1.
(a)
d
1 d 2
ln x = ,
x = 2x
dx
x dx
(seen anywhere)
attempt to substitute into the quotient rule (do not accept product rule)
A1A1
M1
1
x 2 2 x ln x
x
x4
e.g.
correct manipulation that clearly leads to result
A1
x 2 x ln x x (1 2 ln x ) x 2 x ln x
,
, 4
x4
x4
x
x4
e.g.
g ( x ) =
(b)
1 2 ln x
x3
evidence of setting the derivative equal to zero
AG
N0
N2
(M1)
e.g. g(x) = 0, 1 2ln x = 0
ln x =
1
2
A1
x = e2
A1
[7]
2.
gradient of tangent = 8 (seen anywhere)
f(x) = 4kx3 (seen anywhere)
(A1)
A1
recognizing the gradient of the tangent is the derivative
(M1)
setting the derivative equal to 8
e.g. 4kx3 = 8, kx3 = 2
(A1)
substituting x = 1 (seen anywhere)
k=2
(M1)
A1
N4
[6]
3.
(a)
substituting (0, 13) into function
e.g. 13 = Ae0 + 3
13 = A + 3
A = 10
IB Questionbank Maths SL
M1
A1
AG
N0
(b)
substituting into f(15) = 3.49
e.g. 3.49 = 10e15k + 3, 0.049 = e15k
evidence of solving equation
e.g. sketch, using ln
ln0.049
accept
15
k = 0.201
(c)
f(x) = 10e0.201x + 3
f(x) = 10e0.201x 0.201 (= 2.01e0.201x)
(i)
A1
(M1)
A1
N2
A1A1A1
N3
Note: Award A1 for 10e0.201x, A1 for 0.201,
A1 for the derivative of 3 is zero.
(d)
(ii)
valid reason with reference to derivative
e.g. f(x) < 0, derivative always negative
R1
N1
(iii)
y=3
A1
N1
finding limits 3.8953, 8.6940 (seen anywhere)
evidence of integrating and subtracting functions
correct expression
e.g.
8.69
3.90
g ( x) f ( x)dx,
A1A1
(M1)
A1
8.69
3.90
[( x 2 + 12 x 24) (10e 0.201x + 3)]dx
area = 19.5
A2
N4
[16]
4.
(a)
(i)
1.15, 1.15
(ii)
recognizing that it occurs at P and Q
e.g. x = 1.15, x = 1.15
k = 1.13, k = 1.13
IB Questionbank Maths SL
A1A1
(M1)
A1A1
N3
N2
(b)
evidence of choosing the product rule
e.g. uv + vu
(M1)
derivative of x3 is 3x2
(A1)
2x
2
derivative of ln (4 x ) is 4 x
correct substitution
2x
x3
+ ln(4 x 2 ) 3 x 2
2
4x
e.g.
2x 4
+ 3x 2 ln(4 x 2 )
2
g(x) = 4 x
(A1)
A1
AG
N0
(c)
A1A1N2
(d)
w = 2.69, w < 0
A1A2
N2
[14]
5.
(a)
(b)
evidence of valid approach
e.g. f(x) = 0, graph
a = 1.73, b = 1.73 (a = 3 , b = 3 )
(M1)
A1A1
attempt to find max
e.g. setting f(x) = 0, graph
(M1)
c = 1.15 (accept (1.15, 1.13))
(c)
A1
attempt to substitute either limits or the function into formula
e.g. V =
V = 2.16
[ f ( x)]
c
IB Questionbank Maths SL
N3
]
2
dx, x ln(4 x ) ,
1.149...
N2
M1
y dx
A2
N2
(d)
valid approach recognizing 2 regions
e.g. finding 2 areas
(M1)
correct working
e.g.
1.73...
f ( x)dx +
(A1)
1.149...
f ( x)dx;
1.73...
f ( x)dx +
1.149...
f ( x)dx
area = 2.07 (accept 2.06)
A2
N3
[12]
6.
(a)
evidence of using the product rule
M1
f (x) = ex(1 x2) + ex(2x)
A1A1
x
Note: Award A1 for e (1 x ), A1 for e (2x).
f (x) = ex(1 2x x2)
AG
N0
(b)
y=0
A1
N1
(c)
at the local maximum or minimum point
(ex(1 2x x2) = 0)
f (x) = 0
(M1)
1 2x x2 = 0
(M1)
r = 2.41 s = 0.414
(d)
A1A1 N2N2
f(0) = 1
A1
gradient of the normal = 1
A1
evidence of substituting into an equation for a straight line
correct substitution
(M1)
A1
e.g. y 1 = 1(x 0), y 1 = x, y = x + 1
x+y=1
(e)
(i)
AG
intersection points at x = 0 and x = 1 (may be seen as the limits) (A1)
approach involving subtraction and integrals
fully correct expression
(e (1 x ) (1 x ) ) dx ,
1
e.g.
(ii)
N0
area R = 0.5
1
0
f ( x ) dx
(M1)
A2
N4
A1
N1
(1 x ) dx
0
[17]
IB Questionbank Maths SL
