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If \mathrm{g(x)} is a differentiable real valued function satisfying \mathrm{g^{\prime \prime}(x)-3 g^{\prime}(x)>3 \forall x \geq 0} and \mathrm{g^{\prime}(0)=-1} then \mathrm{g(x)+x} for \mathrm{x>0} is

Option: 1

increasing function of x
 


Option: 2

decreasing function of x
 


Option: 3

data insufficient


Option: 4

none of these


Answers (1)

best_answer

\mathrm{\begin{aligned} & \frac{d}{d x}\left(g^{\prime}(x) e^{-3 x}\right)>3 e^{-3 x} \\\\ & \Rightarrow \frac{d}{d x}\left(g^{\prime}(x) e^{-3 x}+e^{-3 x}\right)>0 \end{aligned}}

\mathrm{\Rightarrow\left(g^{\prime}(x)+1\right) e^{-3 x} \text { is increasing function }}

\mathrm{\text { Now }\left(g^{\prime}(x)+1\right) e^{-3 x}>\left(g^{\prime}(0)+1\right) \quad \forall x>0}

\mathrm{\Rightarrow\left(g^{\prime}(x)+1\right)>0}

\mathrm{\Rightarrow \mathrm{g}(\mathrm{x})+\mathrm{x} \text { is an increasing function }}

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Rishi

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