Question

If the sum of the square of the roots of fee equation $x^{2}-\left(sin\alpha-2\right)x -\left(1+sin\alpha\right) = 0$ is least, then $\alpha$ is equal to

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

Answer 1

Answer: (D)

Given equation is
$x^{2}-\left(sin\alpha-2\right)x -\left(1+sin\alpha\right) = 0$
Let $x_{1}$ and $x_{2}$ be two roots of quadraticequation.
$\therefore x_{1}+x_{2} = sin\alpha -2$ and $x_{1}x_{2} = \left(1+sin\alpha \right)$
$\left(x_{1}+x_{2}\right)^{2} = \left(sin\alpha -2\right)^{2} = sin^{2}\alpha +4-4\,sin\alpha$
$\Rightarrow \quad x_{1}^{2}+x_{2}^{2} = sin^{2}\alpha +4-4\,sin\alpha -2x_{1}x_{2}$
$sin^{2}\alpha +4-4\,sin\alpha +2\left(1+sin\alpha\right)$
$sin^{2}\alpha -2\,sin\alpha +6\quad\quad...\left(A\right)$
Now, By putting
$\alpha = \frac{\pi}{6},\,\alpha = \frac{\pi }{4},\,\alpha = \frac{\pi }{3} and \alpha = \frac{\pi }{2} \,in\left(A\right)$
one by one
We get least value of $x_{1}^{2}+x_{2}^{2}$ at $\frac{\pi }{2}$
Hence, $\alpha = \frac{\pi }{2}$