JEE Main 2024 — 5 April, Shift 1 — Question 8

$f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^{3}}$

is continuous at $x=0$

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

$\lim _{x \rightarrow 0} \frac{-\beta+x(3+\alpha)+\frac{9 \beta x^{2}}{\underline{2}}+\left(\frac{-27}{\underline{3}}-\frac{\alpha}{\underline{3}}\right) x^{3} \ldots}{x^{3}}=f(0)$

for exist

$\beta=0,3+\alpha=0,-\frac{27}{\lfloor 3}-\frac{\alpha}{\lfloor 3}=\mathrm{f}(0)$

$\alpha=-3,-\frac{27}{6}-\frac{(-3)}{6}=f(0)$

$f(0)=\frac{-27+3}{6}=-4$