For what value of k is the function f(x) = { (k cos x) / (π - 2x) for x ≠ π/2; 3 for x = π/2 } continuous at x = π/2?
Answer and explanation
Correct answer: C. k = 6
- Step 1: Evaluate limit as x -> π/2: let x = π/2 + h, so as x -> π/2, h -> 0.
- Step 2: cos(π/2 + h) = -sin(h). Denominator: π - 2(π/2 + h) = -2h.
- Step 3: lim_(h -> 0) [k (-sin h)] / (-2h) = (k / 2) lim_(h -> 0) (sin h / h) = k / 2.
- Step 4: For continuity at x = π/2, limit must equal f(π/2) = 3: k / 2 = 3 => k = 6.
Why not the others: The alternative choices (k = 3, k = 2, k = 4) stem from algebraic sign mistakes, misapplying coordinate formulas, or intermediate arithmetic calculation slips.
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