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 $y = {\sin ^p}x.{\cos ^q}x$
(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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