Question

The set of all real values of $\lambda$ for which the function $f\left(x\right)=(1-{\cos }^2\left(x\right))(\lambda +\sin \left(x\right)),x\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right),$ has exactly one maximum and exactly one minimum, is

• A. $\left(-\frac{3}{2},\frac{3}{2}\right)-\left\{0\right\}$
• B. $\left(-\frac{1}{2},\frac{1}{2}\right)-\left\{0\right\}$
• C. $\left(-\frac{3}{2},\frac{3}{2}\right)$
• D. $\left(-\frac{1}{2},\frac{1}{2}\right)$

Answer 1

The correct option is A

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

Explanation for the correct answer.

Step 1: Calculate the first derivative of $f\left(x\right)$

$$\begin{gathered} f\left(x\right)=(1-{\cos }^{2}x)(\lambda +\sin x),x\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right), \\ \Rightarrow f\left(x\right)=\left({\sin }^2\left(x\right)\right)(\lambda +\sin x) \end{gathered}$$

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

$$\begin{gathered} \therefore f\text{'}\left(x\right)=2\lambda \sin x\cos x+3{\sin }^2\left(x\right)\cos \left(x\right) \\ \Rightarrow f\text{'}\left(x\right)=\lambda \sin \left(2x\right)+\frac{3}{2}\sin \left(2x\right)\sin \left(x\right) \end{gathered}$$

Step 2: Find the extremum

For extreme value$f\text{'}\left(x\right)=0$

$$\begin{gathered} \Rightarrow \sin \left(2x\right)=0 \\ \Rightarrow 2x=n\pi \left[n=0,1,2,...\right] \\ \Rightarrow x=\frac{n\pi }{2} \end{gathered}$$

For $n=0$

$\Rightarrow x=0$

Now

$2\lambda +3\sin x=0$

$\Rightarrow \sin x=-\frac{2\lambda }{3}$

$\sin x\in (-1,1)-\left\{0\right\}$

$-1<-\frac{2\lambda }{3}<1$

$\Rightarrow -\frac{3}{2}<\lambda <\frac{3}{2}$

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

Hence, option A is the correct option.