[JEE Main 2004] if u sqrt a 2 cos 2 theta b 2 sin 2 theta sqrt
If $u=\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}+\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$, the difference between the maximum and minimum of $u^2$ is
(a) $(a-b)^2$
(b) $2\sqrt{a^2+b^2}$
(c) $(a+b)^2$
(d) $2(a^2+b^2)$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (a) $(a-b)^2$
$u^2=(a^2+b^2)+2\sqrt{a^2b^2+\frac14(a^2-b^2)^2\sin^22\theta}$. Max $=2(a^2+b^2)$ (at $\sin^22\theta=1$), min $=(a+|b|)^2$; difference $=(a-b)^2$.
JEE Main 2004 · Trigonometry — verified solution by the Vidaara Team.
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