If the quadratic equation $4^{\sec ^2 \alpha} \cdot x^2+2 x+\left(\beta^2-\beta+\frac{1}{2}\right)=0$ has…
If the quadratic equation $4^{\sec ^2 \alpha} \cdot x^2+2 x+\left(\beta^2-\beta+\frac{1}{2}\right)=0$ has real roots, then the value of $\cos ^2 \alpha+\cos ^{-1} \beta$ is
$\frac{\pi}{3}$
$\frac{\pi}{3}+1$
$\frac{\pi}{2}$
$\frac{\pi}{2}-1$
Solution
The quadratic equation,
$
\begin{aligned}
& 4^{\sec ^2 \alpha} x^2+2 x+\left(\beta^2-\beta+\frac{1}{2}\right)=0 \text { have real roots } \\
& \Rightarrow \text { discriminant }=4-4 \cdot 4^{\sec ^2 \alpha}\left(\beta^2-\beta+\frac{1}{2}\right) \geq 0 \\
& \Rightarrow 4^{\sec ^2 \alpha}\left(\beta^2-\beta+\frac{1}{2}\right) \leq 1 \\
& \text { But } 4^{\sec ^2 \alpha} \geq 4, \beta^2-\beta+\frac{1}{2}=\left(\beta-\frac{1}{2}\right)^2+\frac{1}{4} \geq \frac{1}{4}
\end{aligned}
$
So, the equation will be satisfied only
$
\begin{aligned}
& \text { When } 4^{\sec ^2 \alpha}=4 \text { and } \beta^2-\beta+\frac{1}{2}=\frac{1}{4} \\
& \sec ^2 \alpha=1 \text { and }\left(\beta-\frac{1}{2}\right)^2=0 \\
& \cos ^2 \alpha=1 \text { and } \beta=\frac{1}{2} \\
& \therefore \cos ^2 \alpha+\cos ^{-1} \beta=1+\cos ^{-1}(1 / 2) \\
& =1+\pi / 3
\end{aligned}
$