JEE Advanced 2025 · Paper 1, Mathematics question 11

Expand the integrand: $\int_0^x\frac{dt}{1-t^2}=x+\frac{x^3}{3}+O(x^5)$, and $\beta x\cos x=\beta x-\frac{\beta}{2}x^3+O(x^5)$. So
$$\frac{\alpha}{2}\int_0^x\frac{dt}{1-t^2}+\beta x\cos x=\left(\frac\alpha2+\beta\right)x+\left(\frac\alpha6-\frac\beta2\right)x^3+O(x^5).$$

For the limit to exist and equal 2:

• $\frac\alpha2+\beta=0$;
• $\frac\alpha6-\frac\beta2=2$.

Substituting $\beta=-\frac\alpha2$ gives $\frac\alpha6+\frac\alpha4=2$. So $\alpha=\frac{24}5$, $\beta=-\frac{12}5$, and
$$\alpha+\beta=\frac{12}5=2.4.$$