Question:

Step 1: Determine $d(x,[0,1])$ and $d(x,[2,3])$ for different intervals of $x$. For $d(x,[0,1])$: - $x < 0$: $d = -x$ - $0 \leq x \leq 1$: $d = 0$ - $x > 1$: $d = x - 1$ For $d(x,[2,3])$: - $x < 2$: $d = 2 - x$ - $2 \leq x \leq 3$: $d = 0$ - $x > 3$: $d = x - 3$ Step 2: Compute $f(x)$ in each region. - $x < 0$: $f(x) = \dfrac{-x}{-x + (2-x)} = \dfrac{-x}{2-2x}= \dfrac{x}{2x-2}$ (simplify: $f(x) = \dfrac{-x}{2-2x}$) - $0 \leq x \leq 1$: $f(x) = \dfrac{0}{0+(2-x)} = 0$ - $1 < x < 2$: $f(x) = \dfrac{x-1}{(x-1)+(2-x)} = \dfrac{x-1}{1} = x-1$ - $2 \leq x \leq 3$: $f(x) = \dfrac{x-1}{(x-1)+0} = 1$ - $x > 3$: $f(x) = \dfrac{x-1}{(x-1)+(x-3)} = \dfrac{x-1}{2x-4}$ Step 3: Analyze option (a): $f(x)$ is decreasing in $(-\infty,0)$ and increasing in $(3,\infty)$. For $x < 0$: $f(x) = \dfrac{-x}{2-2x}$. Derivative: $f'(x) = \dfrac{-(2-2x) - (-x)(-2)}{(2-2x)^2} = \dfrac{-2+2x-2x}{(2-2x)^2} = \dfrac{-2}{(2-2x)^2} < 0$. So $f$ is decreasing — this part is correct. For $x > 3$: $f(x) = \dfrac{x-1}{2x-4}$. Derivative: $f'(x) = \dfrac{(2x-4)-(x-1)(2)}{(2x-4)^2} = \dfrac{2x-4-2x+2}{(2x-4)^2} = \dfrac{-2}{(2x-4)^2} < 0$. So $f$ is actually decreasing in $(3,\infty)$, not increasing. Option (a) is <b>incorrect</b>. Step 4: Analyze option (b): $f$ is bijective. $f = 0$ on $[0,1]$ (not injective) and $f = 1$ on $[2,3]$, so $f$ is not injective. Option (b) is <b>incorrect</b>. Step 5: Analyze option (c): Number of points where $f(x)$ is non-derivable is 4. Potential non-derivable points: $x = 0, 1, 2, 3$. Check each: - At $x=0$: left derivative $\neq$ right derivative (non-derivable) - At $x=1$: left derivative $\neq$ right derivative (non-derivable) - At $x=2$: left derivative $\neq$ right derivative (non-derivable) - At $x=3$: left derivative $\neq$ right derivative (non-derivable) So there are 4 non-derivable points. Option (c) is <b>correct</b>, meaning it is NOT incorrect. Step 6: Analyze option (d): Number of solutions of $f(x) = \dfrac{1}{2}$ is 2. $f(x) = 1/2$ in region $1 < x < 2$: $x - 1 = 1/2 \Rightarrow x = 3/2$ (1 solution). $f(x) = 1/2$ for $x < 0$: $\dfrac{-x}{2-2x} = 1/2 \Rightarrow -2x = 2-2x \Rightarrow 0 = 2$, no solution. $f(x) = 1/2$ for $x > 3$: $\dfrac{x-1}{2x-4} = 1/2 \Rightarrow 2x-2 = 2x-4 \Rightarrow -2 = -4$, no solution. So only 1 solution, not 2. Option (d) is <b>incorrect</b>. Step 7: The incorrect options are (a), (b), and (d).