Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>Consider the function \( f(x) = |x-2| + |x-5|, \, x \in R \).</p><p><strong>Statement-1:</strong> \( f'(4) = 0 \).</p><p><strong>Statement-2:</strong> \( f \) is continuous in \([2, 5]\), differentiable in \((2, 5)\) and \( f(2) = f(5) \).</p>
<p>Statement-1 is true, statement-2 is false.</p>
<p>Statement-1 is true; statement-2 is true; Statement-2 is a correct explanation for statement-1.</p>
<p>Statement-1 is true; statement-2 is true; Statement-2 is not a correct explanation for statement-1.</p>
<p>Statement-1 is false, statement-2 is true.</p>
Step-by-Step Solution
Key Concept: On the interval (2,5), both |x-2| and |x-5| have constant signs, so f(x) becomes a linear function with constant derivative, not zero. Rolle's theorem requires equality of endpoints, but f(2)=3≠f(5)=3 is false—actually f(2)=f(5)=3, so the statements need careful verification of the derivative.
<p><strong>Step 1: Analyze f(x) = |x-2| + |x-5| by cases</strong></p><p>For x ∈ (2,5): both (x-2)>0 and (x-5)<0</p><p>So f(x) = (x-2) + (5-x) = 3 (constant function)</p><p><strong>Step 2: Find f'(4)</strong></p><p>Since f(x) = 3 on the interval containing x=4, we have f'(4) = 0 ✓</p><p>Statement-1 is <strong>TRUE</strong></p><p><strong>Step 3: Verify Statement-2</strong></p><p>• Continuity on [2,5]: f is continuous (sum of continuous functions) ✓</p><p>• Differentiability on (2,5): f(x)=3 is differentiable with f'(x)=0 ✓</p><p>• f(2) = |2-2| + |2-5| = 0 + 3 = 3</p><p>• f(5) = |5-2| + |5-5| = 3 + 0 = 3</p><p>So f(2) = f(5) = 3 ✓</p><p>Statement-2 is <strong>TRUE</strong></p><p><strong>Step 4: Check logical relationship</strong></p><p>Statement-2 is actually the reason Statement-1 is true (Rolle's theorem applies: all conditions met). However, Statement-2 does NOT directly explain f'(4)=0 as its primary consequence—rather, Rolle's theorem guarantees existence of such a point.</p><p>∴ Answer: <strong>B</strong> (Both statements true, but Statement-2 is not correct explanation of Statement-1, OR both true but independent)</p>
Correct Answer: B