Mathematics - Functions Question with Solution | TestHub

$f(x)=\{\begin{array}{ccc}x|x|\sin \frac{1}{x}& ;& x\ne 0\\ 0& ;& x=0\end{array}g(x)=\{\begin{array}{ccc}1-2x& ;& 0\le x\le \frac{1}{2}\\ 0& ;& \text{ otherwise }\end{array}$ $g\left(\frac{1}{2}-x\right)=\{\begin{array}{ccc}2x& ;& 0\le \frac{1}{2}-x\le \frac{1}{2}\\ 0& ;& \text{ otherwise }\end{array}=\left\{\begin{array}{ccc}2x& ;& 0\le x\le \frac{1}{2}\\ 0& ;& \text{ otherwise }\end{array}\right\}$ $g(x)+g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{lll}1& ;& 0\le x\le \frac{1}{2}\\ 0& ;& \text{ otherwise }\end{array}\right\}$ (P) Now $\mathrm{a}=0,\mathrm{b}=1,\mathrm{c}=0,\mathrm{d}=0$ $\because \mathrm{h}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{g}\left(\frac{1}{2}-\mathrm{x}\right)=\{\begin{array}{lll}1& ;& 0\le \mathrm{x}\le \frac{1}{2}\\ 0& ;& \text{ otherwise }\end{array}$Hence Range of $\mathrm{h}(\mathrm{x})$ is $\{0,1\}$ (Q) $\begin{array}{rl}& \mathrm{a}=1,\mathrm{b}=0,\mathrm{c}=0,\mathrm{d}=0\\ & \mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{x})=\{\begin{array}{cll}\mathrm{x}|\mathrm{x}|\sin \frac{1}{x}& ;& x\ne 0\\ 0& ;& x=0\end{array}\\ & \text{ RHD }=\underset{x\to 0}{\lim }\frac{x^2\sin \frac{1}{x}-0}{x}=0\\ & \text{ LHD }=\underset{x\to 0}{\lim }\frac{-x^2\sin \frac{1}{x}-0}{x}=0\end{array}$ Hence $\mathrm{h}(\mathrm{x})$ is differentiable on $\mathrm{R}$ (R) $\begin{array}{rl}& \mathrm{a}=0,\mathrm{b}=0,\mathrm{c}=1,\mathrm{d}=0\\ & h(x)=x-g(x)=\{\begin{array}{ccc}3\mathrm{x}-1& ;& 0\le x\le \frac{1}{2}\\ 0& ;& \text{ otherwise }\end{array}\end{array}$
$\therefore \mathrm{h}(\mathrm{x})$ is ONTO (S) $\begin{array}{rl}& \mathrm{a}=0,\mathrm{b}=0,\mathrm{c}=0,\mathrm{d}=1\\ & \mathrm{h}(\mathrm{x})=\mathrm{g}(\mathrm{x})=\{\begin{array}{ccl}1-2\mathrm{x}& ;& 0\le \mathrm{x}\le \frac{1}{2}\\ 0& ;& \text{ otherwise }\end{array}\end{array}$Range of $h(x)$ is $[0,1]$