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

If $y = \tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+2x+3}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+5x+7}\right) + \dots + n \text{ terms}$,then $y'(0)$ is

The mean of numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.80$. Then $\operatorname{Tan}^{-1} \frac{1}{a} + \operatorname{Tan}^{-1} \frac{1}{b} =$

If $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $x \in (1, \infty)$,then $f'(x)$ is equal to:

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 -

$\sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module