Question

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

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

Answer 1

The correct option is A $(-\frac{3}{2},\frac{3}{2})-\left\{0\right\}$


$f\left(x\right)=(1-{\cos }^{2}x)(\lambda +\sin x)$
$f\left(x\right)={\sin }^{2}x(\lambda +\sin x)$
$f^{\prime }\left(x\right)=2\sin x\cos x(\lambda +\sin x)+{\sin }^{2}x\left(\cos x\right)$
$=\sin 2x(\lambda +\sin x+\frac{\sin x}{2})$
$=\frac{1}{2}\sin 2x(2\lambda +3\sin x)$
For extreme value $f^{\prime }\left(x\right)=0$
$\Rightarrow \sin 2x=0\Rightarrow \sin x=0$
$\Rightarrow x=0\to$One point
$2\lambda +3\sin x=0$
$\Rightarrow \sin x=\frac{-2\lambda }{3}$
$\sin x\in (-1,1)-\left\{0\right\}$
$-1<\frac{-2\lambda }{3}<1\Rightarrow \frac{-3}{2}<\lambda <\frac{3}{2}$
$\lambda \in (-\frac{3}{2},\frac{3}{2})-\left\{0\right\}$