A) \[\frac{\pi }{p+q}\]
B) \[\frac{2\pi }{p+q}\]
C) \[\frac{\pi }{2(p+q)}\]
D) \[\frac{1}{p+q}\]
Correct Answer: B
Solution :
Given \[\cos p\theta =-\cos q\theta =\cos (\pi +q\theta )\] Þ \[p\theta =2n\pi \pm (\pi +q\theta ),n\in I\] Þ \[\theta =\frac{(2n+1)\pi }{p-q}\] or \[\frac{(2n-1)\pi }{p+q},\,\,n\in I\] Both the solutions form an A.P. \[\theta =\frac{(2n+1)\pi }{p-q}\] gives us an A.P. with common difference \[\frac{2\pi }{p-q}\] and \[\theta =\frac{(2n-1)\pi }{p+q}\] gives us an A.P. with common difference\[=\frac{2\pi }{p+q}\]. Certainly, \[\frac{2\pi }{p+q}<\,\left| \,\frac{2\pi }{p-q}\, \right|\].
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