Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>If the function \(f\) defined on \(\left(\dfrac{\pi}{6},\,\dfrac{\pi}{3}\right)\) by \[f(x) = \begin{cases} \dfrac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \neq \dfrac{\pi}{4} \\ k, & x = \dfrac{\pi}{4} \end{cases}\] is continuous, then \(k\) is equal to:</p>
<p>\(2\)</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(1\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: For continuity at x = π/4, the value k must equal the limit of f(x) as x approaches π/4. Use algebraic simplification and L'Hôpital's rule or trigonometric identities to find this limit.
<p><strong>Step 1:</strong> For continuity at x = π/4, we need k = lim(x→π/4) f(x).</p><p><strong>Step 2:</strong> At x = π/4: numerator = √2·cos(π/4) - 1 = √2·(1/√2) - 1 = 1 - 1 = 0, and denominator = cot(π/4) - 1 = 1 - 1 = 0. This is 0/0 form.</p><p><strong>Step 3:</strong> Rationalize the numerator: multiply by (√2cos x + 1)/(√2cos x + 1):</p><p>Numerator becomes: 2cos²x - 1 = cos(2x)</p><p><strong>Step 4:</strong> Rewrite denominator: cot x - 1 = (cos x - sin x)/sin x</p><p><strong>Step 5:</strong> The expression becomes: f(x) = [cos(2x)·sin x]/[cos x - sin x]</p><p><strong>Step 6:</strong> Since cos(2x) = cos²x - sin²x = (cos x - sin x)(cos x + sin x):</p><p>f(x) = [(cos x - sin x)(cos x + sin x)·sin x]/[cos x - sin x] = (cos x + sin x)·sin x</p><p><strong>Step 7:</strong> Now substitute x = π/4: k = (cos(π/4) + sin(π/4))·sin(π/4) = (1/√2 + 1/√2)·(1/√2) = (2/√2)·(1/√2) = 2/2 = 1</p><p>∴ Answer: <strong>k = 1</strong></p>
Correct Answer: B