A) \[2n\pi \]
B) \[2n\pi +\frac{2\pi }{3}\]
C) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{6}\]
D) \[n\pi +{{(-1)}^{n+1}}\frac{\pi }{4}-\frac{\pi }{3}\]
Correct Answer: B
Solution :
| \[\sqrt{3}\sin x-\cos x=\underset{\lambda \varepsilon R}{\mathop{min}}\,\,\,\{2,{{e}^{2}},\pi ,{{\lambda }^{2}}-4\lambda +7\}\] |
| \[B={{\lambda }^{2}}-4\lambda +7\] |
| \[={{\lambda }^{2}}-2.\,\,2\lambda +4+3\]\[{{(\lambda -2)}^{2}}+3\] |
| \[{{B}_{\min .}}=3\] |
| \[\therefore \sqrt{3}\sin x-\cos x=2\] |
| \[\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x=1\]\[\Rightarrow \sin x.\cos \frac{\pi }{6}=\cos x.\sin \frac{\pi }{6}=1\] |
| \[\sin (x-\frac{\pi }{6})=1,\,\,So\,\,x-\frac{\pi }{6}=2n\pi +\frac{\pi }{2}\] |
| \[x=2n\pi +\frac{\pi }{2}+\frac{\pi }{6}=2n\pi +\frac{2\pi }{3}\] |
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