JEE Advanced 2022 · Paper 2, Mathematics question 5

Expand the numerator to order $x^3$:

• $e^{x^3}=1+x^3+O(x^6)$.
• $(1-x^3)^{1/3}=1-\frac{x^3}3+O(x^6)$.
• $\left((1-x^2)^{1/2}-1\right)\sin x=\left(-\frac{x^2}2+O(x^4)\right)(x+O(x^3))=-\frac{x^3}2+O(x^5)$.

So the numerator is $\left(1+\frac13-\frac12\right)x^3+O(x^5)=\frac56x^3+O(x^5)$.

Denominator. $x\sin^2x=x^3+O(x^5)$.

Answer. $\beta=\frac56$, so $6\beta=5$.