If lim _x → 0 cos (2 x)+a cos (4 x)-b x^4 is finite, then (a+b) is equal to :

$\displaystyle \lim _{x \rightarrow 0} \frac{\cos 2 x+a \cos 4 x-b}{x^4}=$x→0lim​x4cos2x+acos4x−b​= finite

$$\begin{aligned} & L=\frac{\left\{1-\frac{(2 x)^2}{\underline{2}}+\frac{(2 x)^4}{\underline{4} \cdots\}\}+a\left\{1-\frac{(4 x)^2}{\underline{2}}+\frac{(4 x)^4}{\lfloor 4} \ldots \cdot\right\}-b}\right.}{x^4} \\ & L=\frac{(1+a-b)-x^2(2+8 a)+x^4\left(\frac{2}{3}+\frac{32}{3} a\right)+x^6() \ldots}{x^4}\end{aligned}$$​L=x4{1−2​(2x)2​+4​⋯}}+a{1−2​(4x)2​+⌊4(4x)4​…⋅}−b(2x)4​​L=x4(1+a−b)−x2(2+8a)+x4(32​+332​a)+x6()…​​

$$\begin{aligned} & \therefore 1+\mathrm{a}-\mathrm{b}=0 \text { and } 2+8 \mathrm{a}=0 \Rightarrow \mathrm{a}=-\frac{1}{4} \\ & \mathrm{~b}=\mathrm{a}+1 \\ & =-\frac{1}{4}+1=\frac{3}{4} \\ & \therefore \mathrm{a}+\mathrm{b}=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}\end{aligned}$$​∴1+a−b=0 and 2+8a=0⇒a=−41​ b=a+1=−41​+1=43​∴a+b=−41​+43​=21​​