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Let $f(x) = \max \{ \sin A : 0\le A \le x \}$ and $g(x) = \min \{ \sin A : 0\le A \le x \}$
and $h(x)= \lfloor f(x)- g(x)\rfloor$ , where $\lfloor\quad \rfloor$ denotes greatest integer function . Then find the range of $h(x)$.

My Approach : I know that the range of $\lfloor \sin A\rfloor$ is $\{1,-1,0\}$ but I am not understanding how to solve $\max\{\sin A : \ 0\le A \le x\}$ and $g(x) = \min \{\sin A : 0\le A \le x \}$ and make cases of $\sin A$ ..

Please explain ......

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Divide into cases.

  1. If $x\in[0,\pi/2]$, then $f(x)=\sin x$ and $g(x)=0$

  2. If $x\in(\pi/2,\pi]$, then $f(x)=1$ and $g(x)=0$

  3. If $x\in(\pi,3\pi/2]$, then $f(x)=1$ and $g(x)=\sin x$

  4. If $x>3\pi/2$, then $f(x)=1$ and $g(x)=-1$

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