$$If\,\mathop {\lim }\limits_{x \to 0} {{\cos (2x) + a\cos (4x) - b} \over {{x^4}}}is\,finite,\,then\,(a + b)\,is\,equal\,to:$$

<p>To find the value of $a + b$ for which the following limit is finite:</p> <p>$ \lim\limits_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} $</p> <p>we start by expanding the cosine functions using their Taylor series:</p> <p>$ \cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \ldots $</p> <p>$ \cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \ldots $</p> <p>Substitute these expansions into the limit expression:</p> <p>$ \lim\limits_{x \to 0} \frac{\left(1 - \frac{4x^2}{2!} + \frac{(2x)^4}{4!} - \ldots \right) + a\left(1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \ldots\right) - b}{x^4} $</p> <p>To ensure that the limit is finite, the linear and quadratic terms must sum to zero. Let's equate the constant terms first:</p> <p><p>For the limit to be finite: </p> <p>$ 1 + a - b = 0 $</p></p> <p><p>Equating the coefficients of $x^2$:</p> <p>$ -2 - 8a = 0 $</p></p> <p>Solving the second equation, we have:</p> <p>$ 8a = -2 \\ a = \frac{-1}{4} $</p> <p>Substitute $a$ back into the first equation to find $b$:</p> <p>$ 1 + \frac{-1}{4} - b = 0 \\ b = 1 - \frac{1}{4} \\ b = \frac{3}{4} $</p> <p>Thus, the sum of $a$ and $b$ is:</p> <p>$ a + b = \frac{-1}{4} + \frac{3}{4} = \frac{1}{2} $</p> <p>Therefore, the correct value of $a + b$ is $\frac{1}{2}$.</p>