Limits, Continuity & Differentiability
General
Grade 12
Question:
<p>Let <span class="math-inline">\(g(x)=\begin{cases}3x^2-4\sqrt{x}+1 & x<1\\ ax+b & x\ge 1\end{cases}\)</span>. If <span class="math-inline">\(g(x)\)</span> is continuous and differentiable at <span class="math-inline">\(x=1\)</span>, find <span class="math-inline">\(a\)</span> and <span class="math-inline">\(b\)</span>.</p>
a=b=4
a=b=-4
<strong>a=4, b=-4</strong>
a=-4, b=4
Step-by-Step Solution
Key Concept: General
<div class="solution"><p><strong>Continuity at x=1:</strong> <span class="math-block">\[g(1^-)=3(1)-4(1)+1=0,\quad g(1)=a+b\implies a+b=0\quad\cdots(1)\]</span></p><p><strong>Differentiability at x=1:</strong> <span class="math-block">\[g'(x^-)=6x-\frac{2}{\sqrt{x}}\bigg|_{x=1}=6-2=4,\quad g'(x^+)=a\implies a=4\quad\cdots(2)\]</span></p><p>From (1): <span class="math-inline">\(b=-4\)</span>.</p><p><strong>Answer: (C) a=4, b=-4</strong></p><div class="trap-box"><strong>Trap:</strong> Forgetting to differentiate <span class="math-inline">\(\sqrt{x}\)</span> properly: <span class="math-inline">\(\frac{d}{dx}(-4\sqrt{x})=-\frac{2}{\sqrt{x}}\)</span>.</div><div class="key-concept"><strong>Key Concept:</strong> For differentiability: equate LHD = RHD AND continuity simultaneously</div></div>
Correct Answer: 3