1.If limx→03+αsinx+βcosx+loge(1−x)3tan2x=13\displaystyle \lim_{x \to 0} \frac{3 + \alpha \sin x + \beta \cos x + \log_e(1-x)}{3 \tan^2 x} = \frac{1}{3}x→0lim3tan2x3+αsinx+βcosx+loge(1−x)=31, then 2α−β2\alpha - \beta2α−β is equal toLogin to continueOnly logged in users canattempt or see the solution.
