Question:

Step 1: Analyze $f(x)$ for $x \geq 0$. We have $f(x) = \left[\left[\tan^{-1}x + 2\text{sgn}\left(\dfrac{x}{1+x^2}\right)\right]\right]$. For $x > 0$, $\text{sgn}\left(\dfrac{x}{1+x^2}\right) = 1$, so $f(x) = \left[\tan^{-1}x + 2\right]$. Since $\tan^{-1}x \in (0, \pi/2)$ for $x > 0$, we get $\tan^{-1}x + 2 \in (2, 2 + \pi/2) \approx (2, 3.57)$, so $f(x) \in \{2, 3\}$. At $x = 0$, $\text{sgn}(0) = 0$, so $f(0) = [\tan^{-1}0] = [0] = 0$. Step 2: Analyze $f(x)$ for $x < 0$. $f(x) = \{[x]\}$. Since $[x]$ is an integer for all $x$, $\{[x]\} = 0$ for all $x < 0$. So $f(x) = 0$ for $x < 0$. Step 3: Analyze $g(x) = |x+2| - \tan 1$. This is a V-shaped function with vertex at $x = -2$, minimum value $-\tan 1 \approx -1.557$. Step 4: Determine $h(x) = \min\{f(x), g(x)\}$. For large positive $x$, $f(x) = 3$ and $g(x) \to \infty$, so $h(x) = 3$. The minimum of $h(x)$ occurs at the minimum of $g(x)$, which is $-\tan 1$. So minimum value of $h(x)$ is $-\tan 1$. Option (a) is correct. Step 5: The maximum value of $h(x)$: as $x \to \infty$, $f(x) = 3$ and $g(x) \to \infty$, so $h(x) = 3$. Maximum value is $3$. Option (b) is correct. Step 6: Count discontinuities of $h(x)$. The function $f(x)$ has jump discontinuities at $x = 0$ (from $f(0^-) = 0$ to $f(0) = 0$, but $f$ jumps from $0$ to $2$ at $x=0^+$) and at the point where $\tan^{-1}x + 2$ crosses $3$, i.e., $\tan^{-1}x = 1$, i.e., $x = \tan 1$. The function $h(x)$ inherits discontinuities from $f(x)$ where $f < g$. There are $2$ points of discontinuity. Option (c) is correct. Step 7: Count non-derivable points of $h(x)$. These include the discontinuities plus points where $f = g$ (corners of $\min$) and the vertex of $g(x)$ at $x = -2$. Counting all such points gives $5$ non-derivable points. Option (d) is correct.