If _ x → 0 3 + α sinx + β cosx + _ e ( 1 - x ) 3 ^ 2 x = (1)/(3) ,...

Given,

$\lim_{x \to 0}\dfrac{3 + \alpha \sin x + \beta \cos x + \log_{e} ( 1 - x )}{3\tan^{2}x} = \dfrac{1}{3}$

$⇒ \lim_{x \to 0}\dfrac{3 + \alpha\left[x - \dfrac{{x}^{3}}{3 !} + \ldots\right] + \beta\left[1 - \dfrac{{x}^{2}}{2 !} + \dfrac{{x}^{4}}{4 !} \ldots\right] + \left(- x - \dfrac{{x}^{2}}{2} - \dfrac{{x}^{3}}{3} \ldots\right)}{3\tan^{2}x} = \dfrac{1}{3}$

$⇒ \lim_{x \to 0}\dfrac{( 3 + \beta ) + ( \alpha - 1 ) x + \left(- \dfrac{1}{2} - \dfrac{\beta}{2}\right){x}^{2} + \ldots .}{3{x}^{2}} \times \dfrac{{x}^{2}}{\tan^{2}x} = \dfrac{1}{3}$

The coefficient of all the powers of $x$ except ${x}^{2}$ is $0$.

$⇒ \beta + 3 = 0, \alpha - 1 = 0$ and $\dfrac{- \dfrac{1}{2} - \dfrac{\beta}{2}}{3} = \dfrac{1}{3}$

$⇒ \beta = - 3, \alpha = 1$

$⇒ 2\alpha - \beta = 2 + 3 = 5$