Applications Of Derivatives Question 244
Question: One maximum point of $ {{\sin }^{p}}x{{\cos }^{q}}x $ is
[RPET 1997; AMU 2000]
Options:
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) $
Show Answer
Answer:
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}} $ .
BETA
