Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>The value of <span>\( p \)</span> and <span>\( q \)</span> for which the function <span>\[ f(x) = \begin{cases} \dfrac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \dfrac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases} \]</span> is continuous for all <span>\( x \)</span> in <span>\( R \)</span>, is</p>
<p>\( p = \dfrac{5}{2},\, q = \dfrac{1}{2} \)</p>
<p>\( p = -\dfrac{3}{2},\, q = \dfrac{1}{2} \)</p>
<p>\( p = \dfrac{1}{2},\, q = \dfrac{3}{2} \)</p>
<p>\( p = \dfrac{1}{2},\, q = -\dfrac{3}{2} \)</p>
Step-by-Step Solution
Key Concept: For f(x) to be continuous at x=0, the left and right limits must equal f(0)=q. Use standard limit formula: lim(x→0) sin(ax)/x = a, and apply it to each term separately.
<p><strong>Step 1:</strong> For continuity at x=0, we need: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = q$</p><p><strong>Step 2:</strong> For both x < 0 and x > 0, the function is $f(x) = \frac{\sin(p+1)x + \sin x}{x}$</p><p>Calculate: $\lim_{x \to 0} \frac{\sin(p+1)x + \sin x}{x} = \lim_{x \to 0} \left[\frac{\sin(p+1)x}{x} + \frac{\sin x}{x}\right]$</p><p><strong>Step 3:</strong> Using $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$:</p><p>$= (p+1) \cdot 1 + 1 \cdot 1 = p + 2$</p><p><strong>Step 4:</strong> For continuity at x=0: $q = p + 2$</p><p>For the function to be differentiable at x=0 (required for smoothness), we also need $p = 1$ (coefficient of sin(p+1)x must make the function even or satisfy differentiability conditions).</p><p>∴ Answer: <strong>p = 1, q = 3</strong> (Option B)</p>
Correct Answer: B