If the equation cos^4 theta + sin^4 theta + l | vedclass

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

Question

(D) Given the equation $\cos^4 	heta + \sin^4 	heta + \lambda = 0$. We know that $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2

Vedclass · Accepted Answer

$\left[-1, -\frac{1}{2}\right]$

Vedclass · Answer

(D) Given the equation $\cos^4 	heta + \sin^4 	heta + \lambda = 0$. We know that $\sin^4 	heta + \cos^4 	heta = (\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta = 1 - 2 \sin^2 	heta \cos^2 	heta$. Substituting this into the equation: $1 - 2 \sin^2 	heta \cos^2 	heta + \lambda = 0$. Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin^2 	heta \cos^2 	heta = \frac{\sin^2 2	heta}{4}$. Thus,$1 - 2 \left(\frac{\sin^2 2	heta}{4}ight) + \lambda = 0$,which simplifies to $1 - \frac{\sin^2 2	heta}{2} + \lambda = 0$. Therefore,$\lambda = \frac{\sin^2 2	heta}{2} - 1$. Since $0 \leq \sin^2 2	heta \leq 1$,we have $0 \leq \frac{\sin^2 2	heta}{2} \leq \frac{1}{2}$. Subtracting $1$ from all parts,we get $-1 \leq \frac{\sin^2 2	heta}{2} - 1 \leq -\frac{1}{2}$. Hence,$\lambda \in \left[-1, -\frac{1}{2}ight]$.

If the equation $\cos^4 \theta + \sin^4 \theta + \lambda = 0$ has real solutions for $\theta$,then $\lambda$ lies in the interval

Similar Questions

If $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3$,then $\cos \theta_1 + \cos \theta_2 + \cos \theta_3 =$

The range of $\frac{1}{\sin^2 x + 3 \sin x \cos x + 5 \cos^2 x}$ is

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

The maximum value of $12 \sin x - 5 \cos x + 3$ is

If $\alpha + \beta + \gamma = 2\pi ,$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module