Question

The value of $\alpha$, for which the equation $x^{2}-(\sin \alpha-2) x-(1+\sin \alpha)=0$ has roots whose sum of square is least, is

• A. $ \frac{\pi }{3} $
• B. $ \frac{\pi }{4} $
• C. $ \frac{\pi }{2} $
• D. $ \frac{\pi }{6} $

Answer 1

Answer: (C)

Let the roots of the equation be $p, q$.
Let $S=p^{2}+q^{2}$
$=(p+q)^{2}-2 p q \ldots$ (i)
Given equation is
$x^{2}-(\sin \alpha-2) x-(1+\sin \alpha)=0$
$\therefore p+q=(\sin \alpha-2), p q=-(1+\sin \alpha)$ From Eq. (i),
$S=(\sin \alpha-2)^{2}+2(1+\sin \alpha)$
$=\sin ^{2} \alpha-4 \sin \alpha+4+2+2 \sin \alpha$
$=\sin ^{2} \alpha-2 \sin \alpha+6$
$\Rightarrow S=(\sin \alpha-1)^{2}+5$
This is least when $\sin \alpha-1=0$
$\Rightarrow \alpha=\frac{\pi}{2}$