Solve this following


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 :


  1. $\left(-\frac{1}{2}, \frac{1}{2}\right)-\{0\}$

  2. $\left(-\frac{1}{2}, \frac{1}{2}\right)$

  3. $\left(-\frac{3}{2}, \frac{3}{2}\right)$

  4. $\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}$

Correct Option: , 4


$f(x)=\left(1-\cos ^{2} x\right)(\lambda+\sin x)$

$x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

$f(x)=\lambda \sin ^{2} x+\sin ^{3} x$

$f^{\prime}(x)=2 \lambda \sin x \cos x+3 \sin ^{2} x \cos x$

$f^{\prime}(x)=\sin x \cos x(2 \lambda+3 \sin x)$

$\sin x=0, \frac{-2 \lambda}{3},(\lambda \neq 0)$

for exactly one maxima \& minima

$\frac{-2 \lambda}{3} \in(-1,1) \Rightarrow \lambda \in\left(\frac{-3}{2}, \frac{3}{2}\right)$

$\lambda \in\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}$

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