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Question Number 146838 by bobhans last updated on 16/Jul/21

Given 4^{x} + 4^{- x} - 2^{2 - x} + 2^{2 + x} - 7 = 0, x > 0

find 2^{x} + 2^{- x} .

if x \in [- \frac{π}{6}, 0] then minimum value of

function f (x) = \cot (x + \frac{π}{3}) - \tan (\frac{2 π}{3} - x)

when x = ?

Answered by EDWIN88 last updated on 16/Jul/21

(1) let 2^{x} = p \land 2^{- x} = q \Rightarrow pq = 1

\Rightarrow p^{2} + q^{2} - 4 q + 4 p - 7 = 0

\Rightarrow {(p - q)}^{2} + 2 pq + 4 (p - q) - 7 = 0

\Rightarrow {(p - q)}^{2} + 4 (p - q) - 5 = 0

\Rightarrow (p - q + 5) (p - q - 1) = 0

{\begin{cases} p - q = - 5 \\ p - q = 1 \end{cases}

∵ p + q = \sqrt{{(p + q)}^{2}} = \sqrt{p^{2} + q^{2} + 2 pq}

= \sqrt{{(p - q)}^{2} + 4 pq}

= \sqrt{25 + 4} or \sqrt{1 + 4}

= \sqrt{29} or \sqrt{5}

Commented by Rasheed.Sindhi last updated on 16/Jul/21

N i c e!