JEE Advanced 2024 — Paper 2 — Question 5

Note that as $x \to \infty$, $\sin(1/x^2) \sim 1/x^2$. Also $|\sin(x^2)| \le 1$. Thus the limit is equivalent to $\lim_{x\to\infty} \frac{\sin(x^2) (\ln x)^\alpha}{x^{\alpha\beta+2} (\ln(1+x))^\beta}$. Since $|\sin(x^2)|\le 1$, the magnitude is bounded by $\frac{(\ln x)^\alpha}{x^{\alpha\beta+2} (\ln(1+x))^\beta}$. For large $x$, $\ln(1+x) \sim \ln x$, so the bound behaves like $\frac{(\ln x)^{\alpha-\beta}}{x^{\alpha\beta+2}}$. The term $x^{\alpha\beta+2}$ dominates. If $\alpha\beta+2 > 0$, the denominator grows polynomially and the limit is 0. If $\alpha\beta+2 = 0$, the expression is asymptotically $(\ln x)^{\alpha-\beta}$; the limit is 0 only if $\alpha-\beta < 0$. Checking the options: A: $\alpha\beta = -3$ (fails).

B: $\alpha\beta = -1$ (satisfies).

C: $\alpha\beta = -1$ (satisfies).

D: $\alpha\beta = -2$ gives $\alpha\beta+2=0$ and $\alpha-\beta = 3 > 0$, so the limit does not tend to 0. Hence the correct pairs are $(-1,1)$ and $(1,-1)$, i.e., options B and C.