Question

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

• 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,
$4^{\sec ^{2} \alpha} x^{2}+2 x+\left(\beta^{2}-\beta+\frac{1}{2}\right)=0$ have real roots
$\Rightarrow $ discriminant $=4-4 \cdot 4^{\sec ^{2} \alpha}\left(\beta^{2}-\beta+\frac{1}{2}\right) \geq\, 0$
$\Rightarrow \,4^{ sec ^{2} \alpha}\left(\beta^{2}-\beta+\frac{1}{2}\right) \leq \,1$
But $4^{\sec ^{2} \alpha} \geq 4, \beta^{2}-\beta+\frac{1}{2}=\left(\beta-\frac{1}{2}\right)^{2}+\frac{1}{4} \geq \frac{1}{4}$
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}$
$\therefore \, \cos ^{2} \alpha+\cos ^{-1} \beta=1+\cos ^{-1}(1 / 2)$
$=1+\pi / 3$