F.4-5 Maths

1.The floor plan of an apartment which is formed by two squares. If the perimeter and the area of the apartment are 16 m and 13 m 2 respectively, find the lengths of sides of the two squares.

2. Peter pays $45 for x mangoes with unit price $y. If the unit price is reduced by $1, he can buy 2 more mangoes and saves $1. Find the values of x and y.

3.It is given that the line L: x + y + k = 0 intersects the quadratic curve C: y = 2 x 2 - 4 x + k .Find the range of possible values of k.

4. The quadratic curve C: y = px 2 + 2x + 6 touches the line L: y = 8 – 2x at one point A.

(a) Find the value of p.

(b) Find the coordinates of A.

5.

3x-2y-1=0---------(1)

5x^2-y^2=1----------(2)

1 個解答

評分
  • 土扁
    Lv 7
    9 年 前
    最佳解答

    1.

    Let a m and b m be the lengths of sides of the two squares.

    (a > b)

    4a + 4b - 2b = 16 …… [1]

    a² + b² = 13 …… [2]

    From [1]:

    4a + 2b = 16

    2a + b = 8

    b = 8 - 2a …… [3]

    Put [3] into [2]:

    a² + (8 - 2a)² = 13

    a² + 64 - 32a + 4a² = 13

    5a² - 32a + 51 = 0

    (a - 3)(5a - 17) = 0

    a = 3 or a = 17/5

    a = 3 or a = 3.4

    Put a = 3 into [3] :

    b = 8 - 2(3)

    b = 2

    Put a = 3.4 into [3] :

    b = 8 - 2(3.4)

    b = 1.2

    The lengths of the sides of the two square is:

    2 m and 3 m or 1.2 m and 3.4 m

    =====

    2.

    xy = 45 …… [1]

    (x + 2)(y -1) = 45 - 1 …… [2]

    From [1]:

    x = 45/y …… [3]

    Put [3] into [2]:

    [(45/y) + 2](y - 1) = 44

    45 - (45/y) + 2y - 2 = 44

    2y - 1 - (45/y) = 0

    y[2y - 1 - (45/y)] = 0

    2y² - y - 45 = 0

    (y - 5)(2y + 9) = 0

    y = 5 or y = -9/2 (rejected)

    Put y = 5 into [3]:

    x = 45/5

    x = 9

    Hence, x = 9, y = 5

    =====

    3.

    L: x + y + k = 0 …… [1]

    C: y = 2x² - 4x + k …… [2]

    From [1]:

    y = -x - k …… [3]

    [2] = [3]:

    2x² - 4x + k = -x - k

    2x² - 3x + 2k = 0

    Since the equation has two real roots, thus determinant Δ > 0

    (3)² - 4(2)(2k) > 0

    9 - 16k > 0

    -16k > -9

    16k < 9

    k < 9/16

    =====

    4.

    (a)

    C: y = px² + 2x + 6 …… [1]

    L: y = 8 - 2x …… [2]

    [1] = [2]:

    px² + 2x + 6 = 8 - 2x

    px² + 4x - 2 = 0 …… [3]

    Since the equation has double roots, thus determinant Δ = 0

    (4)² - 4(p)(-2) = 0

    16 + 8p = 0

    p = -2

    (b)

    Put p = -2 into [3]:

    -2x² + 4x - 2 = 0

    x² - 2x + 1 = 0

    (x - 1) = 0

    x = 1 (double roots)

    Put x = 1 into [2]:

    y = 8 - 2(1)

    y = 6

    Coordinates of A = (1, 6)

    =====

    5.

    3x - 2y - 1 = 0 …… [1]

    5x² - y² = 1 …… [2]

    From [1]:

    2y = 3x - 1

    y = (3x - 1)/2 …… [3]

    Put [3] into[2]:

    5x² - [(3x - 1)/2]² = 1

    5x² - (3x - 1)²/4 = 1

    20x² - 9x² + 6x - 1 = 4

    11x² + 6x - 5 = 0

    (11x - 5)(x + 1) = 0

    x = 5/11 or x = -1

    Put x = 5/11 into [3]:

    y = [3(5/11) - 1]/2

    y = 2/11

    Put x = -1 into [3]:

    y = [3(-1) - 1]/2

    y = -2

    x = 5/11, y = 2/11 or x = -1, y = -2

    資料來源: 土扁
還有問題嗎?立即提問即可得到解答。