Mathematics - Functions Question with Solution | TestHub

For a fixed $\alpha >1,{\alpha }^n\to \infty$ and ${\alpha }^{-n}\to 0$ as $n\to \infty$, so the ratio tends to $|\sin x|:a=|\sin x|$. For $\alpha <1,{\alpha }^n\to 0,{\alpha }^{-n}\to \infty$, so $b=|\cos x|.c=\frac{\pi }{4}{\int }_0^1\cos \frac{\pi t}{2}dt=\frac{\pi }{4}\cdot \frac{2}{\pi }=\frac{1}{2}$. Since $\max (|\sin x|,|\cos x|)\ge$ $\frac{1}{\sqrt{2}}>\frac{1}{2},f(x)=\max \{|\sin x|,|\cos x|\}$. Its minimum is $\frac{1}{\sqrt{2}}$ (where $|\sin x|=|\cos x|$ ) and maximum is 1 , so the range is $\left[\frac{1}{\sqrt{2}},1\right]$.