Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be functions defined by $f(x)=\left\{\begin{array}{ll}x|x| \sin \left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x=0,\end{array} \quad\right.$ and $g(x)= \begin{cases}1-2 x, & 0 \leq x \leq \frac{1}{2} \\ 0, & \text { otherwise }\end{cases}$ Let $a, b, c, d \in \mathbb{R}$. Define the function $h: \mathbb{R} \rightarrow \mathbb{R}$ by $h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in \mathbb{R}$ Match each entry in List-I to the correct entry in List-II. The correct option is :

• A

  $(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(3) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)$
• B

  $(\mathrm{P}) \rightarrow(5) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(4) \quad(\mathrm{S}) \rightarrow(3)$
• Correct answer
  C

  $(\mathrm{P}) \rightarrow(5) \quad(\mathrm{Q}) \rightarrow(3) \quad(\mathrm{R}) \rightarrow(2) \quad(\mathrm{S}) \rightarrow(4)$
• D

  $(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)$

Solution

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

Hence, the correct answer is option: 'C'.