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. infinite

Answer 1

Answer: (A)

2

Answer 2

The roots of the equation $x^2 - 3x + 2 = 0$ are found by factoring: $(x-1)(x-2) = 0$. The roots are $x=1$ and $x=2$. Let $\alpha$ and $\beta$ be the roots. Without loss of generality, let $\alpha = 1$ and $\beta = 2$.

The function is given by $f(x) = |x - \alpha| + |x - \beta| = |x - 1| + |x - 2|$. We need to find the number of points in the interval $[\alpha, \beta] = [1, 2]$ at which $f$ is not differentiable.

The function $|x-c|$ is not differentiable at $x=c$. The function $f(x) = |x-1| + |x-2|$ is a sum of two absolute value functions. The points where the terms inside the absolute value become zero are $x=1$ and $x=2$. These are the potential points of non-differentiability.

Let's define $f(x)$ piecewise based on the intervals determined by $x=1$ and $x=2$:

1. For $x < 1$: $x-1 < 0$ and $x-2 < 0$.
   $f(x) = -(x-1) - (x-2) = -x + 1 - x + 2 = -2x + 3$.
2. For $1 \le x < 2$: $x-1 \ge 0$ and $x-2 < 0$.
   $f(x) = (x-1) - (x-2) = x - 1 - x + 2 = 1$.
3. For $x \ge 2$: $x-1 \ge 0$ and $x-2 \ge 0$.
   $f(x) = (x-1) + (x-2) = x - 1 + x - 2 = 2x - 3$.

The function is defined as: $f(x) = \begin{cases} -2x + 3 & \text{if } x < 1 \\ 1 & \text{if } 1 \le x < 2 \\ 2x - 3 & \text{if } x \ge 2 \end{cases}$

A function is differentiable at a point if the left-hand derivative (LHD) and the right-hand derivative (RHD) exist at that point and are equal. The derivative of $f(x)$ for $x \ne 1$ and $x \ne 2$ is: $f'(x) = \begin{cases} -2 & \text{if } x < 1 \\ 0 & \text{if } 1 < x < 2 \\ 2 & \text{if } x > 2 \end{cases}$

Now let's check differentiability at $x=1$ and $x=2$.

At $x=1$:

LHD at $x=1$: $f'(1^-) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$. For small $h < 0$, $1+h < 1$.
$f(1+h) = -2(1+h) + 3 = -2 - 2h + 3 = 1 - 2h$.
$f(1) = 1$.
$f'(1^-) = \lim_{h \to 0^-} \frac{(1 - 2h) - 1}{h} = \lim_{h \to 0^-} \frac{-2h}{h} = -2$.

RHD at $x=1$: $f'(1^+) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$. For small $h > 0$, $1 < 1+h < 2$.
$f(1+h) = 1$.
$f(1) = 1$.
$f'(1^+) = \lim_{h \to 0^+} \frac{1 - 1}{h} = \lim_{h \to 0^+} \frac{0}{h} = 0$.

Since $f'(1^-) \ne f'(1^+)$, $f(x)$ is not differentiable at $x = 1$.

At $x=2$:

LHD at $x=2$: $f'(2^-) = \lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h}$. For small $h < 0$, $1 < 2+h < 2$.
$f(2+h) = 1$.
$f(2) = |2-1| + |2-2| = 1 + 0 = 1$.
$f'(2^-) = \lim_{h \to 0^-} \frac{1 - 1}{h} = \lim_{h \to 0^-} \frac{0}{h} = 0$.

RHD at $x=2$: $f'(2^+) = \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$. For small $h > 0$, $2+h > 2$.
$f(2+h) = 2(2+h) - 3 = 4 + 2h - 3 = 1 + 2h$.
$f(2) = 1$.
$f'(2^+) = \lim_{h \to 0^+} \frac{(1 + 2h) - 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.

Since $f'(2^-) \ne f'(2^+)$, $f(x)$ is not differentiable at $x = 2$.

The points where $f$ is not differentiable are $x=1$ and $x=2$. The interval in question is $[\alpha, \beta] = [1, 2]$. Both points $x=1$ and $x=2$ lie within the interval $[1, 2]$. Therefore, there are 2 points in $[\alpha, \beta]$ at which $f$ is not differentiable.