Limits, Continuity & Differentiability
Non-differentiability of composite functions
Grade 12
<p><strong>882.</strong> Let \(f(x) = \begin{cases} \dfrac{ax^3 + bx^2 + cx + d}{x}, & x \neq 0 \\ 2, & x = 0 \end{cases}\) be a continuous function where \(a, b, c, d\) are in arithmetic progression. Then find the number of points where \(|f(|x|)|\) is non derivable.</p>
Step-by-Step Solution
Key Concept: For f(x) to be continuous at x=0, the limit of f(x) as x→0 must equal f(0)=2. Since a,b,c,d are in AP, we can express them using a common difference to find which coefficients must vanish for continuity.
<p><strong>Step 1: Apply AP condition</strong></p><p>Let a, b, c, d be in AP with common difference r. Then: b=a+r, c=a+2r, d=a+3r</p><p><strong>Step 2: Ensure continuity at x=0</strong></p><p>For continuity at x=0: $\lim_{x \to 0} \frac{ax^3 + bx^2 + cx + d}{x} = 2$</p><p>This requires d=0 and $\lim_{x \to 0} \frac{ax^2 + bx + c}{1} = 2$, so c=2</p><p><strong>Step 3: Find coefficients using AP</strong></p><p>From d=a+3r=0 and c=a+2r=2:</p><p>Subtracting: r=-2, so a=6, b=4, c=2, d=0</p><p>Thus: $f(x) = 6x^2 + 4x + 2$ for x≠0, and f(0)=2 ✓</p><p><strong>Step 4: Analyze |f(|x|)|</strong></p><p>Since $f(x) = 6x^2 + 4x + 2 = 6(x+\frac{1}{3})^2 + \frac{4}{3} > 0$ for all x, we have $|f(|x|)| = f(|x|) = 6x^2 + 4|x| + 2$</p><p><strong>Step 5: Find non-derivable points</strong></p><p>The function g(x)=|f(|x|)| has non-derivability where:</p><p>• x=0 (due to |x|)</p><p>• The composition creates a corner at x=0</p><p>Since f(|x|) is an even function, |f(|x|)| is also even. The function is continuous everywhere but has a sharp corner at x=0 due to |x|.</p><p><strong>∴ Answer: 1 point (at x=0)</strong></p>
Correct Answer: 1