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 $$g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text{otherwise}. \end{cases}$$

Let $$a, b, c, d \in \mathbb{R}$$. Define the function $$h : \mathbb{R} \to \mathbb{R}$$ by $$h(x) = af(x) + b\left(g(x) + g\left(\frac{1}{2} - x\right)\right) + c(x - g(x)) + d \cdot g(x)$$, $$x \in \mathbb{R}$$.

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 $$\{0, 1\}$$

The four basic building blocks are

$$f(x)=\begin{cases}x|x|\sin\!\left(\dfrac1x\right),&x\neq0\\[4pt]0,&x=0\end{cases},\qquad g(x)=\begin{cases}1-2x,&0\le x\le\dfrac12\\[4pt]0,&\text{otherwise}\end{cases}$$

and the composite function

$$h(x)=a\,f(x)+b\,[\,g(x)+g\!\left(\tfrac12-x\right)]+c\,[\,x-g(x)]+d\,g(x).$$

Case P (a=0, b=1, c=0, d=0)

Then $$h(x)=g(x)+g\!\left(\tfrac12-x\right).$$
For $$0\le x\le\dfrac12$$ we have $$g(x)=1-2x$$ and, since $$0\le\dfrac12-x\le\dfrac12,$$

$$g\!\left(\tfrac12-x\right)=1-2\!\left(\tfrac12-x\right)=2x.$$

Hence $$h(x)=1-2x+2x=1\quad\text{for }0\le x\le\dfrac12.$$

For $$x\lt0$$ or $$x\gt\dfrac12$$ both arguments of $$g$$ lie outside $$[0,\tfrac12],$$ so $$g=0$$ there and $$h(x)=0.$$ Thus $$h(x)=\begin{cases}1,&0\le x\le\dfrac12\\[4pt]0,&\text{otherwise}\end{cases}.$$

The only attained values are $$0$$ and $$1,$$ so the range is $$\{0,1\}.$$ This matches item (5).

Case Q (a=1, b=0, c=0, d=0)

Here $$h(x)=f(x)=x|x|\sin\!\left(\dfrac1x\right)\;(x\ne0),\;h(0)=0.$$

Differentiability at $$x=0$$:

$$\lim_{h\to0}\dfrac{f(h)-f(0)}h =\lim_{h\to0}|h|\sin\!\left(\dfrac1h\right)=0,$$ because $$|\,\sin(1/h)\,|\le1$$ and $$|h|\to0.$$
For $$x\ne0$$ the formula is a product/composition of differentiable functions, so $$f$$ is differentiable there as well. Hence $$h$$ is differentiable on the whole of $$\mathbb R,$$ giving item (3).

Case R (a=0, b=0, c=1, d=0)

Now $$h(x)=x-g(x).$$ Using the definition of $$g$$:

$$h(x)=\begin{cases} x,&x\lt0\\[4pt] x-(1-2x)=3x-1,&0\le x\le\dfrac12\\[4pt] x,&x\gt\dfrac12 \end{cases}$$

Ranges of the three pieces:

• $$x\lt0:\;h=x\;\Longrightarrow\;(-\infty,0)$$
• $$0\le x\le\dfrac12:\;h=3x-1\;\Longrightarrow\;[-1,\tfrac12]$$
• $$x\gt\dfrac12:\;h=x\;\Longrightarrow\;(\tfrac12,\infty)$$

The union $$(-\infty,0)\cup[-1,\tfrac12]\cup(\tfrac12,\infty)=\mathbb R.$$ Therefore every real number is achieved; $$h$$ is onto, corresponding to item (2).

Case S (a=0, b=0, c=0, d=1)

Then $$h(x)=g(x).$$ For $$0\le x\le\dfrac12,\;h(x)=1-2x,$$ which decreases continuously from $$1$$ to $$0.$$ Hence $$\operatorname{Range}(h)=[0,1],$$ giving item (4).

Collecting all the matches:

P → (5), Q → (3), R → (2), S → (4).

Thus the correct option is
Option C which is: (P) → (5), (Q) → (3), (R) → (2), (S) → (4).