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Question:

If f:R→R is a differentiable function such that f'(x) > 2f(x) for all x∈R, and f(0)=1, then

f(x)>e2x in (0,∞)

f(x) is decreasing in (0,∞)

f'(x)<e2x in (0,∞)

f(x) is increasing in (0,∞)

Solution:

The correct options are C: f(x) is increasing in (0,∞) and D: f(x) > e2x in (0,∞)

Given: f:R→R is a differential function such that f'(x) > 2f(x)

Now, f'(x) > 2f(x) ⇒ f'(x)/f(x) > 2

Integrating both sides, we get
∫f'(x)/f(x) dx > ∫2 dx
⇒ log(f(x)) + c > 2x
⇒ f(x) + C > e2x

Since, f(0) = 1 ⇒ f(0) + C > e0 ⇒ C > 0
⇒ f(x) > e2x
∴ D is correct

Since, e2x increases in the interval (0,∞) ⇒ f(x) is increasing in (0,∞).
Hence, C and D are correct.