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    Graphs/Functions - 2015 Sample Paper 1 JC HL RyanS

    Hello,

    I am having trouble with Questions 7 and 10! Can someone please help?

    Thanks :)

    https://www.examinations.ie/schools/Mathematics_Sample_Paper_P1_-_Higher_Level.pdf

    1. avatar image

      Dazzla16

      Hi RyanS! Here are the worked solutions to those two questions:

      7(a)

      x^2 + (10 - x)^2 = the area of the two shaded squares. We will try different values for x to find the minimum possible value.

      f(1) = (1)^2 + (10 - 1)^2 = 82

      f(2) = (2)^2 + (10 - 2)^2 = 68

      f(3) = (3)^2 + (10 - 3)^2 = 58

      f(4) = (4)^2 + (10 - 4)^2 = 52

      f(5) = (5)^2 + (10 - 5)^2 = 50

      f(6) = (6)^2 + (10 - 6)^2 = 52

      f(7) = (7)^2 + (10 - 7)^2 = 58

      f(8) = (8)^2 + (10 - 8)^2 = 68

      f(9) = (9)^2 + (10 - 9)^2 = 82

      Therefore, the smallest possible value of the total area of the two shaded squares = 50 square units

      (You may also draw the graph of the function f(x) = x^2 + (10 - x)^2 and find the minimum value of f(x). The answer will still be 50)

      (b)

      (before I explain the answer, you have to label the two sides that touches the line d. In my solution, I called them b and c)

      In Pythagoras' theorem: a^2 = b^2 + c^2

      If we apply this to the right angled triangle, then d^2 = b^2 + c^2

      b^2 = area of bigger shaded square

      c^2 = area of smaller shaded square

      Therefore, area of bigger shaded square + area of smaller shaded square = d^2

      10(a)

      3 = (2)^2 + a(2) + b

      3= 4 + 2a + b

      2a + b = -1

      -4 = (-5)^2 + a(-5) + b

      -4 = 25 - 5a + b

      -5a + b = -29

      (b)

      (You have to solve the simultaneous equations)

      2a + b = -1

      - -5a + b = -29

      ____________

      7a = 28

      a = 4

      Sub back in to one of the equations:

      2(4) + b = -1

      8 + b = -1

      b = -9

      (c)

      y = x^2 + 4x - 9

      y = (0)^2 + 4(0) -9

      y = -9

      The co-ordinate is (0,-9)

      (d)

      y = x^2 + 4x - 9

      0 = x^2 + 4x - 9

      x^2 + 4x - 9 = 0

      ax^2 + bx + c = 0, a = 1, b = 4, c = -9

      (Now use the quadratic formula and sub in the values for a, b and c)

      x = -(4) +/- square root((4)^2 - 4(1)(-9)

      ___________________________

      2(1)

      x = -4 +/- square root(16 + 36)

      _____________________

      2

      x = -4 + square root(52)

      ________________

      2

      or

      x = -4 - square root(52)

      ________________

      2

      x = 1.6 or x = -5.6

      The two points are (1.6,0) and (-5.6,0)

      Hope this helps :)

    2. avatar image

      RyanS

      It does help. Thanks a million!! 😃

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      Me

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