Questions

$f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x})$ $x \neq 1$ . Then

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limx→1− f(x)=0 .

limx→1+ f(x) does not exist

limx→1+ f(x)=0

limx→1- f(x) does not exist

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detailed solution

Correct option is A

We have, f(x)=1−x(1+1−x)1−xcos⁡11−x for x<11−x(1−(1−x))−(1−x)cos⁡11−x for x>1⇒ f(x)=1−x(2−x)1−xcos⁡11−x for x<11−x2−(1−x)cos⁡11−x for x>1⇒ f(x)=(1−x)cos⁡11−x for x<1(−1−x)cos⁡11−x for x>1∴ limx→1− f(x)=limx→1 (1−x)cos⁡11−x= 0×(An oscillating number between -1 and 1) =0limx→1+ f(x)=limx→1 −(1+x)cos⁡11−x=-2 (An oscillating number between -1 and 1)⇒ limx→I+ f(x) does not exist .

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