Question:

Step 1: Analyze the given condition. Since $f(a) > f(b)$ whenever $a < b$, the function $f$ is strictly decreasing on $\mathbb{R}$. Step 2: Check option (a). A strictly decreasing function need not be differentiable everywhere (e.g., $f(x) = -x$ for $x \neq 2$ and some non-differentiable point at $x=2$). So the limit $\lim_{h \to 0} \dfrac{f(2+h)-f(2)}{h}$ need not exist. Option (a) is not necessarily correct. Step 3: Check option (b). Since $f$ is strictly decreasing and continuous, it is one-to-one. By the intermediate value theorem, $f(x) = 0$ has at most one solution, and since $f$ is continuous and strictly decreasing over all of $\mathbb{R}$, it has exactly one real root. Option (b) is correct. Step 4: Check option (c). Consider $g(x) = f(x) - f(-x+1)$. Note that $g(x) + g(1-x) = [f(x) - f(1-x)] + [f(1-x) - f(x)] = 0$, so $g(1-x) = -g(x)$. This means $g$ is antisymmetric about $x = 1/2$, so $g(1/2) = 0$, giving exactly one solution $x = 1/2$ (since $g$ is strictly increasing as $f$ is strictly decreasing). Option (c) is correct. Step 5: Check option (d). Suppose $f(x) = f(x+1)$ for some $x$. Since $x < x+1$, strict decrease gives $f(x) > f(x+1)$, a contradiction. So there is no real root of $f(x) = f(x+1)$. Option (d) is correct.