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Question

Let f(x) be continuous and differentiable function satisfying f(x+y)=f(x)f(y) for all x,yϵR.
If f(x) can be expressed as f(x)=1+xp(x)+x2q(x)
where limx0p(x)=a and limx0q(x)=b then f(x) is

A
af(x)
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B
bf(x)
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C
(a+b)f(x)
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D
(a+2b)f(x)
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Solution

The correct option is B af(x)
f(x+y)=f(x)f(y)
Let h0
f(x+h)=f(x)f(h)
f(x)=limh0f(x+h)f(x)h=f(x)f(h)1h=f(x)f(0)
Also
f(x)=1+xp(x)+x2q(x)
f(x)=p(x)+xp(x)+x2q(x)+2xq(x)
limx0f(x)=p(x)+xp(x)+x2q(x)+2xq(x)=p(0)=a
f(0)=a
Thus,
f(x)=f(x)f(0)=af(x)

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