Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow R: f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] \rightarrow R: g(x)=\cos x$. Show that each one of $f$ and $g$ is one one but $(f+g)$ is not one - one.

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Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow R: f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] \rightarrow R: g(x)=\cos x$ . Show that each one of $f$ and $g$ is one one but $(f+g)$ is not one – one.

Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow R: f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] \rightarrow R: g(x)=\cos x$ . Show that each one of $f$ and $g$ is one one but $(f+g)$ is not one – one.

Solution:

$f: \left[0, \frac{\pi}{2}\right] \rightarrow \mathrm{R}$ for given function $\mathrm{f}(\mathrm{x})=\sin$
Recalling the graph for $\sin \mathrm{x}$ , we realise that for any two values on the curve, say $\mathrm{x}_{1}$ and $\mathrm{x}_2$ ,
$\sin x_1 \neq \sin x_2$
As a result, $f(x)=\sin x$ is one to one.

$f: \left[0, \frac{\pi}{2}\right] \rightarrow \mathrm{R}$ for given function $\mathrm{f}(\mathrm{x})=$
$x$
Recalling the graph for $\cos \mathrm{x}$ , we realise that for any two values on the curve, say $\mathrm{x}_1$ and $\mathrm{x}_2$ , $\mathrm{x}_{1} \neq \cos \mathrm{x}_2$
As a result, $f(x)=\cos x$ is one to one.

Now, $f+g=\sin x+\cos x$ mapped from $\left[0, \frac{\pi}{2}\right]$ to $\mathrm{R}$

Recalling the graph for $\sin x+\cos x$ , we realise that for any two values on the curve, say $\mathrm{x}_1$ and $\mathrm{x}_2$ ,
$\sin x_{1}+\cos x_1 \neq \sin x_2+\cos x_2$
As a result, $f(x)=\sin x+\cos x$ is not one to one.
i.e. $\mathrm{f}+\mathrm{g}$ is not one to one, even though $\mathrm{f}(\mathrm{x})$ ar individually.