JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 14

Magnitudes. With $u=12\,\vec a\cdot\vec b$: $|3\vec a+2\vec b|^2=72+u$ and $|3\vec a-2\vec b|^2=72-u$.

Cauchy-Schwarz.
$$3\sqrt{72+u}+4\sqrt{72-u}\le\sqrt{3^2+4^2}\sqrt{144}=60,$$
with equality when $\frac{72+u}{72-u}=\frac9{16}$ ($u=-20.16$, an allowed value since $|u|\le72$).

Maximum 60.

(C)