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Question

The value of f(0) for f(x)=(1+tan2x)12x,so that f(x) is continuous everywhere, is

A
e
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B
12
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C
e12
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D
0
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Solution

The correct option is B e12
Say y=x.
For the function to be continuous at x=0, f(0)=limx0f(x).
limy0(1+tan2y)12y2=limy0e12y2log(1+tan2y)=exp(limy012y2log(1+tan2y))=exp(limy0tan2y2y2log(1+tan2y)tan2y)=exp(121)
(Using limy0tanyy=1,limy0log(1+y)y=1)
This gives the limit, and hence the value of the function at x=0 as e12.

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