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  • If <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{\cos 4 x+a \cos 2 x+b}{x^4}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%200%7D%20%5Cfrac%7B%

If \mathrm{\lim _{x \rightarrow 0} \frac{\cos 4 x+a \cos 2 x+b}{x^4}} is finite, then \mathrm{(a, b)=}
 

Option: 1

(5,-4)
 


Option: 2

(-5,-4)
 


Option: 3

(-4,3)
 


Option: 4

(4,5)


Answers (1)

best_answer

It is given that \mathrm{\lim _{x \rightarrow 0} \frac{\cos 4 x+a \cos 2 x+b}{x^4}} is finite.
Therefore, \mathrm{\lim _{x \rightarrow 0} \frac{\cos 4 x+a \cos 2 x+b}{x^4}} should be of the form \mathrm{\frac{0}{0}} at \mathrm{x=0.}Consequently, the value \mathrm{\cos 4 x+a \cos 2 x+b} must be zero at \mathrm{ x=0}. i.e., \mathrm{1+a+b=0} Using L'Hospital's rule, we have

\mathrm{ \lim _{x \rightarrow 0} \frac{-4 \sin 4 x-2 a \sin 2 x}{4 x^3} \quad\left(\frac{0}{0} \text { form }\right) }

\mathrm{ =\lim _{x \rightarrow 0} \frac{-16 \cos 4 x-4 a \cos 2 x}{12 x^2} \text { [Using L' Hospital's rule] } }

This should be of the form \mathrm{ \frac{0}{0}}

\mathrm{ \therefore-16-4 a=0 }

Solving (i) and (ii), we get \mathrm{ a=-4 \: and \: b=3 }

Hence option 3 is correct.

Posted by

Pankaj Sanodiya

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