Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Let \(f: R \to R\) be a function defined as \[f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a + bx, & \text{if } 1 < x < 3 \\ b + 5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases}\] Then, \(f\) is:</p>
<p>continuous if \(a = 5\) and \(b = 5\)</p>
<p>continuous if \(a = -5\) and \(b = 10\)</p>
<p>continuous if \(a = 0\) and \(b = 5\)</p>
<p>not continuous for any values of \(a\) and \(b\)</p>
Step-by-Step Solution
Key Concept: For f to be continuous at x = 1, the left limit must equal the right limit: lim(x→1⁻) f(x) = lim(x→1⁺) f(x). This gives one equation, and continuity at x = 3 gives another, forming a system to solve for a and b.
<p><strong>Step 1:</strong> Apply continuity at x = 1. We need lim(x→1⁻) f(x) = lim(x→1⁺) f(x).</p><p>Left limit: lim(x→1⁻) f(x) = 5</p><p>Right limit: lim(x→1⁺) f(x) = a + b(1) = a + b</p><p>Therefore: <strong>a + b = 5</strong> ... (Equation 1)</p><p><strong>Step 2:</strong> Apply continuity at x = 3. We need lim(x→3⁻) f(x) = lim(x→3⁺) f(x).</p><p>Left limit: lim(x→3⁻) f(x) = a + b(3) = a + 3b</p><p>Right limit: lim(x→3⁺) f(x) = 21</p><p>Therefore: <strong>a + 3b = 21</strong> ... (Equation 2)</p><p><strong>Step 3:</strong> Solve the system. Subtract Equation 1 from Equation 2:</p><p>(a + 3b) - (a + b) = 21 - 5</p><p>2b = 16 → <strong>b = 8</strong></p><p>From Equation 1: a + 8 = 5 → <strong>a = -3</strong></p><p>∴ Answer: D (a = -3, b = 8)</p>
Correct Answer: D