Let $f:\mathbb{R}\to \; <!--\; \to \; -->\mathbb{R}$ and $g:\mathbb{R}\to \; <!--\; \to \; -->\mathbb{R}$ be functions defined by

$f(x)=\{\begin{array}{cl}x|x|\sin \; <!--\; \; -->\left(\frac{1}{x}\right),& x\ne \; <!--\; \ne \; -->0,\\ 0,& x=0,\end{array}andg(x)=\{\begin{array}{cc}1-\; <!--\; -\; -->2x,& 0\le \; <!--\; \le \; -->x\le \; <!--\; \le \; -->\frac{1}{2},\\ 0,& \text{ otherwise }\end{array}$

Let a, b, c, d$\in \; <!--\; \in \; -->\mathbb{R}$. Define the function $h:\mathbb{R}\to \; <!--\; \to \; -->\mathbb{R}$ by

$h(x)=af(x)+b(g(x)+g(\frac{1}{2}-\; <!--\; -\; -->x))+c(x-\; <!--\; -\; -->g(x))+dg(x),x\in \; <!--\; \in \; -->\mathbb{R.}$

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 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 correct option is

The correct option is:
(P) → (5) (Q) → (3) (R) → (2) (S) → (4)

Let us analyze the given functions and find the mapping for each case step-by-step.

The functions are defined as:

$f(x)=\{\begin{array}{ll}x|x|\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}$

and

$g(x)=\{\begin{array}{ll}1-2x,& 0\le x\le \frac{1}{2}\\ 0,& \text{otherwise}\end{array}$

The general function $h(x)$ is given by:

$h(x)=af(x)+b(g(x)+g(\frac{1}{2}-x))+c(x-g(x))+dg(x)$

Case (P): If $a=0,b=1,c=0,d=0$

Substituting the values, we obtain:
$h(x)=g(x)+g(\frac{1}{2}-x)$

Let's evaluate $h(x)$ on different intervals:
1. If $0\le x\le \frac{1}{2}$:
Then $0\le \frac{1}{2}-x\le \frac{1}{2}$. Thus:
$g(x)=1-2x$
$g(\frac{1}{2}-x)=1-2(\frac{1}{2})=2x$
Hence, $h(x)=(1-2x)+2x=1$.

2. If $x<0$:
Then $g(x)=0$. Also, $\frac{1}{2}-x>\frac{1}{2}$, so $g(\frac{1}{2}-x)=0$. Thus, $h(x)=0$.

3. If $x>\frac{1}{2}$:
Then $g(x)=0$. Also, $\frac{1}{2}-x<0$, so $g(\frac{1}{2}-x)=0$. Thus, $h(x)=0$.

Therefore, the function is:
$h(x)=\{\begin{array}{ll}1,& 0\le x\le \frac{1}{2}\\ 0,& \text{otherwise}\end{array}$

This means the range of $h$ is the set containing only 0 and 1, i.e., $\{0,1\}$.
So, (P) → (5).

Case (Q): If $a=1,b=0,c=0,d=0$

Substituting the values, we get:
$h(x)=f(x)$

For $x\ne 0$, $h(x)$ is clearly differentiable. Let us check the differentiability of $h(x)$ at $x=0$ using the definition of the derivative:

$h^{\prime }(0)=\underset{x\to 0}{lim}\frac{h(x)-h(0)}{x}=\underset{x\to 0}{lim}\frac{x|x|\sin (1/x)-0}{x}=\underset{x\to 0}{lim}|x|\sin \left(\frac{1}{x}\right)$

Since $-1\le \sin (1/x)\le 1$ and $\underset{x\to 0}{lim}|x|=0$, by the Squeeze Theorem, we have:
$\underset{x\to 0}{lim}|x|\sin \left(\frac{1}{x}\right)=0$

Thus, $h^{\prime }(0)=0$, which means $h$ is differentiable on $\mathbb{R}$.
So, (Q) → (3).

Case (R): If $a=0,b=0,c=1,d=0$

Substituting the values, we get:
$h(x)=x-g(x)$

Let's write down $h(x)$ explicitly:
1. If $x<0$:
$h(x)=x-0=x$, which has the range $(-\mathrm{\infty },0)$.

2. If $0\le x\le \frac{1}{2}$:
$h(x)=x-(1-2x)=3x-1$.
As $x$ increases from $0$ to $\frac{1}{2}$, $h(x)$ increases linearly from $-1$ to $\frac{1}{2}$. So the range in this interval is $[-1,\frac{1}{2}]$.

3. If $x>\frac{1}{2}$:
$h(x)=x-0=x$, which has the range $(\frac{1}{2},\mathrm{\infty })$.

Combining the range over all intervals:
$\text{Range}(h)=(-\mathrm{\infty },0)\cup [-1,\frac{1}{2}]\cup (\frac{1}{2},\mathrm{\infty })=\mathbb{R}$

Since the range of $h$ is equal to its codomain $\mathbb{R}$, the function $h$ is onto.
So, (R) → (2).

Case (S): If $a=0,b=0,c=0,d=1$

Substituting the values, we get:
$h(x)=g(x)$

The definition of $g(x)$ tells us that it is non-zero only for $0\le x\le \frac{1}{2}$.
For $x\in [0,\frac{1}{2}]$, $g(x)=1-2x$. The range of this linear segment is:
$[g(1/2),g(0)]=[0,1]$

For all other values of $x$, $h(x)=0$, which is already in the interval $[0,1]$.

Thus, the range of $h$ is $[0,1]$.
So, (S) → (4).