Question

If the equation ${\cos }^4\left(\theta \right)+{\sin }^4\left(\theta \right)+\lambda =0$ has real solutions for $\theta$, then $\lambda$ lies in the interval

• A. $(\frac{-1}{2},\frac{-1}{4}]$
• B. $\left[-1,\frac{-1}{2}\right]$
• C. $\left[\frac{-3}{2},\frac{-5}{4}\right]$
• D. $\left(\frac{-5}{4},-1\right)$

Answer 1

The correct option is B

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

Explanation for correct option

Given equation is ${\cos }^4\left(\theta \right)+{\sin }^4\left(\theta \right)+\lambda =0$

$$\begin{gathered} \Rightarrow \lambda =-\left[{\cos }^4\left(\theta \right)+{\sin }^4\left(\theta \right)\right] \\ \Rightarrow \lambda =-\left[{\left({\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)\right)}^2-2{\cos }^2\left(\theta \right){\sin }^2\left(\theta \right)\right]\left[\because a^2+b^2={\left(a+b\right)}^2-2ab\right] \\ \Rightarrow \lambda =-{\left(1\right)}^2+\frac{1}{2}{\left(2\cos \left(\theta \right)\sin \left(\theta \right)\right)}^2\left[\because {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right] \\ \Rightarrow \lambda =\frac{1}{2}{\sin }^2\left(2\theta \right)-1\left[\because 2\cos \left(\theta \right)\sin \left(\theta \right)=\sin \left(2\theta \right)\right] \end{gathered}$$

We know that $-1\le \sin \left(x\right)\le 1$

Therefore $0\le {\sin }^2\left(x\right)\le 1$

So $\frac{1}{2}{\sin }^2\left(2\theta \right)\in \left[0,\frac{1}{2}\right]$

for $\frac{1}{2}{\sin }^2\left(2\theta \right)=0\Rightarrow \lambda =0-1=-1$

for $\frac{1}{2}{\sin }^2\left(2\theta \right)=\frac{1}{2}\Rightarrow \lambda =\frac{1}{2}-1=\frac{-1}{2}$

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

Hence, Option(B) i.e. $\left[-1,\frac{-1}{2}\right]$ is correct