Question
[Maximum mark: 7] [without GDC]
(a) Express the following numbers to 3 s.f
(i) 0.030473 (ii) 2034999 (iii) 2.3011 [2]
(b) Let \(x=1.3×10^{5}\) and \(y=2×10^{-5}\). Write down
(i) write down the exact value of x + y
(ii) write down the value of \(\frac{x}{y}\) in the standard form \(ax10^{k}\), where
1 ≤ a< 10, \(k ∈ Z\) and \(a\) correct to 3 s.f. [3]
(c) The value of \(A\) given to 3 s.f. is 2.36
Write down the range of the possible values of \(A\). [2]
▶️Answer/Explanation
Ans.
1.
(a) (i) 0.030473 ≅ 0.0305 (ii) 2034999 ≅ 2030000 (iii) 2.3011 ≅ 2.30
(b) \(x=1.3×10^{5}\) = 130000 and \(y=2×10^{-5}\) = 0.00002
(i) x + y =130000.00002 (ii) \(\frac{x}{y}\) =0.65 ×1010 = 6.5 ×109
(c) \(A\) to 3 s.f. = 2.36 hence \(A\) ∈ [2.355,2.365)
Question
a.Prove that the number \(14 641\) is the fourth power of an integer in any base greater than \(6\).[3]
b.For \(a,b \in \mathbb{Z}\) the relation \(aRb\) is defined if and only if \(\frac{a}{b} = {2^k}\) , \(k \in \mathbb{Z}\) .
(i) Prove that \(R\) is an equivalence relation.
(ii) List the equivalence classes of \(R\) on the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.[8]
▶️Answer/Explanation
Markscheme
a.\(14641\) (base \(a > 6\) ) \( = {a^4} + 4{a^3} + 6{a^2} + 4a + 1\) , M1A1
\( = {(a + 1)^4}\) A1
this is the fourth power of an integer AG
[3 marks]
(i) \(aRa\) since \(\frac{a}{a} = 1 = {2^0}\) , hence \(R\) is reflexive A1
\(aRb \Rightarrow \frac{a}{b} = {2^k} \Rightarrow \frac{b}{a} = {2^{ – k}} \Rightarrow bRa\)
so R is symmetric A1
\(aRb\) and \(bRc \Rightarrow \frac{a}{b} = {2^m}\), \(m \in \mathbb{Z}\) and \(bRc \Rightarrow \frac{b}{c} = {2^n}\) , \(n \in \mathbb{Z}\) M1
\( \Rightarrow \frac{a}{b} \times \frac{b}{c} = \frac{a}{c} = {2^{m + n}}\) , \(m + n \in \mathbb{Z}\) A1
\( \Rightarrow aRc\) so transitive R1
hence \(R\) is an equivalence relation AG
(ii) equivalence classes are {1, 2, 4, 8} , {3, 6} , {5, 10} , {7} , {9} A3
Note: Award A2 if one class missing, A1 if two classes missing, A0 if three or more classes missing.
[8 marks]
Question
[Maximum mark: 10] [without GDC]
Rational numbers, by definition, have the form of a fraction \(\frac{a}{b}\) where a ∈ Z and b∈ Z *.
(a) Given that \(x\) and \(y\) are rational numbers, prove that
(i) \(x + y\) is a rational number. (ii) \(xy\) is a rational number. [4]
(b) Given that x and y are irrational numbers, use a counterexample to prove that
(i) \(x + y\) is not always irrational. (ii)\(xy\) is not always irrational. [2]
(c) Given that x is rational and y irrational, use contradiction to prove that
(i) \(x + y\) is irrational (ii)\(xy\) is irrational (provided x ≠ 0 ) [4]
▶️Answer/Explanation
Ans.
(a) Let \(x=\frac{a}{b}\) and \(y=\frac{c}{d}\) Then
(i) \(x+y=\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\) which is rational (as a fraction of integers)
(i) \(xy=\frac{a}{b}\frac{c}{d}=\frac{ac}{bd}\) which is rational (as a fraction of integers)
(b) Given that \(x\) and \(y\) are irrational numbers, use a counterexample to prove that
(i) Let \(x=\sqrt{2}\) and \(y=1-\sqrt{2}\). Then \(x + y =1\) which is rational
(ii) Let \(x=\sqrt{2}\) and \(y=\frac{1}{\sqrt{2}}\). Then \(xy =1\) which is rational
(c) Given that \(x\) is rational and \(y\) irrational, use contradiction to prove that
(i) Suppose that \(x + y\) is rational, then \(y = (x + y) − x\) is also rational, contradiction
(ii) Suppose that \(xy\) is rational, then \(y=xy.\frac{1}{x}\) is also rational, contradiction.