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Question

# A function f:R→R satisfies the equation f(x+y)=f(x)f(y) for all x,y∈R,f(x)≠0. Suppose that the function is differentiable at x=0 and f′(0)=2. Then,

A
f(x)=f(x)
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B
f(x)=2f(x)
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C
2f(x)=f(x)
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D
None of these
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Solution

## The correct option is B f′(x)=2f(x)Given,f(x+y)=f(x).f(y)∀x,y∈R.......(1).Then f(0+0)=f(0).f(0)or, f(0)=1 [ Since f(x)≠0∀x∈R].......(2).Also given,f′(0)=2or, limh→0f(0+h)−f(0)h=2or, limh→0f(0)f(h)−f(0)h=2 [ Using (1)]or, limh→0f(h)−1h=2........(3) [Using (2)].Now,f′(x)=limh→0f(x+h)−f(x)hor, f′(x)=limh→0f(x).f(h)−f(x)h [ Using (1)]or, f′(x)=f(x)limh→0f(h)−1hor, f′(x)=2f(x). [ Using (3)]

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