Let $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ be functions defined by $f(x)=\begin{cases} x|x|\sin\left(\frac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{cases}$ and $g(x)=\begin{cases} 1-2x, & 0\le x\le \frac{1}{2}, \\ 0, & \text{otherwise}. \end{cases}$ Let $a,b,c,d\in\mathbb{R}$. Define the function $h:\mathbb{R}\to\mathbb{R}$ 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. | List-I | List-II | | --- | --- | | (P) If $a=0$, $b=1$, $c=0$, and $d=0$, then | (1) $h$ is one-one. | | (Q) If $a=1$, $b=0$, $c=0$, and $d=0$, then | (2) $h$ is onto. | | (R) If $a=0$, $b=0$, $c=1$, and $d=0$, then | (3) $h$ is differentiable on $\mathbb{R}$ | | (S) If $a=0$, $b=0$, $c=0$, and $d=1$ then | (4) the range of $h$ is $[0,1]$ | | | (5) the range of $h$ is $\{0,1\}$. |

f(x)= x|x|sin(1/x), x≠ 0 0, x=0 g(x)= 1-2x, 0≤ x≤ 1/2 0, otherwise

$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d\,g(x)$

$\text{Match P, Q, R, S to (1)--(5)}$

$\text{To find: the correct option}$

$0\le x\le \tfrac12:\ g(x)=1-2x,\quad g\left(\tfrac12-x\right)=1-2\left(\tfrac12-x\right)=2x$ $\Rightarrow g(x)+g\left(\tfrac12-x\right) = \begin{cases} 1, & 0\le x\le \tfrac12 \\ 0, & \text{otherwise}\end{cases}$ $P:\ h=S$ $Q:\ h=f$ $R:\ h=x-g(x)$ $S:\ h=g$ $P:\ h(x)=\begin{cases}1,& 0\le x\le\tfrac12\\ 0,&\text{otherwise}\end{cases} \Rightarrow \text{range } \{0,1\} \Rightarrow (5)$ $|f(x)| = x^2\left|\sin\tfrac1x\right| \le x^2 \Rightarrow \lim_{x\to0}f(x)=0=f(0)$ $f'(0) = \lim_{x\to0}\dfrac{f(x)-0}{x} = \lim_{x\to0}|x|\sin\tfrac1x = 0$ $\Rightarrow h=f \text{ differentiable on } \mathbb{R} \Rightarrow (3)$ $R:\ h(x)=x-g(x)=\begin{cases} 3x-1, & 0\le x\le\tfrac12 \\ x, & \text{otherwise}\end{cases}$ $\text{range} = (-\infty,0)\cup[-1,\tfrac12]\cup(\tfrac12,\infty)=\mathbb{R} \Rightarrow \text{onto} \Rightarrow (2)$ $S:\ h=g \Rightarrow \text{range} = [0,1] \Rightarrow (4)$ $P\to5,\quad Q\to3,\quad R\to2,\quad S\to4$ $\boxed{\text{option (3)}}$