Question

Let $\alpha ,\beta ,\gamma$ be distinct real numbers such that
$a{\alpha }^2+b\alpha +c=(\sin \theta ){\alpha }^2+(\cos \theta )\alpha$
$a{\beta }^2+b\beta +c=(\sin \theta ){\beta }^2+(\cos \theta )\beta$
$a{\gamma }^2+b\gamma +c=(\sin \theta ){\gamma }^2+(\cos \theta )\gamma$
(where a, b, c, $\in$ R.)
The maximum value of the expression $\frac{a^2+b^2}{a^2+3ab+5b^2}$ is equal to

• A. 1
• B. 2
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

Answer: (B)

The 3 given equation are suggestive that $\alpha ,\beta$ and $\gamma$ are the roots of
$a{x}^2+bx+c=(\sin \theta )x^2+(\cos \theta )x\text{ or }(a-\sin \theta )x^2+(b-\cos \theta )x+c=0$....(1)
$\therefore a=\sin \theta ,b=\cos \theta \text{ and }c=0\text{ (Using concept of identity) }$
$E=\frac{a^2+b^2}{a^2+3ab+5b^2}=\frac{1}{{\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta }$
$=\frac{2}{1-\cos 2\theta +3\sin 2\theta +5(1+\cos 2\theta )}=\frac{2}{6+3\sin 2\theta +4\cos 2\theta }$
$\therefore E_{\max }=\frac{2}{6-5}=2$