Mathematics - Quadratic Equation Question with Solution | TestHub

Given equation is $$x^2-(\sin \alpha -2)x-(1+\sin \alpha )=0$$ Let $x_1$ and $x_2$ be two roots of quadratic equation. $$\begin{array}{rl}& \therefore x_1+x_2=\sin \alpha -2\text{ and }{x}_1{x}_2=-(1+\sin \alpha )\\ & {\left(x_1+x_2\right)}^2=(\sin \alpha -2{)}^2={\sin }^2\alpha +4-4\sin \alpha \end{array}$$ $$\begin{array}{rl}& \Rightarrow x_1^2+x_2^2={\sin }^2\alpha +4-4\sin \alpha -2x_1{x}_2\\ & ={\sin }^2\alpha +4-4\sin \alpha +2(1+\sin \alpha )\\ & ={\sin }^2\alpha -2\sin \alpha +6\end{array}$$ Now, By putting $\alpha =\frac{\pi }{6},\alpha =\frac{\pi }{4},\alpha =\frac{\pi }{3}$ and $\alpha =\frac{\pi }{2}$ in (A) one by one We get least value of $x_1^2+x_2^2$ at $\frac{\pi }{2}$ Hence, $\alpha =\frac{\pi }{2}$