JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 18

Minimum. $f(a)=\tfrac12\left[2(2^a+2^{-a})+(3^a+3^{-a})\right]\ge\tfrac12(4+2)=3$, with equality at $a=0$. So $\alpha=3$.

Integral from $\ln2$ to $\ln3$. Put $t=e^{2x}$:
$$\int\frac{e^{2x}\,dx}{e^{4x}-1}=\int_4^9\frac{dt}{2(t^2-1)}=\frac14\ln\frac{t-1}{t+1}\Big|_4^9=\frac14\left(\ln\frac45-\ln\frac35\right)=\frac14\ln\frac43.$$

(B)