標題:
會考a.maths mock paper 2題[超難]
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發問:
http://life.mingpao.com/htm/hkcee2007/mock/hkcee_amaths_sanford_e.pdf (14,15) [唔好直接打答案,連步驟] 更新: 14(a)(ii)2應該係C1半徑
最佳解答:
[h should be replaced by a] [point P instead of Point A] 14)(a)(i) centre = (2a/2, 0) = (a, 0) radius = 0.5 * root(4a^2 - 8a + 4) = 0.5 * 2 * root(a^2 - 2a + 1) = root((a - 1)^2) = a - 1 14(a)(ii)(1) As L2 is a tangent to C1, we have (3a - 4*0 + 1)^2/25 = (a - 1)^2 9a^2 + 6a + 1 = 25a^2 - 50a + 25 16a^2 - 56a + 24 = 0 a = 3 or 0.5 As a > 1, thus a = 3 sub (a, 0) into L1 2a - 0 - 6 = 2(3) - 6 = 0 Thus, centre of C1 lies on L1 14(a)(ii)(2)radius of C1 = 3-1 = 2units 14)(b)(i)slope L1 = 2, slop L2 = 3/4 tan θ = (2 - 3/4)/(1 + 2 * 3/4) = 1.25/2.5 = 1/2 14)(b)(ii)Point P = Intersection of L1, L2 = (5, 4) Thus, equation of L is x = 5. 14)(c) As C2 touches L, L2 and C1, which is similar to C1, its centre must lies on L1 like C1 Let radius of C2 be R, centre = (h,k) Intersection of L1, L2 = (5, 4) Distance of (5, 4) and (3, 0) = root(2^2+4^2) = 2 * root 5 By simple ratio, R/2 = (2 * root 5 - R - 2)/2 * root 5 R = 0.764, corr 3 sig fig. Also, 2/4 = R/(4 - k) k = 4 - 2R = 2.472 As (h,k) lies on L1, 2h - k - 6 = 0 2h = 6 + k h = 4.236 Thus, centre = (4.236, 2.472) 15)(a)(Int from -6 to a)pi * x^2 dy = (Int from -6 to a)pi * (36 - y^2) dy = pi[(36a - 1/3 * a^3) - (36(-6) - 1/3 * (-6)^3)] = pi(36a-1/3 * a^3 +144) 15)(b)(Int from 0 to b)pi * x^2 dy = (Int from 0 to b)pi * 2y dy = pi * b^2 15)(c)(i) As water depth in vessel 1 is 5, thus, a = -1 Water in vessel 1 = pi(-36-1/3 * -1 +144) = 323pi/3 As water depth in vessel 2 is 5, thus, b = 4 Water vessel 2 = pi * 4^2 = 16pi 16pi + 323pi/3 = pi * h^2 h^2 = 371/3, h = 11.12 15)(c)(ii)V_2 = pi * b^2 dV_2/dt = 2b * pi * db/dt = 2 * 4 * pi * -3 = -12pi V_1 = pi(36a-1/3 * a^3 +144) dV_1 = (-a^2 + 36)pi * da/dt 12pi = (-25 + 36)pi * da/dt da/dt = 12/11
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