Question

If $\underset{x\to 0}{lim}\frac{cos4x+acos2x+b}{x^4}$ is finite, then $(a,b)=$

• A. $(5,-4)$
• B. $(-5,-4)$
• C. $(-4,3)$
• D. $(4,5)$

Answer 1

Answer: (C)

$(-4,3)$

It is given that $\underset{x\to 0}{lim}\frac{cos4x+acos2x+b}{x^4}$ is finite.
Therefore, $\frac{cos4x+acos2x+b}{x^4}$ should be of the form $\frac{0}{0}$ at $x=0$.

So, the numerator also tends to 0 as $x\to 0$
${lim}_{x\to 0}cos4x+acos2x+b=0$
i.e., $1+a+b=0$ ....... (i)
Using L'Hospital's rule the given limit is
$\underset{x\to 0}{lim}\frac{-4sin4x-2asin2x}{4x^3}$
It is of the form $\left(\frac{0}{0}form\right)$
$\underset{x\to 0}{lim}\frac{-16cos4x-4acos2x}{12x^2}$ [Using L' Hospital's rule]
This should be of the form $\frac{0}{0}$.
$\therefore -16-4a=0$ ......... (ii)
Solving (i) and (ii), we get
$a=-4$ and $b=3$.