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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

A
f(x) is increasing in(0,)
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B
f(x) is decreasing in(0,)
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C
f(x)>e2x in (0,)
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D
f(x)<e2x in (0,)
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Solution

## The correct options are A f(x) is increasing in(0,∞) C f(x)>e2x in (0,∞) f′(x)−2f(x)>0 ∀ x∈R ⋯(1) Let y=f(x) Then dydx−2y>0 Integrating function I.F.=e∫−2dx=e−2x Multiplying by e−2x, we get e−2xdydx−2ye−2x>0 ⇒ddx(y.e−2x)=ddx(f(x).e−2x)>0 Suppose g(x)=f(x).e−2x Then g′(x)>0 ∀ x∈R [From (1)] ∴g(x)=f(x).e−2x is strictly increasing for all x∈R Also, g(0)=1 ⇒g(x)>g(0)=1 for all x>0 ⇒f(x).e−2x>1 ∀ x∈(0,∞) Hence f(x)>e2x ∀ x∈(0,∞) As f′(x)>2f(x)⇒f′(x)>2e2x>2 ∀ x∈(0,∞) ∴f(x) is strictly increasing in (0,∞)

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