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F.4 A Maths(1)
1.(a) If f(x)=(a-x)(x-b), find the maximum value of f(x) in terms of a and b.(b) Sketch the graph of y= (5-x)(x-1)(c) Hence, find the range of values of k if the equation(5-x)(x-1) = k has two unequal positive real roots.2.Find the range of values of k for which the equation x^2+(3k+1)x+(k^2-5k-1) = 0 has... 顯示更多 1.(a) If f(x)=(a-x)(x-b), find the maximum value of f(x) in terms of a and b. (b) Sketch the graph of y= (5-x)(x-1) (c) Hence, find the range of values of k if the equation(5-x)(x-1) = k has two unequal positive real roots. 2.Find the range of values of k for which the equation x^2+(3k+1)x+(k^2-5k-1) = 0 has real roots. 3. Find the range of values of values of k if the expression 4x^2+2(k+3)x+9k^2 is positive for all real values of x. thx=]
最佳解答:
1.a. f(x) = (a - x)(x - b) = -x2 + (a + b)x - ab = -[x2 - (a + b)x + [(a + b)/2)]2] + (a + b)2/4 - ab = -[x - (a + b) / 2]2 + (a - b)2 / 4 So, the max. value of f(x) is (a - b)2 / 4. 圖片參考：http://i295.photobucket.com/albums/mm158/Audrey_hepburn2008/A_Hepburn01Aug311128.jpg?t=1220153331 c. The graph has a max when y = 4 So, the range of values of k is 0 <= k < 4 for the equation has two unequal positive real roots. 2. x2 + (3k + 1)x + (k2 - 5k - 1) = 0 For the equation has real roots, discriminant >= 0 i.e. (3k + 1)2 - 4(1)(k2 - 5k - 1) >= 0 5k2 + 26k + 5 >= 0 (5k + 1)(k + 5) >= 0 k <= -5 or k >= -1/5 3. 4x2 + 2(k + 3)x + 9k2 is positive for all real values of x So, 4x2 + 2(k + 3)x + 9k2 = 0 has no real solution Discriminant < 0 [2(k + 3)]2 - 4(4)(9k2) < 0 -35k2 + 6k + 9 < 0 35k2 - 6k - 9 > 0 (5k - 3)(7k + 3) > 0 k < -3/7 or k > 3/5
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F.4 A Maths(1)
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發問:1.(a) If f(x)=(a-x)(x-b), find the maximum value of f(x) in terms of a and b.(b) Sketch the graph of y= (5-x)(x-1)(c) Hence, find the range of values of k if the equation(5-x)(x-1) = k has two unequal positive real roots.2.Find the range of values of k for which the equation x^2+(3k+1)x+(k^2-5k-1) = 0 has... 顯示更多 1.(a) If f(x)=(a-x)(x-b), find the maximum value of f(x) in terms of a and b. (b) Sketch the graph of y= (5-x)(x-1) (c) Hence, find the range of values of k if the equation(5-x)(x-1) = k has two unequal positive real roots. 2.Find the range of values of k for which the equation x^2+(3k+1)x+(k^2-5k-1) = 0 has real roots. 3. Find the range of values of values of k if the expression 4x^2+2(k+3)x+9k^2 is positive for all real values of x. thx=]
最佳解答:
1.a. f(x) = (a - x)(x - b) = -x2 + (a + b)x - ab = -[x2 - (a + b)x + [(a + b)/2)]2] + (a + b)2/4 - ab = -[x - (a + b) / 2]2 + (a - b)2 / 4 So, the max. value of f(x) is (a - b)2 / 4. 圖片參考：http://i295.photobucket.com/albums/mm158/Audrey_hepburn2008/A_Hepburn01Aug311128.jpg?t=1220153331 c. The graph has a max when y = 4 So, the range of values of k is 0 <= k < 4 for the equation has two unequal positive real roots. 2. x2 + (3k + 1)x + (k2 - 5k - 1) = 0 For the equation has real roots, discriminant >= 0 i.e. (3k + 1)2 - 4(1)(k2 - 5k - 1) >= 0 5k2 + 26k + 5 >= 0 (5k + 1)(k + 5) >= 0 k <= -5 or k >= -1/5 3. 4x2 + 2(k + 3)x + 9k2 is positive for all real values of x So, 4x2 + 2(k + 3)x + 9k2 = 0 has no real solution Discriminant < 0 [2(k + 3)]2 - 4(4)(9k2) < 0 -35k2 + 6k + 9 < 0 35k2 - 6k - 9 > 0 (5k - 3)(7k + 3) > 0 k < -3/7 or k > 3/5
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