Limits, Continuity & Differentiability
Differentiability and non-derivable points
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 (ax³+bx²+cx+d)/x as x→0 must equal 2, which forces d=0. Since a,b,c,d are in AP with d=0, we get a=-3b and c=b, making f(x)=ax²+bx+c. The function |f(|x|)| is non-differentiable where f(|x|)=0 or where f(|x|) changes sign.
<p><strong>Step 1: Apply continuity at x=0</strong></p><p>For f to be continuous at x=0: lim(x→0) (ax³+bx²+cx+d)/x = 2</p><p>This limit equals c (coefficient of x⁰ in numerator after removing one x), so c=2.</p><p>Also, for the limit to exist and equal 2, we need d=0.</p><p><strong>Step 2: Use AP condition</strong></p><p>Since a,b,c,d are in AP with d=0 and c=2:</p><p>Common difference = d-c = -2</p><p>So: c=2, b=4, a=6, d=0</p><p>Thus f(x) = 6x² + 4x + 2 for x≠0, and f(0)=2</p><p><strong>Step 3: Analyze |f(|x|)|</strong></p><p>f(x) = 6x²+4x+2 = 6(x+1/3)² + 4/3 > 0 for all x</p><p>So |f(|x|)| = f(|x|) = 6x²+4|x|+2</p><p><strong>Step 4: Find non-differentiable points</strong></p><p>g(x) = 6x²+4|x|+2 is non-differentiable where |x| is non-differentiable:</p><p>• At x=0: The derivative from left is -4, from right is +4 (corner point)</p><p>Since 6x²+4|x|+2 is always positive and smooth away from x=0, the only point of non-differentiability is x=0.</p><p><strong>∴ Answer: 1</strong></p>
Correct Answer: 1