what-minimum-value-of-f-x-9tan-2-x-4cot-2-x-

Question Number 78207 by jagoll last updated on 15/Jan/20

w h a t m i n i m u m

v a l u e o f f (x) = 9 \tan^{2} (x) + 4 \cot^{2} (x)

Commented by jagoll last updated on 15/Jan/20

i^{'} m g o t 24 . i t r i g h t a n s w e r ?

Commented by john santu last updated on 15/Jan/20

s i r M j s i t a n s w e r i s c o r r e c t ?

Commented by MJS last updated on 15/Jan/20

I get 12

Commented by jagoll last updated on 15/Jan/20

h o w s i r ?

Commented by MJS last updated on 15/Jan/20

let \tan x = t

f (t) = 9 t^{2} + \frac{4}{t^{2}}

f^{'} (t) = 18 t - \frac{8}{t^{3}} = \frac{2}{t^{3}} (3 t^{2} - 2) (3 t^{2} + 2) = 0

\Rightarrow t = \pm \frac{\sqrt{6}}{3}

f^{″} (t) = \frac{6}{t^{4}} (3 t^{4} + 4) > 0 \forall t \in R \Rightarrow minima at \pm \frac{\sqrt{6}}{3}

\tan x = \pm \frac{\sqrt{6}}{3} \Rightarrow x = n π \pm \arctan \frac{\sqrt{6}}{3}

f (n π \pm \arctan \frac{\sqrt{6}}{3}) = 12

Commented by jagoll last updated on 15/Jan/20

t h a n k s s i r . m y a n s w e r i s w r o n g

Answered by behi83417@gmail.com last updated on 15/Jan/20

{9 tg}^{2} x . {4 cotg}^{2} x = 36 = cons .

\Rightarrow f (x) is minimum when : {9 tg}^{2} x = {4 cotg}^{2} x

\Rightarrow {tg}^{4} x = \frac{4}{9} = \frac{36}{81} \Rightarrow tgx = \pm \frac{\sqrt{6}}{3}

\Rightarrow f_{\min} = 9 \times \frac{6}{9} + 4 \times \frac{9}{6} = 6 + 6 = 12 .

Commented by jagoll last updated on 15/Jan/20

t h a n k s s i r

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