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

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q17 – JEE-MAIN 2025

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Home Maths Limits and DerivativesQuestion #274

lim_{x \to 0} \frac{\cos (2 x) + a \cos (4 x) - b}{x^{4}}

is finite, then

(a + b)

is equal to :

Clue 1 of 3: Indeterminate Form

For a limit of the form $lim_{x \to 0} \frac{f (x)}{g (x)}$ to be finite when $g (x) \to 0$ , the numerator $f (x)$ must also tend to $0$ .

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Step 1: Apply condition for finite limit✦ Active

For the limit $lim_{x \to 0} \frac{\cos (2 x) + a \cos (4 x) - b}{x^{4}}$ to be finite, the numerator must approach zero as $x \to 0$ . Evaluating the numerator at $x = 0$ :

\cos (0) + a \cos (0) - b = 1 + a - b

Setting this to zero gives the relation between $a$ and $b$ :

1 + a - b = 0 ⟹ b = a + 1

Step 2: Use Taylor series expansion for the numerator○ Expand

Substitute $b = a + 1$ into the numerator and expand $\cos (2 x)$ and $\cos (4 x)$ using the Maclaurin series $\cos (u) = 1 - \frac{u^{2}}{2!} + \frac{u^{4}}{4!} + O (u^{6})$ :

\cos (2 x) = 1 - \frac{(2 x)^{2}}{2!} + \frac{(2 x)^{4}}{4!} + O (x^{6}) = 1 - 2 x^{2} + \frac{2}{3} x^{4} + O (x^{6})

\cos (4 x) = 1 - \frac{(4 x)^{2}}{2!} + \frac{(4 x)^{4}}{4!} + O (x^{6}) = 1 - 8 x^{2} + \frac{32}{3} x^{4} + O (x^{6})

Now, substitute these into the numerator $N (x) = \cos (2 x) + a \cos (4 x) - (a + 1)$ :

N (x) = (1 - 2 x^{2} + \frac{2}{3} x^{4}) + a (1 - 8 x^{2} + \frac{32}{3} x^{4}) - (a + 1) + O (x^{6})

N (x) = (1 + a - (a + 1)) + (- 2 - 8 a) x^{2} + (\frac{2}{3} + \frac{32 a}{3}) x^{4} + O (x^{6})

N (x) = 0 \cdot x^{0} + (- 2 - 8 a) x^{2} + (\frac{2}{3} + \frac{32 a}{3}) x^{4} + O (x^{6})

💡 Teacher's Secret Hint

Ensure to expand up to the $x^{4}$ term, as the denominator is $x^{4}$ .

Step 3: Determine 'a' and calculate (a+b)○ Expand

For the limit $lim_{x \to 0} \frac{N (x)}{x^{4}}$ to be finite, the coefficient of $x^{2}$ in $N (x)$ must be zero. This is because if the $x^{2}$ term were non-zero, the limit would be infinite.

- 2 - 8 a = 0 ⟹ 8 a = - 2 ⟹ a = - \frac{1}{4}

Now, calculate $b$ using the relation from Step 1:

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

Finally, calculate the required value $(a + b)$ :

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

💡 Teacher's Secret Hint

The limit value itself would be the coefficient of $x^{4}$ if the $x^{2}$ term is zero.

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Maths Question 17 – JEE-MAIN 2025

Clue 1 of 3: Indeterminate Form

🧠 Full Solution Path