question_answer

Find the intervals in which the function given by \[f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x},\] \[0\le x\le 2\pi \] is (i) Strictly increasing and (ii) Strictly decreasing.

Answer:

We have,  \[f(x)=\frac{4\sin x-2x-x\,\cos \,x}{2+\cos \,x}\] \[\therefore f'(x)=\frac{\begin{align}   & (2+\cos x)(4\cos x-2-\cos x+x\sin x) \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+(4\sin x-2x-x\cos x)\sin x \\ \end{align}}{{{(2+\cos x)}^{2}}}\] \[=\frac{cos\,x(4-\cos \,x)}{{{(2+\cos \,x)}^{2}}}\] (i) For f(x) to be strictly increasing, we must have             \[f'(x)>0\] \[\Rightarrow \]   \[\frac{\cos x(4-\cos x)}{{{(2+\cos x)}^{2}}}>0\] \[\Rightarrow \]   \[cosx>0\]         \[\left[ \because \frac{4-\cos \,x}{{{(2+\cos \,x)}^{2}}}>0 \right]\] \[\Rightarrow \]   \[x\in \left( 0,\,\,\frac{\pi }{2} \right)\cup \left( \frac{3\pi }{2},\,\,2\pi  \right)\] Hence, f(x) is strictly increasing on \[\left( 0,\,\,\frac{\pi }{2} \right)\cup \left( \frac{3\pi }{2},\,\,2\pi  \right).\] (ii) For f(x) to be strictly decreasing, we must have             \[f'(x)<0\] \[\Rightarrow \] \[\frac{\cos x(4-\cos x)}{{{(2+\cos x)}^{2}}}<0\] \[\Rightarrow \] \[\cos x<0\] \[\Rightarrow \] \[x\in \left( \frac{\pi }{2},\frac{3\pi }{2} \right)\] Hence, r(x) is strictly decreasing on \[\left( \frac{\pi }{2},\frac{3\pi }{2} \right).\]