Question
${\sin ^p}x{\cos ^q}x$ का एक उच्चिष्ठ बिन्दु होगा
$\frac{{dy}}{{dx}} = p{\sin ^{p - 1}}x.\cos x.{\cos ^q}x + q{\cos ^{q - 1}}x.( - \sin x){\sin ^p}x$
$\frac{{dy}}{{dx}} = p{\sin ^{p - 1}}x.{\cos ^{q + 1}}x - q{\cos ^{q - 1}}x.{\sin ^{p + 1}}x$
$\frac{{dy}}{{dx}} = 0$ रखने पर, ${\tan ^2}x = \frac{p}{q}$
$ \Rightarrow $ $\tan x = \pm \sqrt {\frac{p}{q}} $
$\therefore $ उच्चिष्ठ का बिन्दु $ = x = {\tan ^{ - 1}}\sqrt {\frac{p}{q}} $.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.