Q91 / 150 · MHT CET 2024 (15 May Shift 1) · Mathematics
If the equation $\cos ^4 \theta+\sin ^4 \theta+\lambda=0$. has real solutions for $\theta$, then $\lambda$ lies in the interval
- A$\left(-\frac{5}{4},-1\right)$
- B$\left[-\frac{3}{2},-\frac{5}{4}\right]$
- C$\left(-\frac{1}{2},-\frac{1}{4}\right]$.
- Correct answerD$\left[-1,-\frac{1}{2}\right]$
Solution
$\begin{aligned}
& \cos ^4 \theta+\sin ^4 \theta+\lambda=0 \\
& \Rightarrow\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta+\lambda=0 \\
& \Rightarrow 1-2 \sin ^2 \theta \cos ^2 \theta+\lambda=0 \\
& \Rightarrow \lambda=2 \sin ^2 \theta \cos ^2 \theta-1 \\
& \Rightarrow \lambda=\frac{\sin ^2 2 \theta}{2}-1
\end{aligned}$
Since $-1 \leq \sin 2 \theta \leq 1$, $\begin{aligned} & 0 \leq \sin ^2 2 \theta \leq 1 \\ & \Rightarrow 0 \leq \frac{\sin ^2 2 \theta}{2} \leq \frac{1}{2} \\ & \Rightarrow-1 \leq \frac{\sin ^2 2 \theta}{2}-1 \leq-\frac{1}{2} \\ & \Rightarrow \lambda \in\left[-1,-\frac{1}{2}\right] \end{aligned}$
Since $-1 \leq \sin 2 \theta \leq 1$, $\begin{aligned} & 0 \leq \sin ^2 2 \theta \leq 1 \\ & \Rightarrow 0 \leq \frac{\sin ^2 2 \theta}{2} \leq \frac{1}{2} \\ & \Rightarrow-1 \leq \frac{\sin ^2 2 \theta}{2}-1 \leq-\frac{1}{2} \\ & \Rightarrow \lambda \in\left[-1,-\frac{1}{2}\right] \end{aligned}$
Hence, the correct answer is option: 'D'.