Match Functions to Properties of Injectivity Surjectivity Differentiability and Range

To solve the problem, we first analyze the functions $f(x)$, $g(x)$, and $g\left(\dfrac{1}{2}-x\right)$.

The functions are defined as: $f(x) = \begin{cases} x|x|\sin\left(\dfrac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$ $g(x) = \begin{cases} 1-2x, & 0 \le x \le \dfrac{1}{2} \\ 0, & \text{otherwise} \end{cases}$

Now, let's evaluate $g\left(\dfrac{1}{2}-x\right)$: The condition $0 \le \dfrac{1}{2} - x \le \dfrac{1}{2}$ holds if and only if $0 \le x \le \dfrac{1}{2}$.

• For $x \in \left[0, \dfrac{1}{2}\right]$, $g\left(\dfrac{1}{2}-x\right) = 1 - 2\left(\dfrac{1}{2}-x\right) = 2x$.
• Otherwise, $g\left(\dfrac{1}{2}-x\right) = 0$.

Combining $g(x)$ and $g\left(\dfrac{1}{2}-x\right)$: $g(x) + g\left(\dfrac{1}{2}-x\right) = \begin{cases} (1-2x) + 2x = 1, & 0 \le x \le \dfrac{1}{2} \\ 0, & \text{otherwise} \end{cases}$

Now, we analyze each case given in List-I:

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

$h(x) = g(x) + g\left(\dfrac{1}{2}-x\right) = \begin{cases} 1, & 0 \le x \le \dfrac{1}{2} \\ 0, & \text{otherwise} \end{cases}$ The function $h(x)$ takes only two distinct values: $0$ and $1$. Thus, the range of $h$ is $\{0, 1\}$. Hence, $(\text{P}) \to (5)$.

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

$h(x) = f(x) = \begin{cases} x|x|\sin\left(\dfrac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$

• For $x \neq 0$, $h(x)$ is a product of differentiable functions, so it is differentiable for all $x \neq 0$.
• At $x = 0$, we test differentiability using the limit definition: $h'(0) = \lim_{x \to 0} \frac{h(x) - h(0)}{x - 0} = \lim_{x \to 0} \frac{x|x|\sin\left(\dfrac{1}{x}\right)}{x} = \lim_{x \to 0} |x|\sin\left(\dfrac{1}{x}\right)$ Since $\left||x|\sin\left(\dfrac{1}{x}\right)\right| \le |x| \to 0$ as $x \to 0$, by the Sandwich Theorem, $h'(0) = 0$.

Since $h'(x)$ exists for all $x \in \mathbb{R}$, $h$ is differentiable on $\mathbb{R}$. Hence, $(\text{Q}) \to (3)$.

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

$h(x) = x - g(x) = \begin{cases} x - (1-2x) = 3x - 1, & 0 \le x \le \dfrac{1}{2} \\ x, & \text{otherwise} \end{cases}$ Let's find the range of $h(x)$:

• For $x < 0$, $h(x) = x \in (-\infty, 0)$.
• For $0 \le x \le \dfrac{1}{2}$, $h(x) = 3x - 1 \in \left[-1, \dfrac{1}{2}\right]$.
• For $x > \dfrac{1}{2}$, $h(x) = x \in \left(\dfrac{1}{2}, \infty\right)$.

The complete range is: $\text{Range}(h) = (-\infty, 0) \cup \left[-1, \dfrac{1}{2}\right] \cup \left(\dfrac{1}{2}, \infty\right) = (-\infty, \infty) = \mathbb{R}$ Since the range of $h$ equals the co-domain $\mathbb{R}$, $h$ is onto (surjective). Hence, $(\text{R}) \to (2)$.

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

$h(x) = g(x) = \begin{cases} 1 - 2x, & 0 \le x \le \dfrac{1}{2} \\ 0, & \text{otherwise} \end{cases}$

• For $x \in \left[0, \dfrac{1}{2}\right]$, as $x$ increases from $0$ to $\dfrac{1}{2}$, $h(x) = 1-2x$ decreases continuously from $1$ to $0$, covering the entire closed interval $[0, 1]$.
• For $x \notin \left[0, \dfrac{1}{2}\right]$, $h(x) = 0$, which is already inside $[0, 1]$.

Thus, the range of $h$ is $[0, 1]$. Hence, $(\text{S}) \to (4)$.

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Conclusion:

The correct matching is: $(\text{P}) \to (5) \quad (\text{Q}) \to (3) \quad (\text{R}) \to (2) \quad (\text{S}) \to (4)$

This corresponds to Option C.