Question

If the quadratic equation $4^{{\sec }^2\alpha }\cdot x^2+2x+\left({\beta }^2-\beta +\frac{1}{2}\right)=0$ has real roots, then the value of ${\cos }^2\alpha +{\cos }^{-1}\beta$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{3}+1$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{2}-1$

Answer 1

Answer: (B)

The quadratic equation,
$4^{{\sec }^2\alpha }{x}^2+2x+\left({\beta }^2-\beta +\frac{1}{2}\right)=0$ have real roots
$\Rightarrow$ discriminant $=4-4\cdot 4^{{\sec }^2\alpha }\left({\beta }^2-\beta +\frac{1}{2}\right)\ge 0$
$\Rightarrow 4^{se{c}^2\alpha }\left({\beta }^2-\beta +\frac{1}{2}\right)\le 1$
But $4^{{\sec }^2\alpha }\ge 4,{\beta }^2-\beta +\frac{1}{2}={\left(\beta -\frac{1}{2}\right)}^2+\frac{1}{4}\ge \frac{1}{4}$
So, the equation will be satisfied only When $4^{{\sec }^2\alpha }=4$ and ${\beta }^2-\beta +\frac{1}{2}=\frac{1}{4}$
${\sec }^2\alpha =1$ and ${\left(\beta -\frac{1}{2}\right)}^2=0$
${\cos }^2\alpha =1$ and $\beta =\frac{1}{2}$
$\therefore {\cos }^2\alpha +{\cos }^{-1}\beta =1+{\cos }^{-1}(1/2)$
$=1+\pi /3$