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Question

The set of all points where the functionf(x)=1-e-x2 is differentiable is:


A

[0,)

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B

(,)

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C

(,){0}

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D

(1,)

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Solution

The correct option is C

(,){0}


Explanation for the correct option:

Step 1: Solve the right-hand limit for f(x)=1-e-x2

For x0,

f'(x)=12×(-e-x2)(-2x)1-e-x2=xe-x21-e-x2

f'(0+)=limh0+f(h)f(0)h=limh0+1-e-h2h

=limh0e-h2-1-h212=1

Step 2: Solve the left-hand limit for f(x)=1-e-x2

f'(0-)=limh0--e-h2-1-h212=1

As h0- is a negative number,

1-e-h2h=1-e-h2h=1-e-h2h2=e-h2-1-h212

f is not differentiable atx=0.

The differentiable points are (,){0}.

Hence, option C is the correct option.


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