JEE Main 2025 — 2 April, Evening Shift — Question 23

$\lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^{4}}$

$$\lim _{x \rightarrow 0} \frac{\left(1-\frac{4 x^{2}}{2!}+\frac{(2 x)^{4}}{4!}-\ldots .\right)+a\left(1-\frac{(2 x)^{2}}{2!}+\frac{(4 x)^{4}}{4!}-\ldots\right)-b}{x^{4}}$$

For limit to be finite $1+a-b=0$

And $-2-8 a=0$

$\Rightarrow 8 a=-2$

$\Rightarrow a=\frac{-1}{4}$

$\Rightarrow b=1+a$

$$=1-\frac{1}{4}$$ $$b=-\frac{3}{4}$$

$a+b=\frac{-1}{4}+\frac{3}{4}=\frac{1}{2}$

Option (1) is correct.