Question

Let \( f(x) = |x - \alpha| + |x - \beta| \), where \( \alpha, \beta \) are the roots of the equation \( x^2 - 3x + 2 = 0 \). Then the number of points in \( [\alpha, \beta] \) at which \( f \) is not differentiable is:

• A. \( 2 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( \infty \)

Hint:

The function \( |x - a| \) is not differentiable at \( x = a \). The sum of such functions will be non-differentiable at the zeros of the terms inside the absolute values.

Answer 1

The Correct Option is A

Step 1: Find the roots \( \alpha \) and \( \beta \).
The roots of \( x^2 - 3x + 2 = 0 \) are \( x = 1 \) and \( x = 2 \). Let \( \alpha = 1 \) and \( \beta = 2 \). The interval is \( [1, 2] \).
Step 2: Write the function \( f(x) \).
\[ f(x) = |x - 1| + |x - 2| \]
Step 3: Identify potential points of non-differentiability.
Absolute value functions \( |x - a| \) are not differentiable at \( x = a \). Here, potential points are \( x = 1 \) and \( x = 2 \).
Step 4: Analyze differentiability in \( [1, 2] \).
For \( 1<x<2 \), \( f(x) = (x - 1) + (2 - x) = 1 \), \( f'(x) = 0 \). At \( x = 1 \): Left derivative \(\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = -1\), Right derivative \(\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = 0\). Not differentiable.
At \( x = 2 \): Left derivative \(\lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h} = 0\), Right derivative \(\lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h} = 1\). Not differentiable.
The function \( f(x) \) is not differentiable at \( x = 1 \) and \( x = 2 \), both of which are in the interval \( [1, 2] \). There are 2 such points.