Find-the-minimum-value-of-5t-2-8t-5-2-3-t-2-2t-2-3-where-2-3-lt-t-lt-2-3-

Question Number 197882 by CrispyXYZ last updated on 02/Oct/23

Find the minimum value of

\frac{5 t^{2} - 8 t + 5}{(2 + \sqrt{3}) t^{2} - 2 t + 2 - \sqrt{3}} where 2 - \sqrt{3} < t < 2 + \sqrt{3} .

Answered by mr W last updated on 03/Oct/23

\frac{5 t^{2} - 8 t + 5}{(2 + \sqrt{3}) t^{2} - 2 t + 2 - \sqrt{3}} = \frac{1}{k}, s a y

(2 + \sqrt{3} - 5 k) t^{2} - 2 (1 - 4 k) t + 2 - \sqrt{3} - 5 k = 0

Δ = {(1 - 4 k)}^{2} - (2 + \sqrt{3} - 5 k) (2 - \sqrt{3} - 5 k) ⩾ 0

{(1 - 4 k)}^{2} - {(2 - 5 k)}^{2} + 3 ⩾ 0

k (3 k - 4) ⩽ 0

\Rightarrow 0 ⩽ k ⩽ \frac{4}{3} \Rightarrow \frac{3}{4} ⩽ \frac{1}{k} < + \infty

\Rightarrow \frac{5 t^{2} - 8 t + 5}{(2 + \sqrt{3}) t^{2} - 2 t + 2 - \sqrt{3}} ⩾ \frac{3}{4} = m i n i m u n

a t t = \frac{1 - 4 k}{2 + \sqrt{3} - 5 k} = \frac{1 - \frac{16}{3}}{2 + \sqrt{3} - \frac{20}{3}} = \frac{14 + 3 \sqrt{3}}{13}

2 - \sqrt{3} < \frac{14 + 3 \sqrt{3}}{13} < 2 + \sqrt{3}

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