Sejam $\arctan(\large{\frac{1+x}{2}})$ $= \alpha$ e $\arctan(\large{\frac{1-x}{2}})$ $= \beta$ , do enunciado, temos que: $$\alpha + \beta \geq \frac{\pi}{6} \implies \tan{(\alpha + \beta)} \geq \tan{\frac{\pi}{6}} = \frac{1}{\sqrt{3}}$$ $\tan{(\alpha + \beta)} = \Large{\frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha)\tan(\beta)}}$ $=$ $\Large{\frac{\frac{1+x}{2} + \frac{1-x}{2}}{1+\frac{x^2-1}{4}}}$ $=$ $\Large{\frac{4}{3+x^2}}$ , portanto: $\Large{\frac{4}{3+x^2}}$ $\geq$ $\Large{\frac{1}{\sqrt{3}}}$ $\implies$ $x^2 \leq 4\sqrt{3} - 3$ $\implies$ $-\sqrt{4\sqrt{3} - 3} \leq x \leq \sqrt{4\sqrt{3} - 3}$ Como $4\sqrt{3} = \sqrt{48} \cong 7 $ (aproximadamente menor), então: $\sqrt{4\sqrt{3} - 3}$ $\cong$ $2$ (aproximadamente menor que $2$ ). Conclui-se que $x$ com certeza está contido no intervalo $\color{red}{[-2,3]}$ , alternativa $\mathbb{(C)}$