Mathematics - Functions Question with Solution | TestHub

$f(x)=1-{\sin }^2\frac{x}{4}+\sin \frac{x}{4}=-\left\{{\sin }^2\frac{x}{4}-\sin \frac{x}{4}\right\}+1=-\left\{{\left(\sin \frac{x}{4}-\frac{1}{2}\right)}^2-\frac{1}{4}\right\}+1$ $=\frac{5}{4}-{\left(\sin \frac{x}{4}-\frac{1}{2}\right)}^2$

$\mathrm{Maximum}\mathrm{f}(\mathrm{x})=\frac{5}{4}$

Minimum f(x) $=\frac{5}{4}-{\left(-1-\frac{1}{2}\right)}^2=\frac{5}{4}-\frac{9}{4}=-1$

Range of $\mathrm{f}(\mathrm{x})=\left[-1,\frac{5}{4}\right]$