Question:

Step 1: Analyze the two functions $g(x) = |3-x|$ and $h(x) = 3 - x^3$. $g(x) = |3-x|$ has a corner at $x = 3$. $h(x) = 3 - x^3$ is a smooth cubic. Step 2: Find where $g(x) = h(x)$, i.e., $|3-x| = 3 - x^3$. For $x \leq 3$: $3 - x = 3 - x^3 \Rightarrow x^3 - x = 0 \Rightarrow x(x^2-1)=0 \Rightarrow x = 0, \pm 1$. For $x > 3$: $x - 3 = 3 - x^3 \Rightarrow x^3 + x - 6 = 0 \Rightarrow (x-1)(x^2+x+6)=0$, no real root $> 3$. Step 3: Determine which function is larger in each interval to define $f(x) = \max(g(x), h(x))$. At $x = -1, 0, 1$, the two functions intersect, so $f(x)$ switches between the two expressions at these points. At $x = 3$, $g(x)$ has a corner. Step 4: Check continuity. Both $|3-x|$ and $3-x^3$ are continuous everywhere, so their maximum is also continuous $\forall x \in R$. Option (a) is correct. Step 5: Check differentiability. The non-differentiable points occur where the two functions cross (and the max switches) or where one of them is non-differentiable. The crossings at $x = -1, 0, 1$ and the corner of $|3-x|$ at $x = 3$ give potential non-differentiable points. Checking carefully, $f(x)$ is non-differentiable at $x = -1, 1, 3$ (three points). At $x = 0$ the functions meet tangentially so differentiability is preserved there. Thus $f(x)$ is non-derivable at exactly three points. Option (c) is correct.