Application of Derevative($F'(0)=g''(0)-3T'(0)>0$)

Question

If $g(x)$ is a differentiable real valued function satisfying $g′′(x) – 3g′(x) > 3$ $∀ x \ge 0$ and $g′(0) = –1$ then $g(x) + x$ for $x > 0$ is
(A) increasing function of x

(B) decreasing function of x

(C) data insufficient

(D) none of these

My approach is as follow $F'(x) =g''(x)-3(g'(x)+1)>0$ and $T(x)=g(x)+x$

$T'(x)=g'(x)+1$

$T'(0)=g'(0)+1=0$

$F'(x)=g''(x)-3T'(x)>0$

$F'(0)=g''(0)-3T'(0)>0$

$F'(0)=g''(0)>0$

Regarding $g(x)$ I am not able to know its nature hence I cannot proceed from here

Answer 1

Writing the given differential equation in terms of $T(x)$, we get, $$T''(x)-3T'(x)>0\forall x\in \Bbb R\cup \{0\}$$ $$\implies \frac{d}{dx}(e^{-3x}T'(x))>0\forall x\in \Bbb R\cup \{0\}$$ Hence, $e^{-3x}T'(x)$ is increasing on positive reals. It takes value $0$ at $0$, hence, $$T'(x)>0\forall x>0$$ So option $(A)$ is correct.