The value of the derivative of $| x - 1 | + | x - 3 |$ at $x = 2$ is:

2

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0

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Solution

The correct option is B $0$
We know that $| x | = x$ for all $x \geq 0$ and $| x | = - x$ for all $x < 0$ . Therefore,
At $x = 2$ ,
$| x - 1 | = x - 1$ and $| x - 3 | = - (x - 3) = - x + 3$
$\Rightarrow f (x) = (x - 1) + (- x + 3) = 2$
which is a constant function and the derivative of a constant function is always zero. So at $x = 2$ derivative of $f (x)$ is zero.

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