ข้อสอบ PAT 1 - มีนาคม 2556 | ความถนัดทางคณิตศาสตร์

$จากโจทย์ \arctan (1 - x) + arccot (\frac{1}{2 x}) = 2 arcsec \sqrt{1 + 2 x (1 - x)} \to 1$

$กำหนดให้ A = \arctan (1 - x) \tan A = 1 - x กำหนดให้ B = arccot (\frac{1}{2 x}) \cot B = \frac{1}{2 x} \to \tan B = 2 x$

$กำหนดให้ C = arcsec \sqrt{1 + 2 x (1 - x)} \sec C = \sqrt{1 + 2 x (1 - x)} \cos C = \frac{1}{\sqrt{1 + 2 x (1 - x)}} วาดรูปสามเหลี่ยมมุมฉาก$

$หาค่า y จากปีทากอรัส {[\sqrt{1 + 2 x (1 - x)}]}^{2} = y^{2} + 1^{2} 1 + 2 x (1 - x) = y^{2} + 1 y = \sqrt{2 x (1 - x)}$

$จะได้ \tan C = \frac{ข้าม}{ชิด} = \frac{\sqrt{2 x (1 - x)}}{1} = \sqrt{2 x (1 - x)} จาก 1 \arctan (1 - x) + arccot (\frac{1}{2 x}) = 2 arcsec \sqrt{1 + 2 x (1 - x)} แทนค่า A, B, C$
$จะได้ A + B = 2 C (take \tan ทั้งสองข้าง) \tan (A + B) = \tan (2 C) \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} = \frac{2 \tan C}{1 - \tan^{2} C} \frac{(1 - x) + (2 x)}{1 - (1 - x) (2 x)} = \frac{2 (\sqrt{2 x (1 - x)})}{1 - 2 x (1 - x)} \frac{1 + x}{1 - 2 x + 2 x^{2}} = \frac{2 (\sqrt{2 x (1 - x)})}{1 - 2 x + 2 x^{2}}$

$กำหนดให้ D = 1 - 2 x + 2 x^{2} จะได้ \frac{1 + x}{D} = \frac{2 \sqrt{2 x (1 - x)}}{D} D (1 + x) = 2 D \sqrt{2 x (1 - x)} D [(1 + x) - 2 \sqrt{2 x (1 - x)}] = 0; แทนค่า D (1 - 2 x + 2 x^{2}) [(1 + x) - 2 \sqrt{2 x (1 - x)}] = 0$
$จะได้ 2 x^{2} - 2 x + 1 = 0 หรือ (1 + x) - 2 \sqrt{2 x (1 - x)} = 0 2 [x^{2} - x + \frac{1}{4}] + 1 - \frac{1}{2} = 0 1 + x = 2 \sqrt{2 x (1 - x)} 2 {(x - \frac{1}{2})}^{2} + \frac{1}{2} = 0 {(1 + x)}^{2} = 4 [2 x (1 - x)] {(x - \frac{1}{2})}^{2} = - \frac{1}{4} 1 + 2 x + x^{2} = 8 x - 8 x^{2} x = \emptyset 9 x^{2} - 6 x + 1 = 0 {(3 x - 1)}^{2} = 0 x = \frac{1}{3}$