Higher November 2021 Paper 6 Q15 – Westie's Workshop

Higher November 2021 Paper 6 Q15

OCRHigherCurrent spec7 marksIterations

15

(a) Show that the equation \(x^3 - 5x - 1 = 0\) has a solution between \(x = 2\) and \(x = 3\). [3]

(b) Find this solution correct to 1 decimal place.
You must show your working. [4]

Mark scheme (a)

Answer	Marks	Part marks and guidance
\(2^3 - 5 \times 2 - 1 = -3\) \(3^3 - 5 \times 3 - 1 = 11\) Sign change so solution between \(x = 2\) and \(x = 3\)	3	M2 for \(2^3 - 5 \times 2 - 1 = -3\) and \(3^3 - 5 \times 3 - 1 = 11\) or M1 for \(2^3 - 5 \times 2 - 1\) or \(3^3 - 5 \times 3 - 1\) soi by −3 or 11	Accept other values of \(x\) used between 2 and 3 (see table in part (b)) For full marks the two values need to produce a sign change Examples just sufficient for the third mark: change of sign −3 < 0 < 11 \(x = 2\) gives an answer < 0 and \(x = 3\) gives an answer > 0 Example insufficient for the third mark: so \(x\) lies between 2 and 3
		Alternative method After \(x^3 - 5x = 1\) seen M2 for \(2^3 - 5 \times 2 = -2\) and \(3^3 - 5 \times 3 = 12\) A1 for −2 < 1 and 12 > 1 so solution between \(x = 2\) and \(x = 3\) OR M1 for \(2^3 - 5 \times 2\) or \(3^3 - 5 \times 3\) soi by −2 or 12
		Alternative method SC3 for using an iterative equation that converges to a value in the range 2.25 to 2.35 and a concluding statement that 2 < 2.25 to 2.35 < 3 oe or SC2 for using an iterative equation that converges to a value in the range 2.25 to 2.35	If within part (a) candidates refer to their working in part (b), award marks for this final alternative method

Mark scheme (b)

Answer	Marks	Part marks and guidance
Two correct evaluations in the range 2.25 to 2.35, one which gives a positive value and the other giving a negative value	M3	M2 for two correct evaluations between 2 and 3, one positive and one negative or M1 for one correct evaluation between 2 and 3	Likely values: accept rot to 2+ sf (see table below)
and 2.3	A1dep	Dependent on achieving at least M2
Alternative method Rearranges to a correct iterative formula (converging or diverging)	M1		Condone missing subscripts
Attempts first iteration (either substitution seen or found to at least 2 dp (rot))	M1
Continues iteration(s) to reach \(x\) in the range 2.25 to 2.35	M1
2.3	A1	If 0 scored SC1 for answer 2.3 with no worthwhile working	If within part (b) candidates refer to their working in part (a), award up to full marks for part (b)

Likely values

\(x\)	\(x^3 - 5x - 1\)
2.1	−2.239
2.2	−1.352
2.25	−0.859
2.3	−0.333
2.4	0.824
2.5	2.125
2.6	3.576
2.7	5.183
2.75	6.047
2.8	6.952
2.9	8.889

\(x\)	\(x^3 - 5x - 1\)
2.25	−0.859
2.26	−0.757
2.27	−0.653
2.28	−0.548
2.29	−0.441
2.30	−0.333
2.31	−0.224
2.32	−0.113
2.33	−0.001
2.34	0.113
2.35	0.228

‹ Previous30 of 33 in IterationsNext ›