- Home
- Standard 11
- Mathematics
यदि $\left| {\,a\,{{\sin }^2}\theta + b\sin \theta \cos \theta + c\,{{\cos }^2}\theta - \frac{1}{2}(a + c)\,} \right|\, \le \frac{1}{2}k,$ तब ${k^2}$ बराबर है
${b^2} + {(a - c)^2}$
${a^2} + {(b - c)^2}$
${c^2} + {(a - b)^2}$
इनमें से कोई नहीं
Solution
(a) $a{\sin ^2}\theta + b\sin \theta \cos \theta + c{\cos ^2}\theta – \frac{1}{2}(a + c)$
$ = \frac{1}{2}[ – a\cos 2\theta + b\sin 2\theta + c\cos 2\theta ]$
$ = \frac{1}{2}[b\sin 2\theta – (a – c)\cos 2\theta ]$
$|b\sin 2\theta – (a – c)\cos 2\theta |\, \le \sqrt {{b^2} + {{(a – c)}^2}} $
$\therefore $$\left| {\frac{1}{2}\{ b\sin 2\theta – (a – c)\cos 2\theta \} } \right| \le \frac{1}{2}\sqrt {{b^2} + {{(a – c)}^2}} $
==> $\left| {a{{\sin }^2}\theta + b\sin \theta \cos \theta + c{{\cos }^2}\theta – \frac{1}{2}(a + c)} \right|$
$ \le \frac{1}{2}\sqrt {{b^2} + {{(a – c)}^2}} $
