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Let `x _(1) , x _(2), x _(3)` be the points where `f (x) = | 1-|x-4||, x in R` is not differentiable then `f (x_(1))+ f(x _(2)) + f (x _(3))=`

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The correct Answer is:
1

Function is not differentiable at x=3,4,5
So, f(3)+f(4)+f(5)
0+1+0=1
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