A) \[x={{\tan }^{-1}}\sqrt{(p/q)}\]
B) \[x={{\tan }^{-1}}\sqrt{(q/p)}\]
C) \[x={{\tan }^{-1}}(p/q)\]
D) \[x={{\tan }^{-1}}(q/p)\]
Correct Answer: A
Solution :
Let \[y={{\sin }^{p}}x.{{\cos }^{q}}x\] \[\frac{dy}{dx}=p{{\sin }^{p-1}}x.\cos x.{{\cos }^{q}}x+q{{\cos }^{q-1}}x.(-\sin x){{\sin }^{p}}x\] \[\frac{dy}{dx}=p{{\sin }^{p-1}}x.{{\cos }^{q+1}}x-q{{\cos }^{q-1}}x.{{\sin }^{p+1}}x\] Put \[\frac{dy}{dx}=0\], \[\therefore {{\tan }^{2}}x=\frac{p}{q}\]\[\Rightarrow \]\[\tan x=\pm \sqrt{\frac{p}{q}}\] \[\therefore \]Point of maxima \[x={{\tan }^{-1}}\sqrt{\frac{p}{q}}\].
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