4752 Mark Scheme June 2005 Mark Scheme 4752 June 2005 packs/MS_MEI_C2.pdf · 4752 Mark Scheme June 2005 Section A 1 1+ 1 3 2 2 x 1+3 B2 for 1 kx2, or M1 for 3 x2 seen before differentiation

4752 Mark Scheme June 2005

Mark Scheme 4752June 2005


Section A 1 1+

123

2 x 1+3 B2 for 12kx , or M1 for

32x seen before

differentiation or B1 ft their 32x correctly

differentiated

4

2 1170 4 B1 for a = 11 and B1 for d = 5 or 20th

term = 106 and

M1 for 20/2[their (a) + their(106)] or 20/2[2their (a)+ (20-1)×their(d)] OR M1 for 6×20 and M2 for

[ ]205 20 12

⎛ ⎞+⎜ ⎟⎝ ⎠

o.e.

4

3 ±√13/4 3 B2 for (-) √13/4 or

1613

±

or M1 for √13 or sin2θ + cos2θ = 1 used

3

4 x + x-1 soi y′ = 1 − 1/x2

subs x = 1 to get y′ = 0 y′′=2x-3 attempted Stating y′′ > 0 so min cao

B1 B1 B1 M1ftA1

1 – x-2 is acceptable Or solving 1 – x-2 = 0 to obtain x = 1 or checking y′ before and after x = 1 Valid conclusion First quadrant sketch scores B2

5

5 (i) 1 (ii) −2 (iii) 6log x

1 2 2

M1 for 1/9=3-2 or log(1) – log(32) base not requd; M1 for 5 log x or log(x6)

5

6 Correct curve thro’ y axis (0, 1) indicated on sketch or table 5.64

G1 G1 3

y, y′ & y′′ all positive independent B2 for other versions of 5.64(3....) or B1 for other ans 5.6 to 5.7 or M1 for x log 2 = log 50 and M1 for x = log 50 ÷ log 2

5

7 y = 7 − 3/x2 oe 5 B3 for (y =) −3/x2 + c [B1 for each of k/x2, k = − 6/2 and +c] and M1 for substituting (1, 4) in their attempted integration with + c, the constant of integration

5

8 (i) 66° or 66.4 or 66.5…. 293.58 …. to 3 or more sf cao (ii) stretch (one way) parallel to the x-axis sf 0.5

B1 B1 1 1 1

Allow 1.16 or 73.8 Lost for extras in range. Ignore extras outside the range Horizontal, from y axis, in x axis, oe

5

36


Section B 9 i 3x2 − 20x + 12 2 B1 if one error “+c” is an error

2

ii y − 64 = −16(x − 2) o.e. eg y = −16x + 96

4 M1 for subst x = 2 in their y′ A1 for y′ = −16 and B1 for y = 64

4

iii

Factorising f(x) 2)6)(2( −+≡ xx OR Expanding 2)6)(2( −+ xx

B3 M2 E1

or B1 for f(-2) = -8-40-24+72 =0 and B1 for 0)6( =′f and B1dep for f(6)=0

3

iv 4 3210 6 72

4 3x x x x− + +

value at (x = 6) ~ value at (x = −2) 341(.3..) cao

B2 M1 A1

-1 for each error Must have integrated f(x)

4

10 i AB = 7.8(0), 7.798 to 7.799 seen area = 52.2 to 52.3

2 2

M1 for correct use of sine rule For long methods M1A1 for art 7.8 M1 for [2×][0.5 ×] their AB × 11.4 × sin 36°

4

ii tan 0.91 = ST/12.6 ST = 12.6 × tan 0.91 and completion (16.208...) area OSTR = [2×][0.5 ×]12.6 × their(16.2) nb 204. …. area of sector = 0.5 × 12.62 × 1.82 =144.47… Logo = 59.6 to 60.0 arc = 12.6 × 1.82 [=22.9...] perimeter = 55.3 to 55.4

M1 E1 M1 M1 A1 A1 M1 A1

Accept 16.2 if ST is explicit but for long methods with pa check that their explicit expression = 16.2 oe using degrees soi by correct ans Accept 144, 144.5 oe using degrees

8

11 i 81 1 1

ii (1x)3n – 1

1 1

iii (GP with) a = 1 and r = 3 clear correct use GP sum formula

M1 M1

or M1 for = 1+3+9+ ... +3n-1 2

iv

(A) 6 www (B) 243

2 1

M1 for 364 = (3n − 1)/2 3

v their (ii) > 900 (y − 1)log 3 > log 900 y − 1 > log 900 ÷ log 3 y = 8 cao

M1ftM1ftM1 B1

-1 once for = or < seen: condone wrong letter / missing brackets / no base

4

Mark Scheme 4752January 2006

Section A 1 7/9 or 140/180 o.e. 2 B1 for 180° = π rad o.e. or 0.78 or other

approximations 2

2 224 2 M1 for 23 + 33 + 43 + 53 2

3 triangle divided into 2 rt angled tris √3 and 1 indicated 60 indicated

H1 S1 A1

3

4 16.1 overestimate + expn eg sketch

4 1

M3 for ¼{8.2 + 4.2 + 2 (6.4 + 5.5 + 5 + 4.7 + 4.4)} M2 for one slip/error M1 for two slips/errors

5

5 (i)

tan x = ¾ (ii) 36.8 to 36.9 and 216.8 to 216.9

2 M1 A1A1

no numbers required on axes unless more branches shown. G1 for a correct first sweep Allow 37, 217

5

6 y′′ = 2x − 6 y′′ = 0 at x = 3 y′ = 0 at x = 3 showing y′ does not change sign

B1 B1 B1 E1

or that y′′ changes sign

4

7 (i) 5 (ii) 5.646... to 2 sf or more

2 3

M1 for 6 = 1.2r M2 for 2 x 5x sin 0.6 or √(52 + 52 – 2.5.5. cos 1.2) or 5 sin 1.2/sin 0.971 M1 for these methods with 1 error

5

8 32 22

3 3x x c−− + o.e. 5 1 for each element 5

9 (i) log10 y = 0.5x + 3 (ii) y = 100.5x + 3 isw

B3 2

B1 for each term scored in either part o.e. e.g. 1000 10 xy = ×

5

Section B 10 i y′ = 6 − 2x

y′ = 0 used x = 3 y = 16 (0, 7) (−1, 0) and (7,0) found or marked on graph sketch of correct shape

M1 M1 A1 A1 3 1

condone one error 1 each must reach pos. y - axis

8

ii 58.6 to 58.7 3 M1

B1 for 7x + 3x2 − x3/3 [their value at 5] − [their value at 1] dependent on integration attempted

3

iii using his (ii) and 48 1 1

11 i 3x2 − 6 2 1 if one error 2

ii −√2 < x < √2 3 M1 for using their y′ = 0 B1 f.t. for both roots found

3

iii subst x = −1 in their y′ [=−3] y = 7 when x = −1 y + 3x = 4 x3 − 6x + 2 = − 3x + 4 (2, −2) c.a.o.

B1 M1 A1 M1 A1,A1

f.t. f.t. 3 terms f.t.

6

12 i ii

A 23 B 24 C 480 A 11.78 – 11.80 B 5 x 1.1n – 1 > 50 1.1 n – 1 > 10 (n – 1) log 1.1 > 1 n – 1 > 1/log 1.1 n = 26

2 2 2 2 B1 B1 L1 A1 1

M1 for 5, 7, 9 etc or AP with a = 5, d = 2 M1 for 51 = 5 + 2(n − 1) o.e. M1 for attempted use of sum of AP formula eg 20/2[10+19×2] Or other step towards completion (NB answer given) independent

2 2 2



Section A 1 1, 3 1,1

2

2 r = 0.2 3 M1 for 10 = 8/(1 − r), then

M1 dep’t for any correct step 3

3 1/√15 i.s.w. not +/− 3 M2 for √15 seen M1 for rt angled triangle with side 1 and hyp 4, or cos2 θ = 1 − 1/4 2.

3

4 x5/5 −3 x−1/−1 + x [value at 2 − value at 1] attempted 5.7 c.a.o.

B3 M1 A1

1 each term dep’t on B2

5

5 [y =] 3x − x3/3 + c subst of (6, 1) in their eqn with c y = 3x − x3/3 + 55 c.a.o

B1 B1 M1 A1

Dep’t on integration attempt Dep’t on B0B1 Allow c = 55 isw

4

17

6 (i) 3, 8, 13, 18

(ii) use of n/2[2a + (n − 1)d] (S100 = ) 25 050 or (S50 = ) 6275 (S49 = ) 6027 or (S51 = ) 6528 their(S100 − S50 ) dep’t on M1 18 775 cao

B1 M1 A1 M1 A1

Ignore extras Use of a + (n − 1)d u51 = 253 u100 = 498 u50 = 248 u52 = 258 50/2(their(u51 + u100)) dep’t on M1 or 50/2[2 × their(u51) + 49 × 5]

5

7 (i) sketch of correct shape correct period and amplitude period halved for y = cos 2x; amplitude unchanged (ii) 30, 150, 210, 330

G1 G1 G1 B2

Not ruled lines need 1 and −1 indicated; nos. on horiz axis not needed if one period shown B1 for 2 of these, ignore extras outside range.

5

8 √x = x ½ soi 18x2, ½ x – ½

36x Ax−3/2 (from Bx− ½ )

B1 B1B1 B1 B1

−1 if d/dx(3) not = 0 any A,B

5

9 3x log 5 = log 100 3x = log 100/log 5 x = 0.954

M1 M1 A2

allow any or no base or 3x = log5100 dep’t A1 for other rot versions of 0.9537... SC B2/4 for 0.954 with no log wkg SC B1 r.o.t. 0.9537...

4

19


Section B

10 i (A)

5.22 + 6.32 − 2×5.2 × 6.3 ×cos “57” ST = 5.6 or 5.57 cao

M2 A1

M1 for recognisable attempt at cos rule. or greater accuracy

3

i (B)

sin T/5.2 = sin(their 57)/their ST T=51 to 52 or S = 71 to 72 bearing 285 + their T or 408 – their S

M1 A1 B1

Or sin S/6.3 = … or cosine rule If outside 0 to 360, must be adjusted

3

ii 5.2θ , 24 × 26/60 θ = 1.98 to 2.02 θ = their 2 × 180/π or 114.6°... Bearing = 293 to 294 cao

B1B1 B1 M1 A1

Lost for all working in degrees Implied by 57.3

5

11

11 i y′ = 3x2 − 6x use of y′ = 0 (0, 1) or (2, −3) sign of y′′ used to test or y′ either side

B1 M1 A2 T1

condone one error A1 for one correct or x = 0 , 2 SC B1 for (0,1) from their y′ Dep’t on M1 or y either side or clear cubic sketch

5

ii y′ (-1) = 3 + 6 = 9 3x2 − 6x = 9 x = 3 At P y = 1 grad normal = −1/9 cao y − 1 = −1/9 (x − 3) intercepts 12 and 4/3or use of

∫ −12

0 91

34 x dx (their normal)

½ × 12 × 4/3 cao

B1 M1 A1 B1 B1 M1 B1 A1

ft for their y ′ implies the M1 ft their (3, 1) and their grad, not 9 ft their normal (linear)

8

13

12 i log10 P = log10 a + log1010bt

log10 10bt = bt intercept indicated as log10a

B1 B1 B1

condone omission of base 3

ii 3.9(0), 3.94, 4(.00), 4.05, 4.11 plots ft line of best fit ft

T1 P1 L1

to 3 sf or more; condone one error 1 mm ruled and reasonable

3

iii (gradient = ) 0.04 to 0.06 seen (intercept = ) 3.83 to 3.86 seen (a = ) 6760 to 7245 seen P = 7000 x 100.05t oe

M1 M1 A1 A1

7000 × 1.12t

SC P = 10 0.05t + 3.85 left A2

4

iv 17 000 to 18 500 B2 14 000 to 22 000 B1 2

12

7

Mark Scheme 4752January 2007

4752 Mark Scheme Jan 2007

8

Section A

1 325 6

2x×

1+1 - 1 if extra term 2

2 −0.2 3 M1 for

651 r

=−

and M1 dep for correct

constructive step

3

3 √8 or 2√2 not ±√8

3 M1 for use of sin2 θ+ (1/3)2 = 1 and M1for sinθ = √8/3 (ignore ±) Diag.: hypot = 3, one side =1 M1 3rd side √8 M1

3

4 (i) C (ii) B (iii) 2n-1

1 1 1

3

5 (i) −0.93, -0.930, -0.9297… (ii) answer strictly between 1.91 and 2 or 2 and 2.1 (iii) y′ = −8/x3 , gradient = −1

2 B1 M1A1

M1 for grad = (1 – their yB)/(2 − 2.1) if M0, SC1 for 0.93 don’t allow 1.9 recurring

5

6 At least one cycle from (0, 0) amplitude 1 and period 360[°] indicated 222.8 to 223 and 317 to 317.2 [°]

G1 G1dep 2

1 each, ignore extras

4

7 x < 0 and x > 6 3

B2 for one of these or for 0 and 6 identified or M1 for x2 -6x > 0 seen (M1 if y found correctly and sketch drawn)

3

8 a + 6d = 6 correct

( )1030 2 92

a d= + correct o.e.

elimination using their equations a = −6 and d = 2 5th term = 2

M1 M1 M1f.t. A1 A1

Two equations in a and d

5

9 (y =) 2x3 + 4x2 −1 accept 2x3 + 4x2 + c and c = - 1

4 M2 for (y =) 2x3 + 4x2 + c (M1 if one error) and M1 for subst of (1, 5) dep on their y =, +c, integration attempt.

4

10 (i) 3 logax

ii) 1000b

c=

2 2

M1 for 4 logax or − logax; or logx3

M1 for 1000 or 103 seen

4

Section B

4752 Mark Scheme January 2007

9

11 i Correct attempt at cos rule correct full method for C

C = 141.1… bearing = [0]38.8 cao

M1 M1 A1 A1

any vertex, any letter or B4

4

ii ½ × 118 × 82 × sin their C or supp. 3030 to 3050 [m2]

M1 A1

or correct use of angle A or angle B 2

iiiA sin (θ/2) = ( ½ × 189)/130 1.6276 → 1.63

M1 A1

or cosθ = (1302+1302-1892)/(2x130x130) In all methods, the more accurate number to be seen.

2

iiiB 0.5 × 1302 × sin 1.63 0.5 × 1302 × 1.63 their sector − their triangle AOB 5315 to 5340

M1 M1 M1 A1

condone their θ (8435) condone their θ in radians (13770) dep on sector > triangle

4

12 i (2x − 3)(x − 4) x = 4 or 1.5

M1 A1A1

or (11 ±√(121 – 96))/4 if M0, then B1 for showing y = 0 when x = 4 and B2 for x = 1.5

3

ii y′ = 4x − 11 = 5 when x = 4 c.a.o. grad of normal = −1/their y′ y[ − 0 ]=their −0.2 (x − 4) y-intercept for their normal area = ½ × 4 × 0.8 c.a.o.

M1 A1 M1f.t.M1 B1f.t. A1

condone one error or 0 = their (-0.2)x4 + c dep on normal attempt s.o.i. normal must be linear or integrating their f(x) from 0 to 4 M1

6

iii 3 22 113 2 12x x x− + attempt difference between value at 4 and value at 1.5 [−] 5

245 o.e. or [−]5.2(083..)

M1 M1 A1

condone one error, ignore + c ft their (i), dep on integration attempt. c.a.o.

3

13 i log 10 y = log 10 k + log 10 10ax

log 10 y = ax + log 10 k compared to y = mx+c

M1 M1

2

ii 2.9(0), 3.08, 3.28, 3.48, 3.68 plots [tol 1 mm] ruled line of best fit drawn

T1 P1f.t L1f.t.

condone one error

3

iii intercept = 2.5 approx gradient = 0.2 approx y = their 300x 10x(their 0.2)

or y = 10(their 2.5 + their 0.2x)

M1 M1 M1f.t.

or y – 2.7 = m(x – 1) 3

iv subst 75000 in any x/y eqn subst in a correct form of the relationship 11,12 or 13

M1 M1 A1

B3 with evidence of valid working

3

v “Profits change” or any reason for this.

R1 too big, too soon 1


5

4752 Mark Scheme June 2007 1 (i) −√3

(ii) 53 π

1

2

Accept any exact form

accept 53π ,1 ⅔π. M1 π rad = 180° used

correctly

3

2 1 12 2

12

36 or 9 o2

9 o.e.2

y x x

y x−

′ = ×

′′ =

.e.

√36 = 6 used interim step to obtain 3

4

2 1 M1 A1

1 if one error in coeff or power, or extra term f.t. their y′ only if fractional power f.t. their y″ www answer given

5

3 (i) y = 2f(x) (ii) y = f(x − 3)

2 2

1 if ‘y=’ omitted [penalise only once] M1 for y = kf(x), k > 0 M1 for y = f(x + 3) or y = f(x – k)

4

4 (i) 11 27 or ft from their 11 (ii) 20

1 1 2

M1 for 1 × 2 + 2× 3 + 3 × 4 soi, or 2,6,12 identified, or for substituting n = 3 in standard formulae

4

5 θ = 0.72 o.e 13.6 [cm]

2 3

M1 for 9 = ½ × 25 × θ No marks for using degrees unless attempt to convert B2 ft for 10 + 5 × their θ or for 3.6 found or M1 for s = 5 θ soi

5

6 (i) log a 1 = 0, log a a = 1 (ii) showing both sides equivalent

1+1 3

NB, if not identified, accept only in this order M1 for correct use of 3rd law and M1 for correct use of 1st or 2nd law. Completion www A1. Condone omission of a.

5

7 (i) curve with increasing gradient any curve through (0, 1) marked (ii) 2.73

G1 G1 3

correct shape in both quadrants M1 for x log 3 = log 20 (or x=log320) and M1 for x = log 20 ÷ log 3 or B2 for other versions of 2.726833.. or B1 for other answer 2.7 to 2.8

5

8 (i) 2(1 − sin2 θ) + 7 sin θ = 5 (ii) (2 sin θ − 1)( sin θ − 3) sin θ = ½ 30° and 150°

1 M1 DM1 A1 A1

for cos2 θ + sin2 θ = 1 o.e. used 1st and 3rd terms in expansion correct f.t. factors B1,B1 for each solution obtained by any valid method, ignore extra solns outside range, 30°, 150° plus extra soln(s) scores 1

5

6

4752 Mark Scheme June 2007 9 i

y′ = 6x2 − 18x + 12 = 12 y = 7 when x = 3 tgt is y − 7 = 12 (x − 3) verifying (−1, −41) on tgt

M1 M1 B1 M1 A1

condone one error subst of x = 3 in their y′ f.t. their y and y′ or B2 for showing line joining (3, 7) and (−1, −41) has gradient 12

5

ii y′ = 0 soi quadratic with 3 terms x = 1 or 2 y = 3 or 2

M1 M1 A1 A1

Their y′ Any valid attempt at solution or A1 for (1, 3) and A1 for (2,2) marking to benefit of candidate

4

iii cubic curve correct orientation touching x- axis only at (0.2,0) max and min correct curve crossing y axis only at −2

G1 G1 G1

f.t.

3

10 i 970 [m] 4 M3 for attempt at trap rule ½×10×(28+22+2[19+14+11+12+16]) M2 with 1 error, M1 with 2 errors. Or M3 for 6 correct trapezia, M2 for 4 correct trapezia, M1 for 2 correct trapezia.

4

ii concave curve or line of traps is above curve (19+14+11 +11+12+16) × 10 830 to 880 incl.[m]

1 M1 A1

Accept suitable sketch M1 for 3 or more rectangles with values from curve.

3

iii t = 10, vmodel = 19.5 difference = 0.5 compared with 3% of 19 = 0.57

B1 B1f.t.

or 0.5 100 2.619

× ≈

2

iv 28t − ½ t2 + 0.005t3 o.e. value at 60 [− value at 0] 960

M1 M1 A1

2 terms correct, ignore + c ft from integrated attempt with 3 terms

3

11 ai 13 1 1 aii 120 2 M1 for attempt at AP formula ft their a,

d or for 3 + 5 + …+ 21 2

bi 1251296

2

M1 for 31 5

6 6⎛ ⎞×⎜ ⎟⎝ ⎠

2

ii a = 1/6, r = 5/6 s.o.i. 16

561

S∞ =−

o.e.

1+1 1

If not specified, must be in right order

3

iii ( )( )

( )

156

510 106

105

10 6

min

0.006

( 1) log log 0.006log 0.0061

log30

−<

− <

− >

=

n

n

n

n

Or log(1/6)+ log(5/6)n-1 < log0.001 (n -1)log(5/6)< log(0.001/(1/6))

M1 M1

DM1 B1 M1 M1

condone omission of base, but not brackets

NB change of sign must come at correct place

4

7


5

4752 (C2) Concepts for Advanced Mathematics

Section A 1 40x3 2 -1 if extra term 2

2 (i) 3

(ii) 141

1 2

M1 for 9 × (1 + 2 + 3 + 4 + 5) + 1 + 2 +3

3

3 right angled triangle with 1 and 2 on correct sides Pythagoras used to obtain hyp = √5

cos θ = ha

= 5

2

M1 M1 A1

or M1 for sinθ = ½cosθ and M1 for substituting in sin2 θ + cos2θ = 1 E1 for sufficient working

3

4 (i)line along y = 6 with V (1, 6), (2, 2), (3, 6) (ii) line along y = 3 with V (−2,3), (−1,1), (0,3)

2 2

1 for two points correct 1 for two points correct

4

5 46 332 7

4x x x c+ + +

5 1 for 2x6; 2 for

433

4x or 1 for other

43kx ; 1 for 7x;

1 for +c

5

6 (i) correct sine shape through O amplitude of 1 and period 2π shown (ii) 7π/6 and 11π/6

1 1 3

B2 for one of these; 1 for −π/6 found

5

7 (i) 60 (ii) −6 (iii)

2 1 1 1

M1 for 22 + 23 + 24 + 25 o.e. Correct in both quadrants Through (0, 1) shown dep.

5

8 r = 1/3 s.o.i. a = 54 or ft 18 ÷ their r

S = r

a−1

used with -1 < r < 1

S = 81 c.a.o.

2 M1 M1 A1

1 mark for ar = 18 and ar3 = 2 s.o.i. 5

9 (i) 0.23 c.a.o. (ii) 0.1 or 1/10 (iii) 4(3x + 2) or 12x + 8 (iv) [y = ] 103x + 2 o.e.

1 1 1 1

10-1 not sufficient

4

5

2

4

6

2- 0 x

y


6

Section B 10 i h = 120/x2

A = 2x2 + 4xh o.e. completion to given answer

B1 M1 A1

at least one interim step shown

3

ii A′ = 4x − 480/x2 o.e. A′′ = 4 + 960 / x3

2 2

1 for kx-2 o.e. included ft their A′ only if kx-2 seen ; 1 if one error

4

iii use of A′ = 0 x = 3 120 or 4.9(3..) Test using A′ or A′′ to confirm minimum Substitution of their x in A A = 145.9 to 146

M1 A1

T1 M1 A1

Dependent on previous M1

5

11 iA BC2 = 3482 + 3022 − 2 × 348 × 302 × cos 72° BC = 383.86… 1033.86…[m] or ft 650 + their BC

M2 A1 1

M1 for recognisable attempt at Cosine Rule to 3 sf or more accept to 3 sf or more

4

iB sin sin 72302 their

BBC

=

B = 48.4.. 355 − their B o.e. answer in range 306 to307

M1 A1 M1 A1

Cosine Rule acceptable or Sine Rule to find C or 247 + their C

4

ii Arc length PQ = 224 2 120360

π× ×

o.e. or 469.1... to 3 sf or more QP = 222.5…to 3 sf or more answer in range 690 to 692 [m]

M2 B1 A1

M1 for 136 2 120360

π× ×

4

12 iA x4 = 8x (2, 16) c.a.o. PQ = 16 and completion to show ½ × 2 × 16 = 16

M1 A1 A1

NB answer 16 given

3

iB x5/5 evaluating their integral at their co-ord of P and zero [or 32/5 o.e.] 9.6 o.e.

M1 M1 A1

ft only if integral attempted, not for x4 or differentiation c.a.o.

3

iiA 6x2h2 + 4xh3 + h4

2 B1 for two terms correct.

2

iiB 4x3 + 6x2h + 4xh2 + h3 2 B1 for three terms correct 2

iiC 4x3 1 1

iiD gradient of [tangent to] curve 1 1

2


6

4752 (C2) Concepts for Advanced Mathematics Section A 1 210 c.a.o. 2 1 for π rads = 180° soi 2

2

(i) 5.4 × 10-3, 0.0054 or 5000

27

(ii) 6 www

1 2

M1 for S = 5.4 / (1 − 0.1)

3

3 stretch, parallel to the y axis, sf 3

2 1 for stretch plus one other element correct

2

4 [f′(x) =] 12 − 3x2 their f′(x) > 0 or = 0 soi −2 < x < 2

B1 M1 A1

condone −2 ≤ x ≤ 2 or “between -2 and 2”

3

5 (i) grad of chord = (23.1 − 23)/0.1 o.e. = 5.74 c.a.o. (ii) correct use of A and C where

for C, 2.9 < x < 3.1 answer in range (5.36, 5.74)

M1 A1 M1 A1

or chord with ends x = 3 ± h, where 0 < h ≤ 0.1 s.c.1 for consistent use of reciprocal of gradient formula in parts (i) and (ii)

4

6 [y =] kx3/2 [+ c] k = 4 subst of (9, 105) in their eqn with c or c = −3

M1 A1 M1 A1

may appear at any stage must have c; must have attempted integration

4

7 sector area = 28.8 or

5144

[cm2]

c.a.o. area of triangle = ½ × 62 × sin 1.6 o.e. their sector − their triangle s.o.i. 10.8 to 10.81 [cm2]

2

M1 M1 A1

M1 for ½ × 62 × 1.6 must both be areas leading to a positive answer

5

8 a + 10d = 1 or 121 = 5.5(2a+10d) 5(2a + 9d) = 120 o.e. a = 21 s.o.i. www and d = −2 s.o.i. www 4th term is 15

M1 M1 A1 A1 A1

or 121 = 5.5(a + 1) gets M2 eg 2a + 9d = 24

5

9 x log 5 = log235 or

log 235log5

x =

3.39

M1 A2

or x = log5 235 A1 for 3.4 or versions of 3.392…

3

10 2 ( 1 − cos2 θ) = cos θ + 2 − 2 cos2 θ = cos θ s.o.i. valid attempt at solving their quadratic in cos θ cos θ = - ½ www θ = 90, 270, 120, 240

M1 A1 DM1 A1 A1

for 1 - cos2 θ = sin2 θ substituted graphic calc method: allow M3 for intersection of y = 2 sin2 θ and y = cos θ + 2 and A2 for all four roots. All four answers correct but unsupported scores B2. 120 and 240 only: B1.

5

18

18


7

Section B 11 i (x + 5)(x − 2)(x + 2) 2 M1 for a (x + 5)(x − 2)(x + 2) 2

ii [(x + 2)](x2 + 3x − 10)

x3 + 3x2 − 10x + 2x2 + 6x − 20 o.e.

M1 M1

for correct expansion of one pair of their brackets for clear expansion of correct factors – accept given answer from (x + 5)(x2 − 4) as first step

2

iii iv

y′ = 3x2 + 10x − 4 their 3x2 + 10x – 4 = 0 s.o.i. x = 0.36… from formula o.e. (−3.7, 12.6) (-1.8, 12.6)

M2 M1 A1 B1+1 B1+1

M1 if one error or M1 for substitution of 0.4 if trying to obtain 0, and A1 for correct demonstration of sign change accept (−1.9, 12.6) or f.t.( ½ their max x, their max y)

6 2

12 i Area = (-)0.136 seen [m2] www Volume = 0.34 [m3] or ft from their area × 2.5

4 1

M3 for 0.1/2 × (0.14 + 0.16 + 2[0.22 + 0.31 + 0.36 + 0.32]) M2 for one slip; M1 for two slips must be positive

5

ii 2x4 − x3 − 0.25 x2 − 0.15x o.e. value at 0.5 [− value at 0] = −0.1375 area of cross section (of trough) or area between curve and x-axis 0.34375 r.o.t. to 3 or more sf [m3] m3 seen in (i) or (ii)

M2 M1 A1 E1 B1 U1

M1 for 2 terms correct dep on integral attempted must have neg sign

7

13 i log P = log a + b log t www comparison with y = mx + c intercept = log10 a

1 1 1

must be with correct equation condone omission of base

3

ii log t 0 0.78 1.15 1.18 1.20 log P 1.49 1.64 1.75 1.74 1.76 plots f.t. ruled line of best fit

1 1 1 1

accept to 2 or more dp 4

iii gradient rounding to 0.22 or 0.23

a = 101.49 s.o.i. P = 31t m allow the form P = 100.22logt

+1.49

2 1 1

M1 for y step / x-step accept1.47 – 1.50 for intercept accept answers that round to 30 – 32 , their positive m

4

iv answer rounds in range 60 to 63 1 1


6


Section A 1 4x5

1212x

−− + c

1 2 1

M1 for other 12kx

−

4

2 95.25, 95.3 or 95 4 M3 ½×5×(4.3+0+2[4.9+4.6+3.9+2.3+1.2]) M2 with 1 error, M1 with 2 errors. Or M3 for 6 correct trapezia.

4

3 1.45 o.e. 2 M1 for 61

51

41

31

21 ++++ oe 2

4 105 and 165 3 B1 for one of these or M1 for 2x = 210 or 330

3

5 (i) graph along y = 2 with V at (3,2) (4,1) & (5,2) (ii) graph along y = 6 with V at (1,6) (2,3) & (3,6)

2 2

M1 for correct V, or for f(x+2) B1 for (2,k) with all other elements correct

4

6 (i) 54.5 (ii) Correct use of sum of AP formula with n = 50, 20, 19 or 21 with their d and a = 7 eg S50 = 3412.5, S20 = 615 Their S50 − S20 dep on use of ap formula 2797.5 c.a.o.

2 M1 M1 A1

B1 for d = 2.5 or M2 for correct formula for S30 with their d M1 if one slip

5

7 8x − x-2 o.e.

their 0=dxdy

correct step x = ½ c.a.o.

2 M1 DM1A1

B1 each term s.o.i. s.o.i.

5

8 (i) 48 geometric, or GP (ii) mention of |r| < 1 condition o.e. S = 128

1 1 1 2

M1 for 211

192−−

5

9 (i) 1 (ii) (A) 3.5 logax (ii) (B) − logax

1 2 1

M1 for correct use of 1st or 3rd law

4


7

Section B 10 i 7 − 2x

x = 2, gradient = 3 x = 2, y = 4 y – their 4 = their grad (x − 2) subst y = 0 in their linear eqn completion to x = 3

2 (ans given)

M1 A1 B1 M1 M1 A1

differentiation must be used or use of y = their mx + c and subst (2, their 4), dependent on diffn seen

6

ii f(1) = 0 or factorising to (x − 1)(6 − x) or (x − 1)(x − 6) 6 www

1 1

or using quadratic formula correctly to obtain x = 1

2

iii 2 37 1 62 3

x x x− −

value at 2 − value at 1 126

or 2.16 to 2.17

1 4 42 3

× × − their integral

0.5 o.e.

M1 M1 A1 M1 A1

for two terms correct; ignore +c ft attempt at integration only

5

11 i(A) 150 (cm) or 1.5 m 2 M1 for 2.5 × 60 or 2.5 × 0.6 or for 1.5 with no units

2

i(B) ½ × 602 × 2.5 or 4500 ½ × 1402 × 2.5 or 24 500 subtraction of these 20 000 (cm2) isw

M1 M1 DM1A1

or equivalents in m2 or 2 m2

4

ii(A) attempt at use of cosine rule

2 2 23.5 2.8 1.6cos EFP2 2.8 3.5

+ −=

× × o.e.

26.5 to 26.65 or 27

M1 M1 A1

condone 1 error in substitution 3

ii(B) 2.8 sin (their EFP) o.e. 1.2 to 1.3 [m]

M1 A1

2


8

12 i log a + log (bt) www

clear use of log (bt) = t log b dep

B1 B1

condone omission of base throughout question

2

ii (2.398), 2.477, 2.556 , 2.643, 2.724 points plotted correctly f.t. ruled line of best fit f.t.

T1 P1 1

On correct square

3

iii log a = 2.31 to 2.33 a = 204 to 214 log b = 0.08 approx b = 1.195 to 1.215

M1 A1 M1 A1

ft their intercept ft their gradient

4

iv eg £210 million dep 1 their £a million 1

v log1000 their intercept 3 2.32their gradient 0.08

− −≈

= 8.15 to 8.85

M1 A1

or B2 from trials

2


6


Section A 1 using Pythagoras to show that hyp.

of right angled isos. triangle with sides a and a is √2a completion using definition of cosine

M1 A1

www a any letter or a number NB answer given

2

2 62 5x x+ value at 2 − value at 1 131

M2 M1 A1

M1 if one error ft attempt at integration only

4

3 (i) 193 (ii) divergent + difference between terms increasing o.e.

2 1

M1 for 8 + 15 + …+ 63

3

4 (i) 2.4 (ii) 138

2 2

M1 for 43.2 ÷ 18

M1 for their (i) × or

θ = o.e. or for other rot versions of 137.50…

4

5 (i)sketch of cosx ; one cycle, sketch of cos2x; two cycles, Both axes scaled correctly (ii) (1-way) stretch parallel to y axis sf 3

1 1 D1 1 D1

5

6 y′ = 3x2 − 12x − 15 use of y′ = 0, s.o.i. ft x = 5, – 1 c.a.o. x < −1 or x > 5 f.t.

M1 M1 A1 A1 A1

for two terms correct

5

7 use of cos2 θ = 1 − sin2 θ at least one correct interim step in obtaining 4 sin2 θ - sin θ = 0. θ = 0 and 180, 14.(47…) 165 - 166

M1 M1 B1 B1 B1

NB answer given r.o.t to nearest degree or better -1 for extras in range

5


7

8 attempt to integrate 3 x − 5

[y=] 322 5x x c− +

subst of (4, 6) in their integrated eqn

c = 10 or [y=] 322 5 10x x− +

M1 A2 M1 A1

A1 for two terms correct

5

9 (i) 7 (ii) 5.5 o.e.

1 2

M1 for at least one of 5 log10a or ½ log10a or log10a5.5 o.e.

3

Section B 10 i 0.6(0..), 0.8(45..), [1], 1.1(76..)

1.3(0..), 1.6(0..) points plotted correctly f.t. ruled line of best fit

T 1 P1 L1

Correct to 2 d.p. Allow 0.6, 1.3 and 1.6 tol. 1 mm

3

ii b = their intercept a = their gradient -11 ≤ b ≤ -8 and 21 ≤ a ≤ 23.5

M1 M1 A1

3

iii 34 to 35 m 1 1

iv 29 = “22”logt – “9”

t = 10”1.727..”

55 [years] approx

M1 M1 A1

accept 53 to 59

3

v For small t the model predicts a negative height (or h = 0 at approx 2.75) Hence model is unsuitable

1 D1

2


8

11 iA 10+20+30+40+50+60 B1

or ( )6 2 10 5 102

× + × or ( )6 10 602

+ 1

iB correct use of AP formula with a = 10 and d = 10 n (5 + 5n) or 5n (n + 1) or 5 (n2 + n) or (5n2 + 5n) 10n2 + 10n - 20700 = 0 45 c.a.o.

M1 A1 M1 A1

Or better

4

iiA 4 1 1

iiB £2555 2 M1 for 5(1 + 2 + …28) or 5(29 − 1) o.e.

2

iiC correct use of GP formula with a =5, r = 2 5(2n − 1) o.e.= 2621435 2n = 524288 www 19 c.a.o.

M1 DM1 M1 A1

“S” need not be simplified

4

12 i 6.1 2 M1 for

( ) ( )2 23.1 7 3 73.1 3

− − −−

o.e. 2

ii ( )( ) ( )2 23 7 3 7h

h

+ − − −

numerator = 6h + h2 6 + h

M1 M1 A1

s.o.i. 3

iii as h tends to 0, grad. tends to 6 o.e. f.t.from “6”+h

M1 A1

2

iv y − 2 = “6” (x − 3) o.e. y = 6x − 16

M1 A1

6 may be obtained from

2

v At P, x = 16/6 o.e. or ft At Q, x = 7 0.021 cao

M1 M1 A1

3


7

4752 (C2) Concepts for Advanced Mathematics 1 ½ x2 + 3x-1 + c o.e. 3 1 for each term 3

2 (i)

(ii)

5 with valid method 165 www

1 2

eg sequence has period of 4 nos. M1 for 13× (1 + 3 + 5 + 3) + 1 + 3 +5 or for 14 × (1 + 3 + 5 + 3) − 3

3

3 rt angled triangle with √2 on one side and 3 on hyp Pythag. used to obtain remaining side = √7

tan θ = 27

oppadj

= o.e.

1 1

1

or M1 for cos2 θ = 1 − sin2 θ used

A1 for cos θ = 79

A1 for tan θ = sin 2cos 7

θθ

= o.e.

3

4 radius = 6.5 [cm]

3 M1 for ½ × r2 × 0.4 [= 8.45] o.e.

and M1 for r2 = 1694

o.e. [= 42.25]

3

5 (i) (ii)

sketch of correct shape with P (−0.5,2) Q (0,4) and R (2,2) sketch of correct shape with P (−1,0.5) Q (0,1) and R (4,0.5)

2 2

1 if Q and one other are correct 1 if Q and one other are correct

4

6 (i) (ii)

205 253

o.e.

3

2

M1 for AP identified with d = 4 and M1 for 5 + 50 d used

M1 for r = 25

o.e.

5

7 (i) (ii)

sin A sin 795.6 8.4

= s.o.i.

[A =] 40.87 to 41 [BC2 =] 5.62 + 7.82 – 2 × 5.6 × 7.8 × cos (“180-79”) = 108.8 to 108.9 [BC =] 10.4(…)

M1 A1 M1 A1 A1

5

8 123y x−′ =

¾ when x = 16 y = 24 when x = 16 y − their 24 = their ¾ (x − 16) y − 24 = ¾ (x − 16) o.e.

M1 A1 B1 M1 A1

condone if unsimplified

dependent on ddyx

used for m

5


9

(iii)

0.0100 ≤ m < 0.0125 a = 10c or loga = c P = 10c × 10mt or 10mt + c

B2 B1 B1

M1 for stepstep

yx

−−

1.96 ≤ c ≤ 2.02 f.t. their m and a

4

(iv) use of t = 105 1.0 – 2.0 billion approx unreliable since extrapolation o.e.

B1 B1 E1

3

Oxford Cambridge and RSA Examinations

GCE

Mathematics (MEI) Advanced GCE 4752

Concepts for Advanced Mathematics (C2)

Mark Scheme for June 2010


1

SECTION A

1 [1], 1 1 12 3 4, , 2 B1 for [1], 1 1

2 3,

2 (i) 1122 or 25

12 or 2.08(3…) 2

M1 for 1 1 1 11 2 3 4

2 (ii)

6

2( 1)

r

r r o.e.

2 M1 for [f(r) =] r(r + 1) o.e. M1 for [a =] 6

3 (i) 3x2 − 12x − 15 2

M1 if one term incorrect or an extra term is included.

3 (ii) Their ddyx = 0 s.o.i.

x = 5 x = −1

M1 B1 B1

4 crossing x-axis at 0 and 2.5 min at (1.25, −6.25) crossing x-axis at 0 and 5 min at (2.5, −18.75)

1 1 1 1

5 x −

262

x

o.e.

their [5 + 325 ] – [2 + 3

4 ] = 2.37 o.e. c.a.o.

2 M1 A1

M1 for 1 term correct Dependent on at least M1 already earned i.s.w.

6 attempt to integrate

12 26 12x x

[y =] 2x3 + 8x1.5 + c Substitution of (4, 10) [y =] 2x3 + 8a1.5 –182 or c = –182

M1 A2 M1 A1

accept un-simplified; A1 for 2 terms correct dependent on attempted integral with + c term

7 3.5 loga x or k = 3.5

2

B1 for 3 loga x or ½ loga x or loga x3½ seen


2

8 Subst. of 1− cos2 θ or 1 − sin2 θ 5 cos2 θ = 1 or 5 sin2 θ = 4 cos θ = 1

5their or

sin θ = 45their o.e.

63.4, 116.6, 243.4, 296.6

M1 A1 M1 B2

Accept to nearest degree or better; B1 for 2 correct (ignore any extra values in range).

9 log 18 = log a + n log 3 and log 6 = log a + n log 2 log 18 − log 6 = n (log 3 − log 2) n = 2.71 to 2 d.p. c.a.o. log 6 = log a + 2.70951…log 2 o.e. a = 0.92 to 2 d.p. c.a.o.

M1*DM1 A1 M1 A1

or 18 = a × 3n and 6 = a × 2n

3 = 32

n

n = log3log1.5 = 2.71 c.a.o.

6 = a × 22.70951 o.e. = 0.92 c.a.o.

Section A Total: 36 SECTION B

10 (i) ddyx = 4x3

when x = 2, ddyx = 32 s.o.i.

when x = 2, y = 16 s.o.i. y = 32x – 48 c.a.o.

M1 A1 B1 A1

i.s.w.

10 (ii) 34.481 2 M1 for 4 42.1 20.1

10 (iii) (A)

16 + 32h + 24h2 + 8h3 + h4 c.a.o. 3 B2 for 4 terms correct B1 for 3 terms correct

10 (iii) (B)

32 + 24h + 8h2 + h3 or ft 2 B1 if one error

10 (iii) (C)

as h 0, result their 32 from (iii) (B) gradient of tangent is limit of gradient of chord

1 1


3

11 (a) 10.62 + 9.22 − 2×10.6×9.2×cos68°

o.e. QR = 11.1(3…)

sin Qsin 68 sin Rtheir QR 9.2 10.6 or o.e.

Q = 50.01..° or R = 61.98..° bearing = 174.9 to 175°

M1

A1

M1

A1

B1

Or correct use of Cosine Rule 2 s.f. or better

11 (b) (i) (A) ½ × 802 × 2

3

= 64003

M1

A1

6702.(…) to 2 s.f. or more

11 (b) (ii)

DC = 80 sin( 3 ) = 80 3

2 Area = ½×their DA×40√3 or ½×40√3×80×sin(their DCA) o.e. area of triangle = 8003 or 1385.64… to 3s.f. or more

B1

M1

A1

both steps required s.o.i.

11 (b) (iii)

area of ¼ circle = ½ × × (403)2 o.e. “6702” + “1385.6” – “3769.9” = 4300 to 4320

M1

M1

A1

[=3769.9…] i.e. their(b) (i) + their (b) (ii) – their ¼ circle o.e. 933⅓π + 8003


4

12 (i) (A)

1024 2 M1 for number of buds = 210 s.o.i.

12 (i) (B)

2047 2 M1 for 1+2+4+…210 or for 211 – 1 or (their 1024) + 512 + 256 +…+ 1

12 (ii) (A)

no. of nodes = 1 + 2 +..+ 2n-1 s.o.i.

7 2 12 1

n

1 1

no. of leaves = 7 + 14 +…+ 7×2n-1

12 (ii) (B)

7(2n − 1) > 200 000 2n > 200 000

7 + 1 or 200 0077

n log 2 > log ( 200 007

7 ) and completion to given ans [n =] 15 c.a.o.

M1

M1

M1

B1

or log7 + log2n > log200 007

Section B Total: 36

GCE


Unit 4752: Concepts for Advanced Mathematics

Advanced Subsidiary GCE

Mathematics (MEI)

Mark Scheme for January 2011


SECTION A 1 11.4 o.e. 2 M1 for 12/3 + 12/4 + 12/5 + 12/6 o.e.

M0 unless four terms summed

2 161 2

2 4x x c 4

B1 for 612 x , M1 for

12kx , A1 for k = 4

or , B1 for + c dependent on at least one power increased

allow x6 isw,

3 ½ 1.5 (0.6 + 0.7 + 2(2.3+3.1+2.8+1.8)) = 15.975 rounded to 2 s.f. or more

M2

A1

M1 if one error or M2 for sum of 5 unsimplified individual trapezia: 2.175, 4.05, 4.425, 3.45, 1.875

basic shape of formula must be correct. Must be 5 strips. M0 if pair of brackets omitted or h = 7.5 or 1. allow recovery of brackets omitted to obtain correct answer. M0 for other than 5 trapezia isw only if 15.975 clearly identified as cross-sectional area

4 (i) (3, 15)

B2 B1 for each coordinate s.c. B0 for (3, 5)

4 (ii) (1.5, 5)

B2 B1 for each coordinate s.c. B0 for (3, 5)

5 ar = 6 and ar4 = - 48 r = −2 tenth term = 1536

3(1 ( 2) )1 ( 2)

n

o.e.

( 2) 1n

M1 M1 A1

M1

A1

B2 for r = -2 www B3 for 1536 www allow M1 for a = 6÷their r and substitution in GP formula with their a and r

c.a.o.

ignore incorrect lettering such as d =-2 condone the omission of the brackets round “-2” in the numerator and / or the denominator

1


6 a+2d = 24 and a + 9d = 3

d = −3; a = 30 S50 − S20 −2205 cao

M1 A1A1

M1

A1

if M0 , B2 for either, B3 for both ft their a and d;

M1 for S30 = (u21 + u50) o.e. B2 for -2205 www

do not award B2 or B3 if values clearly obtained fortuitously S50 = −2175; S20 = 30 u21 = 30 20 × 3 = 30 u50 = 30 49 × 3 = 117

7 (i) 17 log10 x or log10 x17

B2 M1 for 5log10 x or 12 log10 x or log10 x12 as part of the first step

condone omission of base

7 (ii) −b B2 M1 for loga 1 = 0 or logaa = 1 soi

allow 0 - b

8 substitution of sin2 θ = 1- cos2 θ −5cos2 θ = cos θ θ = 90 and 270, 102 258 101 and 259

M1 A1 A1 A1 A1

SC1

soi or better accept 101.5(…) and 258.(46…) rounded to 3 or more sf; if M0, allow B1 for both of 90 and 270 and B1 for 102 and B1 for 258 (to 3 or more sf)

if the 4 correct values are presented, ignore any extra values which are outside the required range, but apply a penalty of minus 1 for extra values in the range if given in radians deduct 1 mark from total awarded (1.57, 1.77, 4.51, 4.71)

2


9 area sector =

221

2 6 12 rr

area triangle = 2

21 sin2 6 4

aa

½a2×½ = ½×r2 × ×½

2 2

4 24a r

o.e. and completion to

given answer

M1 M1 M1 A1

soi soi soi

allow sin30 no follow through marks available at least one correct intermediate step required, and no wrong working to obtain given answer

Section A Total: 36

3


SECTION B 10 (i) eqn of AB is y = 3x + 1 o.e.

their “3x + 1” = 4x2 (4x + 1) (x – 1) = 0 o.e. so x = −1/4 at C, x = −1/4, y = 4 (−1/4)2 or 3 (− 1/4) + 1[=1/4 as required]

M1 M1 M1 A1

or equiv in y: 214

3yy

or rearranging and deriving roots y = 4 or ¼ condone verification by showing lhs = rhs o.e. or y = ¼ implies x = ¼ so at C x = − ¼

SC3 for verifying that A, B and C are collinear and that C also lies on the curve SC2 for verifying that A, B and C are collinear by showing that gradient of AB = AC (for example) or showing C lies on AB solely verifying that C lies on the curve scores 0

10 (ii) y = 8x at A y = 8 eqn of tgt at A y − 4 = their“8”(x − 1) y = 8x - 4 at C y = 8 −1/4 [=−2] y − ¼ = −2(x − (− ¼)) or other unsimplified equivalent to obtain given result. allow correct verification that (-¼,¼) lies on given line

M1 A1 M1 A1 M1 A1

ft their gradient NB if m = -2 obtained from given answer or only showing that (− ¼, ¼) lies on given line y = −2x − ¼ then 0 marks.

gradient must follow from evaluation of condone unsimplified versions of y = 8x - 4 dependent on award of first M1 SC2 if equation of tangent and curve solved simultaneously to correctly show repeated root

10 (iii) their “8x – 4” = −2x − ¼ y = −1 www

M1 A1

or 144

8 2yy

o.e. [x = 3/8]

4


11 (i)

4 23 3

4 2x xx x

their integral at 3 − their integral at 1 [= 2.25 − 1.75] = −4 isw represents area between curve and x axis between x = 1 and 3 negative since below x-axis

M2

M1

A1

B1

B1

M1 if at least two terms correct dependent on integration attempted

ignore + c M0 for evaluation of 3 23 3x x x or of differentiated version B0 for area under or above curve between x = 1 and 3

11 (ii) y = 3x2 − 6x − 1 their y = 0 soi

x = with a = 3, b = -6 and c = -1 isw

x = or better as final answer 6 48 6 48

6 6x

or ft their

final answer

M1 M1

M1

A1

B1

dependent on differentiation attempted or 3(x – 1)2 - 4 [= 0] or better

eg A1 for 21 33

allow ≤ instead of <

no follow through; NB or better stated without working implies use of correct method A0 for incorrect simplification, eg 1 ± √48 allow B1 if both inequalities are stated separately and it’s clear that both apply allow B1 if the terms and the signs are in reverse order

12 (i) 50% of 25 000 is 12 500 and the population [in 2005] is 12 000 [so consistent]

B1 or 12 000 is 48% of 25 000 so less than 50%[ so consistent]

12 (ii) log 10 P = log10a – kt or

log 10 = -kt o.e. www

B2 condone omission of base; M1 for log 10 P = log10a + log1010- kt or better www

5


6

12 (iii) 4.27, 4.21, 4.13, 4.08 plots ruled line of best fit drawn

B1

B1 B1

accept 4.273…, 4.2108…, 4.130…, 4.079… rounded to 2 or more dp 1 mm tolerance ft their values if at least 4 correct values are correctly plotted

f.t. if at least two calculated values correct must have at least one point on or above and at least one point on or below the line and must cover 0 ≤ t ≤ 25

12 (iv) a = 25000 to 25400 0.01 ≤ k ≤ 0.014 P = a 10-kt or P = 10loga - kt with values in acceptable ranges

B1

B2

B1

allow 104.4..

M1 for -k = using values from table or graph; condone +k B0 if left in logarithmic form

M1 for a correct first step in solving a pair of valid equations in either form A1 for k A1 for a A1 for P = a 10-kt

12 (v) P = a 10-35k 8600 to 9000 comparing their value with 9375 o.e. and reaching the correct conclusion for their value

M1

A1

A1

T heir a and k f.t.

allow logP = loga – 35k

Section B Total: 36

GCE




Mathematics (MEI)


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SECTION A 1 ½x4 + 3x

F[5] – F[2] [=327.5 − 14] =313.5 o.e.

M1 M1

A1

accept unsimplified at least one term correctly integrated, may be implied by A1

ignore + c condone omission of brackets 313.5 unsupported scores 0

2 0.05, 2000, 1.25 × 10-6 or 1 1, 2000, 20 800000

o.e.

divergent

B2

B1

B1 for two correct allow “alternate terms tend to zero and to infinity” o.e.

do not allow “oscillating”, “getting bigger and smaller”, “getting further apart”

3 (i) m =

1 2 4.1 1 2 44.1 4

+ × − + ×−

s.o.i

grad = 9.2 94.1 4

−−

s.o.i

0.3315 cao

M1

M1

A1

no marks for use of Chain Rule or any other attempt to differentiate SC2 for 0.33…. appearing only embedded in equation of chord

3 (ii) selection of value in (4, 4.1) and 4 or of two values in [3.9, 4.1] centred on 4 answer closer to 1/3 than 0.3315(...)

M1

A1

allow selection of 4 and value in (3.9, 4)

4 6 = ab and 3.6 = ab2

a = 10, b = 0.6 c.a.o.

M1

A2

log6 = loga + logb and log3.6 = loga + logb2

A1 each; if M0 then B3 for both, B1 for one

1


5 [ dy

dx= ] 32x3 c.a.o.

substitution of x = ½ in their dydx

grad normal = 14their

−

when x = ½, y = 4 ½ o.e.

1 1 12 ) i.s.w 2 44 (y x− = − −

M1

M1

M1

B1

A1

[= 4]

514 8 o.e. 4y x= − +

must see kx3

their 4 must be obtained by calculus

6 126 2dy x

dx= −

32 2y kx x c= − + o.e. 324 2y x x c+ o.e. = −

correct substitution of x = 9 and y = 4 in their equation of curve

324 2 8 6y x x= − −

M2

A1

M1 dep

A1

M1 for k32x and M1 for -2x + c

dependent on at least M1 already awarded allow A1 for c = −86 i.s.w. if simplified equation for y seen earlier

16x is a mistake, not a misread

“y =” need not be stated at this point, but must be seen

at some point for full marks

must see “+ c”

2


7 sin 2sincos

θ θθ

=

2cos θ – 1 = 0 and sin θ = 0 [θ = ] 0, 180, 360, [θ = ] 60, 300 if 4 marks awarded, lose 1 mark for extra values in the range, ignore extra values outside the range

M1

A1 B1 B1

may be implied by 2cos θ - 1 = 0 or better

or, if to advantage of candidate B4 for all 5 correct B3 for 4 correct B2 for 3 correct B1 for 2 correct if extra value(s) in range, deduct one mark from total do not award if values embedded in trial and improvement approach

8 log log log np s t= + log log logp s n= + t

loglog log[ ] or

log log

pp s sn

t t

− =

[base not required]

M1

M1

A1

or np ts

=

log log pn ts

=

as final answer (i.e. penalise further incorrect simplification)

or A2 for

[ ]logtpns

= [base t needed ] following first M1

9 log16½ or [–] log52 s.o.i.

log(4×75) or log 7525

s.o.i.

x = 12 www

M1 M1 A1

x = 4 7525× implies M1M1

if a = 10 assumed, x = 12 c.a.o. scores B3 www no follow through

10 t1 = − sin θ t2 = sin θ

B1 B1

www www

e.g. sin(θ + 360) = sin θ + sin360 = sinθ B0

Section A Total: 36

3


SECTION B 11 (i) 200 - 2πr2 = 2πrh

2200 22

rhrπ

π−= o.e.

substitution of correct h into V = πr2h V = 100r − πr3 convincingly obtained

M1 M1 M1 A1

100 = πr2 + πrh 100r = πr3 + πr2h 100r = πr3 + V V = 100r − πr3

or

M1 for 2

Vhrπ

=

M1 for 200 = 2πr2 + 2πr× 2

Vrπ

M1 for 200 = 2πr2 + 2 Vr

A1 for V = 100r − πr3 convincingly obtained

sc3 for complete argument working backwards: V = 100r − πr3

πr2h = 100r − πr3

πrh = 100 – πr2

100 = πrh + πr2 200 = A = 2πrh + 2πr2

sc0 if argument is incomplete

11 (ii) dV

dr= 100 − 3πr2

2

2

d Vdr

= −6πr

B2 B1

B1 for each term

allow 9.42(….) r2 or better if decimalised -18.8(…) r or better if decimalised

4


11 (iii) their dV

dr = 0 s.o.i.

r = 3.26 c.a.o. V = 217 c.a.o.

M1 A2 A1

must contain r as the only variable

A1 for 100( )3

rπ

= ± ; may be implied

by 3.25… deduct 1 mark only in this part if answers not given to 3 sf,

there must be evidence of use of calculus

5


12 (i)(A) 390 B2 M1 for 500 – 11 × 10

12 (i)(B) ( )24

24 2 2 500 (24 1) 10= × + − × −S o.e. i.s.w. or ( )24

24 2 500 270S = + o.e. i.s.w. [=9240] (answer given)

B2

nothing simpler than

12(1000 + 23×-10) or 24 (1000 230)2

−

or 12(2 500 230)× − if B2 not awarded, then M1 for use of a.p. formula for S24 with n = 24, a = 500 and d = -10 or M1 for l = 270 s.o.i.

condone omission of final bracket or “(23)-10” if recovered in later work if they write the sum out, all the terms must be listed for 2 marks 12 × (1000 – 230) or 12 × 770 on its own do not score

12 (ii)(A) 368.33(...) or 368.34

B2 M1 for 460 × 0.9811

12 (ii)(B) J20 = 310 M20 = 313.36(...), 313.4, 313.3, 313.37 or 313 J19 = 320 M19 = 319.76(...), 319.8 or 319.7

B3 B3 for all 4 values correct or B2 for 3 values correct or B1 for 2 values correct

values which are clearly wrongly attributed do not score

12 (ii)(C) 8837 to 8837.06

B2 M1 for

( )24

24

460 1 0.98

1 0.98S

−=

− o.e.

12 (ii)(D)

24(1 0.98 ) 9240(1 0.98)

a − =−

o.e.

480.97 to 480.98

M1 A1

f.t. their power of 24 from (ii)C

6


7

13 (i) arc AC = 2.1 × 1.8

= 3.78 c.a.o. area = their 3.78 × 5.5 = 20.79 or 20.8 i.s.w.

M1

A1

M1dep* A1

103 2 2.1 360

π× ×

dependent on first M1

103° or better 3.78 must be seen but may be embedded in area formula

13 (ii) BD = 2.1 cos (π − 1.8) or 2.1cos1.3(4159…..) or 2.1sin0.2(292…) r.o.t to 1 d.p. or more = 0.48

M2

A1

M1 for BDcos( 1.8)2.1

π − = o.e.

allow any answer which rounds to 0.48

M2 for BD = 2.1 cos 76.8675…° or 2.1sin13.1324…rounded to 2 or more sf or M2 for CD = 2.045... r.o.t. to 3 s.f. or better and BD = √(2.12 – 2.0452)

13 (iii) sector area = 3.969 triangle area = 0.487 to 0.491 24.5

M2

M2

A1

M1 for ½ × 2.12 × 1.8 M1 for ½ × 2.1 × their 0.48 × sin (π − 1.8) or ½ × their 0.48 × 2.045.. r.o.t. to 3 s.f. or better allow any answer which rounds to 24.5

or equivalent with degrees for first two Ms N.B. 5.5 × 3.969 = 21.8295 so allow M2 for 21.8295 may be sin 1.8 instead of sin (π − 1.8) N.B. 5.5 × area = 2.6785 to 2.7005 so allow M2 for a value in this range

Section B Total: 36


Question Answer Marks Guidance 1 1

2 21 32

x x oe; isw B3 need not be simplified

B2 for one term correct ignore + c

if B0 allow M1 for either x1/2 or x-1 seen before differentiation deduct one mark for extra term in x

[3] 2 (5), 8, 11, (14),…isw B1 a = 5 and d = 3 soi B1

S50 = 502

(2×5 + (50 – 1) × 3) oe

M1

if M0, award B2 if 3925 is obtained

3925

A1

if M0, SC1 for use of a = 8 and obtaining 4075

from summing individual terms or if unsupported

[4] 3 (i) 9.82 + 6.42 2 × 9.8 × 6.4 × cos 53.4 M1 9.82 + 6.42 74.79... [= 62.2...] M1 for evidence of correct order of operations

used; may be implied by correct answer 6.89 implies M0 262.4368 implies M1 (calc in radian mode), (NB √262.436..=16.199…)

7.887… or 7.89 or 7.9 A1 if M0, B3 for 7.89 or more precise www NB 9.8sin53.4 = 7.87 [3] 3 (ii) ½ × 9.8 × 7.3 × sin (180 53.4) oe seen M1 or sin 53.4 used; may be embedded may be split into height = 9.8 × sin 53.4

then Area = ½ × 7.3 × height 28.716…or 28.72 or 28.7 or 29 isw A1 if M0, B2 for 28.7 or more precise www [2] 4 (i) (6, 9) 2 1 for each co-ordinate SC0 for (6, 3) [2] 4 (ii) (1.5, 3) 2 1 for each co-ordinate SC0 for (6, 3) [2] 5

45 = ½ r2 × 1.6 oe

M1 2 91.673...45

360r

r2 = 90/1.6 oe M1 r = 7.5 or exact equivalent cao A1 or B3 www allow recovery to 7.5 if working in

degrees, but A0 for (eg) 7.49

(their 7.5) × 1.6

M1 91.673...2 (their )

360r

12 implies M1

27 A1 or B2 www [5]

5


Question Answer Marks Guidance 6 gradient = 3 seen

log10 y – 5 = (their 3)( log10 x 1) or using (5, 17) log10 y = 3 log10 x + 2 oe

B1

M1

A1

may be embedded or log10 y = 3 log10 x + c and substitution of (1, 5) or (5, 17) for log10 x and log10 y

condone omission of base throughout NB may recover from eg Y = 3X + 2

103log 210 xy oe

y = 100x3

M1 A1

or 310 10 10log log log 100y x

or 10 3log 2yx

or 310 10log log 100y x

[5] 7 3

2

32

6x

324x

5x + c

M1*

A1

B1

may appear later

B0 if from 12(6 5)y x x c

condone “+ c” not appearing until substitution

substitution of (4, 20) M1dep* [y =] 4x1.5 5x + 8 or c = 8 isw A1 [5] 8 0.775397.. soi M1 or 44.427..°

0.388, 1.18, 3.53, 4.32 in degrees: 22.2, 67.8, 202, 248*

A4

A1 each value if A0 then B1 for at least two of 2.366…, 7.058…, 8.649…for 2θ or all of 135.57…, 404.427…, 495.57…

if any of final answers not given to three sf deduct 1 mark from total A marks *if final answers in degrees deduct 1 from total A marks ignore extra values outside range if four correct answers in degrees or radians, deduct 1 for extra values in range

[5]

6


Question Answer Marks Guidance 9 (i) ½ × 0.2 (0 + 0 + 2(0.5 + 0.7 + 0.75 + 0.7 +

0.5)) [=0.63]

M3 M2 if one error, M1 if two errors condone omission of zeros or M3 for 0.05 + 0.12 + 0.145 + 0.145 + 0.12 + 0.05 may be unsimplified, must be summed

basic shape of formula must be correct must be 6 strips M0 if brackets omitted, but allow recovery M0 if h = 1 or 1.2 Area = 6.3 and 0.53 imply M0

(their 0.63) × 50 M1 31.5 A1 [5] 9 (ii) (A) 4 3 23.8 0.2 – 6.8 0.2 7.7 0.2 – 4.2 0.2

M1 ±0.58032 implies M1 condone one sign error

0.01968 cao isw A1 or B2 if unsupported allow – 0.01968 [2] 9 (ii) (B) 5 4 3 23.8 6.8 7.7 4.2

5 4 3 2x x x x

+ c M2 M1 for two terms correct excluding c

condone omission of c

accept 2.56 to 2.57 for coefficient of x3 allow M1 if all signs reversed

F(0.9) [ F(0)] M1* as long as at least M1 awarded NB F(0.9) = – 0.496... 50 × their ±F(0.9) M1dep* 24.8 to 24.9 cao A1 [5]

7


Question Answer Marks Guidance 10 (i) y´ = 3x2 5 M1 their y´ = 0 M1

(1.3, 4.3) cao

A1 or A1 for 5 3

x oe soi

( 1.3, 4.3) cao A1 allow if not written as co-ordinates if pairing is clear

ignore any work relating to second derivative

[4] 10 (ii) crosses axes at (0, 0) B1 condone x and y intercepts not written as

co-ordinates; may be on graph See examples in Appendix

and ( 5 , 0) B1 ±(2.23 to 2.24) implies ±√5 sketch of cubic with turning points in correct

quadrants and of correct orientation and passing through origin

B1 must meet the x-axis three times B0 eg if more than 1 point of inflection

x-intercepts ±√5 marked B1 may be in decimal form (±2.2…) [4] 10 (iii) substitution of x = 1 in 2f 3 5x x M1 sight of 2 does not necessarily imply

M1: check 2f 3 5x x is correct in part (i)

2 A1 y 4 = (their ) × (x 1) oe f (1) M1* or 4 = 2×(1) + c 2x 2 = x3 5x and completion to given

result www M1dep*

use of Factor theorem in x3 – 3x + 2 with – 1 or ± 2

M1 or any other valid method; must be shown eg long division or comparing coefficients to find (x – 1)(x2 + x – 2) or (x + 2)(x2 – 2 x + 1) is enough for M1 with both factors correct

x = 2 obtained correctly A1 NB M0A0 for x(x2 – 3) = – 2 so x = – 2 or x2 – 3 = – 2 oe

[6]

8


Question Answer Marks Guidance 11 (i) ar = 6 oe B1 must be in a and r

251

ar

oe

B1 must be in a and r

25 61

a

a

M1 or 6 25(1 )r

r

NB assuming a = 10 earns M0

a2 – 25a + 150 [= 0] A1 or 25r2 – 25r + 6 [= 0] All signs may be reversed a = 10 obtained from formula, factorising,

Factor theorem or completing the square A1 r = 0.4 and r = 0.6

a = 15

A1

a = 15

if M0, B1 for r = 0.4 and 0.6 and B1 for a = 15 by trial and improvement

r = 0.4 and 0.6

A1

6 100.6

a oe mark to benefit of candidate

[7] 11 (ii) 10 × (3/5)n – 1 and 15 × (2/5)n – 1 seen M1

15 × 2n – 1 : 10 × 3n-1 or 1 1

1 12 33 : 25 5

n n

n n

M1

may be implied by 3 × 2n – 1 : 2 × 3n – 1

condone ratio reversed

3 × 2n – 1 : 2 × 3n – 1

A1

and completion to given answer www

condone ratio reversed

[3]

9


Appendix: examples for Question 10(ii)

Example 1

3rd B1 BOD inflection

Example 2

3rd B1 BOD Shape

Example 3

3rd B1 BOD Point of inflection on left

Example 4

Clearly does not pass through origin 3rd B0

Example 5

3rd B1

Example 6

3rd B1 condone RHS

10


11

Example 7

3rd B1 condone extreme ends ruled

Example 8

3rd B0 doesn’t meet x- axis 3 times

Example 9

4th B1is earned in spite of the curve not being a cubic

Example 10

x-intercepts: co-ordinates reversed but condone this as candidates who write 2.23 , 2.23 only would not be penalised 4th B1

Example 11

(3rd B0: incorrect orientation) 4th B1

Example 12

4th B1 earned.


5

Question Answer Marks Guidance 1 5

2kx k = 12 + c

M1

A1

A1

[3]

2 (i) converging + valid reason

1 eg converges to 0, r = ½, difference between terms decreasing, sum of terms converges to 6, G.P. with |r| < 1

[1]

2 (ii) neither + valid reason

1 eg divergent oe, A.P., d = 4 oe, convergent and periodic ruled out with correct reasons

[1]

2 (iii) periodic + valid reason

1 eg repeating cycle of terms

[1]

3 (i) (0.8, 2) oe

2 B1 each coordinate SC0 for (4, −2)

[2]

3 (ii) Translation

B1

900

oe

B1 or eg 270 to left

allow B2 for rotation through 180° about (45, 0) oe

[2]


6

Question Answer Marks Guidance 4 (i) 1.2r = 4.2

3.5 cao

M1

A1 or 68.7549... 2 4.2

360r with θ to 3 sf or

better

B2 if correct answer unsupported

[2]

4 (ii) cos 0.6

their3.5d

2.888.. to 2.9

M1

A1

or cos34.377..their3.5

d with θ to 3 sf or

better

or correct use of Sine Rule with 0.9708 (55.623°) or area = 5.709 = 0.5 × h × 3.952, or 3.52 – 1.9762 = d2

[2]

5 gradient = 4 9.5 12

9.5 9

0.6577 to 0.66

M1

A1

or 0.657656...isw

4 38 24 4√38 −24

9 < xC < 9.5

B1 allow 8.53 ≤ xC < 9

[3]

6 6x2 + 18x 24

B1

their 6x2 + 18x 24 = 0 or > 0 or 0 −4 and + 1 identified oe

M1

A1

or sketch of y = 6x2 + 18x 24 with attempt to find x-intercepts

x < 4 and x > 1 cao

A1 or x ≤ 4 and x ≥ 1 if B0M0 then SC2 for fully correct answer

[4]


7

Question Answer Marks Guidance 7 2 2 2105 92 75cosA

2 105 92

oe

0.717598…soi A = 44.14345…° soi [0.770448553…]

M1

A1

A1

or 2 2 275 92 105cos2 75 92

B

oe

0.2220289…soi B = 77.1717719……° soi [1.346901422]

or 2 2 2105 75 92cos

2 105 75C

oe

0.519746…soi C = 58.6847827…° soi [1.024242678…]

ignore minor errors due to premature rounding for second A1 condone A, B or C wrongly attributed

½ × 92 × 105 × sin (their A) 3360 or 3361 to 3365

M1

A1

or ½ × 75 × 92 × sin (their B)

or ½ × 75 × 105 × sin (their C)

or M3 for

136(136 75)(136 105)(136 92) A2 for correct answer 3360 or 3363 - 3364 [5]

8 (i)

M1

A1

for curve of correct shape in both quadrants through (0, 1) shown on graph or in commentary

SC1 for curve correct in 1st quadrant and touching (0,1) or identified in commentary

[2]

x

y

1


8

Question Answer Marks Guidance 8 (ii)

5x 1 = 10

10

log 500000log 3

M1 or 5x 1 = log3500 000 condone omission of base 10 use of logs in other bases may earn full marks

x = 10

10

log 500000( 1)

log 3 ÷ 5

M1

x = (log3500000 + 1)÷5

[x = ] 2.588 to 2.59

A1 oe; or B3 www if unsupported, B3 for correct answer to 3 sf or more www

[3]

9 (i) sin( ) 1coscos

oe

sin θ = cos2 θ and completion to given result

M1

A1

www

[2]

9 (ii) sin2 θ + sin θ 1[ = 0]

1 5[sin ]2

oe may be implied by

correct answers [θ =] 38.17... ,or 38.2 and 141.83..., 141.8 or 142

M1

A1

A1

[3]

allow 1 on RHS if attempt to complete square may be implied by correct answers ignore extra values outside range, A0 if extra values in range or in radians NB 0.6662 and 2.4754 if working in radian mode earns M1A1A0

condone y2 + y – 1 = 0 mark to benefit of candidate ignore any work with negative root & condone omission of negative root with no comment eg M1 for 0.618… if unsupported, B1 for one of these, B2 for both. If both values correct with extra values in range, then B1. NB 0.6662 and 2.4754 to 3sf or more


9

Question Answer Marks Guidance 10 (i) at A y = 3

d 2 4dy xx

B1

B1

their d 2 4 4

dyx

M1*

must follow from attempt at differentiation

grad of normal = –1/their 4

M1dep*

y 3 = (1/4) × (x 4) oe isw

A1

substitution of y = 0 and completion to given result with at least 1 correct interim step www

A1 or substitution of x = 16 to obtain y = 0 correct interim step may occur before substitution

[6]

10 (ii) at B, x = 3

B1 may be embedded

3 24F[ ] 33 2x xx x

M1*

condone one error, must be three terms, ignore + c

F[ 4] F[their 3]

M1* dep

dependent on integration attempted

area of triangle = 18 soi

B1

may be embedded in final answer

area of region = 119

3 oe isw

A1

19.3 or better

[5]


10

Question Answer Marks Guidance 11 (i) (A) 2A + D = 25 oe

4A + 6D = 250 oe B1 B1

condone lower-case a and d

D = 50, A = 12.5 oe

B1 B1

[4]

11 (i) (B) 50 2 492

theirA their D [= 60 625] or

20 2 192

their A their D [= 9250]

their “S50 S20” 51 375 cao

M1

M1

A1

[3]

or a = their A + 20D

3030 ( )2

S a l oe with l = their A + 49D

3030 (2 987.5 29 50)2

S their their


11

Question Answer Marks Guidance 11 (ii) 2 1

=25 1

a r

r

or

4 1250

1a r

r

B1

4

2

( 1)250125( 1)

( 1)

rarrar

oe

and completion to given result www

M1

at least one correct interim step required

allow a(1 + r) as the denominator in the quadruple- decker fraction

use of r4 1 = (r2 1)( r2 + 1) to obtain r2 +1 = 10 www

M1

or multiplication and rearrangement of quadratic to obtain r4 10r2 + 9 = 0 oe with all three terms on one side

r2 = x oe may be used or M1 for valid alternative algebraic approaches eg using a(1 + r) = 25 and ar2 + ar3 = ar2 (1 + r) = 225

r = ± 3

A1 or B2 for all four values correct, B1 for both r values or both a values or one pair of correct values if second M mark not earned

a = 6.25 or 12.5 oe

A1 or A1 for one correct pair of values of r and a

[5] 12 (i) log10p = log10a + log1010kt

log10p = log10a + kt www

M1

A1

condone omission of base;

if unsupported, B2 for correct equation

[2] 12 (ii) 2.02, 2.13, 2.23 B1 allow given to more sig figs 2.022304623…, 2.129657673,

2.229707433 plots correct B1f.t. to nearest half square ruled line of best fit

B1 y-intercept between 1.65 and 1.7 and at least

one point on or above the line and at least one point on or below the line

ft their plots must cover range from x = 9 to 49

[3]


12

Question Answer Marks Guidance 12 (iii) 0.0105 to 0.0125 for k

1.66 to 1.69 for log10a or 45.7 to 49.0 for a

B1

B1

must be connected to k must be connected to a

log10p = their kt + their log10a B1 must be a correct form for equation of line and with their y-intercept and their gradient (may be found from graph or from table, must be correct method)

p = their “47.9 × 100.0115t“or 1.6785 0.011510 t ”

B1 as above, “47.9” and “0.0115” must follow from correct method

[4] 12 (iv) 45.7 to 49.0 million

1 ‘million’ needed, not just the value of p

[1] 12 (v) reading from graph at 2.301..

their 54 2014 cao

M1*

M1dep*

A1

or log10200 = “log10a + kt”

eg for their log 200 1.680.0115

t

if unsupported, allow B3 only if consistent with graph

or 200 = “10loga + kt” oe

or M1 for their

200log47.9

0.0115t

[3]


7

Question Answer Marks Guidance 1 (i) −10x− 6 isw B1

B1 for – 10 for x− 6 ignore + c and y =

if B0B0 then SC1 for −5×2x−5−1 or better soi

[2] 1 (ii) 1

3y x soi 1nkx 231

3x

isw

B1 M1

A1

condone 1

3y x if differentiation follows ft their fractional n ignore + c and y =

allow 0.333 or better

[3] 2 (i) 11.5, 11 and 10.5 oe B1 ignore labelling arithmetic and/or divergent B1 allow AP

ignore references to a, d or n incorrect embellishments such as converging arithmetic…, diverging geometric… do not score. B0 if a choice is given eg AP/GP.

[2] 2 (ii)

n = 30 identified as number of terms in relevant AP

B1

eg 1 + 2 + 3 + …+ 30 is not a relevant AP

30

30 (2 11.5 (30 1) 0.5)2

S

M1 or 3030 (11.5 3)2

S condone one error in a, d or n but do not condone l = −½

127.5 oe

A1

[3]

allow recovery from slip in working (eg omission of minus sign)

SC3 if each term calculated and summed to correct answer or for 127.5 unsupported

3 kx-2 9x-2

+ 2x + c

M1* A1

M1*

may be awarded later c may appear at substitution stage

k ≠ 0 no marks at all for responses based on “mx + c”

substitution of x = 3 and y = 6 in their expression following integration

M1dep on award of either of previous M1s eg 6 = k3−2 + 2×3 + c

c = 1 A1 A0 if spoiled by further working for full marks, must see “y =” at some stage

[5]


8

Question Answer Marks Guidance 4 (i) clear diagram or explanation starting with

equilateral triangle correctly showing 30 as half angle and sides 1 and 2 or multiples of these lengths

B1 units for sides and angle not required

correct use of Pythagoras and adjacent and hypotenuse correctly identified to obtain

given result 3cos302

B1 adjacent and hypotenuse may be identified on diagram

condone abbreviations

[2] 4 (ii) 5 or

6 6

soi

11

6

76

M1

A1

A1

may be implied by correct answer or ±0.523598775…, or may appear on quadrant diagram or graph if A0A0, SC1 for 1.8333333π and 1.16666666π to 3 or more sf or SC1 for 330° and 210° www

condone ±30° or − 150° ignore extra values outside the range if full marks or SC1 awarded, subtract 1 for extra values in the range

[3] 5 (i) ruled line touching curve at x = 2 M1 intent to touch, but must not clearly cut

curve

their 2 1

2 1

y yx x

from their tangent

M1 may be on graph or in working; must use correct points from their line their tangent may be at another point

M0 for reciprocal,

answer in range 2.5 to 3.0 inclusive

A1

both M1s must be awarded

( value is approx 2.773)

[3] 5 (ii) 3.482202253... and 4.59479342... rot to 3 or

more sf

B1

2.78 to 2.7815 or 2.8 B1 mark the final answer 2.781477917.. [2]


9

Question Answer Marks Guidance 6 (i) 2S cao B1 [1] 6 (ii)

21ar

1S

r or 1

1S

r

M1

A1

if M0, SC1 for 2

11

r Sr

oe

[2] 7 h = 1.5

1.5 (2.3 2(2.9 4 4.6 4.2 3) 0)2

all y-values correct and correctly placed in formula 29.775 to 3 sf or better; isw

B1

M1

B1

A1

h = 1.5 basic shape of formula correct, omission of brackets may be recovered later condone omission of outer brackets and/or omission of 0 answer only does not score

allow if used with 6 separate trapezia at least 4 y-values in middle bracket, eg 1.5 (2.3 2(2.9 4 4.6 4.2) 3)2

M0 if any x values used or B1 + B3 if 6 separate trapezia calculated to give correct answer

[4] 8 (i) graph from (1, 1) to (1, 1) to (2, 2) to (3, 0) 2 B1 for three points correct or for all four

points correct but clearly not joined

points must be joined, but not always easy to see, so BOD if in doubt. Accept freehand drawing.

[2] 8 (ii) graph from (2, 3) to (2, 3) to (4, 6) to (6, 0) 2 B1 for three points correct or for all four

points correct but clearly not joined points must be joined, but not always easy to see, so BOD if in doubt. Accept freehand drawing.

[2]


10

Question Answer Marks Guidance 9 (i) 3x2 6x 22

M1 condone one incorrect term, but must be

three terms condone “y =”

their y = 0 soi

M1 at least one term correct in their y may be implied by use of eg quadratic formula, completing square, attempt to factorise

3.89 1.89

A1 A1 if A0A0, SC1 for 3 5 3

3 or 1± 5/√3 or

better, or both decimal answers given to a different accuracy or from truncation

3.886751346 and 1.886751346

[4] 9 (ii) x3 3x2 22x + 24 = 6x + 24 M1 may be implied by x3 3x2 28x [= 0] x3 3x2 28x [= 0]

M1 may be implied by x2 3x 28[= 0]

other point when x = 7 isw A1 dependent on award of both M marks ignore other values of x [3] 9 (iii)

F[x] = 34 23 22 24

4 3 2x x x x

M1*

allow for three terms correct; condone + c

alternative method M1 for 3 2(( 3 22 24) (6 24))x x x x dx

may be implied by 2nd M1

F[0] – F[4]

M1dep

allow 0 – F[−4], condone – F[−4], but do not allow F[−4] only

M1* for F[x] = 4 3 23 28

4 3 2x x x

condone one error in integration

area of triangle = 48

B1

M1dep for F[0] – F[4]

area required = 96 from fully correct working

A1

A0 for – 96, ignore units,

no marks for 96 unsupported

[4]


11

Question Answer Marks Guidance 10 (i) (A) AC2 = 2 212.8 7.5 oe

AC = 14.83543056..

M1

A1

allow correct application of cosine rule or from finding relevant angle and using trig rot to 3 or more sf , or 15

B2 for 14.8 or better unsupported

tan C = 12.8/7.5

or C = 90 − tan −1( 7.5/12.8 ) oe 59.6 to 59.64

M1

A1

or sinC = 12.8/their14.8 or cosC = 7.5/their14.8

or sin sin 9012.8 14.8

Ctheir

or 2 2 214.8 7.5 12.8cos

2 7.5 14.8theirC

their

14.8

sin(155 59.6) sin 35AD theirtheir

oe

25.69 to 25.8

M1

A1

allow B2 for 25.69 ≤ AD < 25.8 unsupported…..but B0 for 25.8 unsupported

M0A0 for 14.8/cos55 = 25.803...

[6]


12

Question Answer Marks Guidance 10 (i) (B) area of ABC = 48 soi

B1 may be implied by correct final answer in

range or by sight of ½ × 12.8 × 7.5 oe condone 48.0…

½×their 14.8…×their 25.7…×sin(their 59.6 – 10)

M1 may be implied by 144.8 to 146

192.8 to 194[m2]

A1

B3 for correct answer in range if unsupported

[3] 10 (ii)

angle HMG = 1.12

or MHG = 0.55 (31.5126°)

B1

or angle EMF or angle MEF

allow 1.02 to 1.021 or 58.487° to 58.5°

HM = 1.7176 to 1.7225

B1 may be implied by final answer

½ × 1.1 × their HM2

or 2

360theirHM

M1 1.63(0661924…) θ =63(.025357… )

check arithmetic if necessary their HM ≠ 0.9 or 1.8

area of triangle EMF = 0.652 to 0.662 B1 or MGH may be implied by final answer or in double this (1.304 to 1.324)

2.95 to 2.952 [m2] cao

A1

full marks may be awarded for final answer in correct range ie allow recovery of accuracy

[5] 11 (i) 65 ×(1 − 0.017)3 oe M1 may be longer method finding decrease year

by year etc

NB use of 3×0.017 leads to 61.685, which doesn’t score

61.7410... showing more than 3 sf

A1 answer 61.7 given

[2] 11 (ii) [d =] 65 × 0.983n oe

B1 eg 63.895 × 0.983n−1 or 61.7 × 0.983n−3

[1]


13

Question Answer Marks Guidance 11 (iii) 65 × 0.983n < 3 or

log10(65 × 0.983n) < logl03 oe log1065 + log100.983n < log103 www

M1*

M1dep

may be implied by eg log1065 + n log100.983 < log103

condone omission of base 10 throughout if M0M0, SC1 for log1065 + n log100.983 < log103 even if < is replaced by eg = or > with no prior incorrect log moves

[log1065 + n log100.983 < log103] n log100.983 < log103− log1065 and

or [log100.983n < log103− log1065]

NB watch for correct inequality sign at each step

completion to 10 10

10

log 3 log 65log 0.983

n AG www

A1

inequality signs must be correct throughout

reason for change of inequality sign not required

n = 180 cao

B1 B0 for n > 180

n > 179.38...

[4] 11 (iv) 63.895 = 65 ×10-k soi

B1 or 65 × 0.983 = 65 × 10-k accept 63.895 rot to 3 or 4 sf;

B1 may be awarded for substitution of t = 1 after manipulation

log10(their 63.895) = log1065 k or k = log10(their 0.983)

M1

their 63.895 must be from attempt to reduce 65 by 1.7% at least once

M1A1A1 may be awarded if other value of t with correct d is used

[ k = ] 7.4 × 10-3 to 7.45 × 10-3

A1 [ k = ] –log100.983 isw

[ d = ] 42.1... to 42.123 [°C] isw

A1 NB B1M1A0A1 is possible; unsupported answers for k and/or d do not score

[4]


3

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


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.

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.


6

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.


7

Question Answer Marks Guidance

1 5

21

kx M1 k is any non-zero constant

7

22x cao A1

+ c A1

[3]

2 (i) 1 1 1 1 121

1 2 2 2 3 2 4 2 5 2 oe

soi

M1

may be implied by correct answer NB 7 + 5.25 + 4.2 + 3.5 + 3

M0 if extra terms or terms missing

22.95 or

459

20 or

19

2022

A1

[2]

2 (ii) a + 45 cao B1 mark the final answer

must be explicitly stated

1045

2a a their

M1 or 10

2 (10 1) 52

a condone wrongly attributed

answers

5(2a + 45) or 10a + 225 cao isw A1 ignore further work attempting to find a B2 if correct answer derived from

adding terms separately

[3]

3 2.4 3.6

2.2 2oe

M1 M1 may be embedded eg in

equation of straight line

− 6 cao A1 B2 if unsupported

ignore subsequent work irrelevant

to finding the gradient

[2]


8


4 (i) (6, 1.5) oe B2 B1 for each value; allow x = 6, y = −1.5 SC0 for (6, − 3)

[2]

4 (ii) (2, 3) B2 B1 for each value; allow x = 2, y = −3 SC0 for (6, − 3)

[2]

5 5.92 + 8.5

2 2 × 5.9 × 8.5 × cos 72

M1

107 – 31 or better M1 76.(.…) or 204.(…) (radians) or 64.(…..) (grad)

NB 6.76cos72 or 2.08(8954882..)

scores M1M0

8.7(2…) A1 if M0M0, B3 for 8.72 or better if

unsupported or 8.7(2…) if obtained

from other valid method

[3]

6

½ × 12.42 ×2.1 (= 161.448)

M1* or 2120.32

π 12.4360

angle in degrees to 3 sf or better

½ × 12.42 × sin2.1 (= 66.3 to 66.4)

or ½ × 21.5(121..) × 6.16(9…)

M1* angle in degrees to 3 sf or better may be implied by

2.81(7168325…) (degrees) or

2.53(5559362) (grad)

their 161.448 – their 66.36 M1dep*

95 to 95.1 A1

if unsupported, B4 for 95.08(446)

r.o.t. to 4 sf or better

[4]


9


7 ar = 24 (i) B1* allow ar2 – 1

= 24

150

1

a

r (ii)

B1*

correct substitution to eliminate one

unknown

M1dep* eg subst. of a = 150(1 – r) or

150

150

ar

in (i)

alternatively, subst. of 24

ar

or 24

ra

in (ii)

if M0, B1 for both values of r and

B1 for both values of a, or B1 for

each pair of correct values

NB 150r2 – 150r + 24 [= 0]

a2 – 150a +3600 [= 0]

r = 0.8 or 0.2 A1 or A1 for each correct pair of values A0 if wrongly attributed

a = 30 or a = 120 A1

ignore incorrect pairing if correct values

already correctly attributed

A0 if wrongly attributed

[5]

8 2sin

sin

cos

or 2cos sin

sin

M1 correct substitution for numerator allow maximum of M1M1 if

M1 correct substitution for denominator 2sin oe substituted

cos θ cao

A1 A0 if follows wrong working

or B3 www or if unsupported

mark the final answer but ignore

attempts to solve for θ

allow recovery from omission of θ

[3]


10


9 71.5(6505118..) soi

M1 or 1.24(9045772..) (rad)

or 79.5(1672353..) (grad)

35.7 to 36 A1 if A0, SC1 for all four answers in radians or

grad r.o.t to 3 or more sf

0.62452286, 2.195319213, 3.76611554,

5.336911867 (rad), but 0 if extra values in

range

39.75836177.., 139.75…, 239.75…

339.75…(grad)

125.78..., 215.78..., 305.78... to 3 or more sf A1 if M1A0A0, SC1 for 251.565…,

431.565…, 611.565…

for second A1, ignore extra values

outside range, A0 if extra values in

range

[3]

10 (x + 1) log 3 = 2x log 5 oe

M1 or x + 1 = 2x log 35

or (x + 1)log53 = 2x

allow recovery from omission of

brackets in later working

log 3 = x(2 log 5 log 3) oe

A1

x(1− 2log35) = − 1 oe

or x(2− log53) = log53 oe

NB 0.477121254= 0.920818754x

− 1.929947041x = − 1

1.317393806x = 0.682606194..

log3

2log5 log3oe

A1 3

1

2log 5 1oe

or 5

5

log 3

2 log 3oe

0.518 cao A1 answer only does not score

[4]


11


11 (i) y′ = 1 + 8x-3

M2 M1 for just 8x-3

or 1 − 8x-3

y′′ = 24x-4

oe A1 but not just

4

24

xas AG

[3]

11 (ii) their y′ = 0 soi M1

x = 2 A1 A0 if more than one x-value x = 2 must have been correctly

obtained for all marks after first M1

y = 3 A1 A0 if more than one y-value

substitution of x = − 2:

4

24

( 2)

M1 or considering signs of gradient either side

of −2 with negative x-values

condone any bracket error

< 0 or = − 1.5 oe correctly obtained isw A1 signs for gradients identified to verify

maximum

must follow from M1 A1 A0 M1 or

better

[5]

11 (iii) y = 5 soi B1

substitution of x = 1 in their y′ M1 may be implied by − 7

grad normal = 1/their − 7 M1* may be implied by eg

1/7

y – their( −5) = their 1/7(x − − 1) M1dep* or their (−5) = their

1/7 × (−1) + c

x + 7y + 34 = 0 oe

A1 allow eg

1 340

7 7y x

must see = 0

do not allow eg 34

7 7

xy

[5]


12


12 (i) h = 3 soi B1 allow if used with 6 separate

trapezia

39 9.1 2(10.7 11.7 11.9 11.0)

2

M1

basic shape of formula correct with their 3;

omission of brackets may be recovered

later;

M0 if any x-values used

(NB y0 = 9 and x3 = 9, so check position)

with 3, 4 or 5 y-values in middle

bracket, eg

39 2(10.7 11.7 11.9) 11.0

2

all y-values correctly placed in formula B1 condone omission of outer brackets

163.05 or 163.1 or 163 isw A1 answer only does not score or B1 + B3 if 5 separate trapezia

calculated to give correct answer

NB 29.55 + 33.6 + 35.4 + 34.35 +

30.15

[4]

12 (ii) (A) −0.001×123 – 0.025×12

2 + 0.6×12 + 9 soi M1 may be implied by 10.872, 10.87 or 10.9 NB allow misread if minus sign

omitted in first term if consistent in

(A) and (B). Lose A1 in this part

only

±0.128[m] or ±12.8cm or ±128mm isw A1 B2 if unsupported appropriate units must be stated if

answer not given in metres

[2]


13


12 (ii) (B) 4 3 20.001 0.025 0.6F[ ] 9

4 3 2

x x xx x

M2 M1 if three terms correct ; ignore + c

F(15) [ F(0) ] soi M1

dependent on at least two terms correct in

F[x]

condone F(15) + 0

161.7 to 162 A1 A0 if a numerical value is assigned to c answer only does not score

NB allow misread if minus sign

omitted in first term if consistent in

(A) and (B). 187.03…

[4]

13 (i) log10h = log10a + bt www B1 condone omission of base

m = b, c = log10a B1

must be clearly stated: linking

equations is insufficient

[2]

13 (ii) 0.15, 0[.00], 0.23, 0.36, 0.56, 0.67, 0.78,

0.91, 1.08, 1.2[0]

B2 B1 if 1 error

plots correct (tolerance half square) B1 condone 1 error – see overlay no ft available for plots

single ruled line of best fit for values of x

from 5 to 50 inclusive

B1 line must not go outside overlay between

x = 5 and x = 50

[4]


14


13 (iii) – 0.3 ≤ y-intercept ≤ − 0.22 B1 may be implied by 0.5 ≤ a ≤ 0.603

valid method to find gradient of line M1

may be embedded in equation;

may be implied by eg m between 0.025 and

0.035

condone values from table;

condone slips eg in reading from

graph

theirtheir 10 bth a

or their log their10 a bth

M1

if B1M1M0, then SC1 for

logh = loga + theirbt isw

0.028 ≤ b ≤ 0.032 and

0.5 ≤ a ≤ 0.603 or – 0.3 ≤ loga ≤ − 0.22

A1 if both values in the acceptable

range for A1

[4]

13 (iv) a1060b

a1050b

their values for a and b

M1 or log b 60 log b 5010 10a a

or their values for loga and b

condone 15.9 as second term

may follow starting with

logh = loga + theirbt

8.0 to 26.1 inclusive A1

NB A0 for estimate without clear

valid method using model;

both marks available even if a or b

or both are outside range in (iii)

[2]

13 (v) comment on the continuing reduction in

thickness and its consequences

B1 eg in long term, it predicts that reduction in

thickness will continue to increase, even

when the glacier has completely melted

[1]

GCE

Mathematics (MEI)





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


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


3


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.


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.

4


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.





5




6


Question Answer Marks Guidance 1 (i) 1

13kx−

oe

234x−

isw cao

M1

A1

k is any non-zero constant ignore + c

allow any equivalent exact simplified form

[2]

1 (ii) 3 1kx− + oe −3x−2 isw + c

M1

A1

A1

k is any non-zero constant

allow any equivalent exact simplified form

[3]

2 2 3 42 2 2

10 10 10, ,2 their 2.5 their 1.6

u u u= = = isw

2 3 42 u u u+ + + soi

10.00625 or 180

1601 or 160

10 cao isw

M1*

M1dep*

A1

must be the sum of 4 terms only B3 if unsupported

NB 2.5, 1.6, 3.90625

or 10 8 125, ,4 5 32

may be implied by eg sight of 3.9 and answer of 10.0 NB 2.5, 1.1, 0.625 scores M0M0

[3]

7


3 a + (10 – 1)d = 11.1 and a + (50 − 1)d = 7.1

M1 may be implied by 40d = ±4 or embedded in attempt to solve

condone one slip in coefficient of d

d = −0.1

A1

if unsupported, B2 for one of these and B3 for both

a = 12

A1

1 50( 7.1)2

their a× + with a > 11.1

M1 or 50 (2 (50 1) )

2a d+ − with a > 11.1 and

d < 0

477.5 or 477½ or 955

2cao

A1

if M0, B2 for any form of correct answer www

[5]

4 27 = ½ r2 × 1.5 oe

M1 or 285.943669...27

360rπ= ×

angle in degrees rounded to 2 sf or more

r = 6 soi

A1 may be embedded in formula for arc length may be implied by later work eg 9 or 21

their r × 1.5

M1 or their 85.943639 2 their

360rπ× ×

if r is incorrect, we must see their r × 1.5 [ + 2r] for M1 if r is correct, M1 may be implied by 9 or 21

21 [cm] cao A1 allow full marks for recovery from working with rounded value of θ in degree form

B4 for 21 unsupported www

[4]

8


5 3x2 − 6 seen

B1

their y′ = 0 or y′ > 0 or y′ ≥ 0 M1 must be quadratic with at least one of only two terms correct

2 and 2 identified−

A1

may be implied by use with inequalities or by ±1.41[4213562] to 3 sf or more

2 x = implies A1

2x < − or 2x ≤ − isw

2>x or 2x ≥

A1

A1

if A1A0A0, allow SC1 for fully correct answer in decimal form to 3 sf or more or A2 for 2 or 2x x> ≥

NB just

2 2 or 2 2x x− > > < < − or 2x > ± implies the first A1 then

A0A0 [5]

6 (i) both curves with positive gradients in 1st and 2nd quadrants; ignore labels for this mark

M1 do not award if clearly not exponential shape; condone touching negative x-axis but not crossing it

consider each curve independently; ignore scales and points apart from (0, 1)

both through (0, 1)

A1 allow if indicated in table of values or commentary if not marked on graph

y = 32x above y = 3x in first quadrant and below it in second

A1 must be clearly labelled, A0 if wrongly attributed or if coincide for negative x from (0, 1)

if M0 allow SC1 for one graph fully correct

[3]

6 (ii) x = 3 3x = 27

B1

B1

[2]

B0 if wrongly attributed B0 if wrongly attributed

allow 33 = 27 with x = 3 stated

9


7 1 − cos2 x = 3cos x – 2 oe

M1*

cos2 x + 3cos x − 3 [= 0]

M1*dep or − cos2

x − 3cos x + 3 = 0

condone one sign error or constant term of – 1 (in LH version) or + 1 (in RH version)

3 21cos their or 2

cos their 0.79 to so0 i.7913

x

x

− +=

=

M1

dependent on award of previous method mark, must be correct for their quadratic

ignore other values (eg −3.79...); condone recovery from x = 0.791287847…but M0 if no recovery

[x =] 0.6578 to 0.66 isw cao A1 A0 for eg 0.66π if 0.66 not seen separately NB x = 0.65788395…

[x =] 5.625 to 5.63 isw cao

A1 if A1A1 extra values in range incur a penalty of 1; ignore extra values outside range if A0A0 allow SC1 for 37.69 to 37.7° and 322 to 322.31° or for (0.209 to 0.21)π and (1.79 to 1.791)π

NB x = 5.625301357… no SC mark available if extra values in range

[5]

8 m = 3 seen log y = mlog x + 2 or log y = mlog x + log100

B1

M1

or log y − 8 = m(log x – 2)

condone lack of base; “c = 2” is insufficient

log y = log x3 + 2 or log y = log x3 + log100 or better y = 100x3 or

33log 2 log 210 or 10x xy y+ += = www isw

M1

A1

or 10log y = 103log x + 2 or 103log x + log100 or better

33log log100 log log10010 or 10x xy y+ += =

condone lack of base, but not bases other than 10 unless fully recovered

[4]

10


9 (i) 2 2 220 13 8[cos ]2 13 20

A + −=

× ×

M1*

or 2 2 28 20 13 2 13 20 cos A= + − × × ×

505[cos ]520

A = oe soi

A = 13.79 to 13.8° or 14°

A1

A1

or 0.971 to 0.9712 or 0.24077 to 0.241 or 0.24 (radians); allow B3 if given to 3sf or more unsupported

or 15.32 (grad)

[Area = ] ½ × 20 × 13 × sin their A

M1dep*

or M1 for eg ½ × 20 × 8 × sin 22.8, as long as angle calculated correctly from their A (other angles are 22.79824…° and 143.40645..° or 36.59355..°)

or

41 41 41 41( 8)( 13)( 20)2 2 2 2

− − −

NB 13sin A = 3.099899192 if ½×b×h used

30.99 to 31.01 isw

or 5 6154

oe isw

A1

allow B2 for unsupported answer within range

[5]

11


9 (ii) h = 4 soi their 4

2× (0 + 0 + 2(1.45+1.56+1.27+1.04))

or their 4

2× (0 + 0 + 2(±0.85± 0.76± 0.55± 0.30))

either 21.28 or ± 9.84 their 21.28 + their 9.84 31.12

B1

M1*

B1

A1

M1dep*

A1

shape of formula correct with 2, 3 or 4 y-values in inner bracket with their h; allow recovery from bracket errors M0 if any non-zero x-values used or if y-values used twice all y-values correctly placed with their h, condone omission of zeros and/or omission of outer brackets ignore subsequent rounding, but A0 if answer spoiled by eg multiplication by 20

eg

( ){ }their 42 1.45 1.04 2 1.56 1.27× + + + ;

signs must be consistent in 2nd alternative or B1 + B3* if area of 2 triangles and 3 trapezia calculated to give correct answer www The final M1dep* A1 may then be earned. NB 2.9 + 6.02 + 5.66 + 4.62 + 2.08 or ± 1.7 ± 3.22 ± 2.62 ± 1.7 ± 0.60 with consistent signs throughout

12


9 (ii) alternatively h = 4 soi attempt to find all y-values 2.3, 2.32, 1.82, 1.34 their 4

2× (0 + 0 + 2(2.3+2.32+1.82+1.34))

31.12

B1

M1

A1

M1

B1FT

A1

yupper – ylower all y-values correct shape of formula correct with 2, 3 or 4 of their y-values in inner bracket with their h; allow recovery from bracket errors M0 if any non-zero x-values used or if y-values used twice all their y-values correctly placed, condone omission of zeros and/or omission of outer brackets ignore subsequent rounding, but A0 if answer spoiled by eg multiplication by 20

M0 if values are added to obtain 0.60, 0.80 etc eg ½ × 4 × {2.3 + 1.34 + 2(2.32+1.82)} or B1M1A1 + B3 if area of 2 triangles and 3 trapezia calculated to give correct answer www NB 4.6 + 9.24 + 8.28 + 6.32 + 2.68

[6]

13


10 (i) ddyx

=

4 × 2 + 3 or 11 isw

M1*

9 = their (4 × 2 + 3 ) × 2 + c

M1dep* or y − 9 = their (4 × 2 + 3 ) × (x − 2)

y = 11x − 13 or y = 11x + c and c = −13 stated isw

A1 or y − 9 = 11(x − 2) isw

[3]

10 (ii) 24 32x x+

M1*

[y =] 2x2 + 3x + c

A1 must see “2” and “ + c”; may be earned later eg after attempt to find c

9 = 2×22 + 3×2 + c M1dep* must include constant, which may be implied by answer

y = 2x2 + 3x − 5 cao

A1

allow first 4 marks for y = 2x2 + 3x + c and c = −5 stated

(1, 0) and (−2.5, 0) oe cao B1 or for x = 1, y = 0 and x = −2.5, y = 0 B0 for just stating x = 1 and x = −2.5

34498

x

y

= −

= −

B1

B1

−6.125 or − 6⅛

[7]

14


10 (iii) substitution to obtain [ y =] f(2x) in polynomial form

M1

f(x) must be the quadratic in x with linear and constant term obtained in part (ii), may be in factorised form

or their x = 1 → their 0.5 and their x = −2.5 → their x = −1.25

y = (2x − 1)(4x + 5) or y = 8x2 + 6x − 5

or 23 492 2

4 8y x −

= +

A1FT

must be simplified to one of these forms, FT their quadratic in x with linear and constant term obtained in part (ii)

hence y = (2x − 1)(4x + 5) FT their x-intercepts from their quadratic in x with linear and constant term obtained in part (ii)

3 49,8 8

− −

oe

B1

or FT their (both non-zero) co-ordinates for minimum point or their quadratic in x with linear and constant term obtained in part (ii)

[3]

11 (i) 3×37 oe 6561

M1

A1

condone 1×37 or B2 if unsupported

do not award if only seen in sum of terms of GP if 0, SC1 for 2187 unsupported

[2]

11 (ii) valid attempt to sum a GP with r = 3 and n = 15

153(3 1)3 1

−−

oe

21 523 359

M1

M1

A1

eg 3 + 32 + …….+ 315 or B2 if M1M0 or B3 if unsupported

must see at least first two terms and last term NB 7 174 453 implies M1 from 1 + 3 + …+ 314

[3]

15


11 (iii) 3(3 1) 10000003 1

n −>

−oe

M1*

M0 for working backwards M0 if = or < used

eg 3n + 1 > 2000003 or 20000003 13

n > +

www correctly taking logs of both sides eg (n + 1) log 3 > log 2000003 or nlog3 > log 2000003 – log 3

eg log 20000031log 3

n + > and completion to

log 2000003 1log 3

n > −

M1dep*

A1

eg log 3n + 1 > log 2000003 www or log 3n + log 3 > log 2000003 www; may be implied by next stage of working without any wrong working

at least one previous progressive interim step needed with no wrong working; M0dep* for log(3n – 1) > … do not allow recovery from bracket errors at any stage

n = 13 seen

B1

B0 for n ≥ 13 or n > 13

[4]

11

(iv)

valid attempt to sum a GP with r = 2 and n = 15 their 21 523 359 – their 65 534

M1*

M1dep*

if correct eg 2 + 22 + …….+ 215 = 65 534 with their 65 534 < their 21 523 359

NB 32767 implies M1 from 1 + 2 +…+ 214

21 457 825 isw A1 allow B3 for 21 457 825 unsupported

[3]

16

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mailto:[email protected]

http://www.ocr.org.uk/


GCE

Mathematics (MEI)

Unit 4752: Concepts for Advanced Mathematics Advanced Subsidiary GCE


2

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 2016


3

Annotations and abbreviations

Annotation in scoris Meaning

and

Benefit of doubt

Follow through

Ignore subsequent working

, Method mark awarded 0, 1

, Accuracy mark awarded 0, 1

, Independent mark awarded 0, 1

Special case

Omission sign

Misread

Highlighting Other abbreviations in mark scheme

Meaning



4

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.


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. 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.


5

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.







6



7


1 (i) 11

2kx

or

1

2kx

seen

1

23

3 or xx

isw

M1

A1

k > 0

A0 for eg

1

23 + x c

B2 for correct answer unsupported

[2]

1 (ii) 2 1kx or 1kx oe seen

1 12 1212 or or x

x x

isw

+ c

M1

A1

A1

for any non-zero k

seen at least once following integration

SC0 for 12 6

or 2x x

A0 for 112 12

or 1

x

x

do not allow MR for integration of

12x2

[3]

2 (i)

(i) [5], 10, 5, [10]

[10 + 5 + 10 =] 25

(ii) 0

M1

A1

B1

[3]

ignore extra terms

not from wrong working

condone wrongly attributed terms

B2 for 25 unsupported

3 1.5 + (4 ‒ 1)d = 12 or better

d = 3.5

M1

A1

or 1.5 × r(4 ‒ 1)

= 12 or better

r = 2

if first M0 B0 allow

B3 for d = 3.5 and r = 2;

B2 for one of these;

may be embedded in calculation of

difference

r = 2 B1

d = 3.5

1.5 × their 29 (1.5 + 9 × their 3.5) oe

M1 M0 for use of their S10 in either term NB 768 ‒ 33

difference = 735 A1 allow ‒735

[5]


8


4 5.62 + 7.2

2 2 × 5.6 × 7.2 × cos 68 seen

53 or 53.0

M1

M1

may be implied by 53 or BC in range

may be implied by BC in range

NB 52.9917243; (allow 47.7 to

47.71 from calculator in radian

mode; may be implied by 6.90 to

6.91)

[BC =] 7.3 or 7.27 to 7.28

7.2 sin 68sin

their C

BC

A1

M1

NB 7.27954…

2 2 2their 5.6 7.2

or [cos ]2 5.6 their

BCC

BC

66 or awrt 66.5

A1 allow 1.2 or awrt 1.16 (radians);

A0 for eg 1.2 degrees

NB sin C = 0.917053…

cos C = 0.398766…

Alternatively

eg if the perpendicular from B to AC, BX, is

used

[5]

eg if perpendicular from C to AB,

CY, is used, mark as follows

7.2 × cos 68 seen

2.7 or 2.697 to 2.70

XC = 5.6 ‒ their AX

BX 7.2 sin 68tan

XC their XCC

66 or awrt 66.5

M1*

A1

M1dep*

M1

A1

[5]

if unsupported, B2 for 2.70 or better

NB 2.902832527

allow 1.2 or awrt 1.19 (radians);

A0 for eg 1.2 degrees

5.6 × cos 68 seen

2.1 or 2.097 to 2.10

BY = 7.2 ‒ their AY

CY 5.6 sin 68tan

BY their BYB

C [= 90 – B] = 66 or awrt 66.5


9


5 (i)

sin kx

y = sin 2x

M1

A1

k > 0 and k ≠ 1

must see “y =” at some stage for A1

condone use of other variable

condone f(x) = sin 2x

[2]

5 (ii)

sketch of sine curve with period 360 and

amplitude 1

sine curve centred on y = ‒ 3 and starting at

(0, 3)

B1

B1

[2]

for 0 ≤ x ≤ 450; ignore curve outside this

range;

do not allow sketch of y = cos x or

y = ‒ sin x for either mark

amplitude, period and centring on

y = ‒ 3 must be clear from correct

numerical scale, numerical labelling

or comment; strokes on axes

insufficient to imply scale:

mark intent

allow full marks if y = sin x and

y = sin x ‒ 3 seen on same diagram

6 (i)

2 2 21 1 1 1 or sin or sin cos seen

2 2 2 2r a a

2 21 1sin

2 2 r a isw oe

M1

A1

do not allow use of variable other than θ

allow eg

2 21 180 or sin seen

2 2r a

oe

[2]

6 (ii) 2 2 21 1 1sin 0.8 12 0.8 sin 0.8

2 2 2a a oe

B1 or eg 2 21 1sin0.8 12 0.8 [ 28.8]

2 4a or

equivalent in degrees NB θ = 45.8366236… ̊

NB 2 57.6

0.717356a = 80.29485

NB θ = 45.83662361… ̊

[a =] 8.96 cao; mark the final answer B1

if unsupported, allow B2 for 8.96 or allow

B1 for 9.0 or 8.96074…to 4 sf or more

NB 1

sin0.8 0.35867..2

[2]


10


7 (i)

substitution of

2 2sintan or 1 sin cos or cos

cos

xx x x x

x

in given LHS

both substitutions seen and completion to

sin x as final answer

M1

A1

[2]

if no substitution, statements must follow a

logical order and the argument must be

clear; if one substitution made correctly,

condone error in other part of LHS

NB AG; answer must be stated

allow consistent use of other variable eg θ

for both marks

condone omission of variable

throughout for M1 only , but allow

recovery from omission of variable

at end

M0 if first move is to square one or

both sides

Simply stating eg sin

tancos

xx

x is

insufficient

Alternatively SC2 for complete

argument eg

sintan

cos

xx

x

[tan x × cos x = sin x ] 2 2sin cos 1x x

2cos 1 sinx x

2

sintan

1 sin

xx

x

2tan 1 sin sinx x x oe

7 (ii) 0, 180, 360

14 or 14.47 to 14.5

166 or awrt 165.5

B1

B1

B1

all 3 required

radians: mark as scheme but deduct one

from total

0, π, 2π;

0.25 or 0.253 or awrt 0.2527;

2.89 or 2.889 or awrt 2.8889

NB sin y = 0 or ¼

ignore extra values outside range

if B3, deduct 1 mark for extra values

within range

[3]


11


8 (i) log 1a = 0 soi or 3mlogaa or logaa‒3m

seen

3m cao

M1

A1

do not condone 3mloga do not allow MR for (logaam)

3

[2]

8 (ii)

3 3(2 1)log 3 log 1000x or 32 1 log 1000x

oe

3log 1000 1[ ]

2x

oe

M1

M1

Or 10 10(2 1)log 3 log 1000x [= 3]

or 10

31

log 3[ ]

2x

oe

condone omission of brackets;

allow omission of base 10 or

consistent use of other base

allow one sign error and / or

omission of brackets

allow recovery from bracket error

for A1

2.64 cao; mark the final answer A1

not from wrong working 0 if unsupported or for answer

obtained by trial and error on

32x + 1

= 1000

[3]

9 (i)

2

h× (0 + 0 + 2[4 + 4.9 + 5 + 4.9 + 4]) oe

all non-zero y-values correctly placed

h = 1 used in formula or consistently with two

triangles and four trapezia

M1

M1

B1

correct formula used with 4, 5 or 6 strips and

numerical value for h; condone omission of

zeros or omission of outer brackets for both

M marks

M0M0 if 1, 2, 3 or 6 used as y-values (these

are x-values)

if M0M0 allow B1 for h = 1 and B2 for 22.8

from area of 4 trapezia and 2 triangles and

B1 for 1140

allow eg

½ × 1 × (4 + 4 + 2[4.9 + 5 + 4.9])

½ × 1 × (4 + 0 + 2[4 +4.9 + 5 + 4.9])

(NB may be implied by 18.8 & 20.8

respectively)

area = 22.8 and volume = 1140 isw cao

A1

ignore units

if M0M0B0 allow SC4 for 22.8 and

1140 obtained correctly by other

method

[4]


12


9 (ii) A substitution of x = 1.2 or 4.8 to find y

M1 allow substitution of 1.2 ≤ x ≤ 1.234 or

4.766 ≤ x ≤ 4.8

or M1 for y = 4.4, x = 1.234 [or

4.766] and

y = 4.35 or 4.352 and correct comparison with

4.4 isw A1

[2]

A1 for comparison of 1.234 with 1.2

or 4.766 with 4.8 [so gap less than

3.6]

9 (ii) B 5

2 3 4

5

5 108 54 12F[ ]

81 2 3 4

xx x x x

oe

eg 2 3 4 510 10 5 1

3 9 27 81x x x x

M2

M1 for 3 correct terms; ignore +c

allow coefficients 3.333333…, 1.11111…,

0.185185.., 0.01234567…r.o.t to 2 sf or

better

or decimal equivalents in numerator:

6.6666…, 3.333333…, 0.74074…,

0.061728… r.o.t to 2 sf or better

condone omission of 5

81;

M0 if 5

81x seen outside bracket

but next M1 is still available;

ignore subsequent attempt to

evaluate c for first M2

F[6] F[0] or 2 × (F[3] ‒ F[0])

M1 dependent on at least two terms correctly

integrated in bracket; condone omission of

F(0)

M0 for non-zero lower limit

24

A1 24 unsupported does not score

1200 B1

[5]

ignore units

10 (i) 2 25.1 10.2 5 10 oe

5.1 5

8.1

M1

A1

[2]

condone omission of brackets

0 for 8.1 unsupported


13


10 (ii) 2

5 2(5 ) their 15 oe

h h

h

25 + 10h + h2 ‒ 10 ‒ 2h oe seen

numerator is 8h + h2

8 + h isw

M1

M1

A1

A1

[4]

condone omission of brackets

allow one sign error

10 (iii)

h → 0

M1

may be embedded; allow eg “tends to 0”

M0 for differentiation of x2 ‒ 2x

M0 for following from part (i)

M0 for h = 0

their 8 A1

[2]

FT their k + h from part (ii)

10 (iv)

y = 8x – 25 isw

B1

or y − 15 = 8 (x − 5) isw

or y = 8x + c and c = ‒ 25 stated isw

non-zero numerical value for x-intercept on their

straight line found

M1

[ x = ] 3.125 oe

A1

may be embedded in calculation for area

1

2× their non-zero y-intercept × their

25

8

M1

condone arithmetic slips in finding values of

intercepts

or integration and evaluation of their 25

8

0

(8 25)dx x ;

lower limit must be 0

625

16 or 1

1639 or 39.0625 isw

A1

[5]

accept rounded to 1 dp or better for A1; but

A0 if final answer negative


14


11 (i) 10 10log log y a bt www B1 B0 for just 10 10 10log log log 10y a bt allow omission of base throughout

question

gradient is b, intercept is 10log a cao B2

[3]

B1 for one correct; award independently of

their equation;

must be stated – linking by arrows etc is

insufficient;

condone m = b and c = log a

ignore t-intercept is 10log a

b

B0 for gradient is bt

11 (ii)

1.58, 1.8[0], 1.98, 2.37, 2.68 B1 allow values which round to these numbers

to 2 dp;

all values must be correct

all values correct and all plotted accurately B1 within tolerance on overlay;

ruled line of best fit for at least 1 ≤ t ≤ 10

B1

within tolerance on overlay: must not cut red

or green line;

line between (1, 0.6) and (1, 1.05) at lower

limit and between (10, 2.3) and (10, 2.75) at

upper limit;

use ruler tool to check if line is ruled

where necessary;

tolerance: one small square

horizontally at each end;

not dependent on correct plots

evaluation of 2 1

2 1

log logy y

t t

or substitution of 1 1, logt y and 2 2, logt y in

log y = bt + log a to obtain a numerical value for

the gradient

0.14 ≤ b ≤ 0.24

M1

A1

1 1, logt y and 2 2, logt y are points on

their line

gradient must be identified as b for A1

condone use of values from table

2.5 ≤ a ≤ 6.3

y = their a × 10theirb×t

or y = 10theirbt + their loga

or 10their log a

× 10theirb×t

oe

a and b or log a and b both in acceptable range

B1

M1

A1

must be identified as a ; not from wrong

working

0.4 ≤ log a ≤ 0.8

if M0A0B0M0 allow SC3 for

substitution directly into given

formula to obtain y = a10bt with a

and b in acceptable range

[8]

11 (iii) 260 or 261 B1

[1]

B0 for non-integer answer

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Documents

4752 Mark Scheme June 2005 Mark Scheme 4752 June 2005 packs/MS_MEI_C2.pdf · 4752 Mark Scheme June 2005 Section A 1 1+ 1 3 2 2 x 1+3 B2 for 1 kx2, or M1 for 3 x2 seen before differentiation