Download Practice Paper C3-D

MEI
Mathematics in Education and Industry
MEI STRUCTURED MATHEMATICS
METHODS FOR ADVANCED MATHEMATICS, C3
Practice Paper C3-D
Additional materials: Answer booklet/paper
Graph paper
List of formulae (MF2)
TIME 1 hour 30 minutes
INSTRUCTIONS
•
•
•
Write your Name on each sheet of paper used or the front of the booklet used.
Answer all the questions.
You are permitted to use a graphical calculator in this paper.
INFORMATION
•
•
•
•
•
The number of marks is given in brackets [] at the end of each question or part-question.
You are advised that an answer may receive no marks unless you show sufficient detail of
the working to indicate that a correct method is being used.
Final answers should be given to a degree of accuracy appropriate to the context.
The total number of marks for this paper is 72.
You are reminded of the need for clear presentation in your answers.
 MEI February 2005
Section A (36 marks)
1
You are given that y2 = 4x + 7.
(i)
Use implicit differentiation to find
(ii)
Make x the subject of the equation.
Find
dy
in terms of y.
dx
[2]
dx
dx 1
and hence show that in this case
=
.
dy
dy dy
dx
[3]
(i)
Expand ( e x + e − x ) .
(ii)
Hence find
(i)
Sketch the graph of y = 3 x − 6 .
[2]
(ii)
Solve the equation 3 x − 6 = x + 4 and illustrate your answer on your graph.
[4]
4
Find
∫ x sin 3x dx .
[4]
5
Make x the subject of t = ln
6
The function f(x) is defined as f(x) =
2
3
2
∫ (e
x
[1]
+ e − x ) dx .
2
[3]
5
.
( x − 3)
P
[4]
lnx
. The graph of the function is shown in Fig. 6.
x
Q
Fig. 6
(i)
Give the coordinates of the point, P, where the curve crosses the x-axis.
[1]
(ii)
Use calculus to find the coordinates of the stationary point, Q, and show that it is a
maximum.
[6]
© MEI February, 2005
MEI Structured Mathematics
Practice Paper C3-D
Page 2
7
An oil slick is circular with radius r km and area A km2. The radius increases with time at a
dr
rate given by
= 0.5 , in kilometres per hour.
dt
dA
(i) Show that
[4]
= πr .
dt
(ii)
Find the rate of increase of the area of the slick at a time when the radius is 6 km.
[2]
Section B (36 marks)
8
Fig. 8 shows the graph of y = x 1 + x . The point P on the curve is on the x-axis.
P
Fig. 8
(i)
Write down the coordinates of P.
(ii)
Show that
[1]
dy 3x + 2
=
.
dx 2 1 + x
[4]
(iii) Hence find the coordinates of the turning point on the curve.
What can you say about the gradient of the curve at P?
[4]
1
1
 32

2
+
=
−
x
1
x
d
x
u
u
 du .
∫−1
∫0 

Evaluate this integral, giving your answer in an exact form.
0
(iv) By using a suitable substitution, show that
(v)
What does this value represent?
[7]
Use your answer to part (ii) to differentiate y = x 1 + x sin2 x with respect to x.
(You need not simplify your result.)
[2]
© MEI February, 2005
MEI Structured Mathematics
Practice Paper C3-D
Page 3
9
The functions f(x) and g(x) are defined by
f(x) = x2 ,
g(x) = 2x − 1,
for all real values of x.
(i)
(ii)
State the ranges of f(x) and g(x).
Explain why f(x) has no inverse.
[3]
Find an expression for the inverse function g-1(x) in terms of x.
Sketch the graphs of y = g(x) and y = g -1(x) on the same axes.
[4]
(iii) Find expressions for gf(x) and fg(x).
[2]
(iv) Solve the equation gf(x) = fg(x).
Sketch the graphs of y = gf(x) and y = fg(x) on the same axes to illustrate your
answer.
(v)
Show that the equation f(x + a) = g 2(x) has no solution if a >
© MEI February, 2005
MEI Structured Mathematics
[4]
1
.
4
Practice Paper C3-D
[5]
Page 4
Qu
Answer
Section A
1
(i)
dy
y 2 = 4 x + 7 ⇒ 2 y. = 4
dx
dy 2
⇒
=
dx y
(ii)
2
x=
3
(
A1
2
)
(i)
e +2+e
(ii)
= ∫ (e + 2 + e
B1
M1
A1
3
B1
−2 x
2x
Comment
M1
1 2
dx 1
y 1
y −7 ⇒
= .2 y = =
4
dy 4
2 2
y
2x
=
Mark
1
−2 x
)dx
B1
B1
B1
1 2x
1
e + 2 x − e −2 x + c
2
2
(i)
B1
B1
One for each
exponential
term, one for
both 2x and
3 constant.
One for two
half lines;
one for
correct
orientation
and meeting
at (2, 0) .
2
At P
−3 x + 6 = x + 4
1
⇒
x=
2
At Q
3x − 6 = x + 4
(ii)
⇒
x=5
The solution is x = 12 , 5 .
As shown on graph
4
∫ x sin 3x dx;
u= x⇒
du
dv
1
= 1,
= sin 3 x ⇒ v = − cos x
dx
dx
3
1
1
= − x cos 3x + ∫ cos 3x dx
3
3
1
1
= − x cos 3x + sin 3x + c
3
9
M1
A1
A1
E1
4
Choice of u
M1
1
− cos 3 x
3
A1
M1
A1
Correct form
c must be
seen
4
 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 2
5
6
t = ln
(i)
5
( x − 3)
1 ( x − 3)
= − ln
2
5
( x − 3)
⇒ −2t = ln
5
x
−
(
3)
⇒ e −2t =
⇒ x = 5e −2t + 3
5
P(1,0)
Rules of
logs
Change to
exponentials
M1
M1
A1
A1
4
B1
1
(ii)
1
x. − 1.ln x
dy
= x 2
x
dx
1 − ln x
=
x2
At Q, gradient is zero, so x = e .
Q is (e, 1e ) .
1
x 2 (− ) − (1 − ln x).2 x
x
x4
−3 + 2 ln x
=
x3
When x = e , this is –ve, so Q is a maximum.
d2 y
=
dx 2
7
(i)
(ii)
dA
= 2π r
dr
dA dA dr 1
⇒
=
. = × 2π r = π r
dt dr dt 2
A = π r2 ⇒
So when r = 6 ,
dA
= 6π (= 15.707...) .
dt
The area increases at 15.7 km2h-1, to 3sf.
quotient rule
M1
A1
M1
A1
=0
M1
Or
equivalent
methods
A1
6
For 1/e.
M1
A1
M1
A1
4
Substituting
M1
A1
2
 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 3
Section B
8
(i)
P(−1,0)
(ii)
B1
1
y = x 1+ x
1
= x(1 + x) 2
1
dy
1
−1
= 1.(1 + x) 2 + x. (1 + x) 2
dx
2
2(1 + x) + x
=
1
2(1 + x) 2
3x + 2
=
2 1+ x
(iii) At a turning point, gradient is zero.
x = − 23 there. Then
⇒
y=−
2 1
3 3
M1
Product rule
A1
Any correct
expression
Combining
fractions
M1
E1
4
M1
A1
2 3
9
These are the coordinates of the TP.
A1
At P the gradient is undefined.
B1
=−
(iv)
∫
0
−1
Accept a
simplified
form with
√3 in the
bottom.
Accept
reference to
infinity or to
4 vertical.
x 1 + xdx
1
= ∫ (u − 1)u 2 du, u = 1 + x
0
du
⇒
=1
dx
1
Integral in u
1
M1
E1
Change of
limits
M1A1
Integrating
= ∫ (u − u )du
3
2
1
2
0
1
2 5 2 3 
=  u2 − u2 
3 0
5
4
=−
15
The magnitude represents the area bounded by the curve and
axis between P and O. It is negative because the curve is
below the axis (except at the end points).
(v)
y = x 1 + x sin 2 x
dy 3 x + 2
sin 2 x + x 1 + x .2 cos 2 x
⇒
=
dx 2 1 + x
A1
B1
B1
One for
geometry
7 and one for
sign.
M1A1
2
 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 4
9
(i)
B1
Range of f is [0, ∞) ,
B1
of g (−∞, ∞) .
f has no inverse because (say) for any value of f > 0 there are
E1
2 corresponding values of x
3
y = 2x −1
(ii)
1
1
y+
2
2
1
1
⇒ g −1 ( x) = x +
2
2
⇒x=
One mark
for one line,
and one
mark for
second
correctly
related
M1
A1
B1
B1
4
(iii)
B1
B1
gf ( x) = 2 x 2 − 1
fg( x) = (2 x − 1) 2
2
(iv)
M1
A1
2 x 2 − 1 = (2 x − 1) 2
⇒ 0 = 2x2 − 4x + 2
⇒ 0 = 2( x − 1) 2
⇒ x =1
B1
y = (2x − 1)2
B1
y = (2x − 1)2
y = (2x − 1)2
4
(v)
f ( x + a) = ( x + a)
2
g 2 ( x) = 2(2 x − 1) − 1 = 4 x − 3
B1
f ( x + a) = g 2 ( x) ⇒ ( x + a)2 = 4 x − 3
M1
⇒ x + (2a − 4) x + a + 3 = 0
⇒ There are two roots to this equation if
M1
(2a − 4) 2 > 4(a 2 + 3)
A1
2
Both fns
correct
Equating
2
i.e. 4a − 16a + 16 > 4a + 12
1
⇒ 16a < 4 ⇒ a <
4
2
Using
b2 − 4ac
Correct
inequality
2
A1
5
Result
 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 5
 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 6