Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If \(f(x) = \begin{cases} x^2\{x\}^2 + x\sin\{x\} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}\), where \(\{x\}\) denotes the fractional part function, then</p>
<p>(a) \(f\) is continuous and differentiable at \(x = 0\)</p>
<p>(b) \(f\) is continuous but not differentiable at \(x = 0\)</p>
<p>(c) \(f\) is continuous and differentiable at \(x = 2\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The fractional part function is bounded by 1, and multiplying by $x^2$ creates a factor that vanishes at the origin, ensuring both continuity and differentiability.
<p><strong>At $x = 0$:</strong> Note that $\{x\} \in [0,1)$, so $\{x\}^2 \leq 1$ and $\sin\{x\} \leq 1$.</p><p><strong>Continuity:</strong> $|f(x)| = |x^2\{x\}^2 + x\sin\{x\}| \leq x^2 + |x| \to 0$ as $x \to 0$. Thus $\lim_{x \to 0} f(x) = 0 = f(0)$, so $f$ is continuous at $x=0$.</p><p><strong>Differentiability:</strong> $f'(0) = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} (h\{h\}^2 + \sin\{h\}) = 0$ since $h\{h\}^2 \to 0$ and $\sin\{h\} \to 0$ as $h \to 0$. Thus $f$ is differentiable at $x=0$.</p><p>∴ Answer is (a).</p>
Correct Answer: a