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 number of solutions of $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$ is

If we consider only the principal values of the inverse trigonometric functions,then the value of $\tan \left( \cos^{-1} \frac{1}{5\sqrt{2}} - \sin^{-1} \frac{4}{\sqrt{17}} \right)$ is

If $\alpha$ and $\beta$ (where $\alpha > \beta$) are two zeroes of the equation $3\cos^{-1}\left(x^2 - 5x - \frac{11}{2}\right) = \pi$,then $(\alpha^2 + \beta^3)$ is -

Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16((\sec^{-1} x)^2 + (\operatorname{cosec}^{-1} x)^2)$ is: (in $\pi^2$)

The number of real solutions of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module