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A function defined as f:R→Rf:R\to R satisfies f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) and f(x)≠0f(x)\ne 0 for all x,y∈Rx,y\in R, where R is set of real numbers. If f(x)f(x) is differentiable at x=0x=0 and f′(0)=2f'(0)=2, then f′(x)f'(x) is equal to

Solution

✅ Correct Option: 4

From the definition, f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0f(x)f(h)−f(x)hf'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\dfrac{f(x)f(h)-f(x)}{h}.

Since f(0)=1f(0)=1, this equals f(x)lim⁡h→0f(h)−f(0)h=f(x)f′(0)f(x)\lim_{h\to 0}\dfrac{f(h)-f(0)}{h} = f(x)f'(0).

With f′(0)=2f'(0)=2, we get f′(x)=2f(x)f'(x) = 2f(x).

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