Question 9. If \lim_{x \to 0} \frac{3+\alpha \sin x+\beta \cos x+\ln(1-x)}{3 \tan^{2} x}=\frac{1}{3} then 2\alpha-\beta is equal to :
(1) 2
(2) 7
(3) 5
(4) 1
Answer (3)
Explanation:
\lim {x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log {e}(1-x)}{3 \tan ^{2} x}=\frac{1}{3} \Rightarrow \lim {x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^{3}}{3!}+\ldots\right]+\beta\left[1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!} \ldots\right]+\left(-x-\frac{x^{2}}{2}-\frac{x^{3}}{3} \ldots\right)}{3 \tan ^{2} x}=\frac{1}{3} \Rightarrow \lim {x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^{2}+\ldots .}{3 x^{2}} \times \frac{x^{2}}{\tan ^{2} x}=\frac{1}{3}\Rightarrow \beta+3=0, \alpha-1=0 and \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3}
\Rightarrow \beta=-3, \alpha=1 \Rightarrow 2 \alpha-\beta=2+3=5