Calculus – Limits – JEE Main 02 April 2025 Shift 2

Solution:

Maclaurin series expansion for cosine:
$$\cos(x) = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \dots$$

Expand $\cos(2x)$ and $\cos(4x)$:
$$\cos(2x) = 1 – \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} – \dots = 1 – 2x^2 + \frac{16x^4}{24} – \dots = 1 – 2x^2 + \frac{2x^4}{3} – \dots$$
$$\cos(4x) = 1 – \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} – \dots = 1 – \frac{16x^2}{2} + \frac{256x^4}{24} – \dots = 1 – 8x^2 + \frac{32x^4}{3} – \dots$$

Substitute these expansions into the limit:
$$\lim_{x\rightarrow0} \frac{\left(1 – 2x^2 + \frac{2x^4}{3} – \dots\right) + a\left(1 – 8x^2 + \frac{32x^4}{3} – \dots\right) – b}{x^4}$$

Group the terms by the powers of $x$:
$$\lim_{x\rightarrow0} \frac{(1 + a – b) – x^2(2 + 8a) + x^4\left(\frac{2}{3} + \frac{32a}{3}\right) – \dots}{x^4}$$

For the limit to be finite as $x \to 0$, the coefficients of the powers of $x$ less than $4$ in the numerator must be zero.

$$1 + a – b = 0 \quad \dots (i)$$
$$-(2 + 8a) = 0 \quad \dots (ii)$$

From equation $(ii)$:
$$8a = -2 \Rightarrow a = -\frac{1}{4}$$

Substitute $a$ into equation $(i)$:
$$1 + \left(-\frac{1}{4}\right) – b = 0$$
$$1 – \frac{1}{4} = b \Rightarrow b = \frac{3}{4}$$

Calculate $(a+b)$:
$$a + b = -\frac{1}{4} + \frac{3}{4} = \frac{2}{4}$$
$$a + b = \frac{1}{2}$$

Ans. (1)