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Question

The third term in the Taylor series expansion of ex about x=1 would be

A
ea(xa)
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B
ea2(xa)2
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C
x22
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D
ea6(xa)3
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Solution

The correct option is B ea2(xa)2
f(x)=ex=exa+a=eaexa
=ea[1+(xa)+(xa)22!+...]
Third term=ea(xa)22

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