If the quadratic equation 4^{sec^2 alpha} x^2 | vedclass

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Question

(B) For the quadratic equation $4^{\sec^2 \alpha} x^2 + 2x + (\beta^2 - \beta + \frac{1}{2}) = 0$ to have real roots,the discriminant $D$ must be $\ge

Vedclass · Accepted Answer

$\frac{\pi}{3} + 1$

Vedclass · Answer

(B) For the quadratic equation $4^{\sec^2 \alpha} x^2 + 2x + (\beta^2 - \beta + \frac{1}{2}) = 0$ to have real roots,the discriminant $D$ must be $\geq 0$. $D = (2)^2 - 4(4^{\sec^2 \alpha})(\beta^2 - \beta + \frac{1}{2}) \geq 0$ $4 - 4(4^{\sec^2 \alpha})(\beta^2 - \beta + \frac{1}{2}) \geq 0$ $4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \leq 1$ Since $\sec^2 \alpha \geq 1$,we have $4^{\sec^2 \alpha} \geq 4^1 = 4$. Also,$\beta^2 - \beta + \frac{1}{2} = (\beta - \frac{1}{2})^2 + \frac{1}{4} \geq \frac{1}{4}$. Multiplying these,$4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \geq 4 	imes \frac{1}{4} = 1$. For the inequality $4^{\sec^2 \alpha}(\beta^2 - \beta + \frac{1}{2}) \leq 1$ to hold,both expressions must be at their minimum values: $4^{\sec^2 \alpha} = 4$ $\Rightarrow \sec^2 \alpha = 1$ $\Rightarrow \cos^2 \alpha = 1$. $(\beta - \frac{1}{2})^2 = 0 \Rightarrow \beta = \frac{1}{2}$. Therefore,$\cos^2 \alpha + \cos^{-1} \beta = 1 + \cos^{-1}(\frac{1}{2}) = 1 + \frac{\pi}{3}$.

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

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