As shown in the figure, if the edge length of the inscribed quadrilateral ABCD is ab = 2, BC = 6, CD = Da = 4, then the area of the quadrilateral ABCD is () A. 163B. 8C. 323D. 83

As shown in the figure, if the edge length of the inscribed quadrilateral ABCD is ab = 2, BC = 6, CD = Da = 4, then the area of the quadrilateral ABCD is () A. 163B. 8C. 323D. 83

Connecting BD, the area of quadrilateral ABCD is s = s △ abd + s △ CBD = 12ab · adsina + 12bc · cdsinc ∵ the quadrilateral ABCD is inscribed in the circle, a + C = 180 ° and Sina = sinc. S = 12ab · adsina + 12bc · cdsinc = 12 (AB · AD + BC · CD) Sina = 12 (2 × 4 + 6 × 4) Sina = 16sina (*) in △ abd, BD2 = AB2 + ad2-2ab · adcosa = 22 + 42-2 × 2 × 4cosa = 20-16cosa can be obtained by cosine theorem. Similarly, in △ CDB, BD2 = CB2 + cd2-2cb · cdcosc = 62 + 42-2 × 6 × 4cosc = 52-48cosc, ｜ 20-16cosa = 52-48cosc, combined with COSC = cos (180 ° - a) = - cosa, 64cosa = - 32 can be obtained, cosa = - 12, ∵ a ∈ (0 ° 180 °), a = 120 ° can be obtained by substituting (*) formula BCD area s = 16sin120 ° = 83, so: D

Given the function f (x) = ax Λ 2 + BX + 1 (a, B are real numbers), X ∈ R, f (x) = {f (x) (x > 0) / - f (x) (x0) and f (x) is even function, judge whether f (m) + F (n) is greater than zero

(1) if f (- 1) = 0, and the range of function f (x) is [0, + ∞), then a > 0, then f (x) = a (x + B / 2a) ^ 2 + 1-B ^ 2 / 4A, f (x) min = f (- 1) = 01-B ^ 2 / 4A = 0-B / 2A = - 1, so a = 1, B = 2F (x) = x ^ 2 + 2x + 1, when x > 0, f (x) = x ^ 2 + 2x + 1, when x0 and f (x) is even function, then f (- x) = f (?)

Let X and Y > 0, and Y ^ 2 / 2 + x ^ 2 = 1, then the maximum value of (1 + y ^ 2) under the root sign of X is?

The known deformation is y ^ 2 = 2-2x ^ 2
The root of X (1 + y ^ 2)
= x radical (3-2x ^ 2)
= radical 2 / 2 · (radical 2) x · radical (3-2x ^ 2)
≤ radical 2 / 2 · {[(radical 2) x] ^ 2 + [radical (3-2x ^ 2)] ^ 2} / 2
= 3 2 / 4