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← JEE Advanced 2024 · Paper 2
MathematicsMultiple correct+4 / -2Question 5 of 51

JEE Advanced 2024 · Paper 2, Mathematics question 5

Let SS be the set of all (α,β)∈R×R(\alpha, \beta) \in \mathbb{R}\times\mathbb{R} such that

lim⁡x→∞sin⁡(x2)(log⁡ex)αsin⁡(1x2)xαβ(log⁡e(1+x))β=0.\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?

Choose one or more options
Answer and worked solution

Official answer: B, C

As x→∞x\to\infty, sin⁡1x2∼1x2\sin\frac{1}{x^2}\sim\frac1{x^2} and ln⁡(1+x)∼ln⁡x\ln(1+x)\sim\ln x. So the expression behaves like
sin⁡(x2) (ln⁡x)α−β x−2−αβ.\sin(x^2)\,(\ln x)^{\alpha-\beta}\,x^{-2-\alpha\beta}.

Since sin⁡(x2)\sin(x^2) is bounded and does not tend to a limit, the limit is 00 exactly when (ln⁡x)α−βx−2−αβ→0(\ln x)^{\alpha-\beta}x^{-2-\alpha\beta}\to0. That holds when:

  • αβ>−2\alpha\beta>-2, or

  • αβ=−2\alpha\beta=-2 and α<β\alpha<\beta.

  • (A) (−1,3)(-1,3): αβ=−3<−2\alpha\beta=-3<-2, so it diverges. ✗

  • (B) (−1,1)(-1,1): αβ=−1>−2\alpha\beta=-1>-2. ✓

  • (C) (1,−1)(1,-1): αβ=−1>−2\alpha\beta=-1>-2. ✓

  • (D) (1,−2)(1,-2): αβ=−2\alpha\beta=-2 but α−β=3>0\alpha-\beta=3>0, giving sin⁡(x2)(ln⁡x)3\sin(x^2)(\ln x)^3 with no limit. ✗

Topics:Limits
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