# The set of all real values of

Question:

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:

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

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

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

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

Correct Option: , 4

Solution:

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

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

$\Rightarrow f^{\prime}(x)=\sin x \cos x[2 \lambda+3 \sin x]=0$

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