Oxford .Cambridge .and .RSA
AS .Level .Further .Mathematics .B .(MEI) .
.Y414/01 . Numerical .Methods .Y414/01 . Numerical
Answer .all .the .questions.
1 You .are .given .the .following .simultaneous .equations.
2x.+.1.8y .= .5
3x.+.2.8y .= .3
The .constants .and .the .coefficients .of .x .are .exact, .but .the .coefficients .of .y .have .been .chopped .to
1 .decimal .place.
(a) Calculate .the .maximum .possible .relative .error .in .each .of .the .coefficients .of .y. [2]
(b) Determine .the .range .of .possible .values .of .y. [4]
(c) Explain .why .this .range .is .so .large. [1]
2 The .table .shows .some .values .of .x .and .the .associated .values .of .y .= .f.(x).
x 1.98 1.99 2 2.01 2.02
f.(x) 1.10311648 1.10514069 1.10714872 1.10914075 1.11111695
dy .
(a) Use .the .central .difference .method .to .calculate .two .approximations .to .the .value .of . at .x .=
.2.
dx
[3]
dy .
(b) State .the .value .of . at .x .= .2 .as .accurately .as .you .can, .justifying .the .precision .quoted. [1]
dx
(c) Calculate .an .approximation .of .the .error .in .using .f.(2) .to .approximate .f.(2.008). [2]
,3 Ali .uses .the .trapezium .rule .with .h .= .1, .h .= .0.5 .and .h .= .0.25 .to .calculate .three .approximations .to
2
f(x).dx.. .Ali‟s .results .are .shown .in .the .table.
1
h n Tn
1 1 0.9462734
0.5 2 0.9645336
0.25 4 0.9691932
y
2.
(a) Use .the .information .in .the .table .to .calculate .two .approximations .to . . f(x) .dx . using
1
Simpson‟s .rule, .giving .your .answers .correct .to .6 .decimal .places. [3]
y
2.
(b) Without .doing .any .further .calculation, .state .the .value .of . . f(x) .dx . as .accurately .as .you .can,
1
justifying .the .precision .quoted. [1]
Ali .states .that .the .graph .of .y .= .f.(x) .is .concave .downwards .between .x .= .1 .and .x .= .2.
(c) Explain .whether .the .information .in .the .table .supports .Ali‟s .statement. [1]
© .OCR
.
, 4 The .equation .3x5 .- .13x2 .+ .11 .= .0 .has .three .roots, .a, .b .and .c .such .that .a .1 .b .1 .c.
Fig. .4.1 .shows .part .of .the .graph .of . y .= .3x5 .-.13x2 .+.11.
y
10
8
6
4
2
x
–1 0 1 2
Fig. .4.1
(a) Explain .why .it .is .not .possible .to .use .the .method .of .false .position .with .initial .values .of .a
.= .1 .and .b .= .1.5 .to .find .b. [1]
Taylor .uses .the .method .of .false .position .to .find .b .using .initial .values .of .a .= .1 .and .b .= .1.2. .The
.associated .spreadsheet .output .is .shown .in .Fig. .4.2.
◢ C D E F G H
3 a f(a) b f(b) xnew f(xnew)
4 1 1 1.2 -0.255040 1.159357 -0.189833
5 1 1 1.159357 -0.189833 1.133933 -0.091284
6 1 1 1.133933 -0.091284 1.122729 -0.035016
7 1 1 1.122729 -0.035016 1.118577 -0.012267
8 1 1 1.118577 -0.012267 1.11714 -0.004160
9 1 1 1.117140 -0.004160 1.116655 -0.001395
10 1 1 1.116655 -0.001395 1.116492 -0.000466
11 1 1 1.116492 -0.000466 1.116438 -0.000156
12 1 1 1.116438 -0.000156 1.116420 -5.19E-05
Fig. .4.2
AS .Level .Further .Mathematics .B .(MEI) .
.Y414/01 . Numerical .Methods .Y414/01 . Numerical
Answer .all .the .questions.
1 You .are .given .the .following .simultaneous .equations.
2x.+.1.8y .= .5
3x.+.2.8y .= .3
The .constants .and .the .coefficients .of .x .are .exact, .but .the .coefficients .of .y .have .been .chopped .to
1 .decimal .place.
(a) Calculate .the .maximum .possible .relative .error .in .each .of .the .coefficients .of .y. [2]
(b) Determine .the .range .of .possible .values .of .y. [4]
(c) Explain .why .this .range .is .so .large. [1]
2 The .table .shows .some .values .of .x .and .the .associated .values .of .y .= .f.(x).
x 1.98 1.99 2 2.01 2.02
f.(x) 1.10311648 1.10514069 1.10714872 1.10914075 1.11111695
dy .
(a) Use .the .central .difference .method .to .calculate .two .approximations .to .the .value .of . at .x .=
.2.
dx
[3]
dy .
(b) State .the .value .of . at .x .= .2 .as .accurately .as .you .can, .justifying .the .precision .quoted. [1]
dx
(c) Calculate .an .approximation .of .the .error .in .using .f.(2) .to .approximate .f.(2.008). [2]
,3 Ali .uses .the .trapezium .rule .with .h .= .1, .h .= .0.5 .and .h .= .0.25 .to .calculate .three .approximations .to
2
f(x).dx.. .Ali‟s .results .are .shown .in .the .table.
1
h n Tn
1 1 0.9462734
0.5 2 0.9645336
0.25 4 0.9691932
y
2.
(a) Use .the .information .in .the .table .to .calculate .two .approximations .to . . f(x) .dx . using
1
Simpson‟s .rule, .giving .your .answers .correct .to .6 .decimal .places. [3]
y
2.
(b) Without .doing .any .further .calculation, .state .the .value .of . . f(x) .dx . as .accurately .as .you .can,
1
justifying .the .precision .quoted. [1]
Ali .states .that .the .graph .of .y .= .f.(x) .is .concave .downwards .between .x .= .1 .and .x .= .2.
(c) Explain .whether .the .information .in .the .table .supports .Ali‟s .statement. [1]
© .OCR
.
, 4 The .equation .3x5 .- .13x2 .+ .11 .= .0 .has .three .roots, .a, .b .and .c .such .that .a .1 .b .1 .c.
Fig. .4.1 .shows .part .of .the .graph .of . y .= .3x5 .-.13x2 .+.11.
y
10
8
6
4
2
x
–1 0 1 2
Fig. .4.1
(a) Explain .why .it .is .not .possible .to .use .the .method .of .false .position .with .initial .values .of .a
.= .1 .and .b .= .1.5 .to .find .b. [1]
Taylor .uses .the .method .of .false .position .to .find .b .using .initial .values .of .a .= .1 .and .b .= .1.2. .The
.associated .spreadsheet .output .is .shown .in .Fig. .4.2.
◢ C D E F G H
3 a f(a) b f(b) xnew f(xnew)
4 1 1 1.2 -0.255040 1.159357 -0.189833
5 1 1 1.159357 -0.189833 1.133933 -0.091284
6 1 1 1.133933 -0.091284 1.122729 -0.035016
7 1 1 1.122729 -0.035016 1.118577 -0.012267
8 1 1 1.118577 -0.012267 1.11714 -0.004160
9 1 1 1.117140 -0.004160 1.116655 -0.001395
10 1 1 1.116655 -0.001395 1.116492 -0.000466
11 1 1 1.116492 -0.000466 1.116438 -0.000156
12 1 1 1.116438 -0.000156 1.116420 -5.19E-05
Fig. .4.2
