Let f(x) = |x - alpha| + |x - beta|, where al | vedclass

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(B) The given equation is $x^2 - 3x + 2 = 0$. Factoring the quadratic equation, we get $(x - 1)(x - 2) = 0$. Thus, the roots are $\alpha = 1$ and $\be

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(B) The given equation is $x^2 - 3x + 2 = 0$. Factoring the quadratic equation, we get $(x - 1)(x - 2) = 0$. Thus, the roots are $\alpha = 1$ and $\beta = 2$. So, $f(x) = |x - 1| + |x - 2|$. For $x \in [1, 2]$, we have $x - 1 \ge 0$ and $x - 2 \le 0$. Therefore, $f(x) = (x - 1) - (x - 2) = x - 1 - x + 2 = 1$. Since $f(x) = 1$ is a constant function on the interval $[1, 2]$, it is differentiable at every point in the interval $[1, 2]$. Thus, the number of points in $[1, 2]$ at which $f$ is not differentiable is $0$.

Let $f(x) = |x - \alpha| + |x - \beta|$, where $\alpha$ and $\beta$ 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

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