JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Application of Derivatives

$f(x)=3\sqrt{x-2}+\sqrt{4-x}$, domain $[2,4]$. Substitute $x-2=2{\sin }^2\theta ,4-x=2{\cos }^2\theta$, $\theta \in [0,\pi /2]$:


$3\sin \theta +\cos \theta$ has amplitude $\sqrt{10}$ (attained at an interior $\theta$ with $\tan \theta =3$, both $\sin \theta ,\cos \theta >0$ — valid on $[0,\pi /2]$), giving the maximum $\beta =\sqrt{2}\cdot \sqrt{10}=\sqrt{20}$.

At the endpoints $\theta =0$: value $\sqrt{2}$; $\theta =\pi /2$: value $3\sqrt{2}$. The smaller of these, $\alpha =\sqrt{2}$, is the minimum (the interior critical point is a maximum, so the minimum sits at an endpoint).