IBDP Maths SL 1.1 Operations with numbers AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) (i) 0.030473 rounded to 3 significant figures: First three non-zero digits are 3, 0, 4, with the next digit 7 (≥ 5), so round up: 0.0305 A1.
(ii) 2034999 rounded to 3 significant figures: First three digits are 2, 0, 3, with the next digit 4 (< 5), so no rounding up: 2030000 A1.
(iii) 2.3011 rounded to 3 significant figures: First three digits are 2, 3, 0, with the next digit 1 (< 5), so no rounding up: 2.30 A1.
[2 marks]
(b) Given \( x = 1.3 \times 10^5 = 130000 \) and \( y = 2 \times 10^{-5} = 0.00002 \).
(i) Exact value of \( x + y = 130000 + 0.00002 = 130000.00002 \) A1.
(ii) Compute \( \frac{x}{y} = \frac{1.3 \times 10^5}{2 \times 10^{-5}} = \frac{1.3}{2} \times 10^{5 – (-5)} = 0.65 \times 10^{10} \).
Convert to standard form \( a \times 10^k \), where \( 1 \leq a < 10 \): \( 0.65 \times 10^{10} = 6.5 \times 10^9 \), with \( a = 6.5 \) (correct to 3 significant figures) M1 A1.
[3 marks]
(c) \( A = 2.36 \) to 3 significant figures. The range of possible values is determined by the rounding interval:
Lower bound: \( 2.355 \), upper bound: \( 2.365 \).
Range: \( [2.355, 2.365) \) A2.
[2 marks]
▶️ Answer/Explanation
(a) Let \( x = \frac{a}{b} \), \( y = \frac{c}{d} \), where \( a, c \in \mathbb{Z} \), \( b, d \in \mathbb{Z}^* \).
(i) \( x + y = \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \). Since \( ad + bc \in \mathbb{Z} \) and \( bd \in \mathbb{Z}^* \), \( x + y \) is rational M1 A1.
(ii) \( xy = \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \). Since \( ac \in \mathbb{Z} \) and \( bd \in \mathbb{Z}^* \), \( xy \) is rational M1 A1.
[4 marks]
(b) (i) Let \( x = \sqrt{2} \) (irrational) and \( y = 1 – \sqrt{2} \) (irrational, as \( 1 – \sqrt{2} \) cannot be written as a fraction). Then \( x + y = \sqrt{2} + (1 – \sqrt{2}) = 1 \), which is rational A1.
(ii) Let \( x = \sqrt{2} \) (irrational) and \( y = \frac{1}{\sqrt{2}} \) (irrational, as \( \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \), and \( \sqrt{2} \) is irrational). Then \( xy = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \), which is rational A1.
[2 marks]
(c) (i) Suppose \( x + y \) is rational, where \( x \) is rational and \( y \) is irrational. Let \( x = \frac{a}{b} \), \( x + y = \frac{c}{d} \), where \( a, b, c, d \in \mathbb{Z} \), \( b, d \neq 0 \). Then \( y = (x + y) – x = \frac{c}{d} – \frac{a}{b} = \frac{bc – ad}{bd} \), which is rational M1. This contradicts the assumption that \( y \) is irrational. Thus, \( x + y \) is irrational A1.
(ii) Suppose \( xy \) is rational, where \( x \neq 0 \) is rational and \( y \) is irrational. Let \( x = \frac{a}{b} \), \( xy = \frac{c}{d} \), where \( a, b, c, d \in \mathbb{Z} \), \( a, b, d \neq 0 \). Then \( y = \frac{xy}{x} = \frac{\frac{c}{d}}{\frac{a}{b}} = \frac{c}{d} \cdot \frac{b}{a} = \frac{bc}{ad} \), which is rational M1. This contradicts the assumption that \( y \) is irrational. Thus, \( xy \) is irrational when \( x \neq 0 \) A1.
[4 marks]