Question

Find the interval on which the following function are strictly increasing & strictly decreasing.
$f\left(x\right)=\frac{4sinx-2x-xcosx}{2+cosx},0\le x\le 2\pi$

Answer 1

$\begin{array}{c}Wehave\\ f\left(x\right)=\frac{4\sin x-2x-x\cos x}{2+\cos x}\\ \sin ce,wehavetofindincra\sin ganddecrea\sin g\\ weconsidertheopeninterval\left(0,2\pi \right)\\ Atfirstfinding{f}^{\prime }\left(x\right)\\ f\left(x\right)=\frac{4\sin x-2x-x\cos x}{2+\cos x}\\ =\frac{4\sin x-x\left(2+\cos x\right)}{2+\cos x}\\ =\frac{4\sin x}{2+\cos x}-\frac{x\left(2+\cos x\right)}{2+\cos x}\\ =\frac{4\sin x}{2+\cos x}-x\\ Therefore,\\ f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{4\sin x}{2+\cos x}-x\right)\\ =\frac{d}{dx}\left(\frac{4\sin x}{2+\cos x}\right)-\frac{d\left(x\right)}{dx}\\ =\frac{d}{dx}\left(\frac{4\sin x}{2+\cos x}\right)-1\\ =\left[\frac{{\left(4\sin x\right)}^{\prime }\left(2+\cos x\right)-{\left(2+\cos x\right)}^{\prime }\left(4sinx\right)}{{\left(2+\cos x\right)}^2}\right]-1\\ =\left[\frac{4\cos x\left(2+\cos x\right)-\left(-\sin x\right)\left(4\sin x\right)}{{\left(2+\cos x\right)}^2}\right]-1\\ =\left[\frac{8\cos x+4{\cos }^{2}x+4{\sin }^{2}x}{{\left(2+\cos x\right)}^2}\right]-1\\ =\frac{8\cos x+4\left({\cos }^{2}x+{\sin }^{2}x\right)}{{\left(2+\cos x\right)}^2}-1\\ =\frac{8\cos x+4}{{\left(2+\cos x\right)}^2}-1\\ =\frac{8\cos x+4-{\left(2+\cos x\right)}^2}{{\left(2+\cos x\right)}^2}\\ =\frac{8\cos x+4-\left(4+{\cos }^{2}x+4\cos x\right)}{{\left(2+\cos x\right)}^2}\\ =\frac{4\cos x-{\cos }^{2}x}{{\left(2+\cos x\right)}^2}\\ Putting{f}^{\prime }\left(x\right)=0\\ \therefore \frac{4\cos x-{\cos }^{2}x}{{\left(2+\cos x\right)}^2}=0\\ so,\\ \cos x\left(4-\cos x\right)=0\\ \cos x=0\\ and4-\cos x=0[but-1\le \cos x\le 1]\\ cosx=4isnotpossible\\ Now,\\ x=\left(2n+1\right)\frac{\pi }{2},n\in Z\\ puttingn=1\\ x=\left(2\left(1\right)+1\right)\frac{\pi }{2}=\frac{3\pi }{2}\\ Puttingn=2\\ x=\left(2\left(2\right)+1\right)\frac{\pi }{2}\\ =\frac{5\pi }{2}\\ since,x\in \left(0,2\pi \right)\\ So,valueofxare\frac{\pi }{2}and\frac{\pi }{2}\\ Thus,wedividetheinterval\left(0,2\pi \right)intothreedisjointinterval\\ \left(0,\frac{\pi }{2}\right),\left(\frac{\pi }{2},\frac{3\pi }{2}\right)and\left(\frac{3\pi }{2},2\pi \right)\\ Hence,\\ f\left(x\right)isstrictilyincrea\sin gon\left(0,\frac{\pi }{2}\right)\&\left(\frac{3\pi }{2},2\pi \right)\\ and,\\ f\left(x\right)isstrictlydecrea\sin gon\left(\frac{\pi }{2},\frac{3\pi }{2}\right)\end{array}$