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IB Mathematics: Applications and Interpretation SL · 12.5: Functions · 12.5.6
Functions: exam review Exam review Worked solutions
Section A: short questions
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The diagram shows the graphs of \(f(x)=e^x\) and \(g(x)=\ln x\), and the line \(y=x\). The point A, with x-coordinate 2, lies on the graph of f. [4]
(a) Write down the coordinates of A. Give the y-coordinate to 3 significant figures. [1]
- A1 (2, 7.39)
(b) The point A is reflected in the line \(y=x\). Write down the coordinates of its image. [1]
- A1 (7.39, 2)
(c) Write down the value of \(g(f(2))\). [1]
- A1 2 (g undoes f)
(d) Write down how the functions f and g are related. [1]
- R1 f and g are inverse functions (their graphs are reflections in y = x)
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The function f is defined by \(f(x)=\dfrac{5x}{x-2}\), \(x\ne 2\). [5]
(a) Write down the equation of each asymptote of the graph of f. [2]
- A1 x = 2
- A1 y = 5
(b) The graph of \(g(x)=\dfrac{bx}{x-c}\) has asymptotes that meet at the point \((4, 3)\). Find b and c. [2]
- A1 c = 4 (vertical asymptote x = c)
- A1 b = 3 (horizontal asymptote y = b)
(c) Write down the range of f. [1]
- A1 f(x) ≠ 5
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The diagram shows the graphs of \(f(x)=10^x\) and \(g(x)=\log_{10}x\), and the line \(y=x\). The point A, with x-coordinate 0.5, lies on the graph of f. [4]
(a) Write down the coordinates of A. [1]
- A1 (0.5, 3.16) (\(10^{0.5}=3.16\))
(b) Write down the coordinates of the point B on the graph of g that is the reflection of A, and the equation of the mirror line. [2]
- A1 B(3.16, 0.5)
- A1 y = x
(c) Write down the value of \(g(3.16)\). Give your answer to 3 significant figures. [1]
- A1 0.500 (B is on the graph of g, so \(g(3.16)=0.5\))
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The number of members N of a club, t months after it opens, is modelled by \(N(t)=\dfrac{450t}{t+3}\), for \(t\ge0\). [4]
(a) Write down the equations of the vertical and horizontal asymptotes of the graph of \(N\). [2]
- A1 t = −3
- A1 N = 450
(b) Say what the horizontal asymptote tells you about the number of members. [1]
- A1 N tends to 450 as t grows, but never reaches it
(c) Find the number of members after 6 months. [1]
- A1 300 (\(450\times6\div9\))
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A photo shop prints holiday photos. The cost \(C\) euros of an order of n photos is modelled by \(C(n)=2.5+0.35n\). [9]
(a) Find \(C(40)\), and find the value of n for which \(C(n)=23.5\). [3]
- A1 C(40) = 16.50
- M1 for 2.5 + 0.35n = 23.5
- A1 n = 60
(b) The shop accepts at most 120 photos in one order. Write down a reasonable domain and a reasonable range for this model. [2]
- A1 domain 0 ≤ n ≤ 120, n a whole number (you cannot order part of an item)
- A1 range 2.5 ≤ C ≤ 44.50
(c) The function g is defined by \(g(x)=\sqrt{x-4}+3\). Write down the domain and the range of g. [2]
- A1 x ≥ 4 (the number under the square root cannot be negative)
- A1 g(x) ≥ 3 (a square root is at least 0)
(d) The function h is defined by \(h(x)=x^{2} - 6 x + 5\), for \(0\le x\le 5\). Use your calculator to sketch the graph of h, and write down the range of h. [2]
- M1 for the graph of h on the calculator with its end points and its lowest point
- A1 −4 ≤ h(x) ≤ 5
Section B: long questions
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A boat travels towards a harbour that is 18 km from where it starts. Its distance D km from the start after t hours is modelled by \(D=a+bk^{-t}\), where \(k>1\). [10]
(a) Explain why a is equal to 18. [1]
- R1 as t gets large, k^(−t) tends to 0, so D approaches a, the whole 18 km
(b) Find b. [2]
- M1 for D(0) = 0, so a + b = 0
- A1 b = −18
(c) Given that \(D(1)=5.4\), find the value of k. [3]
- M1 for 18 − 18k^(−1) = 5.4
- M1 for k^(−1) = 0.700 (12.6/18)
- A1 k = 1.43
(d) Find the distance still to travel after 3 hours. [2]
- M1 for 18 − D(3) or 18k^(−3)
- A1 6.17
(e) Find the time at which the boat is 1 km from the end. Give your answer in hours to 3 significant figures. [2]
- M1 for solving 18 − D(t) = 1, or 18k^(−t) = 1
- A1 8.10
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A freight train travels towards a depot that is 45 km from where it starts. Its distance D km from the start after t hours is modelled by \(D=a+bk^{-t}\), where \(k>1\). The train starts at \(D=0\), and after 1 hour it has travelled 20 km. [9]
(a) Explain why a is the total distance to the depot, and write down the value of a. [2]
- R1 as t increases, \(k^{-t}\) tends to 0, so D tends to a
- A1 a = 45
(b) Find the value of b. [1]
- A1 b = −45 (\(D(0)=0\): \(45+b=0\))
(c) Find the value of k. [2]
- M1 for \(45-45k^{-1}=20\)
- A1 k = 1.80
(d) Find how much further the train has to travel after 3 hours to reach the depot. Give your answer to 3 significant figures. [2]
- M1 for \(45\times1.80^{-3}\) (the distance remaining is \(45-D\))
- A1 7.72
(e) Use your calculator to find the time, to the nearest hour, when the train has travelled 44.5 km. [2]
- M1 for solving \(45-45\times1.80^{-t}=44.5\)
- A1 8 (t = 7.66, to the nearest hour)
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A bakery uses four models in its planning. [16]
(a) The profit \(P\) euros from selling loaves at x euros each is modelled by \(P(x)=- 3 x^{2} + 36 x - 60\). Use your calculator to find the y-intercept, the zeros and the vertex of the graph of P, and say what the vertex tells you about the profit. [4]
- A1 y-intercept −60
- A1 zeros x = 2 and x = 10
- A1 vertex (6, 48)
- R1 the greatest profit is 48 euros, made when the price is 6 euros
(b) The temperature of a loaf taken from the oven is modelled by \(T(t)=22+178e^{-0.06t}\), where T is in °C and t is in minutes after it leaves the oven. Write down the equation of the horizontal asymptote of the graph of T, say what it means in this context, and find T after 15 minutes. Give your answer to 3 significant figures. [3]
- A1 T = 22
- R1 the temperature of the loaf falls towards 22 °C (the temperature of the kitchen) but never goes below it
- A1 94.4 °C
(c) The number N of loaves sold in a day is inversely proportional to the price p euros, so \(N=\dfrac{a}{p}\). When \(p=2.5\), \(N=80\). Find a, and find N when \(p=4\). Explain why the graph of N against p has the vertical axis as an asymptote. [4]
- M1 for a = 80 × 2.5
- A1 a = 200
- A1 N = 50
- R1 as p tends to 0, N becomes very large, so the graph gets close to the vertical axis but never meets it
(d) The cost per loaf is modelled by \(C(n)=\dfrac{4n+60}{n+5}\), where n is the number made in a day. Find the y-intercept of the graph of C, and write down the equation of each asymptote. [3]
- A1 y-intercept 12 (C(0) = 60/5)
- A1 vertical asymptote n = −5
- A1 horizontal asymptote C = 4
(e) The amount y of dough after h hours is measured: when \(h=0, 1, 2\), y is 5, 10, 20. Say whether a linear, a quadratic or an exponential model fits better, and justify your choice. [2]
- A1 exponential
- R1 y is multiplied by 2 each hour (a constant ratio), not increased by a constant amount
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