Question 17

1. Analyze $f(x)$: $f(x) = x|x|\sin(1/x)$ for $x \neq 0$ and $f(0) = 0$. $f'(0) = \lim_{k \to 0} \frac{k|k|\sin(1/k) - 0}{k} = \lim_{k \to 0} |k|\sin(1/k) = 0$. Thus, $f(x)$ is differentiable on $\mathbb{R}$. Also, $\lim_{x \to \pm\infty} f(x) = \pm\infty$, so the range of $f$ is $\mathbb{R}$ (onto).

2. Analyze $g(x)$: $g(x) = 1-2x$ for $x \in [0, 1/2]$ and $0$ otherwise. $g(x)$ is discontinuous at $x=0$ (LHL=0, RHL=1) and continuous at $x=1/2$.

3. Match (P): $a=0, b=1, c=0, d=0 \Rightarrow h(x) = g(x) + g(1/2 - x)$. If $x \in [0, 1/2]$, then $1/2 - x \in [0, 1/2]$. $h(x) = (1-2x) + (1-2(1/2-x)) = 1-2x + 2x = 1$. If $x < 0$, $g(x) = 0$ and $1/2 - x > 1/2 \Rightarrow g(1/2-x) = 0$. $h(x) = 0$. If $x > 1/2$, $g(x) = 0$ and $1/2 - x < 0 \Rightarrow g(1/2-x) = 0$. $h(x) = 0$. Range of $h$ is $\{0, 1\}$. Thus (P) $\to$ (5).

4. Match (Q): $a=1, b=0, c=0, d=0 \Rightarrow h(x) = f(x)$. As shown, $f$ is differentiable on $\mathbb{R}$. Thus (Q) $\to$ (3).

5. Match (R): $a=0, b=0, c=1, d=0 \Rightarrow h(x) = x - g(x)$. $h(x) = \begin{cases} x & x < 0 \\ 3x-1 & 0 \leq x \leq 1/2 \\ x & x > 1/2 \end{cases}$ Range of $h$: $(-\infty, 0) \cup [-1, 1/2] \cup (1/2, \infty) = \mathbb{R}$. Thus $h$ is onto. (R) $\to$ (2).

6. Match (S): $a=0, b=0, c=0, d=1 \Rightarrow h(x) = g(x)$. Range of $g$ is $[0, 1]$. Thus (S) $\to$ (4).

Correct Option: C