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Let f(x) = ((1 - x(1+ |1-x | )) /(|1-x|)...

Let `f(x) = ((1 - x(1+ |1-x | )) /(|1-x|)) cos(1/(1-x))` for `x!=1`

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The correct Answer is:
C, D
`f(1^(+))=underset(hto0)lim(1-(1+h)(1+h))/(h)"cos"(1)/(h)`
`=underset(hto0)lim(-h^(2)-2h)/(h)"cos"(1)/(h)=underset(hto0)lim(-h-2)"cos"(1)/(h)`
Thus, `underset(xto1^(+))limf(x)` does nto exist.
`f(1^(-))=underset(hto0)lim(1-(1-h))/(h)"cos"(1)/(h)`
`=underset(hto0)lim(-1(1-h^(2)))/(h)"cos"(1)/(h)``=underset(hto0)lim(h^(2))/(h)"cos"(1)/(h)=underset(hto0)lim h"cos"(1)/(h)=0`
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