Limits, Continuity & Differentiability
Continuity and Differentiability of Piecewise Functions
Grade 12
Question:
<p>If the function <br/><br/>\[f(x) = \begin{cases} x^3, & -2 \leq x < -1 \\ -1, & -1 \leq x < 0 \\ 0, & 0 \leq x < 1 \\ \frac{f(x)}{x}, & x \geq 1 \end{cases}\]<br/><br/>where \(f(x) = \sin(x-2) + a\cos(x-2)\)<br/><br/>is continuous and differentiable in \((4, 6)\), then find the range of \(a\).</p>
<p>(a) \(a \in [8, 64]\)</p>
<p>(b) \(a \in (0, 8]\)</p>
<p>(c) \(a \in [64, \infty)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For a piecewise function involving trigonometric expressions to be differentiable, the amplitude must be sufficiently large so that the function becomes constant in the given interval.
<p><strong>Step 1:</strong> For $x \in (4, 6)$, we have $0 < x-2 < 4$, which gives $\frac{8}{a} \leq (x-2)^3 \leq \frac{64}{a}$ (since $a > 0$).</p><p><strong>Step 2:</strong> For $f(x)$ to be continuous and differentiable in $(4, 6)$, the expression $\frac{\sin(x-2) + a\cos(x-2)}{a}$ must attain a constant value for all $x \in (4, 6)$.</p><p><strong>Step 3:</strong> This is only possible when $a \geq 64$.</p><p><strong>Step 4:</strong> In this case, $f(x) = a\cos(x-2)$, which is continuous and differentiable.</p><p>∴ Answer is (c) $a \in [64, \infty)$.</p>
Correct Answer: C