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Question

Tìm giá trị lớn nhất, nhỏ nhất :

$P = \frac{2\tan ^{2}x + 12\tan x}{\tan ^{2}x + 2\tan x + 3}$

trilac2013 · Answer 1 · 8/31/2014 12:16:19 PM

TXD:D=R\{ ${\frac{\pi }{2}+k\pi}$ }

giả sử có 1 giá trị P sao cho $x\in D$

Ta có P= $\frac{2(\tan x)^{2}+12\tan x}{(\tan x)^{2}+2\tan x+3}$

<=>(P-2) $(\tan x)^{2}$ +(2P-12) $\tan x$ +3P=0

$\Delta$ = $(P-6)^{2}$ -3P(P-2)= $-2P^{2}$ -6P+36 $\geq$ 0

<=> $-6\leq P\leq 3$

MaxP=3<=> $x=\arctan(3)+k\pi$

MinP=-6<=> $x=\arctan (-\frac{3}{2})+k\pi$

Trả lời 31-08-14 12:16 PM

trilac2013
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