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If f(x)={{:(x,olexle1),(2-x,1lexle2'):} ...

If `f(x)={{:(x,olexle1),(2-x,1lexle2'):}` then Rolle's theorem is not applicable to f(x) because

A

f(x) is not defined everywhere on [0,2]

B

f(x) is not continuous on [0,2]

C

f(x) is not differentiable on (1,2)

D

f(x) is not differentiable on (0,2)

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