1(i) M1A1A1A1 [4] M1A1 M1A1 [4] A1A1 A1 M1 B1 M1 ... packs/MS_MEI_NM.pdf6 -0.00218 sign change so root is correct to 3 dp [M1A1] [TOTAL 8] 3 h M T S 2 3.46410 2 3.65028 2 3.52616 2

1(i) h 0.4 0.2 0.1 est f '(2) 3.195 2.974 2.871 M1A1A1A1 [4] (ii) differences -0.221 -0.103 M1A1 differences approximately halving so extrapolate to 2.871 - 0.103 = 2.768. Last figure unreliable so 2.77. Accept argument to 2.8. M1A1 [4] 2(i) E.g. 2/3 rounded to 0.666 666 7, chopped to 0.666 666 6 E1 [1] (ii) 2/3 stored as 0.666 666 7 Absolute error 0.000 000 033... A1A1 mpe is 0.000 000 05 A1 mpre is greatest when x is least M1 mpre is 0.000 000 05 / 0.1 = 5 * 10^

-7 5E-07 M1A1 [6]

3 x f(x) 1.4 -0.82176 1.5 1.09375 root in the interval (1.4, 1.5) B1 r Xr f(Xr) 0 1.4 -

0.82176

1 1.5 1.09375 M1 2 1.4429 -

0.07436 M1A1

3 1.446535 -0.00609

A1

4 1.446859 3.88E-05

M1A1

Root at 1.447 seems secure. B1 [8] 4 x f(x) 2 1.553774 M = 3.305783 M1A1 3 1.652892 T = 3.285825 A1 4 1.732051 S = (2*M + T) / 3

= 3.299130 M1A1

S(h=2) 3.299130 diffs S(h=1) 3.299231 0.0001 S(h=0.5) 3.299238 7 E -06 Differences reducing very rapidly. 3.29924 seems secure. M1A1A1 [8] 5 Computations of this type contain rounding errors E1 The rounding errors will be different when the two sums are computed E1 Adding from large to small loses precision (the small number is lost) E1 Adding from small to large allows each number to contribute to the sum E1 Hence the second sum is likely to be more accurate E1

6(i) x f(x) Δf(x) Δ²f(x) Δ³f(x) 1 4 –3 2 1 6 3 a – 13 3 4 a – 7 a – 4 87 – 3a 4 a 80 – 2a A4 76 – a (-1 each 5 76 error) [4] (ii)

87 – 3a = a – 13 gives a = 25 M1

f(x) = 4 - 3(x-1) + 6(x-1)(x-2)/2 + 12(x-1)(x-2)(x-3)/6 M1A1A1A1A1 = 4 - 3x + 3 + 3x2 - 9x + 6 + 2x3 - 12x2 + 22x -12 A1 = 2x3 - 9x2 + 10x + 1 A1 [8] Algebra may appear in

(iii)

(iii)

f '(x) = 6x2 - 18x + 10 = 0 rather than (ii) for full credit

M1

x = 2.26 (2.26376) A1 f(2.26...) = 0.718 A1 [3] (iv) f(x) = 4 (x - 2)(x - 3)(x - 5)/(1 - 2)(1 - 3)(1 - 5) + three similar terms M1A1A1 [3] Total

18 7(i) 0

4 3

0.785398 2.828427 2.214602 1.570796 2.45E-16 1.429204 G1 2.356194 -2.82843 0.643806 G1 3.141593 -4 -0.14159 3.926991 -2.82843 -0.92699 4.712389 -7.4E-16 -1.71239 5.497787 2.828427 -2.49779 shows 6.283185 4 -3.28319 two roots E1 [3]

(ii) E.g.: r 0 1 2 3 4 5 6 7 Xr 1 1.04720 1.06077 1.06465 1.06576 1.06608 1.06617 1.06620 M1A1A1 alpha = 1.066 correct to 3 decimal places A1 [4] (iii) 0 1 2 3 4 5 6 1 1.04720 1.06077 1.06465 1.06576 1.06608 1.06617 diffs 0.04720 0.01357 0.00388 0.00111 0.00032 0.00009 ratio of diffs 0.28756 0.28615 0.28575 0.28564 0.28561 M1A1 ratios (approx) constant so first order convergence. E1 [3] (iv) Obtain N-R iteration (beware printed answer) M1A1 E.g.: r 0 1 2 3 4 Xr 5 4.35177 4.36435 4.36432 4.36432 M1A1 beta = 4.3643 correct to 4 decimal places A1 [5] (v) 0 1 2 3 4 5 4.35177 4.36435 4.36432 4.36432 diffs -0.64823 0.01258 -

0.00002 0.00000

ratio of diffs -0.01940 -0.00184

0.00000 M1A1

ratios getting (much) smaller so faster than first order E1 [3] Total

18

4776 Mark Scheme January 2006 MEI Numerical Methods 4776

1 Use binomial expansion of (1 + r)-1 or sum of GP, or 1 - r2 with r2 taken to be zero to obtain given result. [M1A1] Relative error in reciprocal is of same magnitude but opposite sign [E1E1] E.g. 10 is approx 2% greater than 9.8 1/10 = 0.1 is approx 2% less than 1/9.8 = 0.10204 [M1A1]

[TOTAL

6] 2(i) xr+1 = 1/sin(xr) [M1] r 0 1 2 3 10 11 12 xr 1 1.188395 1.077852 1.135147 1.113855 1.114323 1.114067 [M1A1] root is 1.11 to 3 sf. [A1]

[subtotal

4](ii) x 2.7725 2.7735

1/x - sin(x) -8.4E-05 0.000719 change of sign, so 2.773 correct to 3 dp [M1A1A1]

[subtotal

3]

[TOTAL

7] 3 h M T 2 2.60242 2.44866 1 2.56982 T2 = (M1 + T1)/2 = 2.52554 [M1A1] T4 = (M2 + T2)/2 = 2.54768 [A1]

S1 = (2M1 + T1)/3 = 2.55117 [M1A1]

S2 = (2M2 + T2)/3 = 2.55506 [A1]

I = 2.56 (or 2.555) is justified [A1]

[TOTAL

7] 4 x 1 2 3 4 5 f(x) 3 4.5 5.4 6.2 6.7 (i) (ii) (i) h f '(3) min max [M1A1A1] 2 0.925 0.9 0.95 (ii) 1 0.85 0.8 0.9 [M1A1A1] Larger h gives smaller interval (or 0.9 is the only common value) [E1] f '(3) = 0.9 is the value that seems justified. [E1] (Or 0.8 seems to be the limit the process is tending to. [E1E1])

[TOTAL 8]

4776 Mark Scheme January 2006

5 x 1 3 4 g(x) 4 1 11 L(2) = 4(2-3)(2-4)/(1-3)(1-4) + 1(2-1)(2-4)/(3-1)(3-4) + 11(2-1)(2-3)/(4-1)(4-3) [M1A1A1A1] = -11/3 [A1] x g(x) Δg(x) Δ2g(x) 1 4 -5.33333 7.666666

2 -

1.33333 2.333333 7.666667 2nd differences constant 3 1 10 so correct for a quadratic [M1A1E1] 4 11 [TOTAL 8] 6 (i) y ' = 10x9 - 10 = 0 only when x = 1, hence at most one turning point [M1A1] tenth degree polynomial is positive as x tends to plus or minus infinity [E1] (hence exactly one turning point) (or other methods) x 0 1 2 f(x) 1 -8 1005 [M1A1] changes of sign so roots in [0,1] and [1,2] [E1] since only one turning point cannot be any more roots [E1] [subtotal 7]

(ii) NR: xr+1 = xr - (xr10 - 10xr + 1)/(10xr9 - 10) [M1A1A1] r 0 1 2 3 4 5 xr 1.2 1.315589 1.284353 1.280004 1.279928 1.279928 [M1A1A1] 1.2799 [subtotal 6]

(iii) xr+1 = (xr10 + 1)/10 [M1A1]

If x0 = 0.1 then x1 = (0.110 + 1)/10 = 0.111 + 0.1 [M1A1] This is so close to 0.1 that further iterations are unnecessary [E1] [subtotal 5] [TOTAL 18]


7 (i) Mid-point rule with h=1 and 4 strips to obtain given result. [M1A1]

loge(5) = 1.60943

8

Mid-pt

= 1.57460

3 error is (-) 0.034835 [M1A1A1]

[subtotal 5]

(ii) N 10 20 40 80

loge(N) 2.30258

5 2.995732 3.688879 4.382027 [M1A1A1] [subtotal 3] (iii)(iv) N Mid-pt ln(N) est k diffs ratio of estimates

10 2.26651

1 2.302585 0.036074 diffs

[M1A1A1]

20 2.95934

6 2.995732 0.036386

0.000312

[subtotal 3]

40 3.65241

6 3.688879 0.036463

7.72E-050.247231

diffs [M1A1]

80 4.34554

3 4.382027 0.036484

2.02E-050.261472

ratio [M1A1] (approx 0.25) extrapolating: 0.036489 5.05E-06 (or equivalent) 0.036490 1.26E-06 0.036490 3.15E-07 [M1A1] 0.03649(0) seems secure [A1] [subtotal 7]

[TOTAL

18]

4776 Mark Scheme June 2006

123

1 Sketch, with explanation. [M1A1E1] f '(x) = 1/(2√x) [M1A1]

hence mpe is approx 0.05 / (2√2.5) = 0.01581

1 (0.016) [M1A1]

(or 0.05 / 2√2.45 = 0.015972 or 0.05 / 2√2.55 =

0.015656 )

[TOTAL 7] 2 x 0 1 f(x) 1 -3 change of sign so root [M1A1] a b f(a) f(b) x f(x) 0 1 1 -3 0.25 -0.24902 [M1A1]

0 0.25 1 -0.249020.20015

6 -0.00046 [A1] 0.200 to 3 dp [A1] x 0.1995 0.2005

f(x) 0.00281

6 -0.00218 sign change so root is correct to 3 dp [M1A1] [TOTAL 8] 3 h M T S

2 3.46410

2 3.65028

2 3.52616

2 values

1 3.51041

1 3.55719

2 3.52600

4 [A1A1A1A1A1]

evidence of efficient formulae for T and S

[M1M1] 3.526(0) appears to be justified [A1] [TOTAL 8] 4 h 0 0.1 0.01 0.001 f(2 + h) 1.4427 1.3478 1.4324 1.4416

est f '(2)

-0.949 -1.03 -1.1 [M1A1A1A1]

Clear loss of significant figures as h is reduced [E1] Impossible to know which estimate is most accurate [E1] [TOTAL 6] 5 x g(x) Δg Δ

2g 1 3.2 9.6 6 table

2

12.8 15.6 6.2second differences nearly constant [M1A1]

3 28.4 21.8 5.9 so approximately quadratic [E1] 4 50.2 27.7 6 5 77.9 33.7 6 111.6


124

g(1.5) = 3.2 + 0.5*9.6 + 0.5*(-0.5)*6/2 = 7.25 [M1A1A1A1] [TOTAL 7]

6 (i) x 4.6 4.7 NB: 3 pi /2 =4.71 (not reqd)

x2-tan(x) 12.2998

3 -58.6228 change of sign, so root [M1A1] a b sign f(a) sign f(b) x sign f(x) mpe 4.6 4.7 1 -1 4.65 1 0.05 [M1A1] 4.65 4.7 1 -1 4.675 -1 0.025 [M1A1]

4.65 4.675 1 -1 4.66250.012

5 [M1A1]

root is 4.6625 with mpe 0.0125 [A1]

[subtotal 9]

(ii) x 7.7 7.9

x2-tan(x) 52.8471

3 84.1251

1 no change of sign, so no evidence of root [M1A1] Sketch showing asymptote for tan(x) at 5pi/2 = 7.854 [G2] So x2 curve is above tan(x) at both end points [E1] [subtotal 5] (iii) best possible estimate is 7.8 [A1]

x 7.75 7.85 [M1] x2-tan(x) 50.4801 -189.529 change of sign so 7.8 is correct to 1 dp [A1E1] [subtotal 4] [TOTAL 18] 7 (i) D = (36 - 8) / (4 - 2) = 14

[M1A1]

I = 0.5 (-3 + 8) + (8 + 36) = 46.5 [M1A1] [subtotal 4]

(ii) q(x) = -3 (x-2)(x-4)/(1-2)(1-4) + 8 (x-1)(x-4)/(2-1)(2-4) + 36 (x-1)(x-2)/(4-1)(4-2) [M1A1A1A1] = - (x2-6x+8) - 4 (x2-5x+4) + 6 (x2-3x+2) [A1] = x2 + 8x - 12 [A1]

q'(x) = 2x + 8 so D = 12 [M1A1] ∫ q(x) dx = x3/3 + 4x2 - 12x so I = 45 [M1A1A1] [subtotal 11] (iii) Large relative difference between estimates of D [E1]

Small relative difference in estimates of I [E1] To be expected as integration is a more stable process than differentiation [E1] [subtotal 3] [TOTAL 18]

4776 Mark Scheme Jan 2007

90

MEI Numerical Methods (4776) January 2007 Mark scheme 1 mpe: 0.000 000 05 x 10

98 = 5 x 1090 [M1A1] mpre: 0.000 000 05 / 1.7112245 = 2.92 x 10

-8 [M1A1] Extra digits are used internally so that rounding errors will not (usually) [E1] show in the displayed answer [TOTAL 5] 2 (i) tan 0.2=

0.202710 approx =

0.202667 [A1A1]

error: -4.3E-05 rel error: -0.00021 [A1A1] [subtotal 4]

(ii) k 0.2^5= 4.34E-05 hence k= 0.13552

8 accept 0.13 or 0.14 [M1A1A1] [subtotal 3] [TOTAL 7] 3 r 0 1 2 3

xr 0.35 0.354767 0.35646

20.35706

7

Differences 0.004767 0.00169

50.00060

5 [M1A1]

Ratio of differences 0.35557

00.35693

2 [M1A1]

root = 0.35706

7 +0.000605 (0.356932 + 0.3569322 + …) [M1A1]

= 0.35740

3 [A1] 0.3574 seems justified [A1] [TOTAL 8] 4 Graph of y = cos x and y = x2 showing one intersection for x > 0. (Or equivalent.) [G2] x 0.7 0.9 change of sign so root

cos x -x2 0.27484

2 -0.18839 [M1A1] r 0 1 2 3 4 root

xr 0.7 0.9 0.81866

30.82390

90.82413

3 0.824 [M1A1A1]

f(x) 0.27484

2 -0.18839 0.01298

90.00053

1 -1.6E-06 to 3dp [A1] [TOTAL 8]5 x 0 0.25 0.5 f(x) 1.1105 1.2446 1.4065 h 0.5 0.25 f '(0) 0.5920 0.5364 [M1A1A1] poor accuracy: estimates very different, at most 1 dp reliable [E1] [subtotal 4] h 0.25 f '(0.25) 0.5920 [M1A1] accuracy likely to be better because of the use of central difference formula; [E1]

4776 Mark Scheme Jan 2007

91

nothing more than 1 dp because there is nothing to compare the answer with. [E1] [subtotal 4] [TOTAL 8] 6 x 0.9 1.1 1.2 1.4 1.5 f(x) -0.43 -0.09 0.15 0.78 1.15 (i) y = -0.09 (x - 1.2) / (1.1 - 1.2) + 0.15 (x - 1.1) / (1.2 - 1.1) = 2.4 x - 2.73 [M1A1A1A1]

Estimate of α: 1.1375 [A1] Using values -0.085 and 0.155 gives α as 1.1354 [M1A1] Using values -0.095 and 0.145 gives α as 1.1396 [M1A1] Hence quote 1.14 [A1] [subtotal 10] (ii) y = -0.09 (x - 1.2) (x - 1.4) / (1.1 - 1.2) (1.1 - 1.4) + two similar terms [M1A1A1A1]

y = -3 (x2 - 2.6x + 1.68) - 7.5 (x2 - 2.5x +1.54) + 13 (x2 - 2.3x + 1.32) [A1] = 2.5 x2 - 3.35x + 0.57 [A1] y = 0 gives α = 1.14 (reject other root) [M1A1] [subtotal 8] [TOTAL 18] 7 (i) x x^-x M T S 1 1 2 0.25 0.625

1.5 0.54433

1 0.544331 0.58466

60.57122

1 (h=0.5) [M1A1A1]

1.25 0.75659

3

1.75 0.37556

4 0.566078 0.57537

20.57227

4(h=0.25)

[M1A1A1A1] [subtotal 7]

(ii) a = 0.57122

1

b = 0.57227

4 b - a = 0.00105

3

c = 0.57234

4 c - b = 7E-05ratio = 0.06647

7

[M1A1A1] Theoretically 1/16 (= 0.0625): good agreement with theory. [E1E1] [subtotal 5] (iii) 0.572344 + 0.0000699 (1/16 + 1/162 + …) [M1A1]

= 0.572349 [A1] 0.57235 appears completely secure from the rate of convergence [A1E1] but there may be rounding errors in the 6th dp [E1] [subtotal 6] [TOTAL 18]


1 x f(x) 1 2.414214 < 3 1.4 3.509193 > 3 change of sign hence root in (1, 1.4) [M1A1] 1.2 2.92324 < 3 root in (1.2, 1.4) est 1.3 mpe 0.1 1.3 3.206575 > 3 root in (1.2, 1.3) est 1.25 mpe 0.05 1.25 3.0625 > 3 root in (1.2, 1.25) est 1.225 mpe 0.025 [M1A1A1] mpe [A1] mpe reduces by a factor of 2, 4, 8, … Better than a factor of 5 after 3 more iterations [M1A1] [TOTAL 8] 2 x 1/(1+x^4) values: [A1] 0 1 M = 0.498054 [A1] 0.25 0.996109 T = 0.485294 [A1] 0.5 0.941176 S = (2M + T) / 3 = 0.493801 [M1] h S ΔS 0.5 0.493801 0.25 0.493952 0.000151 / one term enough [M1] Extrapolating: 0.493952 + 0.000151 (1/16 + 1/162 +...) = 0.493962 [M1A1] 0.49396 appears reliable. (Accept 0.493962) [A1] [TOTAL 8] 3 Cosine rule: 5.204972 [M1A1] Approx formula: 5.205228 [A1] Absolute error: 0.000255 [B1] Relative error: 0.000049 [B1]

[TOTAL 5] 4(i) r represents the relative error in X [E1] (ii) X

n = xn(1 + r)n ≈ xn(1 + nr) for small r [A1E1] hence relative error is nr

(iii) pi = 3.141593 (abs error: 0.001264 ) [M1] 22/7 = 3.142857 rel error: 0.000402 [A1] approx relative error in π2 (multiply by 2): 0.000805 (0.0008) approx relative error in sqrt(π) (multiply by 0.5): 0.000201 (0.0002) [M1M1A1] [TOTAL 8] 5 x f(x) -1 3 f(x) = 3 x (x-4) / (-1)(-5) + [M1A1] 0 2 2 (x+1)(x-4) / (1)(-4) + [A1] 4 9 9 (x+1) x / (5)(4) [A1] f(x) = 0.55 x2 - 0.45 x + 2 [A1] f '(x) = 1.1 x - 0.45 [B1] Hence minimum at x = 0.45 / 1.1 = 0.41 [A1] [TOTAL 7]

4777 Mark Scheme June 2007 6(i) Sketch showing curve, root, initial estimate, tangent, intersection of tangent with x-axis as improved estimate [E1E1E1]

[subtotal 3]

(ii) 0 0 0.35 -0.33497 Sketch showing root, α [G2] 0.7 -0.55771 1.05 -0.35668 E.g. starting values just to the 1.4 2.997884 left of the root can produce an [M1] x1 that is the wrong side of the asymptote [E1]

E.g. starting values further left [M1] can converge to zero. [E1] [subtotal 6] (iii) Convincing algebra to obtain the N-R formula [M1A1] r 0 1 2 3 4 xr 1.2 1.169346 1.165609 1.165561 1.165561 [M1A1A1] root is 1.1656 to 4 dp [A1] differences from root -0.03065 -0.00374 -4.8E-05 Accept diffs of successive terms ratio of differences 0.1219 0.012877 [M1A1] ratio of differences is decreasing (by a large factor), so faster than first order [E1] [subtotal 9]

[TOTAL 18] 7 (i) x g(x) Δg Δ2g 1 2.87 2 4.73 1.86 3 6.23 1.50 -0.36

4 7.36 1.13 -0.37 5 8.05 0.69 -0.44 [M1A1A1]

Not quadratic [E1] Because second differences not constant [E1] [subtotal 5] (ii) x g(x) Δg Δ2g 1 2.87 3 6.23 3.36 5 8.05 1.82 -1.54 [B1] Q(x) = 2.87 + 3.36 (x - 1)/2 - 1.54 (x - 1)(x - 3)/8 [M1A1A1A1] = 0.6125 + 2.45 x - 0.1925 x2 [A1A1A1] [subtotal 8] (iii) x Q(x) g(x) error rel error Q: [A1A1] 2 4.7425 4.73 0.0125 0.002643 errors: [A1] 4 7.3325 7.36 -0.0275 -0.00374 rel errors: [M1A1] [subtotal 5] [TOTAL 18]


64

4776 Numerical Methods

1 x 2 3 root = (2 x 0.03 - 3 x 0.24) / (0.03 - 0.24) [M1A1] f(x) 0.24 0.03 = 3.142857 [A1] Eg: graph showing turning point at x = 3 with root some way [G2] to the left or the right. [TOTAL 5] 2 x f(x) 0 1 1 0.333333 T1 = 0.666667 [M1A1]

0.5 0.477592 M = 0.477592 [M1A1] hence T2 = (T1 + M)/2 = 0.572129 [M1A1] and S = (T1 + 2*M)/3 = 0.540617 [M1A1] [TOTAL 8] 3 x 0 1 3 f(x) 2 2.57 3.85 3 terms: [M1] form: [M1] use x=2: [M1] f(2) = 2(2-1)(2-3)/(0-1)(0-3) + 2.57(2-0)(2-3)/(1-0)(1-3) + 3.85(2-0)(2-1)/(3-0)(3-1) [A1A1A1] = 3.186667 (3.19) [A1] [TOTAL 7] 4 x 1.5 2 x3(2-x)-1 0.6875 -1 change of sign, so root (may be implied) [M1A1] a b x x

3(2-x)-1 mpe 1.5 2 1.75 0.339844 0.25 [M1A1] 1.75 2 1.875 -0.17603 0.125 [A1] 1.75 1.875 1.8125 0.0625 [A1] 4 further iterations reqd: mpe 0.0325, 0.015625, 0.0078125, 0.00390625 [M1A1] [TOTAL 8] 5 Sketch showing curve, tangent, chord, h. Makes clear that tangent and chord have substantially different gradients. [G3] h 0 0.1 0.01 0.001 g(2 + h) 3.61 3.849 3.633 3.612 est g '(2) 2.39 2.3 2 [M1A1A1A1] Clear loss of significant figures as h is reduced [E1] [TOTAL 8]


65

6 x f(x) Δf Δ2f Δ3f

(i) 3 1 4 3 2 5 -1 -4 -6 6 -10 -9 -5 1 [M1A1A1] quadratic = 1 + 2(x-3) - 6(x-3)(x-4)/2 [M1A1] = 1 + 2x-6 - 3x2+21x-36 [A1] = -3x2 +23x -41 [A1] q'(x) = -6x + 23 = 0 at x = 23/6 (= 3.833...) [M1A1]

q(x) = 0 at x = 4.847(127); also at 2.81954 - not reqd. [M1A1] q(6) = -11 (or point out that the second differences not constant) [A1]

[subtotal

12]

(ii) cubic est = 1 + 2(4.5-3) - 6(4.5-3)(4.5-4)/2 + 1(4.5-3)(4.5-4)(4.5-5)/6 [M1A1A1] = 1.6875 [A1] S = 1.5/3 (1 + 4x1.6875 -10) = -1.125 [M1A1] [subtotal 6] [TOTAL 18] 7 (i) mpe 0.000 000 5 [B1]

mpre 0.000 000 5 / 2.506 628 = 1.99 x 10-7 [M1A1]

[subtotal 3]

(ii) mpe 1000 x 0.000 000 5 = 0.000 5 [M1A1] In practice the positive and negative errors will tend to cancel out [E1]

[subtotal 3]

(iii) mpe 1000 x 0.000 001 = 0.001 [M1A1] In practice 1000 x 0.000 000 5 = 0.000 5 [M1A1] because average error in chopping will be 0.000 000 5 [E1] [subtotal 5]

(iv) L to R: 1 (or 1.000 000) [B1] R to L: 1.000 001 [B1] L to R requires 8 sf, (R to L doesn't) [E1] [subtotal 3]

(v) Reverse order more accurate [E1] as that way allows the very small terms at the end of the series to contribute to the sum. [E1] The spreadsheet is likely to work to greater accuracy [E1] The spreadsheet works to more sf than are displayed [E1] [subtotal 4] [TOTAL 18]


1 x 3 3.5 root = (3 x (-0.8) - 3.5 x 0.5) / (-0.8 - 0.5) [M1A1A1] f(x) 0.5 -0.8 = .192308 (3.192, 3.19) [A1] (-) mpe is 3.5 - 3.192308 = 0.307602 (0.308, 0.31) [M1A1] [TOTAL 6]

2 1 2 3 1 -1 5 5 4 5

7 k k-5 k-9 k-14 9 2 2-k 7-2k 16-3k [M1A1A1A1] 16-3k = k-14 hence k = 7.5 [M1A1] [TOTAL 6]

3 h f(2+h) f(2-h) f '(2) derivatives 0.2 .494507 .867869 .566594 [M1A1A1A1] 0.1 .323418 .010586 .564163 -0.00243 differences 0.05 .241636 .085281 .563555 -0.00061 [M1A1] differences reducing by a factor 4 so next estimate about 1.56340. [M1] 1.563 secure to 3 dp. [B1] [TOTAL 8]

4 f(x) = x3-25 f '(x) = 3x2 [M1A1A1] xr+1 = xr - (xr3-25)/3xr2 (a.g.)

r 0 1 2 3 xr 4 3.1875 .945197 2.92417 [M1A1] diffs -0.8125 -0.2423 -0.02103 [B1] ratios .298219 .086783 [B1] differences reducing at an increasing rate (hence faster than first order) [E1] [TOTAL 8]

5 (i) 0.001 369 352 (accept 0.001 369 4) [B1]

(ii) sin 86o = 0.997 564

sin 85o = 0.996 195

[B1B1]

sin 86o - sin 86o = 0.001 369 [A1]

(iii) 2 x 0.0784591 x 0.008 726 54 [M1] = 0.00136935 [A1]

(iv) Rounding has different effects in the two expressions (may be implied) [E1] First method involves subtraction of nearly equal numbers and so loses accuracy [E1] [TOTAL 8]

6 (i) h M T 2 2.763547 2.425240 1 2.677635 2.594393 mid-point: [M1A1A1] 0.5 2.656743 2.636014 trapezium: [M1A1A1A

1] [subtotal 7]

(ii) M: 2.763547 diffs 2.677635 -0.08591 2.656743 -0.02089 reducing by a factor 4 (may be implied) [M1A1E1]

Differences in T reduce by a factor 4, too [B1] [subtotal 4]

(iii) M T S 2.763547 2.425240 2.650778 [M1] 2.677635 2.594393 2.649888 -0.00089033 S values: [A1A1] 2.656743 2.636014 2.649833 -0.000054333 diffs [A1] Differences in S reducing fast e.g by a factor of (about) 16 [E1] How this leads to an answer, e.g: Next difference about -0.0000034 and/or next answer about 2.649830 [E1] Accept 2.6498 or 2.64983 [A1] [subtotal 7] [TOTAL 18]

7 (i) Eg: graph of x2 and 4 + 1/x for x > 0 showing single intersection [G2] Change of sign to find interval (2,3) - i.e. a = 2 [B1]

r 0 1 2 3 4 5 xr 2.5 2.097618 2.115829 2.114859 2.11491 2.114907 [M1A1A1] 2.1149 secure to 4 dp [A1] [subtotal 7]

(ii) The iteration gives positive values only. [E1] r 0 1 2 3 4 5 xr -2 -1.87083 -1.86158 -1.86087 -1.86081 -1.86081 [M1A1A1] -1.8608 secure to 4 dp [A1] [subtotal 5]

(iii) Eg r 0 1 2 3 4 xr -0.5 -1.41421 -1.81463 -1.85713 -1.86052 not converging to required root (converging to previous root) [M1A1] Eg xr+1 = 1 / (xr2 - 4) [M1] r 0 1 2 3 4 5 xr -0.5 -0.26667 -0.25452 -0.25412 -0.2541 -0.2541 [M1A1] -0.2541 secure to 4 dp [A1] [subtotal 6] [TOTAL 18]


66


1(i) x y 1st diff 2nd diff -3 -16 -1 -2 14

1 4 6 -8 [M1A1] 3 2 -2 -8 2nd difference constant so quadratic fits [E1] (ii) f(x) = -16 + 14(x + 3)/2 - 8(x + 3)(x + 1)/8 [M1A1A1A1] = -16 + 7x + 21 - x2 - 4x - 3 = 2 + 3x - x2 [A1] [TOTAL 8] 2(i) Convincing algebra to demonstrate result [M1A1] (ii)(A) Direct subtraction: 0.0022 [B1] (B) Using (*): 1/(223.6090+223.6068) = 0.002236057 [M1A1] Second value has many more significant figures ("more accurate") -- may be implied [E1] Subtraction of nearly equal quantities loses precision [E1] [TOTAL 7] 3(i) x f(x) 0 1 0.8 0.819951 T1 = 0.72798 [M1] 0.4 0.994867 M1 = 0.795893 [M1] hence S1 = 0.773256 [M1] all values [A1] (ii) T2 = 0.761937 [B1] M2 = 0.784069 so S2 = 0.776692 [M1A1] S2 will be much more accurate than S1 so 0.78 or 0.777 would be justified [A1] [TOTAL 8]

4(i) x cosx 1 - 0.5x2

error rel error 0.3 0.955336 0.955 -0.000336 -0.000352 condone signs here [M1A1A1A1] but require correct (ii) want k 0.34 = 0.000336 sign for k [M1] gives k = 0.041542 (0.0415, 0.042, 1/24) [A1] [TOTAL 6] 5 r 0 1 2 xr 3 3 3 xr 2.99 2.9701 2.911194 [M1A1A1] xr 3.01 3.0301 3.091206 Derivative is 2x - 3. Evaluates to 3 at x = 3 [M1A1] 3 is clearly a root, but the iteration does not converge [E1] Need -1 < g'(x) < 1 at root for convergence [E1] [TOTAL 7]


67

6(i) Demonstrate change of sign (f(a), f(b) below) and hence existence of root [B1] a b f(a) f(b) x mpe f(x)

0.2 0.3 -0.06429 0.021031 0.25 0.05 -0.01827 [M1] 0.25 0.3 -0.01827 0.021031 0.275 0.025 0.002134 [M1] 0.25 0.275 0.2625 0.0125 -0.00787 [A1A1A1] [subtotal 6] (ii) r xr fr 0 0.2 -0.06429 1 0.3 0.021031 2 0.275352 0.00241 [M1A1] 3 0.272161 -0.0001 [M1A1] accept 0.27 or 0.272 as secure [A1] secant method much faster [E1] [subtotal 6] (iii) r xr er er+1/er2 0 1.4 0.101496 e col: [M1A1] 1 1.314351 0.015847 1.538329 e/e2 col: [M1A1]

2 1.298887 0.000383 1.525122 3 1.298504 = root equal values show 2nd order convergence [E1] second order convergence: each error is proportional to the square of the previous error [E1] [subtotal 6] [TOTAL 18] 7(i) fwd diff: h 0.4 0.2 0.1 f '(0) 0.444758 0.473525 0.48711 [M1A1A1] diffs 0.028768 0.013585 approx halved [B1] [subtotal 4]

(ii) cent diff: h 0.4 0.2 0.1 f '(0) 0.491631 0.498315 0.50008 [M1A1A1] diffs 0.006684 0.001765 reduction greater than [B1]

for forward difference [subtotal 4]

(iii) (D2 - d) = 0.5 (D1 - d) convincing algebra to d = 2D2 - D1 [M1A1] (D2 - d) = 0.25 (D1 - d) convincing algebra to d = (4D2 - D1)/3 [M1A1A1] [subtotal 5] (iv) fwd diff: 2(0.48711) - 0.473525 = 0.500695 [M1A1] cent diff: (4(0.50008) - 0.498315) / 3 = 0.500668 [M1A1] 0.5007 seems secure [E1] [subtotal 5] [TOTAL 18]


99


1(i) f(x) = 1.6(x - 0.4)(x - 1)/(-0.4)(-1) + 2.4x(x - 1)/0.4(0.4 - 1) + 1.8x(x - 0.4)/1(1 - 0.4) [M1A1,1,1] = 4(x

2 - 1.4x + 0.4) - 10(x2 - x) + 3(x2 - 0.4x) [A1] = - 3x

2 + 3.2x + 1.6 [A1] (ii) Newton's formula requires equally spaced data [E1] [TOTAL 7] 2 x 1 2 x2 + 1/x - 3 -1 1.5 (change of sign so root) [M1A1]

f(x) = x2 + 1/x - 3 so f '(x) = 2x - 1/x2 hence NR formula [M1A1] r 0 1 2 3 xr 1.5 1.532609 1.532089 1.532089 1.53209 [M1A1A1] [TOTAL 7] 3(i) term X X+Y X - Y 10X + 20Y

mpe 0.0005 0.001

0.001 0.015 [B1B1B1B

1] (ii) term X Y XY X/Y

mpre 0.000184 0.000159 0.000343 0.000343

[B1B1B1B1]

[TOTAL 8] 4(i) to 6 dp: sin A sin B LHS RHS

0.84683

2 0.841471

0.5361 0.536088 [B1B1] (ii) It is an approximate equality. LHS involves subtraction of nearly equal numbers. LHS involves 2 trig functions, RHS just 1. [E1E1] (iii) Subtraction of nearly equal quantities is a bigger problem as the difference decreases. RHS involves no such problem. [E1E1] [TOTAL 6] 5

0 0 1 r xr

0.25 0.25 0.996094 0 0.6 [G2] 0.5 0.5 0.9375 1 0.8704 0.75 0.75 0.683594 2 0.426048 [M1A1A1] 1 1 0 3 0.967052 cobweb diagram showing [M1A1A1] spiralling out from root [TOTAL 8]


100

6(i) x f(x)

0 1.732051 0.8 1.777639 T1 = 1.403876 M [M1] 0.4 1.8 M1 = 1.44 T2 = 1.421938 T [M1]

0.2 1.777639

values [A1,1,1,1]

0.6 1.8 M2 = 1.431056

[subtotal

6] (ii) S1 = 1.427959 (a.g.) [M1] S2 = 1.428016 [M1A1]

[subtotal

3] (iii) S4 = (2 M4 + T4) / 3 = 1.428020 [M1A1]

[subtotal

2] (iv) M 1.44 1.431056 1.428782 diffs -0.00894 -0.00227

ratio 0.254186 approx 0.25 [M1A1] S 1.427959 1.428016 1.428020 diffs 5.77E-05 3.99E-06 ratio 0.069037 (approx 0.0625) [M1A1A1] Reasoning to: integral is secure as 1.42802(0) [M1B1]

[subtotal

7]

[TOTAL

18] 7(i) x f(x) 1st diff 2nd diff 1 0.6 1.2 -0.1 -0.7 1.4 0.4 0.5 1.2 [M1A1]

f(x) = 0.6 + (-0.7)(x - 1) / 0.2 + 1.2(x - 1)(x - 1.2) /(2 (0.2)2)

[M1A1A1A

1 = 0.6 - 3.5x + 3.5 +15x

2 - 33x + 18 = 15x

2 - 36.5x + 22.1 [M1A1]

[subtotal

8]

(ii) f '(x) = 30x - 36.5 f '(1.2) = 36 - 36.5 = -0.5 [M1A1] Central difference: (0.4 - 0.6)/(1.4 - 1) = -0.2/0.4 = -0.5 [M1A1] Suggests central difference is accurate for quadratics. [E1]

[subtotal

5] (iii) f '(1) = 30 - 36.5 = -6.5 [B1]


101

Forward difference: (-0.1 - 0.6)/(1.2 - 1) = -0.7/0.2 = -3.5 [M1A1] Shows that forward difference is not exact for quadratics. [E1] Quadratic estimate (-6.5) is likely to be more accurate. (Allow comments [E1]

saying that we cannot be sure.) [subtotal

5]

[TOTAL

18]


102

1(i) -1 < g'(α) < 1 [B1] E.g. Multiply both sides of x = g(x) by λ and add (1 - λ)x to both sides. [M1A1]

Derivative of rhs set to zero at root: λg'(α) + 1 - λ = 0 [M1A1]

algebra to obtain given result [A1] In practice use an initial estimate x0 in place of α [A1]

[subtotal 7]

(iii) x x 3sinx - 0.5 0 0 -0.5 0.5 0.5 0.938277 1 1 2.024413 1.5 1.5 2.492485 2 2 2.227892 2.5 2.5 1.295416 3 3 -0.07664 [G3] Roots approximately 0.25, 2.1 [B1B1] Eg: r xr xr xr xr xr xr 0 0 0.2 0.4 2 2.2 2.4

1 -0.5 0.096008 0.668

255 2.2278921.92

5489 1.52639

2 -1.93828 -0.21242 1.358

852 1.8753082.31326 2.497043

3 -3.29971 -1.13247 2.432

871 2.361981.71

0416 1.302517

4 -0.02763 -3.21639 1.452

591 1.6090122.47

0807 2.392685

5 -0.58289 -0.2758 2.479

066 2.497811.36

4805 1.542517

6 -2.15131 -1.31696 1.345

334 1.3006762.43

6576 2.498801

7 -3.00855 -3.40387 2.424

072 2.3912171.44

4139 1.298298

8 -0.89795 0.277847 1.472

555 1.5457412.47

5969 2.389305

9 -2.84615 0.322857 2.485

535 2.4990581.35

2649 1.549934

10 -1.37349 0.451832 1.329

994 1.2976792.42

89 2.499347 No convergence in each case [M1A1A1] Let g(x) = 3 sinx - 0.5

Then g'(x) = 3 cosx

So λ = 1 / (1 - 3 cosα) [M1A1]


103

Smaller root: λ = -0.52446

Larger root: λ = 0.397687

(approx -0.5) (approx 0.4) [M1A1A1] r xr r xr

0 0.25 NB: must 0 2.1

1 0.253894 be using 1

2.095851

2 0.254078 relaxation 2

2.095866

3 0.254087 32.09586

6

4 0.254088 42.09586

6 [M1A1]

5 0.254088 52.09586

6 [M1A1]

[subtotal

17]

[TOTAL

24] 2(i) f(x) = 1 2h = 2a + b [M1A1]

f(x) = x, x3 give 0 = 0 [M1A1]

f(x) = x2 2h3/3 = 2aα2 [A1]

f(x) = x4 2h5/5 = 2aα4 [A1]

Convincing algebra to verify given results [A1A1]

[subtotal

8] (ii) L R m h x1 x2

0 0.785398 0.392

6990.3926

990.08851

60.69688

2

function values 1.189

207 1.04343

1 1.35535 setup:

weights

0.349066

0.218166

0.218166 [M3A3]

integral

0.415112

0.227641

0.295691

0.938444 [A1]

L R m h x1 x2

0 0.392699 0.196

350.1963

50.04425

80.34844

1


949 1.02190

31.16758

9

weights

0.174533

0.109083

0.109083

integral

0.191105

0.111472

0.127364

0.429941

0.392699 0.785398 0.589

0490.1963

50.43695

7 0.74114


58 1.21122

61.38390

1 repeat: weig 0.174 0.10908 0.10908 [M2]


104

hts 533 3 3

integral

0.225423

0.132124 0.15096

0.508508

0.9384

49 [A1] Either repeat with h halved to verify that 0.938449 is correct to 6 dp [M1A1]

Or observe that the method is converging so rapidly that 0.938449 will be correct to 6dp or [E1A1]

[subtotal

12] (iii) Use routine known to deliver 6dp and vary k:

k = 1.4657

2 L R m h x1 x2

0 0.392699 0.196

350.1963

50.04425

80.34844

1


464 1.03194

61.23791

8

weights

0.174533

0.109083

0.109083

integral

0.19835

0.112568

0.135036

0.445954

0.392699 0.785398 0.589

0490.1963

50.43695

7 0.74114 modify


898 1.29791

81.53016

4 [M1A1]

weights

0.174533

0.109083

0.109083

integral

0.24555

0.141581

0.166915

0.554046

1.0000

00 k 1.465 1.466 1.467 find k

integral 0.999908 1.000

0361.0001

63 [M1A1] hence k = 1.466

[subtotal

4]

[TOTAL

24]


105

3(i) Use central difference formulae for 2nd and 1st derivatives to obtain first given result [M1A1A1

] Hence obtain y1 = h2 - y-1 [M1A1] Use central difference to obtain y1 - y-1 = 2h [M1A1] Hence given result for y1 [M1]

[subtotal

8] (ii)

h x y

0.1 0 0

0.1 0.105 0.2 0.216472 0.3 0.332426 0.4 0.450961 0.5 0.570174 0.6 0.68815 0.7 0.802981 0.8 0.912793 0.9 1.015786 1 1.11027 1.1 1.194705 1.2 1.26774 1.3 1.328248 1.4 1.375354 1.5 1.40846 1.6 1.42726 1.7 1.431751 1.8 1.42223 1.9 1.399287 2 1.363785 2.1 1.316838 2.2 1.259773 2.3 1.194096 2.4 1.121445 2.5 1.04354 2.6 0.962141 2.7 0.878993 2.8 0.79578 2.9 0.714082 3 0.635337 3.1 0.560807 3.2 0.491549 3.3 0.428404 3.4 0.371982 3.5 0.322662 3.6 0.280597 3.7 0.245729 3.8 0.217808 3.9 0.196416 4 0.180999 4.1 0.170894 4.2 0.165365 4.3 0.163635 4.4 0.164915 4.5 0.168435 4.6 0.173469 4.7 0.179352 4.8 0.185502


106

4.9 0.191424 5 0.196725

setupnumb

ers graph [M3] [A3] [A3]

[subtotal

9] (ii) Obtain formula y1 = ah + 0.5h2 [M1A1] Modify routine [M1A1]

Trial on a to obtain a = -1.4 or -1.5 [M1A1G1

] h x y

0.1 0 0

a 0.1 -0.135 -

1.4 0.2 -0.25582 0.3 -0.36107 0.4 -0.44993 0.5 -0.5219 0.6 -0.57677 0.7 -0.6146 0.8 -0.63565 0.9 -0.64047 1 -0.6298 1.1 -0.60462 1.2 -0.56614 1.3 -0.51572 1.4 -0.45494 1.5 -0.3855 1.6 -0.3092 1.7 -0.22792 1.8 -0.14356 1.9 -0.05802 2 0.026884 2.1 0.109408 2.2 0.187962 2.3 0.26113 2.4 0.327696 2.5 0.386672 2.6 0.437316 2.7 0.479135 2.8 0.51189 2.9 0.535589 3 0.550471 3.1 0.556986 3.2 0.555768 3.3 0.547604 3.4 0.533401 3.5 0.514147 3.6 0.490876 3.7 0.464631 3.8 0.43643 3.9 0.40724 4 0.377942 4.1 0.349319 4.2 0.322033 4.3 0.296623 4.4 0.27349


107

4.5 0.252909 4.6 0.235026 4.7 0.219875 4.8 0.207386 4.9 0.197404 5 0.189706

[subtotal

7]

[TOTAL2

4] 4(i) Diagonal dominance: the magnitude of the diagonal element in any row is greater than or equal to the sum of the magnitudes of the other element. [E1] | a | > | b | + 2 will ensure convergence. ( > required as dominance has to be strict) [E1E1]

[subtotal

3] (ii) a b 4 1 2 1 1 4 2 1 4 1 2 0 2 1 4 1 0 1 2 1 4 0 0 0 0 0 0.25 -0.0625 -0.10938 -0.00391 setup 0.321289 -0.05103 -0.14691 -0.01808 [M3A3] 0.340733 -0.03941 -0.15599 -0.02648 0.344469 -0.03388 -0.15715 -0.02989 0.344515 -0.0319 -0.15681 -0.03098 0.344124 -0.03134 -0.15648 -0.03124 0.343886 -0.03123 -0.15633 -0.03127 0.343789 -0.03123 -0.15627 -0.03127 0.343758 -0.03124 -0.15625 -0.03126 0.34375 -0.03125 -0.15625 -0.03125 0.343749 -0.03125 -0.15625 -0.03125 0.34375 -0.03125 -0.15625 -0.03125 values 0.34375 -0.03125 -0.15625 -0.03125 [A3] a b 2 1 4 1 1 2 4 1 2 1 4 0 4 1 2 1 0 1 4 1 2 0 0 0 0 0 0.5 -0.25 -0.875 0.6875 2.03125 -1.95313 -3.42969 4.605469 6.033203 -10.5127 -9.11279 22.56519 12.69934 -46.9236 -13.2195 94.10735 3.347054 -183.278 37.89147 345.9377 -156.613 -632.515 456.5137 1115.079 values -1153.81 -1881.51 2690.835 2994.509 [A3] -5937.67 -4365.6 12560.88 5419.593

[subtotal

12]


108

(iii) No convergence when a = 2, b = 0 [M1A1] Indicates that non-strict diagonal dominance is not sufficient [E1E1]

[subtotal

4] (iv) Use RHSs 1,0,0,0 0,1,0,0 0,0,1,0 0,0,0,1 [M1] to obtain inverse as 0.34375 -0.03125 -0.15625 -0.03125 [A1] -0.03125 0.34375 -0.03125 -0.15625 [A1] -0.15625 -0.03125 0.34375 -0.03125 [A1] -0.03125 -0.15625 -0.03125 0.34375 [A1]

[subtotal

5]

[TOTAL

24]


64

4776 Numerical Methods 1 x LHS 1.3 2.868415 < 3 1.5 3.181981 > 3 [M1A1]

mpe (may be implied) 1.4 3.017945 0.1 [M1] 1.35 2.941413 0.05 [A1] 1.375 2.979232 0.025 [A1] 1.3875 0.0125 finishing at this point: [A1] mpe: 0.00625 0.003125 0.001563 0.000781 < 0.001 so 4 more iterations [M1A1] [TOTAL 8] 2 h M T S 1 2.579768 2.447490 2.535675 T [M1A1] 0.5 2.547350 2.513629 2.536110 S [M1A1A1] 2.536 secure by comparison of S values. [E1A1]

[TOTAL 7]

3(i) f '(x) = 3 x2 – 2 x so f '(0.5) = –0.25 [B1B1] f(0.5) = 0.875 hence given result [B1] (ii) Require –0.0005 < 0.25 h < 0.0005 [M1A1] Hence –0.002 < h < 0.002 [A1] And so 0.498 < x < 0.502 [B1]

[TOTAL 7]

4(i) Convincing algebra to given result [M1A1] (ii) Eg k = 1000 correct evaluation to 2 [B1] k = 1000000 incorrect evaluation to zero (NB some will need larger k) [B1] Mathematically equivalent expressions do not always evaluate equally [E1] (because calculators do not store (large) numbers exactly) Subtraction of nearly equal quantities often causes problems [E1]

[TOTAL 6]


65

5(i) x f(x) Δf(x) Δ2f(x) Δ3f(x) 0 1.883 1st diff: [M1A1] 1 2.342 0.459 2nd, 3rd [F1] 2 2.874 0.532 0.073 3 3.491 0.617 0.085 0.012 4 4.206 0.715 0.098 0.013 3rd diffs almost constant [E1] (ii) f(1.5) = 1.883 + 0.459 × 1.5 + 0.073 × 1.5 × 0.5 / 2! + 0.012 × 1.5 × 0.5 × (–0.5) / 3! [M1A1A1] f(1.5) = 2.598125 or 2.598 to 3 dp [A1]

[TOTAL 8]

6 (i) Sketch of smooth curve and its tangent. [G1]

Forward and central difference chords. [G1G1] Clear statement or implication that the central difference chord has gradient closer to that of the tangent [E1]

[subtotal 4]

(ii) h tan 60º tan (60 + h)º derivative

2 1.732051 1.880726 0.074338 [M1A1] 1 1.732051 1.804048 0.071997 [A1] 0.5 1.732051 1.767494 0.070886 [A1]

[subtotal 4]

(iii) h tan (60 + h)º tan (60 – h)º derivative

2 1.880726 1.600335 0.070098 [M1A1] 1 1.804048 1.664279 0.069884 [A1] 0.5 1.767494 1.697663 0.069831 [A1]

[subtotal 4]

(iv) forward difference: derivative diffs ratio

of diffs 0.074338 0.071997 –0.00234

0.070886 –0.00111 0.474407 (about 0.5, may be implied) [M1A1A1] central difference: derivative diffs ratio of diffs 0.070098 0.069884 –0.00021 0.069831 –5.3E–05 0.24896 (about 0.25, less than [M1A1E1] forward difference, hence faster) [subtotal 6]

[TOTAL 18]


66

7 (i) Sketch showing y = 3 sin x and y = x with intersection in (½π, π) [G1G1] State or show that there is only one other non-zero root [E1] [subtotal 3]

(ii) r 0 1 2 3 4 5 xr 2 2.727892 1.206001 2.80259 0.997639 2.52058 [M1A1A1]

clearly not converging [B1] Cobweb diagram to illustrate process [G3]

[subtotal 7] (iii) Convincing algebra to given result. [M1]

r 0 1 2 3 4 5 xr 2 2.242631 2.277768 2.278844 2.278862 2.278863 [M1A1A1] Root appears to be 2.27886 to 5 dp [A1] x sin x + ⅔ x 2.278855 < 2.2788625 2.278865 > 2.2788627 hence result is correct to 5 dp [M1A1E1] [subtotal 8] [TOTAL 18]

Oxford Cambridge and RSA Examinations

GCE

Mathematics (MEI) Advanced GCE 4776

Numerical Methods

Mark Scheme for June 2010


1

1(i) x LHS RHS 1 1 < 2 (Change of sign implies root.) 2 0.5 > -1 (or equivalent) [M1A1]

r 0 1 2 3 4 5 6 xr 1.5 1.333333 0.818182 0.429078 0.355127 0.347961 0.347352 [M1A1] State or clearly imply convergence outside the interval (1, 2) [E1] (ii) E.g. xr+1 = √(3 - 1/x) E.g. xr+1 = 3/x - 1/x2 [B1] r 0 1 2 3 0 1 2 3 xr 1.5 1.527525 1.531452 1.532 1.5 1.555556 1.515306 1.544287 4 5 4 5 [M1A1] 1.532077 1.532087 1.523326 1.538438 [TOTAL 8] 2(i) Forward difference: (0.9996 - 0.9854)/0.2 = 0.071 [M1A1] Central difference: (0.9996 - 0.9508)/0.4 = 0.122 [M1A1] Central difference expected to be more accurate. [E1] (ii) Forward difference maximum: (0.99965 - 0.98535)/0.2 = 0.0715 [B1] Central difference maximum: (0.99965 - 0.95075)/0.4 = 0.12225 [B1] [TOTAL 7] 3(i) r is the relative error (in X as an approximation to x) [E1] Xn = xn (1 + r)n (1 + r)n = 1 + nr (provided r is small) [M1M1A1] (ii) G2 (= 0.332 929, not required) is about 0.08% smaller than g2 √G (= 0.795 605, not required) is about 0.02% smaller than √g [M1A1A1] [TOTAL 7] 4(i) x sin + tan 2x error rel error accept: +ve, +ve 0.2 0.401379 0.4 -0.00138 -0.00344 -ve, +ve [M1A1A1A1] 0.1 0.200168 0.2 -0.00017 -0.00084 -ve, -ve (ii) 2 × 0.23 / k = 0.00138 gives k = 11.59 Either of these (or other methods) [M1A1] 2 × 0.13 / k = 0.00017 gives k = 11.76 to suggest k = 12 [B1] [TOTAL 7] 5 Data not equally spaced in x [E1] f(x) = - 10(x - 3)(x - 6) / (1 - 3)(1 - 6) - 12(x - 1)(x - 6) / (3 - 1)(3 - 6) + 30(x - 1)(x - 3) / (6 - 1)(6 - 3) [M1A1A1A1] f(x) = - (x2 - 9x +18) + 2(x2 - 7x + 6) + 2(x2 - 4x + 3) [A1] = 3x2 - 13x [A1] [TOTAL 7]


2

6(i) h M T S

0.8 1.547953 1.611209 1.569038 M: [M1A1A1] 0.4 1.563639 1.579581 1.568953 T: [M1A1] 0.2 1.567619 1.571610 1.568949 S: [M1A1] [subtotal 7](ii) 1.56895 appears justified Comparison of last two S values, e.g.: [B1] last change in S is -0.000004; next change negligible [E1] [subtotal 2](iii) h M error T error 0.8 -0.02100 0.04226 accept consistent 0.4 -0.00531 0.01063 use of other sign 0.2 -0.00133 0.00266 convention [M1A1A1] (A) M errors are about half the T errors so M is twice as accurate as T [E1A1] (B) Errors for both T and M reduce by a factor of 4 as h is halved so [E1] the rates of convergence are the same, both second order [A1A1] [subtotal 8] [TOTAL 17] 7(i) f(0) = 5, f(1) = -2. (Change of sign implies root.) [M1A1] f '(x) = 5x4 - 8 hence N-R formula [M1A1] r 0 1 2 3 4 xr 0.5 0.634146 0.638232 0.638238 0.638238 [M1A1A1] differences 0.134146 0.004086 5.98E-06 1.29E-11 [A1] ratios 0.030457 0.001462 2.17E-06 [M1A1] The ratios of differences are decreasing (fast) so process is faster than first order [E1]

[subtotal

11]

(ii) r 0 1 2 3 4 xr 1.4 1.5 1.458054 1.462741 1.46312 f(xr) -0.82176 0.59375 -0.0747 -0.00559 5.99E-05 [M1A1A1] root is 1.46 correct to 3 sf [A1] differences 0.1 -0.04195 0.004687 0.000379 [A1] ratios -0.41946 -0.11175 0.080876 [M1A1] The ratios of differences are decreasing (fast) so process is faster than first order [E1] accept 'second order' [subtotal 8] [TOTAL 19]

GCE


Mathematics (MEI) Advanced Subsidiary GCE

Unit 4776: Numerical Methods

Mark Scheme for January 2011


Question Answer Marks Guidance 1 (i) x LHS RHS 1 2 > 1.557408 no explicit explanation required 1.2 2.2 < 2.572152

M1 A1

[2] (ii) r 0 1 2 3 4 xr 1.1 0.96476 0.442927 –0.52564 –1.58007

M1 A1 r = 3 required

[2] (iii) e.g. re-arrange to x = arctan(1 + x) B1 r 0 1 2 3 4 5 xr 1.1 1.126377 1.131203 1.132076 1.132233 1.132261 M1 A1 1.132 A1 [4] 2 h M T Simpson's rule (2M + T) / 3 2 1.987467 1.354440 1.776458 T: M1A1A1 1 1.830595 1.670954 1.777381 S: M1A1A1 Lose 1 for any additional 'answer'(s) 0.5 1.750774 but do not penalise extrapolation Reference to justification/accuracy : 1.777 or 1.78 E1 A1 [8] 3 (i) h = 1 g′(0) = (2.0100 – 1.4509)/1 = 0.5591 B1 h = 0.5 g′(0) = (1.6799 – 1.4509)/0.5 = 0.458 B1 Estimate with smaller h (0.458) likely to be more accurate: B1 smaller h is more accurate (provided there is no great loss of significant

figures) E1

[4] (ii) h = 0.5 g′(0.5) = (2.0100 – 1.4509)/1 = 0.5591 M1 This estimate, central diff, likely to be more accurate than either of the

forward diffs E1

[2]

1


Question Answer Marks Guidance 4 (i) Max poss loss: 365 (or 366) times 0.01 pence: = 3.65 (or 3.66) pence B1 Arises if each daily amount would round up but gets chopped down E1 Average loss 1.825 (or 1.83) pence, because average is half of max. B1 E1 [4] (ii) £150 000 divided by 1.825 pence: about 8.2 million (8 million) accounts M1 A1 [2] 5 x P(x) ΔP(x) Δ2P(x) Δ3P(x) –1 –11 (i) bold: 1 –10 1 Diff table M1 A1 3 3 13 12 3rd diffs constant E1 5 44 41 28 16 so cubic B1 7 129 85 44 16 (ii) italic: 9 274 145 60 16 working forwards M1 A1 11 495 221 76 16 working backwards M1 A1 [4] + [4] 6 (i) x f g h abs err g rel err g abs err h rel err h

0.2 0.013351 0.013333 0.013423–

0.0000179 –0.0013424 0.0000716 0.0053600abs M1 f, g, h values may be implied

0.1 0.003334 0.003333 0.003339

–0.0000011 –0.0003339 0.0000045 0.0013350

rel M1

A1 A1 A1 A1 A1 A1 A1 [9]

(ii) Errors in g and h are of opposite sign; g is about 4 times as accurate as h. E1 E1 x f (4g + h)/5 abs err rel err 0.2 0.013351 0.013351 –2.5E–08 –1.9E–06 M1 0.1 0.003334 0.003334 –4E–10 –1.2E–07

A1 A1 A1

[6]

(iii) x / sin x ≈ 1.000 000 002 ≈ 1 B1 g(10-4) = 3.33 × 10-9 B1 Subtraction of nearly equal quantities E1 [3]

2


3

Question Answer Marks Guidance 7 (i) f(0) = –1 f(1) = 1 (hence root) B1 f '(x) = 7x6 + 5x4 which is zero only at x = 0. M1 A1 Convincing argument that this is not a turning point B1 No turning points implies no other roots. E1

-2

-1

0

1

2

3

4

5

6

0 0.5 1 1.5

G2

[7] (ii) NR iteration: xr+1 = xr – (xr7 + xr5 – 1) / (7xr6 + 5xr4) B1 r 0 1 2 xr 0.6 1.51756 1.289164 A1 A1 On graph: tangent at 0.6, intersection at 1.5, ordinate & tangent, intersection

at 1.3 G4

[7] (iii) r 0 1 2 xr 0.3 22.1703 19.00128 A1 Comment: e.g. converging but initially very slow (or difficult to tell with only

2 iter'ns) E1

r 0 1 2 xr 0.9 0.890174 0.889891 A1 Comment: e.g. almost converged, root very close to 0.89 E1 [4]

GCE


Mathematics (MEI) Advanced Subsidiary GCE




1(i) Best estimate: 1.1 Maximum possible error: 0.7 [B1B1] Bisecting mpe: 0.7 → 0.35 → 0.175 → 0.0875 → 0.04375 so 4 iterations [M1A1]

(ii) False position: estimate is (0.4 × 0.5 – 1.8 × (–0.2) / (0.5 – (–0.2)) BoD for 0.8 alone [M1A1] = 0.8 [A1] [TOTAL 7] 2(i) (3.02 – 2.66) / (2.20 – 1.80) = 0.9 [M1A1] (ii) max: (3.025 – 2.655) / (2.20 – 1.80) = 0.925 [M1A1] min: (3.015 – 2.665) / (2.20 – 1.80) = 0.875 [A1] (iii) max: (3.025 – 2.655) / (2.195 – 1.805) = 0.94872 [M1] either denominator [M1A1] min: (3.015 – 2.665) / (2.205 – 1.795) = 0.85366 correct. Max [2] if 3 dp [A1] [TOTAL 8] 3(i) Linear interpolation: Q(0) = –8 write down or any method [B1] (ii) Lagrange: Q(x) = (–4)(x – 1)(x – 5) / (–1 – 1)(–1 – 5) + Lagrange form [M1] (–12)(x + 1)(x – 5) / (1 + 1)(1 – 5) + three terms [DM1] 20(x + 1)(x – 1) / (5 + 1)(5 – 1) terms [A1,1,1] (iii) Hence by substitution Q(0) = –10 cao [M1A1] [TOTAL 8] 4(i) x 1 – x4 0.7 < 0.7599 0.8 > 0.5904 (hence root) [M1A1] (ii) Derivative of 1 – x4 is –4x3 [M1] x abs(–4x3) 0.7 1.372 > 1 [M1] for grad anywhere 0.8 2.048 > 1 (so for all [0.7, 0.8] abs gradient or abs RHS > 1) in interval [M1A1E1] [TOTAL 6] 5(i) x X abs err rel err do not insist on sign 3.162278 3.162 –0.00028 –0.0000878 but do require consistency [M1A1A1] between parts (i) and (ii) (ii) x

4 X4 (abs err rel err 100 99.96488 –0.03512) –0.0003512 allow multiplying answer in (i) by 4 [M1A1A1] (iii) Relative error will be approximately k × (–)0.0000878 [B1] [TOTAL 7]

5


6

6(i) x f(x) 2 3 M1 = 3.0801558 Lose A1, 2, 3, 4 [M1A1] 2.4 3.850195 T1 = 3.1163298 if 6, 5, 4, 3 dp [M1A1] 2.8 4.790825 T2 = 3.0982428 (=(M1+T1)/2) [M1A1] S1 = 3.0922138 (=(2M1+T1)/3) [M1A1] [subtotal 8](ii) x f(x) 2.2 3.412917 M2 = 3.0891620 [M1A1] 2.6 4.309988 S2 = 3.0921890 (=(2M2+T2)/3) [B1] or ... 889 [subtotal 3] (iii) M4 = 3.0914298 T4 = 3.0937024 [B1] S4 = 3.0921873 [B1] S1 = 3.0922138 diffs ratio of S2 = 3.0921890 –2.5E–05 diffs S4 = 3.0921873 –1.6E–06 0.0648518 approx 1/16 so fourth order [M1A1E1] (FT their precision) Consider rate of convergence of S: conclude I = 3.09219 (or 3.092187 if using extrapolation) [E1A1] [B2] for 3.09219 alone [subtotal 7] [TOTAL 18] 7(i) x f(x) Δf(x) Δ2f(x) 1 4 3 –2 –6 5 10 12 18 [M1A1] f(x) = 4 + (–6)(x – 1) / 2 + 18(x – 1)(x – 3) / (22 × 2!) [M1A1A1A1] [subtotal 6](ii) f(2) = –1.25 likely to be more accurate (interpolation) f(6) = 22.75 than this (extrapolation) [A1A1E1]

[subtotal 3](iii) x f(x) Δf(x) Δ2f(x) Δ3f(x) 1 4 3 –2 –6 5 10 12 18 7 11 1 –11 –29 ← extend table [M1A1] new f(x) = old f(x) + (–29)(x – 1)(x – 3)(x – 5) / (23 × 3!) FT incorrect old f(x) here [M1A1] f(2) = –1.25 + (–1.8125) = –3.0625 [M1] for substituting [M1A1] f(6) = 22.75 + (–9.0625) = 13.6875 cao for each [A1] [A1] [subtotal 7](iv) Absolute change greater in f(6), relative change greater in f(2) Must have all 4 values correct [E1E1] [E1], [E1] for intelligent comments on absolute, relative changes [subtotal 2] [TOTAL 18]


Question Answer Marks Guidance 1 y = 3 x (x 2) / (1) (3) + 6 (x + 1) (x 2) / (1) (2) 4 (x + 1) x / (3) (2) B1B1B1 y = (x2 2x) 3 (x2 x 2) ⅔ (x2 + x) A1 y = ⅓ (8x2 + x + 18) A1 y ' = ⅓ (16x + 1) = 0 when x = 1/16 M1A1 M1 for diffn, A1 cao [7] 2 (i) rearrange convincingly to r = (X x) / x M1A1 No further explanation required [2] 2 (ii) (1 + r)n = 1 + nr + n(n1)r2 / 2 + … M1 Need to see correct r2 term result given E1 For saying r2 negligible [2] 2 (iii) (A) n=3 6% B1 (B) n=1 2% B2 B1 for 2% [3] 3 (i) M1A1 No explanation required

x LHS RHS 1 1 < 3 2 32 > 18

Or equivalent

[2] 3 (ii) f(x) = x5 x4 – 2 f '(x) = 5x4 4x3 hence N-R formula M1A1 May be implied by subsequent work M1A1A1 M1 1st application, A1 2nd, A1 for 2

successive values agreeing to 6 dp

r 0 1 2 3 4 xr 1.5 1.455026 1.451113 1.4510851 1.4510851

hence root is 1.451085 to 6 decimal places A1 [6] 4 (i)

& h g'(5)

(ii) 0.4 0.37203 M1A1A1 Estimates 0.2 0.39065 0.01863 [3] 0.1 0.40040 0.00975 0.52349 (or 'the differences approx. halve') M1A1 M1 diffs, A1 ratio (may be implied) approx 0.5 (as h is halved) so first order E1

5


Question Answer Marks Guidance [3] 4 (iii) Some way off convergence, so accept 0.4 A1for 0.4 or for 0.41 with (or an argument by extrapolation to 0.41) E1A1 extrapolation, E1 for a reason [2] 5 (i) It means 5.5 × 10-17 E1 It is not zero because of rounding errors in the representation of numbers like 0.6, 0.4, 0.2 (or in the calculations) E2,1,0 (Or 0.6 was not exact) [3] 5 (ii) Cell B4 displays 1 either because the spreadsheet does not show enough dp E1 or because adding 1 pushes the error beyond the sf the spreadsheet stores. E1 The zero in C4 shows that the error must have been lost in B4 E1 [3] 6 (i) M1A1 Any T value A1 2 further T values A1 All T values B1 Ratios

T1 2.357070 diffs ratiosT2 2.394855 0.037785 T4 2.404253 0.009397 0.248710T8 2.406599 0.002346 0.249667

both approximately 0.25 E1 indicates 2nd order method E1 explanation [7] 6 (ii) Using ⅓ (4*T2n Tn) to obtain Simpson's rule estimates M1 (Or via M values) A1 Any S value A1 All S values B1 Ratio

S1 2.407450 diffs ratiosS2 2.407385 6.5E05 S4 2.407381 4.2E06 0.064103

approximately 1/16 E1 indicates 4th order method E1 Explanation [6] 6 (iii) Convergence of the Simpson's rule estimates suggests 2.40738 (or even 2.407381 by extrapolation) A1E1 Either answer If data are rounded to 5 dp, there is an error in the range ± 0.000 005 in B1 Sight of f(x) + or 0.000 005 each value Since the integral is over a range of 2 units the correct value will lie within ± 0.000 01 of the value given previously. A1E1 [5]

6


7

Question Answer Marks Guidance 7 (i) M1A1A1

x LHS RHS 1 0 > -0.84147 0 1 < 0 1 0 < 0.841471 or equivalent 2 3 > 0.909297 no explicit explanation reqd. [3]

7 (ii) M1A1 M1A1 M1 is for correct end-points A1

a b ) f( f(a b) x f(x) 1 0 0.841471 1 0.54304 0.18836 1 0.54304 0.841471 0.18836 0.62662 0.02093 1 0.62662 0.841471 0.02093 0.63568

The sequence of values of x has not converged: 0.6 is the only safe E1A1 estimate. [7] 7 (iii) M1A1A1A1

x 1 2 1.286979 1.377411 1.412046f(x) 0.84147 2.090703 0.30368 0.0841 0.006448

The sequence of values of x has not converged, but 1.4 appears safe E1A1 [6] 7 (iv) The rules converge at about the same rate

The rules converge slowly Reward other appropriate comments

Secant rule involves less calculation at each step (but may be harder to E2,1 implement on a calculator) [2]


5

Question Answer Marks Guidance 1 (i)

x LHS 0 1 1 −2 2 9

B1 E1

Values Reference to 2 changes of sign

[2] 1 (ii)

N-R formula 4

41 3

( 4 1)(4 4)r

r rr

x xx xx

M1 A1

B1 for derivative only

r 0 1 2 3xr 1.5 1.493421 1.493359 1.493359

M1 A1

1.4934 to 4dp A1 [5] 2 (i)

r exact approx error rel error1 1 0.723607 −0.27639 −0.276396 8 8.024922 0.024922 0.003115

B1 M1 A1 A1

Approximations Errors Errors: condone positive Relative errors: must be opposite sign

[4] 2 (ii) 20 6765.0000296 hence 6765 M1 21 10945.9999817 hence 10946 A1 Approximate values A1

dep integers

[3] 3 (i)

h f '(x) 0.2 (1.569−1.464)/0.2 0.525 0.1 (1.516−1.464)/0.1 0.52

M1 A1 A1

Number of sf reduces as h decreases (subtracting nearly equal quantities)

E1

[4]


6

Question Answer Marks Guidance 3 (ii) Change in f '(x) is −0.005. If this is half the error in the first estimate

then a better estimate is 0.52 − 0.005 = 0.515 M1 A1 E1

Or sum a gp

[3] 4 (i)

(ii)

h T M S 1 1.332375 1.377495 1.362455

0.5 1.354935 1.366179 1.362431

M1 A1 [2]

For T

M1 A1 A1

For S

[3] 4 (iii) I = 1.3624 with comment that second S more accurate than first B1

E1

OR I = 1.36243 with comment on the next difference in S [2] 5 (i)

x g(x) ∆ ∆2 ∆3

1 −15 2 −14 1 3 k k + 14 k + 134 54 54 − k 40 − 2k 27 – 3k5 145 91 37 + k 3k − 3

M1 A1

27 – 3k = 3k − 3 M1 k = 5 A1 [4] 5 (ii)

x g(x) ∆ ∆2 ∆3

0 −10 −5 6 12

M1

g(0) = −10 A1 [2]


7

Question Answer Marks Guidance 5 (iii)

2 5 3 14

5 14

M1

= 2.736842 A1 [2] 6 (i) 7.5( 2)( 4) 9.0( 1)( 4) 2.2( 1)( 2)

f( )( 3)( 5) (3)( 2) (5)(2)x x x x x x

x

M1 A1 A1 A1

20.78 1.28 9.56x x M1 A1 A1

Convincing algebra First coefficient Second coefficient

f(0) 9.56 A1 CAO Solve f( ) 0x M1 Positive root is 4.4(16295) A1 Interpolation (f(0)) generally more reliable than extrapolation (f(x) = 0) E1 [11] 6 (ii) 0.5 3(7.5 9.0) 0.5 2(9.0 2.2)L RT T M1 = 35.95 A1 [2] 6 (iii) Simpson's rule requires equal spacing in x E1 f(1.5) 9.725 M1

A1 1.5 used

13 2.5(7.5 4 9.725 2.2)S M1 = 40.5 A1 [5] 7 (i)

0

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5 6 7

G2 G1 E1

One graph Second graph Comment on roots

[4]


8

Question Answer Marks Guidance

7 (ii) Eg x0 x1 x2 x3 x4 x5

0.5 0.675938 0.615146 0.634084 0.627967 0.6299210.629921 0.629295 0.629495 0.629431 0.629451 0.629445

M1 A1

0.629 to 3dp A1 [3] 7 (iii)

x LHS RHS 3.9 0.25641 < 0.312234 4.1 0.24390 > 0.181723

B1

Eg x0 x1 x24 4.111884 5.715937

M1

May offer more iterations, or just x1 with explanation. A1

a b m f(m) mpe3.9 4.1 4 0.006802 0.13.9 4 3.95 −0.02365 0.05

3.95 4 3.975 0.025

B1 M1 A1 A1

Mpe s One iteration m

[7] 7 (iv)

x 5.2 5.4 5.328615 5.334433 5.334677f(x) 0.075762 −0.04205 0.003732 0.00015 −5.9E−07

M1 A1 A1

5.33 to 3sf A1 [4]


7

Question Answer Marks Guidance 1 (i) Convincing sketches of x2 and cos x.

Single intersection. G2 G1 for each graph

Estimate of root in [0.5, 1] B1 Accept π/4. Accept an interval in [0.5, 1] [3] 1 (ii)

Iteration 0.51 (cos )r rx x+ = M1 A1

For any valid rearrangement For writing it as an iteration (soi)

r 0 1 2 3 4 5 xr 0.8 0.83469 0.819395 0.826235 0.823195 0.82455

M1 A1

Max 1 for a diverging iteration A1 requires agreement to 2 dp

0.82 correct to 2 dp A1 Dependent on previous A1 [5] 2 (i)

n exact approx error rel error 5 252 258.3688 6.36877 0.025273

10 184756 187079 2322.973 0.012573

M1 Requires method for abs and rel error

B1 B1 B1

Approximations Errors Relative errors

errors increase but relative errors decrease with n B1 [5] 2 (ii) 110 79.5548

0.01257k = =

M1

8k = to nearest integer A1 Must be an integer OR 15 39.5726

0.0257k = =

M1

8k = to nearest integer A1 Must be an integer [2] 3 (i) x f(x) ∆ ∆2

0.1 1.641 0.2 1.990 0.349 0.3 1.840 -0.150 -0.499 these almost equal 0.4 1.192 -0.648 -0.498 (so approx qdratic)

M1 A1 E1

[3]


8

Question Answer Marks Guidance 3 (ii)

2

0.349(0.15 0.1) 0.499(0.15 0.1)(0.15 0.2)(1.5) 1.6410.1 2(0.1)

− − −= + −f

M1 A1 A1

For recognizable attempt at correct formula Either second or third term correct All three terms correct. Accept cubic.

2 out 3 for formula with x and no 0.15

= 1.878 to 3dp A1 Accept any awrt 1.878 [4] 4 (i) Sketch or convincing argument to an increasing function G2/E2 hence a single root

x 0.7 3.782174 < 4 function 0.8 4.149326 > 4

(Hence root)

M1 A1

If comparing with zero: -0.2178, 0.1493 Max 1 if function sign but not values

[4] 4 (ii)

a f(a) b f(b) x f(x) 0.7 -0.21783 0.8 0.149326 0.759329 -0.00431

0.759329 -0.00431 0.8 0.149326 0.760469

M1 A1 M1

Allow a maximum of 3 out 4 for a solution which goes wrong but self corrects For correct interval and calculating x

0.76 to 2dp A1 Must follow from false position [4] 5 (i)

h g(-h) g(h) g '(0) 0.2 1.1292 1.2745 0.36325 0.1 1.1766 1.2489 0.3615

0.05 1.1974 1.2335 0.361

M1 A1 A1 A1

Full marks for h = 0.15, 0.1, 0.05. Max 3 if other values of h used

h = 0.15 gives 0.361667

[4] 5 (ii) 0.36 because last figure still changing and so unreliable A1 E1

Any sensible comment or attempt to analyse

Or 0.361 if some argument about convergence or extrapolation is used errors [2] 6 (i)

x f(x) T M S 0 1

0.5 1.243504 0.25 1.12042 0.560876 0.560210 0.560432

M1 M Award these marks for a correct answer

M1 S or a correct method with wrong answer

M1 T Do not penalise no. of sf

[3]


9

Question Answer Marks Guidance 6 (ii)

x f(x) T M S 0.125 1.060969 0.375 1.18052 0.560543 0.560372 0.560429

0.0625 1.030816 0.1875 1.090747 0.3125 1.150255 0.4375 1.211499 0.560458 0.560415 0.560429

A6 Values Lose 1 for each error Lose 1 overall if no. of sf is not 6 FT sensible but incorrect M and/or T to S

0.560429 is justified (for information only: 0.5604289 is justified if more sf used)

A1

[7] 6 (iii)

T diffs ratio M diffs ratio 0.560876 0.560210 0.560543 -0.000333 0.560372 0.000162 0.560458 -0.000085 0.256788 0.560415 0.000043 0.262091

M1 A1 T Allow small errors that still give ratios …

M1 A1 M … of approximately 0.25

Ratios about 0.25 in each case; indicates both have 2nd order convergence

E1 E1

Allow correct explanations using 0.25 if the ratios come out wrong

But M is more accurate than T; smaller differences so nearer the correct answer

E1 E1

Allow correct statements about M and T even if not supported by the numbers

[8] 7 (i) In the first 100 terms the positive rounding errors exceed the negative

rounding errors E1 Allow E1 for an incomplete explanation that

shows some understanding

The opposite occurs in the first 200 terms. E1 [2] 7 (ii) Chopping will reduce the sum E1 by an average of 0.00005 per term ie by 0.005 and 0.01 in S100 and S200 M1A1 M1 for 0.00005, A1 rest Hence estimate as 18.5846 (18.585) and 26.8493 (26.85) A1 [4]


10

Question Answer Marks Guidance 7 (iii) 0.50.5

0.5 0.5

1 d 2kk

k kx x

x

++

− − = ∫ M1

= RHS A1 Answer given Midpoint rule M1 Must be convincing Gives LHS A1 Answer given [4] 7 (iv)

approx exact error S100 18.63572 18.5896 0.046124 S200 26.90539 26.8593 0.046091

B1 B1

Approximations

M1 A1

Errors

Errors almost exactly equal E1 [5] 7 (v)

approx assumed

error estimate S1000 61.84715 0.046 61.80115

(61.801 or 61.80)

B1 Approx

M1 A1

Correction using 0.046 (or similar) Penalize more dp

(For information, correct sum is 61.80101 to 5dp) [3]


1

1. Annotations and abbreviations

Annotation in scoris Meaning

Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response.

and BOD Benefit of doubt

FT Follow through

ISW Ignore subsequent working

M0, M1 Method mark awarded 0, 1

A0, A1 Accuracy mark awarded 0, 1

B0, B1 Independent mark awarded 0, 1

SC Special case

^ Omission sign

MR Misread

Highlighting

Other abbreviations in mark scheme

Meaning

E1 Mark for explaining

U1 Mark for correct units

G1 Mark for a correct feature on a graph

M1 dep* Method mark dependent on a previous mark, indicated by *

cao Correct answer only

oe Or equivalent

rot Rounded or truncated

soi Seen or implied

www Without wrong working


2

2. Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand a Annotations should be used whenever appropriate during your marking.

The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.

c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


3

E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.

f Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. 2 or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader.

g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.


4

If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


5


1 (i) a b f(a) f(b) x f(x)

0.2 0.3 0.04 –1.57667 0.202474 –0.02010

0.2 0.202474 0.04 –0.02010 0.201647

M1A1

M1A1

0.201 640 suggests that 0.2016 is secure, accept 0.20164 B1 (Root is 0.201640 to 6 dp)

[5]

1 (ii) Initial estimate 2.1 with mpe 0.1 B1

mpe: 0.05, 0.025, 0.0125, 0.00625, 0.003125 M1 soi

So 5 further steps required A1 cao

[3]

2 (i) h M T S

0.5 0.536650 0.594027 0.555776

0.25 0.551860 0.5653385 0.556353

0.125 0.555282 0.558599 0.556388

B1 Missing T

M1A1 Missing M

M1A1A1A1 S values

[7]

2 (ii) Agreement suggests 0.5564; extrapolation gives 0.556 39 B1 Accept either

[1]

3 (i) k exact approximate relative error

1 1024 1000 –0.0234375

2 1048576 1000000 –0.0463257

3 1073741824 1000000000 –0.0686774

B1 For exact and approximate values soi

B1B1B1

Each relative error

Condone consistent missing minus

signs on this occasion

[4]

3 (ii) k = 2 and k = 3 represent the square and cube of value when k = 1 E1

The relative errors have doubled and tripled respectively E1

[2]


6


4 (i) fwd diff h est

0.1 7.49638

0.05 7.12152

M1

A1

Correct formula for either value

Both values correct

[2]

(ii) cent diff h est

0.1 6.82034

0.05 6.78451

M1

Correct formula for either value

A1

Both values correct

[2]

4 (iii) 6.8 looks secure B1 (Extrapolation gives 6.77...)

Using central difference values (as the method is more accurate) E1 Seen or implied

[2]

5 (i) g(1.14) = 1.234 281 + 0.4 (1.287 500 – 1.234 281) = 1.255 568 6 M1A1 Accept 5-7 dp

[2]

5 (ii) 1 1.188395

1.1 1.234281 0.045886

1.2 1.287500 0.053219 0.007333

B1

g(x) = 1.188395 + 0.045886 (x – 1) / 0.1 M1 formula

+ 0.007333 (x – 1) (x – 1.1) / ((0.1)2 2!) A1A1 A1 for 2 terms, A1A1 for all 3

[4]

5 (iii) Substitute x = 1.14 to get g(1.14) = 1.254 688 6 M1A1 Accept 5-7 dp

[2]


7


6 (i) Exhibit changes of sign, e.g. M1 Accept signs instead of numbers

x –1 0 0.5 1

f(x) –6 1 –0.5625 2

A1

A1

One interval

Rest

[3]

6 (ii) E.g. starting at –0.5 iterates are:

–0.73072 –0.67800 –0.69558 –0.69018 –0.69188

M1A1A1 M1 1st term, A1 next 2, A1 rest

Root is –0.69 to 2 dp A1 cao

[4]

6 (iii) Start at 0.34:

0.340231 0.340042 0.340197 0.340070

converging to a value close to 0.34

M1A1

E1

Derivative of RHS is 15x4 – 3x B1

At x = 0.34 this is –0.82 B1

Approaching 1 in magnitude, hence slow convergence E1

[6]

6 (iv) Req’d derivative is 30x4 – 6x – 2 Hence NR formula M1A1 NR formula soi

NR iterations eg

0.75 1.005222 0.911575 0.876798 0.872091 0.872009

1 0.909091 0.876258 0.872074 0.872009 0.872009

A1A1A1 A1 2 it’ns, A1 rest, A1 0.8720

[5]


8


7 (i) T = 2.08300 M1A1 Accept 2.083 or 2.0830

[2]

7 (ii)

G1

G1

Points

Curve

Draw lines on graph. T will be an overestimate; G1E1

[4]

7 (iii) T + S = 0.77450 + 1.30167 = 2.07617 M1A1 A1 each new correct T or S value

S + T = 1.43350 + 0.63525 = 2.06875 M1A1

Trapezium rule seems like a better fit over (1, 1.5) than (0, 0.5) E1

So second estimate likely to be more accurate A1

[6]

7 (iv) 4-point rule gives 2.06719 M1A1

Errors: T only: 0.01581 so it is an overestimate A1E1 Error soi

T + S: 0.00898

S + T: 0.00156 so S + T is more accurate than T + S A1E1

Accept relative errors to 3dp.

[6]


GCE

Mathematics (MEI)


Advanced Subsidiary GCE


OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2015


3

1. Annotations and abbreviations

Annotation in scoris Meaning and BOD Benefit of doubt

FT Follow through

ISW Ignore subsequent working

M0, M1 Method mark awarded 0, 1

A0, A1 Accuracy mark awarded 0, 1

B0, B1 Independent mark awarded 0, 1

SC Special case

^ Omission sign

MR Misread

Highlighting

Other abbreviations in mark scheme

Meaning

E1 Mark for explaining

U1 Mark for correct units

G1 Mark for a correct feature on a graph

M1 dep* Method mark dependent on a previous mark, indicated by *

cao Correct answer only

oe Or equivalent

rot Rounded or truncated

soi Seen or implied

www Without wrong working


4

2. Subject-specific Marking Instructions for GCE Mathematics (MEI) Pure strand

a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.

c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


5

E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

Documents

1(i) M1A1A1A1 [4] M1A1 M1A1 [4] A1A1 A1 M1 B1 M1 ... packs/MS_MEI_NM.pdf6 -0.00218 sign change so root is correct to 3 dp [M1A1] [TOTAL 8] 3 h M T S 2 3.46410 2 3.65028 2 3.52616 2