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Question

Let f:RR be a differentiable function satisfying f(3)+f(2)=0. Then limx0(1+f(3+x)f(3)1+f(2x)f(2))1/x is equal to :

A
e
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B
1
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C
e1
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D
e2
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Solution

The correct option is B 1
L=limx0(1+f(3+x)f(3)1+f(2x)f(2))1/x [1 form]

L=exp(limx01x[1+f(3+x)f(3)1+f(2x)f(2)1])

=exp(limx01x[f(3+x)f(3)f(2x)+f(2)1+f(2x)f(2)])

=exp(limx0f(3+x)f(2x)+f(2)f(3)x)

Applying L'Hopital rule, we get
L=exp(limx0f(3+x)+f(2x)1)

=exp(f(3)+f(2))

=exp(0)=1

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