> GCE Further Mathematics B (MEI) Y434/01: Numerical methods Advanced GCE Mark Scheme for Autumn 2021

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Mark Scheme October 2021 3 Question Answer Marks AOs Guidance 1 (a) i 1.414214−√2 √2 or 1.4142142−2 2 oe soi 0.000000309449 isw 0.000000618898 isw M1 A1 A1 1.1a 1.1 1.1 ignore... modulus signs to 2 sf or more to 2 sf or more [3] 1 (a) ii the second relative error is double the first relative error oe B1 2.2a [1] 1 (b) Ben is wrong because the spreadsheet stores 1.414214 to a higher precision than is displayed (and so when the square of this number is calculated, 2 is returned) isw B1 2.4 or 1.414214 is an approximation to √2 so 1.4142142 ≠ 2 oe [1] 2 (a) x f(x) Δ Δ² 1 ‒0.65 0.3 2 ‒0.35 1.82 2.12 3 1.77 1.82 3.94 4 5.71 1.82 5.76 5 11.47 M1 A1 1.1 1.1 finds 4 Δ values, allow one error all correct [2]Y434/01 Mark Scheme October 2021 4 Question Answer Marks AOs Guidance 2 (b) the second differences are constant oe B1 1.1 allow the 3rd differences are zero [1] 2 (c) ‒0.65 + 0.3(x ‒ 1) + 1.82×(??−1)(??−2) 2! [P2(x) =] 0.91x² ‒ 2.43x + 0.87 M1 A1 A1 1.1 1.1 1.1 must be correct form; allow 1 substitution error two of three terms correct all correct [3] 3 (a) sinh x² ‒ x³ ‒ 2 = 0 B1 1.1 must see = 0 [1] 3 (b) =IF(H5>0,G5,E5) B1 1.1 or =IF(H5<0,E5,G5) must see = [1] 3 (c) 1.48719×17.2899−2×‒0.77825 17.2899−−0.77825 oe awrt 1.50928 awrt 1.52603 M1 A1 A1 3.1a 1.1 1.1 may be implied by 1.509… NB f(1.50928) = ‒0.6111 to 4 sf [3]Y434/01 Mark Scheme October 2021 5 Question Answer Marks AOs Guidance 3 (d) the ratios are decreasing which suggests the convergence is (slightly) faster than 1st order the ratios are close to 1 which suggests the convergence is slow B1 B1 2.2b 2.2b allow between 1st and 2nd order do not allow eg not first order [2] 4 (a) 4.2472072‒4 0.1 or 4.0239468‒4 0.01 or 4.0023871‒4 0.001 or 4.0002386‒4 [Show More]

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