JEE Advanced 2024 Paper 2 Q05 Mathematics Differentiation & Applications Limits & Continuity Hard

JEE Advanced 2024 Paper 2 · Q05 · Limits & Continuity

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $$\lim_{x \to \infty} \frac{\sin(x^{2})\,(\log_{e} x)^{\alpha}\,\sin\!\left(\dfrac{1}{x^{2}}\right)}{x^{\alpha\beta}\,(\log_{e}(1+x))^{\beta}} = 0.$$ Then which of the following is (are) correct?

  1. A. $(-1, 3) \in S$
  2. B. $(-1, 1) \in S$
  3. C. $(1, -1) \in S$
  4. D. $(1, -2) \in S$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C

Solution

As $x \to \infty$, $\sin(1/x^{2}) \sim 1/x^{2}$, and $\log(1+x) \sim \log x$. The expression behaves like $\dfrac{\sin(x^{2})\,(\log x)^{\alpha-\beta}}{x^{\alpha\beta+2}}$. With $|\sin(x^{2})| \le 1$, the limit is $0$ iff the magnitude $\to 0$, which (for any $\alpha-\beta$) holds when $\alpha\beta + 2 > 0$, i.e. $\alpha\beta > -2$. Check each: (A) $(-1,3)$: $\alpha\beta = -3 \not> -2$ — FAILS. (B) $(-1,1)$: $\alpha\beta = -1 > -2$ — OK. (C) $(1,-1)$: $\alpha\beta = -1 > -2$ — OK. (D) $(1,-2)$: $\alpha\beta = -2$, not strictly greater; the $(\log x)^{\alpha-\beta} = (\log x)^{3}$ blows up, so limit is not $0$ — FAILS. Correct: (B), (C).

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