IB Mathematics SL 2.2 Concept of a function, domain, range and graph AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Let the function h(x) represent the height in centimetres of a cylindrical tin can with diameter x cm.
\[ h(x) = \frac{640}{x^2} + 0.5 \quad \text{for } 4 \leq x \leq 14. \]
(a) Determine the range of h. [3]
The function h-1 is the inverse function of h.
(b)(i) Determine h-1(10). [2]
(b)(ii) In the context of the question, interpret your answer to part (b)(i). [1]
(b)(iii) Determine the range of h-1. [1]
▶️ Answer/Explanation
Markscheme
(a)
Evaluate \( h(x) = \frac{640}{x^2} + 0.5 \) at the boundaries \( x = 4 \) and \( x = 14 \):
At \( x = 4 \): \( h(4) = \frac{640}{4^2} + 0.5 = \frac{640}{16} + 0.5 = 40 + 0.5 = 40.5 \).
At \( x = 14 \): \( h(14) = \frac{640}{14^2} + 0.5 = \frac{640}{196} + 0.5 \approx 3.265306 + 0.5 \approx 3.765306 \). M1
Since \( h(x) \) is decreasing, the range is \( 3.765306 \leq h(x) \leq 40.5 \).
Rounded: \( 3.77 \leq h(x) \leq 40.5 \). A1 A1
Note: Award A1 for correct endpoints, A1 for correct interval.
[3 marks]
Evaluate \( h(x) = \frac{640}{x^2} + 0.5 \) at the boundaries \( x = 4 \) and \( x = 14 \):
At \( x = 4 \): \( h(4) = \frac{640}{4^2} + 0.5 = \frac{640}{16} + 0.5 = 40 + 0.5 = 40.5 \).
At \( x = 14 \): \( h(14) = \frac{640}{14^2} + 0.5 = \frac{640}{196} + 0.5 \approx 3.265306 + 0.5 \approx 3.765306 \). M1
Since \( h(x) \) is decreasing, the range is \( 3.765306 \leq h(x) \leq 40.5 \).
Rounded: \( 3.77 \leq h(x) \leq 40.5 \). A1 A1
Note: Award A1 for correct endpoints, A1 for correct interval.
[3 marks]
(b)(i)
Solve \( h(x) = 10 \): \( \frac{640}{x^2} + 0.5 = 10 \). M1
Simplify: \( \frac{640}{x^2} = 9.5 \), \( x^2 = \frac{640}{9.5} \approx 67.368421 \).
Thus: \( x = \sqrt{\frac{640}{9.5}} \approx 8.20782 \).
Alternatively: \( h^{-1}(x) = \sqrt{\frac{640}{x – 0.5}} \), so \( h^{-1}(10) = \sqrt{\frac{640}{10 – 0.5}} \approx 8.20782 \).
Rounded: \( x \approx 8.21 \, \text{cm} \). A1
[2 marks]
Solve \( h(x) = 10 \): \( \frac{640}{x^2} + 0.5 = 10 \). M1
Simplify: \( \frac{640}{x^2} = 9.5 \), \( x^2 = \frac{640}{9.5} \approx 67.368421 \).
Thus: \( x = \sqrt{\frac{640}{9.5}} \approx 8.20782 \).
Alternatively: \( h^{-1}(x) = \sqrt{\frac{640}{x – 0.5}} \), so \( h^{-1}(10) = \sqrt{\frac{640}{10 – 0.5}} \approx 8.20782 \).
Rounded: \( x \approx 8.21 \, \text{cm} \). A1
[2 marks]
(b)(ii)
A tin can with a height of 10 cm has a diameter of approximately 8.21 cm. A1
Note: Accept the converse interpretation.
[1 mark]
A tin can with a height of 10 cm has a diameter of approximately 8.21 cm. A1
Note: Accept the converse interpretation.
[1 mark]
(b)(iii)
The range of \( h^{-1} \) is the domain of \( h \): \( 4 \leq h^{-1} \leq 14 \). A1
Note: Accept \( 4 \leq y \leq 14 \).
[1 mark]
The range of \( h^{-1} \) is the domain of \( h \): \( 4 \leq h^{-1} \leq 14 \). A1
Note: Accept \( 4 \leq y \leq 14 \).
[1 mark]
Total Marks: 7
Question
A function is defined by \(f(x) = 2 – \frac{12}{x + 5}\) for \( -7 \leq x \leq 7 \), \( x \neq -5 \)
(a) Determine the range of f. [3]
(b) Determine the value of f−1(0). [2]
▶️ Answer/Explanation
Markscheme
(a)
Rewrite the function: \( f(x) = 2 – \dfrac{12}{x + 5} \).
As \( x \to -5^+ \), \( x + 5 \to 0^+ \), so \( \dfrac{12}{x + 5} \to +\infty \), hence \( f(x) \to -\infty \).
As \( x \to -5^- \), \( x + 5 \to 0^- \), so \( \dfrac{12}{x + 5} \to -\infty \), hence \( f(x) \to +\infty \).
Near \( x = -5 \), \( f(x) \) takes all real values except possibly some gaps.
Check the endpoints:
At \( x = -7 \): \( f(-7) = 2 – \dfrac{12}{-7 + 5} = 2 – \dfrac{12}{-2} = 2 + 6 = 8 \).
At \( x = 7 \): \( f(7) = 2 – \dfrac{12}{7 + 5} = 2 – \dfrac{12}{12} = 2 – 1 = 1 \). A1
The function is continuous on \( [-7, -5) \) and \( (-5, 7] \).
On \( [-7, -5) \): As \( x \) increases from -7 to \( -5^- \), \( x + 5 \) increases from -2 to \( 0^- \) (negative).
So, \( \dfrac{12}{x + 5} \) is negative and decreases (becomes more negative) as \( x \to -5^- \).
Thus, \( f(x) = 2 – (\text{negative}) = 2 + \text{positive} \), and it increases to \( +\infty \) as \( x \to -5^- \).
At \( x = -7 \), \( f(-7) = 8 \).
So, on \( [-7, -5) \), \( f(x) \) ranges from 8 to \( +\infty \Rightarrow [8, +\infty) \).
On \( (-5, 7] \): As \( x \) increases from \( -5^+ \) to 7, \( x + 5 \) increases from \( 0^+ \) to 12.
So, \( \dfrac{12}{x + 5} \) is positive and decreases from \( +\infty \) to 1.
Thus, \( f(x) = 2 – (\text{positive}) \), and it increases from \( -\infty \) to 1.
At \( x = 7 \), \( f(7) = 1 \).
So, on \( (-5, 7] \), \( f(x) \) ranges from \( -\infty \) to \( 1 \Rightarrow (-\infty, 1] \).
Overall, \( f(x) \) takes all real values except those between 1 and 8 (exclusive).
Range: \( (-\infty, 1] \cup [8, +\infty) \). A1 A1
[3 marks]
Rewrite the function: \( f(x) = 2 – \dfrac{12}{x + 5} \).
As \( x \to -5^+ \), \( x + 5 \to 0^+ \), so \( \dfrac{12}{x + 5} \to +\infty \), hence \( f(x) \to -\infty \).
As \( x \to -5^- \), \( x + 5 \to 0^- \), so \( \dfrac{12}{x + 5} \to -\infty \), hence \( f(x) \to +\infty \).
Near \( x = -5 \), \( f(x) \) takes all real values except possibly some gaps.
Check the endpoints:
At \( x = -7 \): \( f(-7) = 2 – \dfrac{12}{-7 + 5} = 2 – \dfrac{12}{-2} = 2 + 6 = 8 \).
At \( x = 7 \): \( f(7) = 2 – \dfrac{12}{7 + 5} = 2 – \dfrac{12}{12} = 2 – 1 = 1 \). A1
The function is continuous on \( [-7, -5) \) and \( (-5, 7] \).
On \( [-7, -5) \): As \( x \) increases from -7 to \( -5^- \), \( x + 5 \) increases from -2 to \( 0^- \) (negative).
So, \( \dfrac{12}{x + 5} \) is negative and decreases (becomes more negative) as \( x \to -5^- \).
Thus, \( f(x) = 2 – (\text{negative}) = 2 + \text{positive} \), and it increases to \( +\infty \) as \( x \to -5^- \).
At \( x = -7 \), \( f(-7) = 8 \).
So, on \( [-7, -5) \), \( f(x) \) ranges from 8 to \( +\infty \Rightarrow [8, +\infty) \).
On \( (-5, 7] \): As \( x \) increases from \( -5^+ \) to 7, \( x + 5 \) increases from \( 0^+ \) to 12.
So, \( \dfrac{12}{x + 5} \) is positive and decreases from \( +\infty \) to 1.
Thus, \( f(x) = 2 – (\text{positive}) \), and it increases from \( -\infty \) to 1.
At \( x = 7 \), \( f(7) = 1 \).
So, on \( (-5, 7] \), \( f(x) \) ranges from \( -\infty \) to \( 1 \Rightarrow (-\infty, 1] \).
Overall, \( f(x) \) takes all real values except those between 1 and 8 (exclusive).
Range: \( (-\infty, 1] \cup [8, +\infty) \). A1 A1
[3 marks]
(b)
Solve \( f(x) = 0 \) for \( x \) (since \( f^{-1}(0) \) is the \( x \)-value mapped to 0).
\( 0 = 2 – \dfrac{12}{x + 5} \Rightarrow \dfrac{12}{x + 5} = 2 \Rightarrow x + 5 = 6 \Rightarrow x = 1 \). M1
Because \( 0 \in (-\infty, 1] \) (the right-hand branch’s range) and \( f \) is increasing there, the solution is unique and valid (\( 1 \in (-5, 7] \)).
Answer: \( f^{-1}(0) = 1 \). A1
[2 marks]
Solve \( f(x) = 0 \) for \( x \) (since \( f^{-1}(0) \) is the \( x \)-value mapped to 0).
\( 0 = 2 – \dfrac{12}{x + 5} \Rightarrow \dfrac{12}{x + 5} = 2 \Rightarrow x + 5 = 6 \Rightarrow x = 1 \). M1
Because \( 0 \in (-\infty, 1] \) (the right-hand branch’s range) and \( f \) is increasing there, the solution is unique and valid (\( 1 \in (-5, 7] \)).
Answer: \( f^{-1}(0) = 1 \). A1
[2 marks]
Total Marks: 5