Question

Find the intervals in which f(x) is increasing or decreasing:

(i) f(x) = x|x|, x $\in$R

(ii) f(x) = sinx + |sinx|, 0 < x $\le 2\mathrm{\pi }$

(iii) f(x) = sinx(1 + cosx), 0 < x < $\frac{\mathrm{\pi }}{2}$
[CBSE 2014]

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)f\left(x\right)=x\left|x\right|,x\in \mathbf{R} \\ \mathrm{Case}\mathrm{I}:\mathrm{When}x\ge 0 \\ f\left(x\right)=x\left|x\right|=x\left(x\right)=x^2 \\ \Rightarrow f\text{'}\left(x\right)=2x\ge 0\forall x\ge 0 \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{for}x\ge 0. \\ \mathrm{Case}\mathrm{II}:\mathrm{When}x<0 \\ f\left(x\right)=x\left|x\right|=x\left(-x\right)=-x^2 \\ \Rightarrow f\text{'}\left(x\right)=-2x\ge 0\forall x<0 \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{for}x<0. \\ \mathrm{Hence},f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{for}x\in \mathbf{R}. \\ \left(\mathrm{ii}\right)f\left(x\right)=\sin x+\left|\sin x\right|,0<x\le 2\mathrm{\pi } \\ \mathrm{Case}\mathrm{I}:\mathrm{When}x\in \left(0,\mathrm{\pi }\right) \\ f\left(x\right)=\sin x+\sin x=2\sin x \\ \Rightarrow f\text{'}\left(x\right)=2\cos x \\ \mathrm{As},\cos x>0\mathrm{for}x\in \left(0,\frac{\mathrm{\pi }}{2}\right)\mathrm{and}\cos x<0\mathrm{for}x\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right) \\ \mathrm{So},f\text{'}\left(x\right)>0\mathrm{for}x\in \left(0,\frac{\mathrm{\pi }}{2}\right)\mathrm{and}f\text{'}\left(x\right)<0\mathrm{for}x\in \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right) \\ \therefore f\left(x\right)\mathrm{is}\mathrm{increaing}\mathrm{on}\left(0,\frac{\mathrm{\pi }}{2}\right)\mathrm{and}f\left(x\right)\mathrm{is}\mathrm{decreasing}\mathrm{on}\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right). \\ \mathrm{Case}\mathrm{II}:\mathrm{When}x\in \left(\mathrm{\pi },2\mathrm{\pi }\right) \\ f\left(x\right)=\sin x\mathit{-}\sin x=0 \\ \Rightarrow f\text{'}\left(x\right)=0 \\ \mathrm{So},f\left(x\right)\mathrm{is}\mathrm{neither}\mathrm{increaing}\mathrm{nor}\mathrm{decreasing}\mathrm{on}\left(\mathrm{\pi },2\mathrm{\pi }\right). \\ \left(\mathrm{iii}\right)f\left(x\right)=\sin x\left(1+\cos x\right),0<x<\frac{\mathrm{\pi }}{2} \\ \Rightarrow f\left(x\right)=\sin x+\sin x\cos x \\ \Rightarrow f\text{'}\left(x\right)=\cos x+\sin x\left(-\sin x\right)+\cos x\cos x \\ \Rightarrow f\text{'}\left(x\right)=\cos x-{\sin }^{2}x+{\cos }^{2}x \\ \Rightarrow f\text{'}\left(x\right)=\cos x+{\cos }^{2}x-1+{\cos }^{2}x \\ \Rightarrow f\text{'}\left(x\right)=2{\cos }^{2}x+\cos x-1 \\ \Rightarrow f\text{'}\left(x\right)=2{\cos }^{2}x+2\cos x-\cos x-1 \\ \Rightarrow f\text{'}\left(x\right)=2\cos x\left(\cos x+1\right)-1\left(\cos x+1\right) \\ \Rightarrow f\text{'}\left(x\right)=\left(2\cos x-1\right)\left(\cos x+1\right) \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{increasing},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)>0 \\ \Rightarrow \left(2\cos x-1\right)\left(\cos x+1\right)>0 \\ \mathrm{This}\mathrm{is}\mathrm{only}\mathrm{possible}\mathrm{when} \\ \left(2\cos x-1\right)>0\mathrm{and}\left(\cos x+1\right)>0 \\ \Rightarrow \left(2\cos x-1\right)>0\mathrm{and}\left(\cos x+1\right)>0 \\ \Rightarrow \cos x>\frac{1}{2}\mathrm{and}\cos x>-1 \\ \Rightarrow x\in \left(0,\frac{\mathrm{\pi }}{3}\right)\mathrm{and}x\in \left(0,\frac{\mathrm{\pi }}{2}\right) \\ \mathrm{So},x\in \left(0,\frac{\mathrm{\pi }}{3}\right) \\ \therefore f\left(\mathrm{x}\right)\mathrm{is}\mathrm{increasing}\mathrm{on}\left(0,\frac{\mathrm{\pi }}{3}\right). \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{be}\mathrm{decreasing},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow \left(2\cos x-1\right)\left(\cos x+1\right)<0 \\ \mathrm{This}\mathrm{is}\mathrm{only}\mathrm{possible}\mathrm{when} \\ \left(2\cos x-1\right)<0\mathrm{and}\left(\cos x+1\right)>0 \\ \Rightarrow \left(2\cos x-1\right)<0\mathrm{and}\left(\cos x+1\right)>0 \\ \Rightarrow \cos x<\frac{1}{2}\mathrm{and}\cos x>-1 \\ \Rightarrow x\in \left(\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{2}\right)\mathrm{and}x\in \left(0,\frac{\mathrm{\pi }}{2}\right) \\ \mathrm{So},x\in \left(\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{2}\right) \\ \therefore f\left(\mathrm{x}\right)\mathrm{is}\mathrm{decreasing}\mathrm{on}\left(\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{2}\right). \end{gathered}$$