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Question

Let f(x+y)=f(x)f(y)x,yϵR and f(x)=1+xg(x) where limx0g(x)=1. Then:

A
f(x) is discontinuous at x=0
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B
f’(0) = 1
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C
f(x)=f(x)x
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D
f(x)=f"(x)x
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Solution

The correct option is C f(x)=f(x)x
f(x+y)f(x)f(y)f(0)=f(0)2
f(x)=1+xg(x)f(0)=1f(0)=0or1(1)
Let f(0)=1.At x=a(a is arbitary)
f(a)=limh0f(a+h)f(a)h
=limh0f(a)f(h)f(a)h
=f(a)limh0f(h)1h
=f(a)limh0f(h)1h
=f(a)limh01+hg(h)1h
=f(a)limh0=g(h)
=f(a)×1=f(a).
Hence f(a)=f(a)af(x)=f(x)x.

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