Math Studies 2: Calculus Review Questions
1.
The cost per person, in euros, when x people are invited to a party can be determined by the function
C(x) = x +
(a)
100
x
Find C′ (x).
C′(x) = 1 
100
x2
(A1)(A1)(A1)3
Notes: (A1) for 1, (A1) for −100 , (A1) for x2 as
denominator or x−2 as numerator.
Award a maximum of (A2) if an extra term is seen.
(3)
(b)
Show that the cost per person is a minimum when 10 people are invited to the party.
For studying signs of the derivative at either side of x = 10
For saying there is a change of sign of the derivative
(M1)
(M1)(AG)
OR
For putting x = 10 into C′ and getting zero
For clear sketch of the function or for mentioning that the
function changes from decreasing to increasing at x = 10
(M1)
(M1)(AG)
OR
For solving C′(x) = 0 and getting 10
(M1)
For clear sketch of the function or for mentioning that the
function changes from decreasing to increasing at x = 10
(M1)(AG)2
Note: For a sketch with a clear indication of the minimum or for a
table with values of x at either side of x = 10 award (M1)(M0).
(2)
(c)
Calculate the minimum cost per person.
C(10) = 10 
C(10) = 20
100
10
(M1)
(A1)(G2)2
(2)
(Total 7 marks)
1
2.
A closed rectangular box has a height y cm and width x cm. Its length is twice its width. It has a fixed
2
outer surface area of 300 cm .
y cm
x cm
2x cm
(a)
2
Show that 4x + 6xy = 300.
Area = 2(2x)x + 2xy + 2(2x)y
Notes: (M1) for using the correct surface area
formula (which can be implied if numbers in the
correct place).
(A1) for using correct numbers.
300 = 4x2 + 6xy
Note: Final line must be seen or previous (A1) mark is lost.
(M1)(A1)
(AG)2
(2)
(b)
Find an expression for y in terms of x.
6xy = 300 – 4x2
y=
(M1)
300  4 x 2 150  2 x 2
or
6x
3x
(A1)
(2)
(c)
Hence show that the volume V of the box is given by V =100x 
4 3
x .
3
Volume = x (2x) y
 300  4 x 2
V = 2x2 
6x

= 100 x 
(M1)




4 3
x
3
Note: Final line must be seen or previous (A1) mark is lost.
(A1)(ft)
(AG)2
(2)
2
(d)
Find
dV
.
dx
dV
12 x 2
 100 
or 100 – 4x2
dx
3
Note: (A1) for each term.
(A1)(A1)2
(2)
(e)
(i)
Hence find the value of x and of y required to make the volume of the box a maximum.
For maximum
dv
= 0 or 100 − 4x2 = 0
dx
x=5
y=
=
(ii)
333
(M1)
(A1)(ft)
2
 150  252
300  45
or 

65
35





20
3
(M1)
(A1)(ft)
Calculate the maximum volume.
1
cm3 (333 cm3)
3
(A1)(ft)5
(UP)
Note: (ft) from their (e)(i) if working for volume is seen.
(5)
(Total 13 marks)
3
3.
The function f (x) is such that f ′ (x)  0 for 1  x  4. At the point P (4, 2) on the graph of f (x) the
gradient is zero.
(a)
Write down the equation of the tangent to the graph of f (x) at P.
y = 2. Award (A1) for y =…, (A1) for 2.
Accept f (x) =2 and y = 0x + 2
(A1)(A1)
(C2)
(2)
(b)
State whether f (4) is greater than, equal to or less than f (2).
Less (than).
(A2)(C2)
(2)
(c)
Given that f (x) is increasing for 4  x  7, what can you say about the point P?
Local minimum (accept minimum, smallest or equivalent)
Award (A1) for stationary or turning point mentioned.
Note: No mark is awarded for gradient = 0 as this is given
in the question.
(A2)
(C2)
(2)
(Total 6 marks)
4.
Consider the function f (x) =
(a)
1 3
x  2 x 2  3.
2
Find f ′ (x).
3x 2
 4x
2
(A1)(A1)
Note: Award (A1) for each correct term and
no extra terms; award (A1)(A0) for both terms
correct and extra terms; (A0) otherwise.
(C2)
(2)
(b)
Find f ′′ (x).
3x − 4
(A1)(ft)(A1)(ft)
accept 3x1 − 4x0
(C2)
(2)
(c)
Find the equation of the tangent to the curve of f at the point (1, 1.5).
y = −2.5x + 4 or equivalent
Notes: Award (A1)(ft) on their (a) for –2.5x
(must have x), (A1) for 4 or equivalent correct
answer only.
Accept y − 1.5 = −2.5 (x − 1)
(A1)(ft)(A1)
(C2)
(2)
(Total 6 marks)
4
5.
2
The figure below shows the graphs of functions f1(x) = x and f2 (x) = 5 − x .
6
f 2 (x )
4
f1 ( x )
2
–3
–2
–1
0
1
2
x
–2
–4
(a)
(i)
Differentiate f1(x) with respect to x.
f1(x) = 1
(ii)
(A1)
Differentiate f2 (x) with respect to x.
f2(x) = –2x
(A1)(A1)
(C3)
Note: (A1) for correct differentiation of each term.
(b)
Calculate the value of x for which the gradient of the two graphs is the same.
1 = –2x
x
(c)
(M1)
1
2
(A1)(ft)
Draw the tangent to the curved graph for this value of x on the figure, showing clearly the
property in part (b).
Note:(A1) is for the tangent drawn at x = 
1
and reasonably parallel to the
2
line f1 as shown.
5
(C2)
(A1) (C1)
(Total 6 marks)
6
6.
The diagram below shows the graph of a line L passing through (1, 1) and (2, 3) and the graph P of the
2
function f (x) = x − 3x − 4
y
L
P
0
(a)
x
Find the gradient of the line L.
for attempt at substituted
ydistance
xdistance
gradient = 2
(M1)
(A1)(G2)
2
(2)
(b)
Differentiate f (x).
2x – 3
(A1)(A1)
2
(A1) for2x, (A1) for –3
(2)
(c)
Find the coordinates of the point where the tangent to P is parallel to the line L.
for their 2x – 3 = their gradient and attempt to solve
(M1)
x = 2.5
(A1)(ft)
y = –5.25 ((ft) from their x value)
(A1)(ft)(G2) 3
(3)
(d)
Find the coordinates of the point where the tangent to P is perpendicular to
the line L.
for seeing
1
1
solving 2x – 3 =  (or their value)
2
their(a)
(M1)(M1)
x = 1.25
(A1)(ft)(G1)
y = – 6.1875
(A1)(ft)(G1) 4
7
8
(4)
(e)
Find
(i)
the gradient of the tangent to P at the point with coordinates (2, −6);
22–3=1
(ii)
((ft) from (b))
(A1)(ft)(G1)
the equation of the tangent to P at this point.
y = mx + c or equivalent method to find c  –6 = 2 + c
y=x–8
(M1)
(A1)(ft)(G2) 3
(3)
(f)
State the equation of the axis of symmetry of P.
x = 1.5
(A1)
1
(1)
(e)
Find the coordinates of the vertex of P and state the gradient of the curve
at this point.
for substituting their answer to part (f) into the equation of the parabola
(1.5, – 6.25) accept x = 1.5, y = –6.25
gradient is zero (accept
dy
0 )
dx
(M1)
(A1)(ft)(G2)
(A1) 3
(3)
(Total 18 marks)
8.
(a)
2
Expand the expression 2x(x −1).
9
2x3 – 2x
(b)
(A1)
(C1)
2
Hence differentiate f (x) = 2x(x −1) with respect to x.
f  (x) = 6x2 – 2
(A1)
(C2)
(ft)(A1)(ft)
Note: Award (A1) for each term.
(c)
Find the gradient of the tangent to the curve y = f (x) at the point where x = −1.
gradient = f (–1)
= 6(–1)2 – 2
=4
(M1)
(d)
(A1)(ft) (C2)
If the angle between the x-axis and the tangent in part (c) is , write down the value of tan .
tan = 4
(A1)(ft)
(C1)
(Total 6 marks)
7.
A football is kicked from a point A (a, 0), 0  a  10 on the ground towards a goal to the right of A.
10
The ball follows a path that can be modelled by part of the graph
2
y = − 0.021x +1.245x −6.01, x 
, y  0.
x is the horizontal distance of the ball from the origin
y is the height above the ground
Both x and y are measured in metres.
(a)
Using your graphic display calculator or otherwise, find the value of a.
a = 5.30 (3s.f.) (Allow (5.30, 0) but 5.3 receives an (AP).)
(A1) 1
(1)
(b)
Find
dy
.
dx
dy
  0.042 x 1.245
dx
Note: (A1) for each term.
(A1)(A1)
(2)
(c)
(i)
Use your answer to part (b) to calculate the horizontal distance the ball has travelled from A
when its height is a maximum.
Maximum value when f(x) = 0,
– 0.042x + 1.245 = 0,
Note: (M1) is for either of the above but at least one must be
seen.
(M1)
(x = 29.6.)
Football has travelled 29.6 – 5.30 = 24.3 m (3s.f.) horizontally.
Note: For answer of 24.3 m with no working or for correct
subtraction of 5.3 from candidate’s x-coordinate at the
maximum (if not 29.6), award (A1)(d).
(ii)
(A1)(ft)
Find the maximum vertical height reached by the football.
Maximum vertical height, f (29.6) = 12.4 m
(M1)(A1)(ft)(G2) 4
(UP)
Note: (M1) is for substitution into f of a value seen in part
(c)(i). f(24.3) with or without evaluation is awarded (M1)(A0).
For any other value without working, award (G0). If lines are
seen on the graph in part (d) award (M1) and then (A1) for
candidate’s value 0.5(3s.f. not required.)
(4)
11
2
(d)
Draw a graph showing the path of the football from the point where it is kicked to the point where
it hits the ground again. Use 1 cm to represent 5 m on the horizontal axis and 1 cm to represent 2
m on the vertical scale.
(A1)(A1)
(A1)(ft)
(A1)(ft)
Note: Award (A1) for labels (units not required) and scale,
(A1)(ft) for max (29.6,12.4), (A1)(ft) for x-intercepts at 5.30
and 53.9, (all coordinates can be within 0.5), (A1) for welldrawn parabola ending at the x-intercepts.
(4)
The goal posts are 35 m from the point where the ball is kicked.
(e)
At what height does the ball pass over the goal posts?
f (40.3) = 10.1 m (3s.f.).
(M1)(A1)(ft)(G2)
(UP)
Notes: Follow through from (a).
If graph used, award (M1) for lines drawn and (A1) for
candidate’s value 0.5. (3s.f. not required).
(2)
(Total 13 marks)
9.
The function g (x) is defined by g (x) =
1 4 9 2
x +
x –5x + 7, x  0.
8
4
12
(a)
Find g (2).
1
9
g (2)  (2) 4  (2) 2  5(2)  7
8
4
(M1)
=8
(A1)(G2)
(2)
(b)
Calculate g′ (x).
1
9
g( x)  x 3  x  5
2
2
(A1)
3
(A1)(A1)
Notes: Award (A1) for each correct term.
An extra constant means that the –5 is incorrect.
(3)
The graph of the function y =g (x) has a tangent T1 at the point where x = 2.
(c)
(i)
Show that the gradient of T1 is 8.
1
9
g(2)  (8)  (2)  5  8
2
2
For use of their derivative function
(M1
)
For substitution of 2 leading to an answer of 8 only
(A1)
(AG
)
Note: beware that g(2) = 8 also. This receives no marks in this
part.
(ii)
Find the equation of T1. Write the equation in the form y = mx+ c.
y = 8x + c
(A1)
Note: Award (A1) for 8x seen. Must be 8.
Use of point (2, 8)
(M1
)
Note: Follow through with y value from part (a)
y = 8x – 8
(A1)
(ft)(G3)
Note: Answer must be an equation. Allow T1 = 8x – 8
OR
8x – y = 8(2) – 8
(A1)
13
(M1)
y = 8x – 8
(A1)
5
(ft)(G3)
(5)
(d)
 35 
 . T2 has zero gradient.
 8
The graph has another tangent T2 at the point 1,
Write down the equation of T2.
y
35
8
For y = Allow T2 =
For
(A1)
35
(4.375)
8
(A1)
2
(2)
(e)
(i)
Sketch the graph of y = g (x) in the region 0  x  3, 0  y  22.
Note: The window must clearly have used x values from 0 to 3
and y values from 0 to 22. Axes labels are not required.
Some indication of scale must be present. This need not be
a formal scale. eg tick marks or single numbers on axes are
adequate.
(ii)
(A1)
Correct shape
(A1)
Minimum in approximately correct position
(A1)
Add the two tangents T1 and T2 to your sketch, in the correct positions.
Each tangent drawn at correct point.
(A1)
14
(ft)(A1)
Note: Lines must be straight and must be tangents.
5
(5)
(Total 17 marks)
15