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請問數學多題問題 急急急...
1.k為實數，如果點P﹝3k+1分之k+1,k-2分2k+1﹞在第二象限內，下列何者正確?(A)-2分之1<3<-3分之1(B)-1
1. 答案是： (B) -1 < k < -1/2 在第二象限內，x 坐標為負值，y 坐標為正值。 (k + 1)/(3k + 1) < 0 及 (2k + 1)/(k - 2) > 0 -1 < k < -1/3 及 (k <-1/2 或 k > 2) 取重疊範圍： -1 < k< -1/2 2. 答案是： (B) 0 f(x) = 5x2 + 4x + 1 f(x) = 5[x2 + (4/5)x + (2/5)2] - 5(2/5)2 + 1 f(x) = 5[x + (2/5)]2 + (1/5) f(x) 在 x = -2/5 時有最小值 1/5 a = -2/5, b = 1/5 a + 2b = (-2/5) + 2(1/5) = 0 3. 答案是： (B) √5 PQ = √[(1 - 2)2 + (3 - 5)2] = √5 4. 答案是： (C) -1/5 tanx + cotx = 25/12 tanx + (1/tanx) = 25/12 (tan2x + 1)/tanx = 25/12 12tan2x - 25tanx + 12 = 0 (4tanx - 3)(3tanx - 4) = 0 tanx = 3/4 或 tanx = 4/3 (捨去，因 x > π/4，不合題意) tanx = 3/4 sinx = 3/5 及 cosx =4/5 sinx - cosx = (3/5) - (4/5) = -1/5 5. 答案是： (B) -1/7 tan(α - β) = (tanα - tanβ)/(1 + tanα*tanβ) = (2 - 3)/(1 + 2*3) = -1/7 6. 答案是： (B) π/3 2sin2θ + 11cosθ - 7 = 0 2(1 - cos2θ) + 11cosθ - 7 = 0 2 - 2cos2θ + 11cosθ - 7 = 0 2cos2θ - 11cosθ + 5 = 0 (2cosθ - 1)(cosθ - 5) = 0 cosθ =1/2 或 cosθ = 5 (捨去) θ = π/3 7. 答案是： (D) 49π/3 b2 = a2 + c2 - 2*a*c*cosB (餘弦定律) b2 = 52 + 32 - 2*5*3*cos120° b = 7 b/sinB = 2R (正弦定律) 7/sin120° = 2R 外接圓半徑 R = 7/√3 外接圓面積 = πR2 = π(7/√3)2 = 49π/3 8. 答案是： (D) 136 f(1) = 2(1)^7 - 15(1)^5 + 52(1)^4 - 65(1)^3 + 18(1) + 10 = 2 f(2) = 2(2)^7 - 15(2)^5 + 52(2)^4 - 65(2)^3 + 18(2) + 10 = 134 f(1) + f(2) = 2 + 134 = 136 9. 答案是： (B) 4 餘式 = 4(1/2)^4 + 4(1/2)^3 + (1/2)^2 + 3 = 4 10. 答案是： (A) 11 log2x + log2y = 1 log2xy = 1 xy = 2 ...... [1] x2 + y2 = 117 ...... [2] [1]*2 + [2] : x2 + 2xy + y2 = 121 (x + y)2 = 121 x + y = 11 或 x + y = -11 (捨去)
1.P(k+1/3k+1,2k+1/k-2)在第二象限內=>k+1/3k+1<0，且2k+1/k-2>0=>(k+1)?(3k+1)<0，且(2k+1)?(k-2)>0=>-1
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