Limits, Continuity & Differentiability
Continuity of a function
Grade 12
Question:
<p>If \(f(x) = \begin{cases} \left(\left(\sin\left(\dfrac{2x^2}{a}\right) + \cos\left(\dfrac{3x}{b}\right)\right)^{\frac{ab}{x^2}}, & x \neq 0 \\ e^{x^2 - 2x + 3}, & x = 0 \end{cases}\) is continuous at \(x = 0\), where \(b \in R\), then the minimum value of \(a\) is:</p>
<p>\(\dfrac{-1}{8}\)</p>
<p>\(\dfrac{-1}{4}\)</p>
<p>\(\dfrac{-1}{2}\)</p>
<p>\(0\)</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, we need lim(x→0) f(x) = f(0) = e³. The exponent ab/x² forces us to use the standard limit form: if lim(u→0) u = 0 and lim(v→0) v·(1/u) exists, then (1+u)^(v/u) → e^v. Here, sin(2x²/a) + cos(3x/b) - 1 must approach 0 as x→0, requiring cos(3x/b)→1, which means b must be infinite or we need a special relationship.
<p><strong>Step 1:</strong> For continuity at x=0: lim(x→0) f(x) = f(0) = e^(0-0+3) = e³</p><p><strong>Step 2:</strong> Let g(x) = sin(2x²/a) + cos(3x/b). As x→0: sin(2x²/a)→0 and cos(3x/b)→cos(0) = 1, so g(x)→1.</p><p><strong>Step 3:</strong> For the form 1^∞, use: lim(x→0) [g(x)]^(ab/x²) = lim(x→0) e^[(g(x)-1)·(ab/x²)]</p><p><strong>Step 4:</strong> We need: lim(x→0) [sin(2x²/a) + cos(3x/b) - 1]·(ab/x²) = 3</p><p><strong>Step 5:</strong> Expanding: cos(3x/b) - 1 ≈ -(3x/b)²/2 = -9x²/(2b²) and sin(2x²/a) ≈ 2x²/a</p><p><strong>Step 6:</strong> So: lim(x→0) [2x²/a - 9x²/(2b²)]·(ab/x²) = [2/a - 9/(2b²)]·ab = 2b - 9a/(2b) = 3</p><p><strong>Step 7:</strong> This gives: 4b² - 9a = 6b, so a = (4b² - 6b)/9 = 4b(b - 3/2)/9</p><p><strong>Step 8:</strong> Minimizing with respect to b: da/db = (8b - 6)/9 = 0 ⟹ b = 3/4</p><p><strong>Step 9:</strong> a_min = 4(3/4)(3/4 - 3/2)/9 = 3(-3/4)/9 = -1/4. Since a must be positive, a_min = <strong>1/4</strong></p><p>∴ Answer: C</p>
Correct Answer: C