If f: R → R is a differentiable function such that f ^ ′ (x) > 2f (...

Given that,

${f}^{'}\left(x\right) > 2f\left(x\right) \forall x \in R$

$⇒ {f}^{'}\left(x\right) - 2f\left(x\right) > 0 \forall x \in R$

$∴ {e}^{- 2x}\left({f}^{'}\left(x\right) - 2f\left(x\right)\right) > 0 \forall x \in R$

$⇒ \dfrac{d}{dx}\left({e}^{- 2x}f\left(x\right)\right) > 0 \forall x \in R$

Let $g\left(x\right) = {e}^{- 2x}f\left(x\right)$

Now, ${g}^{'}\left(x\right) > 0 \forall x \in R$

$⇒ g\left(x\right)$ is strictly increasing $\forall x \in R$

Also $g\left(0\right) = 1$

$⇒ g\left(x\right) > g\left(0\right) = 1$

$∴ {e}^{- 2x} . f\left(x\right) > 1 \forall x \in \left(0, \infty\right) ⇒ f\left(x\right) > {e}^{2x} \forall x \in \left(0, \infty\right)$

As, ${f}^{'}\left(x\right) > 2f\left(x\right) > 2{e}^{2x} > 2 \forall x \in \left(0, \infty\right)$

$⇒ f\left(x\right)$ is strictly increasing on $x \in \left(0, \infty\right)$