Limits, Continuity and Differentiability – Limits – JEE Main 22 Jan 2026 Shift 2

Solution:

We use the standard series expansions for the functions involved:
$$ e^x = 1 + x + \frac{x^2}{2!} + \dots $$
$$ \cos x = 1 – \frac{x^2}{2!} + \dots $$
$$ \log(1+x) = x – \frac{x^2}{2} + \frac{x^3}{3} – \dots $$

**Denominator Expansion:**
$$ D = x\left(1 – \frac{x^2}{2} + \dots\right) – \left(x – \frac{x^2}{2} + \frac{x^3}{3} – \dots\right) $$
$$ D = x – \frac{x^3}{2} – x + \frac{x^2}{2} – \frac{x^3}{3} \approx \frac{x^2}{2} $$

**Numerator Expansion:**
$$ N = \left[1 + (a-1)x + \frac{(a-1)^2 x^2}{2}\right] + 2\left[1 – \frac{b^2 x^2}{2}\right] + (c-2)\left[1 – x + \frac{x^2}{2}\right] $$
Grouping terms by powers of $x$:
Constant term: $1 + 2 + c – 2 = c + 1$
Coefficient of $x$: $(a-1) – (c-2) = a – c + 1$
Coefficient of $x^2$: $\frac{(a-1)^2}{2} – b^2 + \frac{c-2}{2}$

For the limit to exist and be finite, the constant and $x$ terms in the numerator must be zero.
1. $c + 1 = 0 \Rightarrow c = -1$
2. $a – c + 1 = 0 \Rightarrow a – (-1) + 1 = 0 \Rightarrow a + 2 = 0 \Rightarrow a = -2$

Now, equate the ratio of $x^2$ coefficients to the limit value 2:
$$ \frac{\text{Coeff of } x^2 \text{ in } N}{\text{Coeff of } x^2 \text{ in } D} = 2 $$
$$ \frac{\frac{(a-1)^2}{2} – b^2 + \frac{c-2}{2}}{1/2} = 2 $$

Substitute $a = -2$ and $c = -1$:
$$ \frac{\frac{(-3)^2}{2} – b^2 + \frac{-3}{2}}{1/2} = 2 $$
$$ \frac{\frac{9}{2} – b^2 – \frac{3}{2}}{1/2} = 2 $$
$$ \frac{3 – b^2}{1/2} = 2 $$
$$ 2(3 – b^2) = 2 \Rightarrow 3 – b^2 = 1 \Rightarrow b^2 = 2 $$

Finally, calculate $a^2 + b^2 + c^2$:
$$ a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 $$
$$ = 4 + 2 + 1 = 7 $$

Ans. (3)