Question

If $|a{\sin }^2\theta +b\sin \theta \cos \theta +c{\cos }^2\theta -\frac{1}{2}(a+c\left)\right|\le \frac{1}{2}k$, then $k^2$ is equal to

• A. $b^2+(a-c{)}^2$
• B. $a^2+(b-c{)}^2$
• C. $c^2+(a-b{)}^2$
• D. $a^2+b^2+c^2$

Answer 1

The correct option is A $b^2+(a-c{)}^2$

$f\left(\theta \right)=|a{\sin }^2\theta +b\sin \theta \cos \theta +c{\cos }^2\theta -\frac{1}{2}a-\frac{c}{2}|$

$[2{\sin }^2\theta -1=-\cos 2\theta ,2{\cos }^2\theta -1=\cos 2\theta ]$

$\therefore f\left(\theta \right)=\left|\frac{a}{2}\right(-\cos 2\theta )+\frac{c}{2}\cos 2\theta +\frac{b}{2}\sin 2\theta |$

$f\left(\theta \right)=\frac{1}{2}\left|\right(c-a)\cos 2\theta +b\sin 2\theta |$

$a\sin \theta +b\cos \theta$ lies in between it's maximum and minimum value as given by

$\sqrt{(a^2-b^2)}\le a\sin \theta +b\cos \theta \le \sqrt{(a^2+b^2)}$

Max. value of $f\left(\theta \right)$ is $\frac{1}{2}\sqrt{(c-a{)}^2+b^2}$

$\therefore k^2=(c-a{)}^2+b^2$