Question

If a ${\sin }^2\theta +b\sin \theta \cdot \cos \theta +c{\cos }^2\theta \le q\forall \theta \in R$ then value of q is

• A. $\frac{1}{2}[a+c+\sqrt{b^2+(a-c{)}^2}]$
• B. $\frac{1}{2}[a+c-\sqrt{b^2+(a-c{)}^2}]$
• C. $\frac{1}{2}[a+c+\sqrt{b^2+(a+c{)}^2}]$
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{2}[a+c+\sqrt{b^2+(a-c{)}^2}]$

${\mathrm{a\; sin}}^2\theta +b\sin \theta \cdot \cos \theta +c{\cos }^2\theta \le q$

$\mathrm{LHS}(1-{\cos }^2\theta )+\frac{b}{2}\left(2\sin \theta \cos \theta \right)+{\cos }^2\theta$

${\cos }^2\theta (c-a)+\frac{\sin 2\theta b}{2}+a$

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

$\frac{(c-a)\cos 2\theta }{2}-\frac{c-a}{2}+\frac{b\sin 2\theta }{2}+a$

$\frac{1}{2}\left[\cos 2\theta \right(c-a)+b\sin 2\theta ]+\frac{c}{2}+\frac{a}{2}$

$\begin{array}{cc}k& \le \sqrt{b^2+(c-a{)}^2}\\ \frac{k}{2}& \le \sqrt{b^2+(a-c{)}^2}\end{array}$

$\begin{array}{cc}\frac{k}{2}-\frac{c}{2}+\frac{3a}{2}& \le \frac{a}{2}+\frac{c}{2}+\sqrt{b^2+(a-c{)}^2}\\ & \le \frac{1}{2}[a+c+\sqrt{b^2+(a-c{)}^2}]\end{array}$

option a is correct.