JEE Mains · Maths · STD 12 - 6. Application of derivatives
The set of all real values of \(\lambda\) for which the function \(f(x)=\left(1-\cos ^{2} x\right) \cdot(\lambda+\sin x)\) \(x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),\) has exactly one maxima and exactly one minima, is
- A \(\left(-\frac{1}{2}, \frac{1}{2}\right)-\{0\}\)
- B \(\left(-\frac{1}{2}, \frac{1}{2}\right)\)
- C \(\left(-\frac{3}{2}, \frac{3}{2}\right)\)
- D \(\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}\)
Answer & Solution
Correct Answer
(D) \(\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\left(1-\cos ^{2} x\right)(\lambda+\sin x)\) \(x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) \(f(x)=\lambda \sin ^{2} x+\sin ^{3} x\) \(f^{\prime}(x)=2 \lambda \sin x \cos x+3 \sin ^{2} x \cos x\) \(f^{\prime}(x)=\sin x \cos x(2 \lambda+3 \sin x)\)…
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