Let $$S$$ be the set of all $$(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$$ such that

$$\lim_{x \to \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin\left(\frac{1}{x^2}\right)}{x^{\alpha\beta}(\log_e(1+x))^\beta} = 0$$

Then which of the following is (are) correct?

Write the given expression as $$f(x)=\dfrac{\sin(x^{2})\;(\log_e x)^{\alpha}\;\sin\!\left(\dfrac{1}{x^{2}}\right)}{x^{\alpha\beta}\;(\log_e(1+x))^{\beta}}$$ and study $$\displaystyle\lim_{x\to\infty}f(x)$$.

Step 1 : Approximate each elementary factor for large $$x$$
(i) $$\sin(x^{2})$$ is always bounded: $$|\sin(x^{2})|\le1$$.
(ii) For very small argument, $$\sin\theta\approx\theta$$, so $$\sin\!\left(\dfrac{1}{x^{2}}\right)\approx\dfrac{1}{x^{2}}$$ when $$x\to\infty$$.
(iii) $$\log_e(1+x)\sim\log_e x$$ as $$x\to\infty$$.

Step 2 : Construct the asymptotic form of $$f(x)$$
Using (i)-(iii):

$$f(x)\sim \dfrac{(\log_e x)^{\alpha}\;\dfrac{1}{x^{2}}}{x^{\alpha\beta}\;(\log_e x)^{\beta}} =\dfrac{(\log_e x)^{\alpha-\beta}}{x^{\,2+\alpha\beta}}\,.\tag{-1}$$

Set $$L(x)=\dfrac{(\log_e x)^{\alpha-\beta}}{x^{\,2+\alpha\beta}}$$. Since $$|\sin(x^{2})|\le 1$$, the original limit equals $$\displaystyle\lim_{x\to\infty}L(x)$$.

Step 3 : Decide when $$L(x)\to 0$$
Compare the polynomial factor $$x^{2+\alpha\beta}$$ with the logarithmic factor.

Case 1: $$2+\alpha\beta\gt0$$ Then the denominator grows like a positive power of $$x$$. Because any power of $$\log_e x$$ grows slower than any positive power of $$x$$, the fraction tends to $$0$$.

Case 2: $$2+\alpha\beta=0$$ Here $$L(x)\sim(\log_e x)^{\alpha-\beta}$$. • If $$\alpha-\beta\lt0$$, the logarithmic power is negative, so $$L(x)\to0$$.
• If $$\alpha-\beta\ge0$$, the limit is $$\ge1$$ and not zero.

Case 3: $$2+\alpha\beta\lt0$$ Now the denominator behaves like $$x^{\text{negative power}}$$, which tends to $$0$$, so $$L(x)\to\infty$$ and the limit is not zero.

Therefore the limit is $$0$$ exactly when $$\boxed{2+\alpha\beta\gt0}\quad\text{or}\quad\boxed{2+\alpha\beta=0\ \text{and}\ \alpha-\beta\lt0}\,.\tag{-2}$$

Step 4 : Test each option

Option A : $$(\alpha,\beta)=(-1,3)$$ $$\alpha\beta=-3,\;2+\alpha\beta=-1\lt0\;(\text{Case 3})\implies$$ limit ≠0. Hence not in $$S$$.

Option B : $$(\alpha,\beta)=(-1,1)$$ $$\alpha\beta=-1,\;2+\alpha\beta=1\gt0\;(\text{Case 1})\implies$$ limit $$=0$$. Thus $$(-1,1)\in S$$.

Option C : $$(\alpha,\beta)=(1,-1)$$ $$\alpha\beta=-1,\;2+\alpha\beta=1\gt0\;(\text{Case 1})\implies$$ limit $$=0$$. Thus $$(1,-1)\in S$$.

Option D : $$(\alpha,\beta)=(1,-2)$$ $$\alpha\beta=-2,\;2+\alpha\beta=0\ (\text{Case 2})$$ and $$\alpha-\beta=1-(-2)=3\gt0$$, so the required condition $$\alpha-\beta\lt0$$ is violated. Hence limit ≠0 and the pair is not in $$S$$.

Conclusion
The correct statements are
Option B which is: $$(-1,1)\in S$$,
Option C which is: $$(1,-1)\in S$$.