Name: ___________________________
Date: ______________________
IB SL2 Semester Review
1*. Evaluate the following limits:
a.
3𝑥 2 + 1
lim
𝑥→−∞ 2𝑥 2 + 3𝑥 − 2
b.
c.
2*.
lim
7−6𝑥 5
𝑥→∞ 𝑥+3
lim
𝑥 2 − 25
𝑥→−5 𝑥+5
Gas is being pumped into a spherical balloon at the rate of 8in³/sec. Find the rate of change of the
radius of the balloon when the radius of the balloon is 6in.
3*. A particle P moves in a straight line with velocity function v(t) = t2 -7t + 10 ms-1. Find the displacement of
P after 3 seconds and the total distance travelled after 3 seconds.
4*. Evaluate the following integrals:

a)

b)

3

4
c)
3
0
1
1
sin xdx
( x 2  2)dx
3
(2 x  )dx
x
5*. Find the area of the region enclosed by y = x+2 and y = x2 + x – 2.
6*. Integrate cos3xsinx with respect to x.
IB Questionbank Maths SL
2
1.
The following diagram shows part of the graph of the function f(x) = 2x2.
diagram not to scale
The line T is the tangent to the graph of f at x = 1.
(a)
Show that the equation of T is y = 4x – 2.
(5)
(b)
Find the x-intercept of T.
(2)
(c)
The shaded region R is enclosed by the graph of f, the line T, and the x-axis.
(i)
Write down an expression for the area of R.
(ii)
Find the area of R.
(9)
(Total 16 marks)
IB Questionbank Maths SL
3
2.
Given the function f (x) = x2 – 3bx + (c + 2), determine the values of b and c such that f (1) = 0 and
f (3) = 0.
(Total 4 marks)
x
3.
The diagram shows part of the graph of y = e 2 .
y
x
y = e2
P
ln2
(a)
x
Find the coordinates of the point P, where the graph meets the y-axis.
(2)
The shaded region between the graph and the x-axis, bounded by x = 0
and x = ln 2, is rotated through 360° about the x-axis.
(b)
Write down an integral which represents the volume of the solid obtained.
(4)
(c)
Show that this volume is .
(5)
(Total 11 marks)
IB Questionbank Maths SL
4
4.
Let g(x) =
(a)
ln x
x2
, for x > 0.
Use the quotient rule to show that g ( x) 
1  2 ln x
x3
.
(4)
(b)
The graph of g has a maximum point at A. Find the x-coordinate of A.
(3)
(Total 7 marks)
5.
Consider the function f with second derivative f′′(x) = 3x – 1. The graph of f has a minimum point
 4 358 
at A(2, 4) and a maximum point at B   ,
.
 3 27 
(a)
Use the second derivative to justify that B is a maximum.
(3)
(b)
Given that f′ =
3 2
x – x + p, show that p = –4.
2
(4)
(c)
Find f(x).
(7)
(Total 14 marks)
IB Questionbank Maths SL
5
6.
The acceleration, a m s–2, of a particle at time t seconds is given by
a=
1
+ 3sin 2t, for t ≥ 1.
t
The particle is at rest when t = 1.
Find the velocity of the particle when t = 5.
(Total 7 marks)
7.
The function f is such that f  (x) = 2x – 2.
When the graph of f is drawn, it has a minimum point at (3, –7).
(a)
Show that f (x) = x2 – 2x – 3 and hence find f (x).
(6)
(b)
Find f (0), f (–1) and f (–1).
(3)
(c)
Hence sketch the graph of f, labelling it with the information obtained in part (b).
(4)
(Note: It is not necessary to find the coordinates of the points where the graph
cuts the x-axis.)
(Total 13 marks)
IB Questionbank Maths SL
6
8.
Let f(x) =
(a)
x . Line L is the normal to the graph of f at the point (4, 2).
Show that the equation of L is y = –4x + 18.
(4)
(b)
Point A is the x-intercept of L. Find the x-coordinate of A.
(2)
In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L.
(c)
Find an expression for the area of R.
(3)
(d)
The region R is rotated 360° about the x-axis. Find the volume of the solid formed, giving
your answer in terms of π.
(8)
(Total 17 marks)
IB Questionbank Maths SL
7
9.
A farmer wishes to create a rectangular enclosure, ABCD, of area 525 m2, as shown below.
The fencing used for side AB costs $11 per metre. The fencing for the other three sides costs $3
per metre. The farmer creates an enclosure so that the cost is a minimum.
Find this minimum cost.
(Total 7 marks)
IB Questionbank Maths SL
8
10.
The diagram shows part of the graph of the curve with equation
y = e2x cos x.
y
P(a, b)
0
(a)
Show that
x
dy
= e2x (2 cos x – sin x).
dx
(2)
(b)
Find
d2 y
.
dx 2
(4)
There is an inflexion point at P (a, b).
(c)
Use the results from parts (a) and (b) to prove that:
(i)
tan a = 3 ;
4
(3)
(ii)
the gradient of the curve at P is e2a.
(5)
(Total 14 marks)
IB Questionbank Maths SL
9
11.
The function f is given by f(x) = 2sin(5x – 3).
(a)
Find f′′(x).
(4)
(b)
Write down
 f ( x)dx .
(2)
(Total 6 marks)
12.
Let f′(x) = –24x3 + 9x2 + 3x + 1.
(a)
There are two points of inflexion on the graph of f. Write down the x-coordinates of these
points.
(3)
(b)
Let g(x) = f″(x). Explain why the graph of g has no points of inflexion.
(2)
(Total 5 marks)
IB Questionbank Maths SL
10
Answer Key
1*. a) 3/2 (tie, coefficients = 3/2)
b) -∞ (top bigger, so ∞ or −∞; top -, bottom +, so −∞)
c) -10 (reduce, then plug in -5 to get -10)
2*.
1
18𝜋
in/sec
3*. 9.83m or 59/6m
4*. a) ½
2
b) 123
c) 15 + 6ln2
5*. 10
2
3
units2 (example 5 page 480)
1
6*. -4cos4x + c (example 19a page 467)
1.
(a)
f (1) = 2
(A1)
f ′(x) = 4x
A1
evidence of finding the gradient of f at x =1
M1
e.g. substituting x =1 into f ′(x)
finding gradient of f at x =1
A1
e.g. f ′(1) = 4
evidence of finding equation of the line
M1
e.g. y – 2 = 4(x –1), 2 = 4(1) + b
y = 4x – 2
(b)
AG
appropriate approach
N0
5
N2
2
(M1)
e.g. 4x – 2 = 0
x=
(c)
(i)
1
2
A1
bottom limit x = 0 (seen anywhere)
(A1)
approach involving subtraction of integrals/areas
(M1)
e.g. ∫ f (x) – area of triangle, ∫ f – ∫l
correct expression
e.g.
(ii)
1

0
2 x 2dx 
A2
1
0.5 4x  2dx, 0 f xdx  , 0
1
1
0.5
2
2 x 2dx 
N4
 f x  4x  2dx
1
0.5
METHOD 1 (using only integrals)
correct integration
IB Questionbank Maths SL
(A1)(A1)(A1)
11

2 x 2 dx 
2x 3
,
3
 4 x  2 dx  2 x
2
 2x
substitution of limits
(M1)
1 2
1 1 
  2  2     1
12 3
 12 2 
e.g.
1
6
area =
A1
N4
METHOD 2 (using integral and triangle)
1
2
(A1)
correct integration
(A1)
area of triangle=

2 x 2 dx 
2x 3
3
substitution of limits
(M1)
2 3 2 3 2
1  0 ,  0
3
3
3
e.g.
correct simplification
(A1)
2 1

3 2
e.g.
1
6
area =
A1
N4
9
[16]
2.
f (1) = 12 – 3b + c + 2 = 0
(M1)
f (x) = 2x – 3b,
f (3) = 6 – 3b = 0
(M1)
3b = 6, b = 2
(A1)
1 – 3(2) + c + 2 = 0, c = 3
(A1)
Note: In the event of no working shown, award (C2) for 1 correct
answer.
[4]
3.
(a)
y = ex/2 at x = 0 y = e0 = 1 P(0, 1)
(b)
V=

ln 2
0
( e x / 2 ) 2 dx
(A1)(A1)
2
(A4)
4
Notes: Award (A1) for 
(A1) for each limit
(A1) for (ex/2)2.
(c)
V=

ln 2
0
e x dx
2
=  [e x ] ln
0
IB Questionbank Maths SL
(A1)
(A1)
12
= [eln2 – e0]
= [2 – 1] = 
=
(A1)
(A1)(A1)
(AG)
5
[11]
4.
(a)
d
1 d
ln x  , x 2  2 x (seen anywhere)
dx
x dx
attempt to substitute into the quotient rule (do not accept product rule)
A1A1
M1
1
x 2    2 x ln x
 x
e.g.
x4
correct manipulation that clearly leads to result
e.g.
x  2 x ln x x1  2 ln x  x 2 x ln x
,
, 4 
x4
x4
x
x4
g x  
(b)
A1
1  2 ln x
AG
x3
evidence of setting the derivative equal to zero
N0
4
N2
3
(M1)
e.g. g′(x) = 0, 1– 2ln x = 0
ln x 
1
2
A1
1
x  e2
A1
[7]
5.
(a)
(b)
substituting into the second derivative
 4
e.g. 3 ×     1
 3
 4
f″    = –5
 3
since the second derivative is negative, B is a maximum
setting f′(x) equal to zero
M1
A1
R1
N0
(M1)
4

evidence of substituting x = 2  or x   
3

e.g. f′(2)
correct substitution
(M1)
A1
2
e.g.
3
3  4  4
( 2) 2  2  p ,         p
2
2  3  3
correct simplification
8 4
e.g. 6 – 2 + p = 0,  + p = 0, 4 + p = 0
3 3
p = –4
(c)
evidence of integration
IB Questionbank Maths SL
A1
AG
N0
(M1)
13
f(x) =
1 3 1 2
x  x  4x  c
2
2
A1A1A1
 4 358 
substituting (2, 4) or   ,
(M1)
 into their expression
 3 27 
correct equation
A1
1 3 1 2
1
1
e.g.  2   2  4  2  c  4,  8   4 – 4  2  c  4, 4  2  8  c  4
2
2
2
2
1
1
A1
f ( x)  x 3  x 2  4 x  10
2
2
N4
[14]
6.
evidence of integrating the acceleration function
(M1)
1

e.g.   3 sin 2t dt
t

3
correct expression ln t –
cos 2t + c
A1A1
2
evidence of substituting (1, 0)
(M1)
3
e.g. 0 = ln 1 –
cos 2 + c
2
3
 3

c = –0.624   cos 2  ln 1 or cos 2 
(A1)
2
2


3
3
3
3
3


v = ln t – cos 2t  0.624   ln t  cos 2t  cos 2 or ln t  cos 2t  cos 2  ln 1 (A1)
2
2
2
2
2


v(5) = 2.24 (accept the exact answer ln 5 – 1.5 cos 10 + 1.5 cos 2)
A1

N3
[7]
7.
(a)
f (x) = 2x – 2
f (x) = x2 – 2x + c
= 0 when x = 3
 0 =9–6+c
c = –3
f (x) = x2 – 2x – 3
x3
f (x) =
– x2 – 3x + d
3
When x = 3,



(b)
f (x) = –7
–7 = 9 – 9 – 9 + d
d =2
x3
f (x) =
– x2 – 3x + 2
3
(M1)(M1)
(A1)
(AG)
(M1)
(M1)
(A1)
f (0) = 2
(A1)
1
f (–1) = – – 1 + 3 + 2
3
2
=3
3
f (–1) = 1 + 2 – 3
=0
(A1)
IB Questionbank Maths SL
(A1)
6
3
14
(c)
2

f (–1) = 0    1, 3  is a stationary point
3

y
–1, 3 23
2
x
(3, –7)
(A4)
4
Note: Award (A1) for maximum, (A1) for (0, 2)
(A1) for (3, –7), (A1) for cubic.
[13]
8.
(a)
finding derivative
e.g. f′(x) =
1
2
1

x 2
,
(A1)
1
2 x
correct value of derivative or its negative reciprocal (seen anywhere)
e.g.
1
1
2 4 4
,
gradient of normal = 
e.g. 
(b)
(c)
A1
1
(seen anywhere)
gradient of tangent
A1
1
 4,  2 x
f (4)
substituting into equation of line (for normal)
e.g. y – 2 = –4(x – 4)
M1
y = –4x + 18
AG
N0
recognition that y = 0 at A
e.g. –4x + 18 = 0
18  9 
x=
 
4  2
(M1)
A1
N2
splitting into two appropriate parts (areas and/or integrals)
correct expression for area of R
4
4.5
4
1
e.g. area of R =
xdx  (4 x  18)dx,
xdx   0.5  2 (triangle)
0
4
0
2
(M1)
A2
N3



Note: Award A1 if dx is missing.
IB Questionbank Maths SL
15
(d)
correct expression
for the4volume from
x = 0 to x = 4
4
4
2
2
π f ( x) dx, π x dx, πxdx
e.g. V = 0
0
0
 
 

(A1)
4
1

V =  πx 2 
2
0
1 
1
V = π 16   0 
2 
2
V = 8π
A1
(A1)
A1
finding the volume from x = 4 to x = 4.5
EITHER
recognizing a cone
1
e.g. V = πr2h
3
(M1)
1
1
π(2) 2 
3
2
2π
=
3
V=
total volume is 8π +
(A1)
A1
2
π
3
 26 
π

 3 
A1
N4
OR
V= π
=


4.5
4
4.5
4
(4 x  18) 2 dx
(M1)
π(16 x 2  144 x  324)dx
4.5
16

= π  x 3  72 x 2  324 x
3
4
2π
=
3
total volume is 8π +
2
π
3
 26 
π

 3 
A1
A1
A1
N4
[17]
9.
METHOD 1
correct expression for second side, using area = 525
525
e.g. let AB = x, AD =
x
(A1)
attempt to set up cost function using $3 for three sides and $11 for one side
e.g. 3(AD + BC + CD) + 11AB
(M1)
correct expression for cost
525
525
525
525
3150
e.g.
3
 3  11x  3x,
3
 3  11AB  3AB,
 14 x
x
x
AB
AB
x
A2
EITHER
sketch of cost function
identifying minimum point
e.g. marking point on graph, x = 15
IB Questionbank Maths SL
(M1)
(A1)
16
minimum cost is 420 (dollars)
A1
N4
OR
correct derivative (may be seen in equation below)
 1575  1575
e.g.C′(x) =

 14
x2
x2
(A1)
setting their derivative equal to 0 (seen anywhere)
 3150
e.g.
 14  0
x2
minimum cost is 420 (dollars)
(M1)
A1
N4
METHOD 2
correct expression for second side, using area = 525
e.g. let AD = x, AB =
(A1)
525
x
attempt to set up cost function using $3 for three sides and $11 for one side
e.g. 3(AD + BC + CD) + 11AB
(M1)
correct expression for cost
525  525
525  525
7350


e.g. 3 x  x 
11, 3 AD  AD 
11, 6 x 


x 
x
AD  AD
x


A2
EITHER
sketch of cost function
identifying minimum point
e.g. marking point on graph, x = 35
(M1)
(A1)
minimum cost is 420 (dollars)
A1
N4
OR
correct derivative (may be seen in equation below)
7350
e.g. C′(x) = 6 –
x2
(A1)
setting their derivative equal to 0 (seen anywhere)
7350
0
e.g. 6 –
x2
(M1)
minimum cost is 420 (dollars)
A1
N4
[7]
10.
(a)
(b)
y = e2x cos x
dy
= e2x (–sin x) + cos x (2e2x)
dx
= e2x (2 cos x – sin x)
d2 y
= 2e2x (2 cos x – sin x) + e2x (–2 sin x – cos x)
2
dx
= e2x (4 cos x – 2 sin x – 2 sin x – cos x)
= e2x (3 cos x – 4 sin x)
IB Questionbank Maths SL
(A1)(M1)
(AG)
2
(A1)(A1)
(A1)
(A1)
4
17
(c)
(i)
(ii)
d2 y
=0
dx 2
 3 cos x = 4 sin x
3
 tan x = 4
(M1)
3
At P, x = a, ie tan a = 4
(A1)
At P,
The gradient at any point e2x (2 cos x – sin x)
Therefore, the gradient at P = e2a (2 cos a – sin a)
3
4
3
When tan a = , cos a = , sin a =
4
5
5
(by drawing a right triangle, or by calculator)
8 3
Therefore, the gradient at P = e2a   
5 5
2a
=e
(R1)
(M1)
(A1)(A1)
(A1)
(A1)
8
[14]
11.
(a)
(b)
Using the chain rule
f′(x) = (2 cos(5x – 3))5 (= 10 cos(5x – 3))
f′′(x) = –(10 sin(5x – 3))5
= –50sin(5x – 3)
Note: Award A1 for sin (5x – 3), A1 for –50.
(M1)
A1
A1A1
2
 f ( x)dx   5 cos(5x  3)  c
N2
A1A1
N2
2
Note: Award A1 for cos(5x – 3), A1 for  .
5
[6]
12.
(a)
valid approach
e.g. f″(x) = 0, the max and min of f′ gives the points of inflexion on f
–0.114, 0.364 (accept (–0.114, 0.811) and (0.364, 2.13))
(b)
R1
A1A1 N1N1
METHOD 1
graph of g is a quadratic function
a quadratic function does not have any points of inflexion
R1
R1
N1
N1
R1
R1
N1
N1
R1
R1
N1
N1
METHOD 2
graph of g is concave down over entire domain
therefore no change in concavity
METHOD 3
g″(x) = –144
therefore no points of inflexion as g″(x) ≠ 0
[5]
IB Questionbank Maths SL
18