<p>Let \(f(x) = \cot^{-1}\left(\dfrac{x^{2018}+5}{(x-5)(x-10)}\right)\), then:</p>
<p>\(\lim_{x \to 5^-} f(x) = 0\)</p>
<p>\(\lim_{x \to 5^+} f(x) = \pi\)</p>
<p>\(\lim_{x \to 10^-} f(x) = \pi\)</p>
<p>\(\lim_{x \to 10^+} f(x) = 0\)</p>
Step-by-Step Solution
Key Concept: Analyze the behavior of the argument of cot⁻¹ as x → 5⁻, 5⁺, 10⁻, 10⁺ by examining the sign and magnitude of the rational function (x²⁰¹⁸+5)/((x-5)(x-10)). The range of cot⁻¹ is (0,π), and cot⁻¹(y) → π as y → -∞ and cot⁻¹(y) → 0 as y → +∞.
<p><strong>Step 1: Identify points of discontinuity.</strong> The denominator (x-5)(x-10) = 0 at x = 5 and x = 10. The numerator x²⁰¹⁸ + 5 > 0 for all real x.</p><p><strong>Step 2: Analyze behavior at x = 5.</strong><br>• As x → 5⁻: (x-5) → 0⁻ and (x-10) → -5, so denominator → 0⁺, making the argument → +∞, thus f(x) → cot⁻¹(+∞) = 0.<br>• As x → 5⁺: (x-5) → 0⁺ and (x-10) → -5, so denominator → 0⁻, making the argument → -∞, thus f(x) → cot⁻¹(-∞) = π.<br>Left and right limits exist but are unequal: f is discontinuous at x = 5.</p><p><strong>Step 3: Analyze behavior at x = 10.</strong><br>• As x → 10⁻: (x-5) → 5 and (x-10) → 0⁻, so denominator → 0⁻, making the argument → -∞, thus f(x) → cot⁻¹(-∞) = π.<br>• As x → 10⁺: (x-5) → 5 and (x-10) → 0⁺, so denominator → 0⁺, making the argument → +∞, thus f(x) → cot⁻¹(+∞) = 0.<br>Left and right limits exist but are unequal: f is discontinuous at x = 10.</p><p><strong>Step 4: Conclusion.</strong> f has jump discontinuities at both x = 5 and x = 10. The function is continuous and differentiable on ℝ \ {5, 10}. Both discontinuities are of the jump type (limits exist but don't match).</p><p>∴ Answer: Depends on given options (typically: f is discontinuous at x = 5 and x = 10; both are jump discontinuities; f is continuous on (-∞,5)∪(5,10)∪(10,∞))</p>
Correct Answer: A,B,C,D