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IB Mathematics: Applications and Interpretation HL · 12.4: Solving Equations with Technology · 12.4.8

Solving Equations with Technology: exam review Exam review: second paper Worked solutions

Section A: short questions

  1. Three equations are \(3x + 2y = 4z - 4\), \(x - y + 2z = -3\) and \(2x + 9 = 3y - z - 3\). [5]

    1. (a) Write the equations in the form \(ax + by + cz = d\). [2]

      • A1 \(3x + 2y - 4z = -4\)
      • A1 \(2x - 3y + z = -12\) (the second equation is already in this form)

    2. (b) Use your calculator to find the values of \(x\), \(y\) and \(z\). [3]

      • M1 for entering three equations in three unknowns in the simultaneous equation solver
      • A1 for two correct values
      • A1 \(x = −2,\ y = 3,\ z = 1\)

  2. Use your calculator in this question. Give answers to 3 significant figures. [6]

    1. (a) Show that \(3x + \dfrac{4}{x} = 9\) can be written as \(3x^2 - 9x + 4 = 0\). [2]

      • M1 for multiplying every term by \(x\): \(3x^2 + 4 = 9x\)
      • A1 for moving the right-hand side across: \(3x^2 - 9x + 4 = 0\)

    2. (b) Hence find the possible values of \(x\). [2]

      • M1 for entering the coefficients 3, −9, 4 in the polynomial solver
      • A1 \(x = 0.543\) or \(x = 2.46\)

    3. (c) Find the real solution of \(2x^3 + x + 8 = 0\). [2]

      • M1 for entering the coefficients 2, 0, 1, 8 (0 for the missing \(x^2\) term)
      • A1 \(x = −1.48\) (the other two solutions are complex)

  3. Solve each equation. Give real solutions to 3 significant figures. [6]

    1. (a) Solve \(16x^2 - 24x + 9 = 0\). [2]

      • M1 for entering the coefficients 16, −24, 9
      • A1 \(x = 0.75\) (the solver gives one value: the root is repeated)

    2. (b) Show that \(x^2 + 6x + 13 = 0\) has no real solutions. [2]

      • M1 for entering the coefficients 1, 6, 13
      • R1 the solutions are \(−3 \pm 2i\): complex, so there are no real solutions

    3. (c) Find the real solutions of \(x^4 - 6x^2 + x + 3 = 0\). [2]

      • M1 for entering the coefficients 1, 0, −6, 1, 3 (0 for the missing \(x^3\) term)
      • A1 \(x = −2.43, −0.649, 0.855, 2.22\)

  4. The graph shows \(y = f(x)\), where \(f(x) = x^3 + 3x^2 - 9x\). Use your calculator. [6]

    −4−22−30−20−10102030xy

    1. (a) Find the \(x\)-intercepts of the graph. [2]

      • M1 for solving \(f(x) = 0\) (polynomial solver, coefficients 1, 3, −9, 0) or reading the intercepts from the graph
      • A1 \(x = −4.85, 0, 1.85\)

    2. (b) Solve \(f(x) = 10\). [2]

      • M1 for solving \(x^3 + 3x^2 - 9x = 10\), or for the intersections of the curve with the line \(y = 10\)
      • A1 \(x = −4.51, −0.917, 2.42\)

    3. (c) The line \(y = c\) meets the curve at exactly two points. Find the possible values of \(c\). [2]

      • M1 for the line touching the curve at a turning point: the calculator gives a maximum of height \(27\) and a minimum of height \(−5\)
      • A1 \(c = −5\) or \(c = 27\)

  5. Consider the equations \(5x + 2y = 4\) and \(15x + 6y = k\), where \(k\) is a constant. [4]

    1. (a) Find the value of \(k\) for which the system has infinitely many solutions. [2]

      • M1 for the second equation being 3 times the first, so \(k = 3 \times 4\)
      • A1 \(k = 12\)

    2. (b) Explain, using the graphs of the two equations, why the system has no solution when \(k = 15\). [2]

      • R1 both lines have gradient \(−\frac{5}{2}\), so they are parallel
      • R1 for \(k = 15\) the lines are different (not the same line), so they never meet and there is no solution

  6. Three numbers \(x\), \(y\) and \(z\) satisfy \(x \le y \le z\). Their mean is 10, their median is 9 and their range is 11. [5]

    1. (a) Write down three equations in \(x\), \(y\) and \(z\). [3]

      • A1 \(x + y + z = 30\) (mean 10 of three numbers)
      • A1 \(y = 9\) (the middle value is the median)
      • A1 \(z - x = 11\)

    2. (b) Find the three numbers. [2]

      • M1 for entering the three equations in the simultaneous equation solver
      • A1 \(x = 5,\ y = 9,\ z = 16\)

  7. A curve has equation \(y = ax^2 + bx\). It passes through the point \((3,\ 9)\) and its gradient at that point is 9. [7]

    1. (a) Write down two equations in \(a\) and \(b\). [2]

      • M1 for substituting \(x = 3,\ y = 9\): \(9a + 3b = 9\)
      • A1 \(6a + b = 9\) (from \(\frac{dy}{dx} = 2ax + b = 9\) at \(x = 3\))

    2. (b) Find the values of \(a\) and \(b\). [2]

      • M1 for solving the two equations with the solver
      • A1 \(a = 2,\ b = −3\)

    3. (c) The line \(y = 2x + 3\) meets the curve at the points P and Q. Use your calculator to find the distance PQ. [3]

      • M1 for \(2x^2 - 3x = 2x + 3\) or graphing the curve \(y = 2x^2 - 3x\) and the line
      • A1 \(x = −0.5\) and \(x = 3\) (so P and Q are \((−0.5,\ 2)\) and \((3,\ 9)\))
      • A1 PQ = 7.83

Section B: long questions

  1. A gardener mixes three fertilisers to make 300 kg of mixture. Fertiliser A contains 4% nitrogen, fertiliser B contains 6% nitrogen and fertiliser C contains 10% nitrogen. There is 20 kg less of fertiliser B than of fertiliser A. The mixture contains 18.8 kg of nitrogen. Let \(a\), \(b\) and \(c\) be the masses, in kg, of fertilisers A, B and C. [8]

    1. (a) Write down an equation for the total mass of the mixture. [1]

      • A1 \(a + b + c = 300\)

    2. (b) Write down an equation for the mass of nitrogen in the mixture. [2]

      • M1 for changing the percentages to decimals and multiplying by the amounts
      • A1 \(0.04a + 0.06b + 0.1c = 18.8\)

    3. (c) Write down an equation using the fact that there is 20 kg less of fertiliser B than of fertiliser A. [1]

      • A1 \(a - b = 20\)

    4. (d) Find the mass of fertiliser C in the mixture. [2]

      • M1 for entering the three equations in the simultaneous equation solver
      • A1 \(c = 80\): 80 kg of fertiliser C

    5. (e) Find the percentage of the nitrogen that comes from fertiliser C. Give your answer to 3 significant figures. [2]

      • M1 for \(\frac{0.1 \times 80}{18.8} \times 100\) (nitrogen from C is 8 kg)
      • A1 42.6%

  2. A school kitchen makes pies, quiches and tarts. Each pie uses 200 g of flour and 80 g of butter and sells for $4.50. Each quiche uses 150 g of flour and 90 g of butter and sells for $5.00. Each tart uses 100 g of flour and 60 g of butter and sells for $3.50. One day the kitchen uses 14.5 kg of flour and 7.1 kg of butter, and everything made is sold for a total of $400. Let \(p\), \(q\) and \(t\) be the numbers of pies, quiches and tarts made. [7]

    1. (a) Write down three equations in \(p\), \(q\) and \(t\). [3]

      • A1 \(200p + 150q + 100t = 14500\) (14.5 kg = 14 500 g)
      • A1 \(80p + 90q + 60t = 7100\) (7.1 kg = 7 100 g)
      • A1 \(4.5p + 5q + 3.5t = 400\)

    2. (b) Find the number of quiches made. [2]

      • M1 for entering the three equations in the simultaneous equation solver
      • A1 \(q = 30\) (with \(p = 40\), \(t = 20\))

    3. (c) Find the mean price at which an item is sold. Give your answer in dollars to 2 decimal places. [2]

      • M1 for \(\frac{400}{40 + 30 + 20}\)
      • (A1)(ft) $4.44

  3. A rectangular piece of fabric measures 60 cm by 40 cm. A square of side \(x\) cm is cut from each corner and the sides are sewn up to make an open storage bag. [8]

    60 − 2xx40 − 2xxEFGHIJKLMNOP

    1. (a) Show that the volume of the bag is \(V = 4x^3 - 200x^2 + 2400x\) cm³. [2]

      • M1 for length \(60 - 2x\), width \(40 - 2x\) and height \(x\): \(V = x(60 - 2x)(40 - 2x)\)
      • A1 expanding: \(V = 4x^3 - 200x^2 + 2400x\)

    2. (b) Explain why \(x\) must be less than 20. [1]

      • R1 the width \(40 - 2x\) must be positive, so \(x < 20\)

    3. (c) The volume of the bag is 7000 cm³. Find the possible values of \(x\). [3]

      • M1 for \(4x^3 - 200x^2 + 2400x - 7000 = 0\)
      • A1 the solver gives \(x = 4.37, 11.9, 33.8\)
      • A1 reject \(x = 33.8\) because it is greater than 20: \(x = 4.37\) or \(x = 11.9\)

    4. (d) Find the length and the width of the bag when \(x\) takes the smaller possible value. [2]

      • A1 length \(60 - 2 \times 4.37 = 51.3\) cm
      • (A1)(ft) width \(40 - 2 \times 4.37 = 31.3\) cm

  4. A farmer records the mass of apples, in kg, picked on days 1, 2, 3, … of a harvest. The masses form an arithmetic sequence. On day 6 she picks 29 kg. In the first 8 days she picks 172 kg in total. In a second orchard the masses picked on days 1, 2 and 3 form a geometric sequence with first term 6. The total picked over these three days is 42 kg. [10]

    1. (a) Find the mass picked on day 1 and the common difference. [3]

      • M1 for \(u_1 + 5d = 29\)
      • M1 for \(\frac{8}{2}(2u_1 + 7d) = 172\), so \(2u_1 + 7d = 43\)
      • A1 \(u_1 = 4,\ d = 5\)

    2. (b) The total mass picked in the first \(n\) days is 664 kg. Find \(n\). [3]

      • M1 for \(\frac{n}{2}(2 \times 4 + (n - 1) \times 5) = 664\), which gives \(5n^2 + 3n - 1328 = 0\)
      • A1 the solver gives \(n = −16.6\) or \(n = 16\)
      • A1 \(n = 16\) (a number of days is positive)

    3. (c) In the second orchard, find the possible values of the common ratio \(r\). [3]

      • M1 for \(6 + 6r + 6r^2 = 42\), or for the sum formula with \(r \ne 1\)
      • A1 \(6r^2 + 6r - 36 = 0\)
      • A1 \(r = 2\) or \(r = −3\)

    4. (d) State which value of \(r\) is possible for masses, giving a reason. [1]

      • R1 \(r = 2\): with \(r = −3\) the mass on day 2 would be −18 kg, which is negative

  5. A theatre is hired for a show. The income from \(x\) tickets sold is \(R = 95x - 0.4x^2\) dollars and the cost is \(C = 2000 + 15x\) dollars. [8]

    1. (a) Find the break-even quantities. [2]

      • M1 for graphing \(R\) and \(C\) (or solving \(95x - 0.4x^2 = 2000 + 15x\)) and finding the intersections
      • A1 \(x = 29.3\) and \(x = 171\) (170.7)

    2. (b) Find the smallest and the largest whole number of tickets that give a profit. [2]

      • M1 for the profit lying between the break-even values
      • A1 30 and 170 (round into the profit range: 29.3 gives 30 and 170.7 gives 170)

    3. (c) Find the greatest profit and the number of tickets that gives it. [2]

      • M1 for finding the maximum of \(P = 80x - 0.4x^2 - 2000\), or the greatest gap between the curve and the line
      • A1 $2000 at \(x = 100\)

    4. (d) The organisers want a profit of at least $1500. Find the range of whole numbers of tickets that achieves this. [2]

      • M1 for the intersections of \(P\) with \(y = 1500\): \(x = 64.6\) and \(x = 135.4\)
      • A1 65 to 135 tickets

  6. A stone is thrown upwards from a cliff top. Its height above the sea after \(t\) seconds is \(h = −4.9t^2 + 12t + 1.5\) metres, for \(t \ge 0\). [7]

    1. (a) Write down the height of the cliff top. [1]

      • A1 1.5 m (the value at \(t = 0\))

    2. (b) Find the greatest height of the stone and the time at which it is reached. [2]

      • M1 for the maximum of the graph
      • A1 8.85 m at \(t = 1.22\) s

    3. (c) Find the time at which the stone hits the sea. [2]

      • M1 for the positive root of \(h = 0\) (the other root, −0.119, is rejected)
      • A1 \(t = 2.57\) s

    4. (d) Find the length of time for which the stone is higher than 5 m above the sea. [2]

      • M1 for the intersections of \(h\) with \(h = 5\): \(t = 0.338\) and \(t = 2.11\)
      • A1 \(2.1105 − 0.3384 = 1.77\) s

jmaths.xyz · AI HL · 12.4.8 Unit review