Differentiation and Integration Mark Scheme - Core 2
1.(a)B11
(b)M1
Allow all M marks for equivalent work with (n odd)
a =A12
Accept unsimplified
(c)M1
b =A12
(d)Substitution of x = 4 into M1
A1F2
ft wrong coeff of
[7]
2.(a)B11
Accept k = 2.5
(b)y = M1
Good attempt at integration;
condone absence of ‘+c’.
1 = m1
Valid start to find c
y = oeA1ft3
ft on their fractional value of k from (a)
must have y = … somewhere
eg k = 1.5 y = 2.8x2.5 – 1.8
[4]
3.(a)y = B11
(b)p = 5B1F
ft wrong coeff of giving integer p
Use of M1
OE; PI
Conclusion fully justifiedA1F3
ft one error
[4]
4.(a)(i)B1
PI by sight of
M1
One term correct
OEA13
(ii)When x = 2, A11
CSO AG (be convinced)
(b)When x = 2, y = 3 B1
For y = 3
gradient of normal = − 2M1
= 1 used
Equation normal M1
OE
Award at 1st correct formA14
[8]
5.= 2x + M1
Attempt at derivative
x = 1 = 4A1 cso
Tangent y = 4x – 5B1 ft
x3 – 4x2 + 5x – 2 = 0M1 A1 cao
(x – 1)2(x – 2) = 0M1
(use of factor theorem of division etc)
and no other values
x = 2A1 cso
y = 3A1 cso
[5]
6.(a)M1
k = 8A1
A1F3
Condone use of decimals once in a or b ft wrong k
(b)M1
k = 6A1
At x = 2, A1F3
Use of decimal points penalised here if wrong k
[6]
7.(a)YP = 4B11
(b)B2,1,02
(B1ifonlyoneerrorintheexpansion)
ForB2thelastlineofthecandidate‘s
solutionmustbecorrect
y = 1 + 4x–1 + 4x–2
(c)M1
Indexreducedby1afterdifferentiatingx
toanegativepower
Atleast1terminxcorrectftonexpnA1ft
CSOFullcorrectsolution.ACFA13
(d)When x = 2, M1
Attempttofindy′(2).
Gradient = 1 1 = 2A12
AG(beconvinced-noerrorsseen)
(e) 2 m ′ = 1M1
m1m2=1OEstatedorused.PI
y 4 =m(x − 2)M1
C‘syPfrompart(a)ifnotrecovered;
mmustbenumerical.
A1ft
Ftoncandidate‘syPfrompart(a)ifnot recovered.
x 2y + 6 = 0A14
CAOMustbethisor0=x2y+6
[12]
8.(a)DifferentiationM1
at least one term correct
y = 2x + 2x–3A1
Accept unsimplified
... 4 when x = 2A1F3
ft numerical or sign error
(b)IntegrationM1
at least one term correct
A12
Accept unsimplified
[5]
9.h = 1B1
PI
f(x) =
Area ≈ h/2{…}
{…}= f(0) + f(3) + 2[f(1) + f(2)]M1
OEsummingofareasofthe‘trapezia’..
{…}= 1 + √8 + 2(√2 + 2)A1
OE
(Area ≈) 5.3284… = 5.328 (to 3dp)A14
CAOMustbe5.328
[4]
10.(a)(i)When x = 4, B11
AG Be convinced
(ii)B11
Accept k = –2
(iii) – 0M1
A power decreased by 1
A1;
A1ft3
candidate’s negative integer k
[−1 for >2 term(s)]
(iv)When x = 4, M1
Attempt to find y′′(4) reaching as far as two
simplified terms
Minimum since y′′(4) > 0E1ft2
candidate’s sign of y′′(4)
[Alternative: Finds the sign of y′(x) either side of the point
where x = 4, need evidence rather than just a statement: (M1)
Correct ft conclusion with valid reason E1ft]
[In both, condone absent statement y′(4)=0]
(b)(i)At P(1,8), = 12B11
AG Be convinced
(ii)Gradient of normal = M1
Use of or stating
m × m′ = 1
Equation of normal is M1
Can be awarded even if m = 12
12y – 96 = – x + 1
A13
Any correct form of the equation
(c)(i)dx =
…….. = M1
One power correct.A2,1,0ft3
A1 if 2 of 3 terms correct
candidate’s negative integer k
Condone absence of “+ c”
(ii) (*)B1ft
y = candidate’s answer to (c)(i) with tidied coefficients
and with ‘+c’.
(‘y =’PI by next line)
When x = 1, y = 8 8 = 2 – 16 – 7 + cM1
Substitute. (1,8) in attempt to find constant of integration
A13
Accept c = 29 after (*), including y =, stated
[17]
11.(a)y(0) = 6, y(1) = –1B1B1
Sign change, so root betweenE13
(b)(i)y = 2 …M1A1
M1for
… = 9B1
y = 3M1A15
M1forasderivof1stterm
(ii)At SP = 9M1
OrB1forx=9verified,thenB1fory=–27
So x = 9A1F
f.t.numericalerroriny
and y = –27A13
(iii)At SP y = B1
This is positive, so minimumE1F2
f.t.wrongvalueforyatSP
[13]
12.h = 0.5B1
Integral = h/2{…}
{…}= [f(1) + f(2.5) + 2 (f(1.5) +f(2))]M1
Where f(x) = 2x – 1.
={1+(4√2 –1) +2[(2√2 –1) + 3]}A1
{4√2 –1=4.65685..} {2√2 –1=1.82842…}
All 4 terms correct.[accept 3 dp or better for each term or 15.31(37…) seen or 3.82(8…) seen if index or surd form not given]
Integral to 3sf = 3.83A14
cao Must be 3.83
[4]
13.(a) (+c)M1A12
M1 for the correct power of x
(b)Substitution of x = 2m1
A1F
ft wrong coeff ; decimals not Allowed
... = A1F3
ditto
[5]
14.(a)(i)M1A12
M1 if index correct or for example ;
Condone for
(ii)Substitution of x = 4m1
= 12.8A1F2
ft wrong coefficient of
(b)Required Area = area of – 12.8M1
Condone eg area of = 4 × 5
= = 3.2A1F2
ft wrong answer to (a)(ii) provided answer > 0
[6]
15.(a)y = 1 – 8x–3M1A12
(b)At SP, 8x–3 = 1m1
SP is (2, 3)A1A13
NMS x = 2, B1 y = 3, B1
(c)y = 24x–4m1A1F
m1 if index correct;
ft numerical error or y = –8x–3
...= at SPA1F
ft wrong coefficient of x–4
so SP is minE1F4
ft wrong (non-zero) value of y at SP;
allow ‘y = 24x–4 > 0’ without a value
[9]
16.(a)Use of y = nx n–1M1
coeff or index right or both approx right
y = A12
(b)(i)M1
Coeff or index right or consistent with each other
A12
Accept unsimplified
(ii)Substituting x = 8 in integralM12
not in y or y’
A1F
ft wrong coeff of x; allow decimals
[6]
17.(a)B11
(b)M1A1
M1 for with c’s non-integer value of p
... = 67 ½ A1F3
f.t. wrong coefficient of
[4]
18.y=M1A1
x=8y=A1F
ft num error; accept 2.67 OB; NMS 2 out of 3
[3]
19.(a)B11
(b)Integn of 3 terms of correct degreeB1B1B13
Accept unsimp; condone no +c
[4]
20.(a)Deriv of is kx½M1
y = 1–½A1
y(0) =1, y(1) = – ½A1F
ft error in coeff of x½
Conclusion drawn (AG)E14
Magnitudes (OE) must be mentioned
(b)(i)M1A12
At least one term correct for M1 Accept unsimp; condone no +c
(ii)Substitution of x = 1m1
Area =A1F2
ft num error leading to positive answer
[8]
21.(a)arc = x B1
Accept if seen on a diagram
5 sides + arc 5x + x = 10B12
ag convincingly shown
(b)(i)Area of triangle = x2 sin60°M1
= x2A1
Area of sector = x2M1
= x [10 – 5x]A1
A = x2 – x2 + 5xA15
ag convincingly shown
(ii) = x – 5x + 5M1
At least one term correct or answer a multiple of
correct answer
A max = 0m1
Putting = 0 and first step to solve
5x – x = 5
x = oe {= 1.209489...}A13
Accept 1.21
oe must not have fractions in the numerator or denominator
(iii){= –4.13...}B1ft
Onlyft if one slip in finding A'
...... < 0 maximumE12
ft sign of
[12]
22.(a)y = 2x – 8x– ³M1A1
M1 if at least one term correct
Attempt to solve y = 0m1
x4 = 4A1F
ft one error in coeff
x = ±2A1F
both needed; accept ; ft error in coeff
y = 4A16
Allow even if only one x found dec approx penalise once only
NMS 2/6 for (±1.4, 4), 1/6 for (1.4, 4), 1/6 for x = ±1.4
(b)(i)Result convincingly establishedM1A12
M1 for reasonable attempt
(ii)u² – 5u + 4 = 0B1
NMS 1/3 for 4 correct values of x
u = 1 or 4B1
x = ±1 or ±2B13
All four needed
(iii)B1B1
B1 for each term
Evaluation of M1.
where a, b are possible values found in (ii); OE
... = A1F
ft error in coeffs (not in limits of integration)
Area = M1
independent of previous M1
... = A16
convincingly established (AG)
[17]
South Wolds Comprehensive School1