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The value of the derivative of \mathrm{|x-1|+|x-3|} at x=2 is.

Option: 1

-2


Option: 2

0


Option: 3

2


Option: 4

Not defined.


Answers (1)

best_answer

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

Since f(x) = 2 for \mathrm{1 \leq x<3 }. Therefore f '(x) = 0 for all \mathrm{x \in(1,3) }.

Hence, f '(x)=0 at x=2.

Posted by

himanshu.meshram

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