Let α, β ∈ R be such that _ x → 0 x ^ 2 ( ) - sin x = 1 . Then 6(α...

If $\alpha \ne 1,$ then $\lim_{x \to 0}\dfrac{xsin\left(\betax\right)}{\left(\dfrac{\alphax - \sin x}{x}\right)} = 0$
$∴$ $\alpha = 1$
$⇒$ $\lim_{x \to 0}\dfrac{\beta{x}^{3}\left(\dfrac{\sin \left(\betax\right)}{\betax}\right)}{{x}^{3}\left(\dfrac{x - \sin x}{{x}^{3}}\right)} = \dfrac{\dfrac{\beta}{1}}{6}$

Using,$\lim_{x \to 0}\dfrac{x - \sin x}{{x}^{3}} = \dfrac{1}{6} \text{ and } \lim_{x \to 0}\dfrac{\sin x}{x} = 1$
$⇒ 6\beta = 1 ⇒ \beta = \dfrac{1}{6}$
$⇒ 6\left(\alpha + \beta\right) = 7$