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Question

The function f(x)=1x2e2x1,x0, is continuous at x=0. Then

A
f(0)=1
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B
f(x) is differentiable at x=0
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C
f(x) is not differentiable at x=0
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D
f(0)=13
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Solution

The correct options are
A f(0)=1
B f(x) is differentiable at x=0
f(0)=limx01x2e2x1
By L hospitals rule twice we get
f(0)=limx04e2x4e2x+x4e2x=1
For f(x) to be differentiable at x=0 limx0f1(x) should exist i.elimx0f(x+h)f(x)h should exist.

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