A) \[\left( -1,1 \right)\]
B) \[\left( -\frac{3}{2},0 \right)\cup \left( 0,\frac{3}{2} \right)\]
C) \[\left( 0,\infty \right)\]
D) \[\left( -\frac{3}{2},-\frac{1}{2} \right)\cup \left( \frac{1}{2},\frac{3}{2} \right)\]
Correct Answer: B
Solution :
| \[f'\left( x \right)=\sin x\cos x\left( 3\sin x+2\lambda \right)\] |
| For maximum and minimum \[f'\left( x \right)=0\Rightarrow \sin x=0\] or \[\cos x=0\] or \[\sin x=-\frac{2\lambda }{3}\] |
| \[\therefore \]critical points in \[\left( -\frac{\pi }{2},\frac{\pi }{2} \right)\]are \[x=0,\]\[{{\sin }^{-1}}\left( -\frac{2\lambda }{3} \right)\] |
| One of these is point of minima and other is point of maxima, provided\[-1<-\frac{2\lambda }{3}<1\]\[\Rightarrow -\frac{3}{2}<\lambda <\frac{3}{2}\] |
| But if \[\lambda =0\], then \[\sin x=0\]which gives only one critical points |
| \[\therefore \lambda \in \left( -\frac{3}{2},\frac{3}{2} \right)-\{0\}\] |
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