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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

The value of $\cos^{-1}(\cos 12) - \sin^{-1}(\sin 14)$ is

The derivative of $\tan^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} \right)$ with respect to $\sin^{-1} \left( \frac{2x}{1+x^2} \right)$ is...

If $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=k$ then $\sin ^{-1}\left(\sqrt{\frac{k}{2}}\right)+\cos ^{-1}\left(\frac{k}{3}\right)=$

If $\tan^{-1} (x^2 + 3|x|-4) = \tan^{-1} (4\pi + \sin^{-1}(\sin 14))$,then the value of $\cos^{-1}(\cos 3|x|)$ is equal to

For any $y \in R$,let $\cot ^{-1}(y) \in(0, \pi)$ and $\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the sum of all the solutions of the equation $\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}$ for $0 < |y| < 3$,is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module