JEE Main · 2004 · 24 Apr 2004 · Offline · Application of Derivatives

Write $p=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }$ and $q=\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$, so $u=p+q$ and $u^2=p^2+q^2+2pq$. Notice $p^2+q^2=a^2+b^2$ is constant, so $u^2=a^2+b^2+2pq$ and only $pq$ varies. A short computation gives $p^2{q}^2=a^2{b}^2+(a^2-b^2{)}^2{\cos }^2\theta {\sin }^2\theta$, and since ${\cos }^2\theta {\sin }^2\theta$ runs between $0$ and $\frac{1}{4}$, the product $pq$ runs from $|ab|$ at $\theta =0$ up to $\frac{a^2+b^2}{2}$ at $\theta =\frac{\pi }{4}$. Hence $u^2$ is least when $2pq=2|ab|$, giving $u_{\min }^2=a^2+b^2+2|ab|=(a+b{)}^2$, and greatest when $2pq=a^2+b^2$, giving $u_{\max }^2=2(a^2+b^2)$. Their difference is $2(a^2+b^2)-(a+b{)}^2=(a-b{)}^2$.