Limits, Continuity & Differentiability
General
Grade 12
Question:
<p><span class="math-inline">\(f(x)=\begin{cases}|x-3| & x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4} & x<1\end{cases}\)</span> is:</p>
<strong>cont. at x=1</strong>
<strong>diff. at x=1</strong>
<strong>cont. at x=3</strong>
diff. at x=3
Step-by-Step Solution
Key Concept: General
<div class="solution"><p><strong>At x=1:</strong><br>f(1⁺)=|1-3|=2; f(1⁻)=1/4-3/2+13/4=1-6+13)/4=8/4=2. <strong>Continuous</strong> ✓ (A)</p><p>f'(1⁻)=(x/2-3/2)|₁=1/2-3/2=-1; f'(1⁺)=derivative of |x-3| at x=1: since 1<3, |x-3|=3-x, f'=-1. LHD=RHD=-1. <strong>Differentiable</strong> ✓ (B)</p><p><strong>At x=3:</strong> |x-3| has corner. <strong>Continuous</strong> but <strong>not differentiable</strong> ✓ (C), (D) false.</p><p><strong>Answer: (A),(B),(C)</strong></p><div class="key-concept"><strong>Key Concept:</strong> Check both junction points separately</div></div>
Correct Answer: A,B,C