One maximum point of ^p x ^q x is - Vidyadip

MCQ

One maximum point of ${\sin ^p}x{\cos ^q}x$ is

✓
$x = {\tan ^{ - 1}}\sqrt {(p/q)} $
B
$x = {\tan ^{ - 1}}\sqrt {(q/p)} $
C
$x = {\tan ^{ - 1}}(p/q)$
D
$x = {\tan ^{ - 1}}(q/p)$

✓

Answer

Correct option: A.

$x = {\tan ^{ - 1}}\sqrt {(p/q)} $

a
(a) 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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