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If f(x)={x^(2{e^(1/x)},)x != 0 k, x=0 is...

If `f(x)={x^(2{e^(1/x)},)x != 0 k, x=0` is continuous at `x=0,` where {*} represents fractional part function, then

A

f(x) is differentiable at x=0.

B

k=1

C

f(x) is continuous but not differentiable at x=0

D

f(x) is continuous every where in its domain.

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(xto0)Limf(x)=0(becauseunderset(xto0)Limx^(2)=0and{e^(1//x)}" is a bounded function")`
`:." "k=0`
Now, `f'(0)=underset(xto0)Lim(f(0+h)-f(0))/(h)=underset(xto0)Limh{e^(1//h)}=0rArrf'(0)=0`
Note that f(x) is discontinuous at `x=pm(1)/(ln2),pm(1)/(ln3)and` so on.
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