June 2024 Paper 2 Q11

AQACurrent spec6 marksProof

(a) A student states that 3 is the smallest value of \(k\) in the interval \(3 \lt k \lt 4\)

Explain the error in the student’s statement. [1 mark]

(b) The student’s teacher says there is no smallest value of \(k\) in the interval \(3 \lt k \lt 4\)

The teacher gives the following correct proof:

Step 1:	Assume there is a smallest number in the interval \(3 \lt k \lt 4\) and let this smallest number be \(x\)
Step 2:	let \(y = \dfrac{3 + x}{2}\)
Step 3:	\(3 \lt y \lt x\) which is a contradiction.
Step 4:	Therefore, there is no smallest number in interval \(3 \lt k \lt 4\)

(i) Explain the contradiction stated in Step 3 [1 mark]

(ii) Prove that there is no largest value of \(k\) in the interval \(3 \lt k \lt 4\) [4 marks]

Scheme	Marks	AO
Explains that 3 is not in \(3 \lt k \lt 4\) OE	E1	2.4
	(1)

Mark scheme (b)

Scheme	Marks	AO
(i) Explains that \(3 \lt y \lt x\) which contradicts the definition of \(x\) as the smallest value OE	E1	2.4
	(1)
(ii) Assumes there is a largest value in (3, 4) OE	B1	2.1
Constructs a value “\(y\)” in (3, 4) which is greater than their “\(x\)” Must have referenced their “\(x\)” before this step	B1	2.2a
States that \(x \lt y \lt 4\) which is a contradiction OE	E1	2.4
Concludes that there is no largest value in (3, 4) OE CSO	R1	2.1
	(4)
	(6 marks)

(i)

\(y\) is between 3 and \(x\) which contradicts the definition of \(x\) as the smallest value in (3, 4)

(ii)

Step 1: Assume there is a largest number in the interval \(3 \lt k \lt 4\) and let this largest number be \(x\)

Step 2: let \(y = \dfrac{x + 4}{2}\)

Step 3: \(x \lt y \lt 4\) which is a contradiction.

Step 4: Therefore, there is no largest value in \(3 \lt k \lt 4\)