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

Option: 1

-2


Option: 2

0


Option: 3

0


Option: 4

does not exist


Answers (1)

best_answer

We have,
\mathrm{f(x)=|x-1|+|x-3|=\left\{\begin{array}{rr} -2 x+4, & x<1 \\ 2, & 1 \leq x<3 \\ 2 x-4, & x \geq 3 \end{array}\right.}

Clearly, f(x) being a constant function between 1 and 3 , is differentiable at x=2 such that \mathrm{f^{\prime}(2)=0.}

Posted by

himanshu.meshram

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