Question:
If the equation \(\cos^4\theta + \sin^4\theta + \lambda = 0\) has real solutions for \(\theta\), then \(\lambda\) lies in the interval :
If the equation \(\cos^4\theta + \sin^4\theta + \lambda = 0\) has real solutions for \(\theta\), then \(\lambda\) lies in the interval :
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Write \(\cos^4\theta+\sin^4\theta = 1-2\sin^2\theta\cos^2\theta\) and find its range; then \(\lambda\) is the negative of that range.
Updated On: Nov 5, 2026
- \(\left[-\frac{3}{2}, -\frac{5}{4}\right]\)
- \(\left(-\frac{1}{2}, -\frac{1}{4}\right]\)
- \(\left(-\frac{5}{4}, -1\right]\)
- \(\left[-1, -\frac{1}{2}\right]\)
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The Correct Option is D
Solution and Explanation
\(\lambda = -(\cos^4\theta+\sin^4\theta)\)
\(\cos^4\theta+\sin^4\theta = (\cos^2\theta+\sin^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1-\frac{1}{2}\sin^2 2\theta\)
Since \(0\le \sin^2 2\theta\le 1\), the expression takes all values in \(\left[\frac{1}{2}, 1\right]\).
Hence \(\lambda = -\left(1-\frac{1}{2}\sin^2 2\theta\right)\in\left[-1, -\frac{1}{2}\right]\).
Answer: (4) \(\left[-1, -\frac{1}{2}\right]\)
\(\cos^4\theta+\sin^4\theta = (\cos^2\theta+\sin^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1-\frac{1}{2}\sin^2 2\theta\)
Since \(0\le \sin^2 2\theta\le 1\), the expression takes all values in \(\left[\frac{1}{2}, 1\right]\).
Hence \(\lambda = -\left(1-\frac{1}{2}\sin^2 2\theta\right)\in\left[-1, -\frac{1}{2}\right]\).
Answer: (4) \(\left[-1, -\frac{1}{2}\right]\)
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