Consider a function $f:\left[0, \frac{\pi}{2}\right]$ $ \rightarrow$ $R$ given by $f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] $ $\rightarrow$ $R$ given by $g(x)=\cos x .$ Show that $f$ and $g$ are one-one, but $f\,+\,g$ is not one-one.
Consider a function $f:\left[0, \frac{\pi}{2}\right]$ $ \rightarrow$ $R$ given by $f(x)=\sin x$ and $g:\left[0, \frac{\pi}{2}\right] $ $\rightarrow$ $R$ given by $g(x)=\cos x .$ Show that $f$ and $g$ are one-one, but $f\,+\,g$ is not one-one.
since for any two distinct elements $x_{1}$ and $x_{2}$ in $\left[0, \frac{\pi}{2}\right]$, $\sin x_{1} \neq \sin x_{2}$ and $\cos x_{1} \neq \cos x_{2},$ both $f$ and $g$ must be one-one. But $(f\,+\,g)(0)=\sin 0+\cos 0=1$ and $(f+g)\left(\frac{\pi}{2}\right)$ $=\sin \frac{\pi}{2}+\cos \frac{\pi}{2}=1 .$ Therefore, $f\,+\,g$ is not one-one.
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{ - 2x + {{\log }_2}({b^2} - 2),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, > 1\,\,\,\,\,\,\,\,\,\,\,\,}
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