Question:

Step 1: Analyze the argument $g(x) = \dfrac{x^{2018}+5}{(x-5)(x-10)}$. Note $x^{2018}+5 > 0$ always. Step 2: As $x \to 5^-$: $(x-5) \to 0^-$, $(x-10) \to -5 < 0$, so $(x-5)(x-10) \to 0^+$. Thus $g(x) \to +\infty$. So $f(x) = \cot^{-1}(+\infty) = 0$. Option (a) is correct. Step 3: As $x \to 5^+$: $(x-5) \to 0^+$, $(x-10) \to -5 < 0$, so $(x-5)(x-10) \to 0^-$. Thus $g(x) \to -\infty$. So $f(x) = \cot^{-1}(-\infty) = \pi$. Option (b) is correct. Step 4: As $x \to 10^-$: $(x-5) \to 5 > 0$, $(x-10) \to 0^-$, so $(x-5)(x-10) \to 0^-$. Thus $g(x) \to -\infty$. So $f(x) = \cot^{-1}(-\infty) = \pi$. Option (c) is correct. Step 5: As $x \to 10^+$: $(x-5) \to 5 > 0$, $(x-10) \to 0^+$, so $(x-5)(x-10) \to 0^+$. Thus $g(x) \to +\infty$. So $f(x) = \cot^{-1}(+\infty) = 0$. Option (d) is correct.