If x 0 (2x) + a (4x) - bx^4 is finite, then (a + b) is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2025Limits

If $\lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4}$ is finite, then $\left(a + b\right)$ is equal to:

Answer

Correct answer:A

Standard Method

Given: $\lim_{x \to 0} \frac{\cos(2x) + a\cos(4x) - b}{x^4}$ is finite.

Find: $a+b$ .

Use the Taylor series expansions about $x=0$ :

\cos(2x)=1-\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}+O(x^6)

\cos(4x)=1-\frac{(4x)^2}{2!}+\frac{(4x)^4}{4!}+O(x^6)

Substituting into the numerator,

\cos(2x)+a\cos(4x)-b =\left(1+a-b\right)-\left(\frac{(2x)^2}{2!}+a\frac{(4x)^2}{2!}\right)+\left(\frac{(2x)^4}{4!}+a\frac{(4x)^4}{4!}\right)+O(x^6)

For the limit after division by $x^4$ to remain finite, the constant term and the $x^2$ term in the numerator must both vanish. Hence,

1+a-b=0

and

\frac{(2x)^2}{2!}+a\frac{(4x)^2}{2!}=0

which gives

2+8a=0

a=-\frac{1}{4}

Then from $1+a-b=0$ ,

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

Therefore,

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

So, the correct option is A.

Setting only the constant term to zero is incomplete. Since the denominator is $x^4$ , the $x^2$ term in the numerator must also vanish; otherwise the quotient behaves like $\frac{1}{x^2}$ and the limit is not finite.
Using the cosine expansion incorrectly, especially the sign of the $x^2$ term, leads to a wrong equation for $a$ . Remember that $\cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} + \cdots$ .
Substituting into $1+a-b=0$ incorrectly after finding $a$ can give the wrong value of $b$ . Rearrange carefully: $b=1+a$ .

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