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 minima, is:
Correct Option: , 4
$f(x)=\left(1-\cos ^{2} x\right)(\lambda+\sin x)=\sin ^{2} x(\lambda+\sin x)$
$\Rightarrow f(x)=\lambda \sin ^{2} x+\sin ^{3} x$
$\Rightarrow f^{\prime}(x)=\sin x \cos x[2 \lambda+3 \sin x]=0$
$\Rightarrow \sin x=0$ and $\sin x=-\frac{2 \lambda}{3} \Rightarrow x=\alpha$ (let)
So, $f(x)$ will change its sign at $x=0, \alpha$ because there is
exactly one maxima and one minima in $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
Now, $\sin x=-\frac{2 \lambda}{3}$
$\Rightarrow-1 \leq-\frac{2 \lambda}{3} \leq 1 \Rightarrow-\frac{3}{2} \leq \lambda \leq \frac{3}{2}-\{0\}$
$\because$ If $\lambda=0 \Rightarrow f(x)=\sin ^{3} x($ from (i))
Which is monotonic, then no maxima/minima
So, $\lambda \in\left(-\frac{3}{2}, \frac{3}{2}\right)-\{0\}$