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 $x_1{x}_2=\left(1+sin\alpha \right)$
${\left(x_1+x_2\right)}^2={\left(sin\alpha -2\right)}^2=si{n}^2\alpha +4-4sin\alpha$
$\Rightarrow x_1^2+x_2^2=si{n}^2\alpha +4-4sin\alpha -2x_1{x}_2$
$si{n}^2\alpha +4-4sin\alpha +2\left(1+sin\alpha \right)$
$si{n}^2\alpha -2sin\alpha +6...\left(A\right)$
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, $\alpha =\frac{\pi }{2}$