ is an increasing function for the given values of $x$. Hence it is one-one function
Option D -$f\left(x\right)$ is an even function. So it is not one-one for the given values of $x$