JEE Main 2024 — 27 January, Shift 2 — Question 9

$\lim _{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _{e}(1-x)}{3 \tan ^{2} x}=\frac{1}{3}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^{3}}{3!}+\ldots\right]+\beta\left[1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!} \ldots\right]+\left(-x-\frac{x^{2}}{2}-\frac{x^{3}}{3} \ldots\right)}{3 \tan ^{2} x}=\frac{1}{3}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^{2}+\ldots}{3 x^{2}} \times \frac{x^{2}}{\tan ^{2} x}=\frac{1}{3}$

$\Rightarrow \beta+3=0, \alpha-1=0$ and $\frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3}$

$\Rightarrow \beta=-3, \alpha=1$

$\Rightarrow 2 \alpha-\beta=2+3=5$