Additional HL Practice
1. [17 marks]
A geometric sequence
, with complex terms, is defined by
and
(a) Find the fourth term of the sequence, giving your answer in the form
(b) Find the sum of the first 20 terms of
and
, giving your answer in the form
.
.
where
are to be determined.
A second sequence
(c) (i) Show that
is defined by
.
is a geometric sequence.
(ii) State the first term.
(iii) Show that the common ratio is independent of k.
A third sequence
is defined by
(d) (i) Show that
is a geometric sequence.
.
(ii) State the geometrical significance of this result with reference to points on the complex plane.
Markscheme
(a)
(A1)
M1
A1
[3 marks]
(b)
(M1)
(M1)
Note: Only one of the two M1s can be implied. Other algebraic methods may be seen.
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(A1)
A1
[4 marks]
(c) (i) METHOD 1
M1
A1
this is the general term of a geometrical sequence R1AG
Notes: Do not accept the statement that the product of terms in a geometric sequence is also geometric
unless justified further.
If the final expression for
is
award M1A1R0.
METHOD 2
M1
A1
this is a constant, hence sequence is geometric R1AG
Note: Do not allow methods that do not consider the general term.
(ii)
A1
(iii) common ratio is
(which is independent of k) A1
[5 marks]
(d) (i) METHOD 1
M1
2
M1
A1
this is the general term for a geometric sequence R1AG
METHOD 2
M1
A1
A1
this is the general term for a geometric sequence R1AG
Note: Do not allow methods that do not consider the general term.
(ii) distance between successive points representing
in the complex plane forms a geometric
sequence R1
Note: Various possibilities but must mention distance between successive points.
[5 marks]
Total [17 marks]
2a. [6 marks]
Find three distinct roots of the equation
giving your answers in modulus-
argument form.
Markscheme
3
METHOD 1
M1(A1)
(A1)
M1
,
,
. A2
Note: Accept
as the argument for
.
Note: Award A1 for correct roots.
Note: Allow solutions expressed in Eulerian
form.
Note: Allow use of degrees in mod-arg (r-cis) form only.
METHOD 2
so
is a factor
Attempt to use long division or factor theorem: M1
A1
Attempt to solve quadratic: M1
A1
4
,
,
. A2
Note: Accept
as the argument for
.
Note: Award A1 for correct roots.
Note: Allow solutions expressed in Eulerian
form.
Note: Allow use of degrees in mod-arg (r-cis) form only.
METHOD 3
Substitute
M1
and
A1
Attempt to solve simultaneously: M1
A1
,
,
. A2
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Note: Accept
as the argument for
.
Note: Award A1 for correct roots.
Note: Allow solutions expressed in Eulerian
form.
Note: Allow use of degrees in mod-arg (r-cis) form only.
[6 marks]
2b. [3 marks]
The roots are represented by the vertices of a triangle in an Argand diagram.
Show that the area of the triangle is
.
Markscheme
EITHER
Valid attempt to use
M1
A1A1
Note: Award A1 for correct sides, A1 for correct sin
OR
Valid attempt to use
M1
A1A1
Note: A1 for correct height, A1 for correct base.
THEN
AG
[3 marks]
Total [9 marks]
6
.
3a. [2 marks]
The wingspans of a certain species of bird can be modelled by a normal distribution with mean
and standard deviation
cm
cm.
According to this model,
of wingspans are greater than
cm.
Find the value of .
Markscheme
(M1)
A1
[2 marks]
3b. [3 marks]
In a field experiment, a research team studies a large sample of these birds. The wingspans of each bird
are measured correct to the nearest
cm.
Find the probability that a randomly selected bird has a wingspan measured as
cm.
Markscheme
(M1)(A1)
A1
[3 marks]
Total [5 marks]
4a. [3 marks]
The number of complaints per day received by customer service at a department store follows a
Poisson distribution with a mean of
.
On a randomly chosen day, find the probability that
(i) there are no complaints;
(ii) there are at least three complaints.
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Markscheme
(i)
A1
(ii)
(M1)
A1
[3 marks]
4b. [2 marks]
In a randomly chosen five-day week, find the probability that there are no complaints.
Markscheme
EITHER
using
(M1)
OR
using
(M1)
THEN
A1
[2 marks]
4c. [3 marks]
On a randomly chosen day, find the most likely number of complaints received.
Justify your answer.
Markscheme
(most likely number of complaints received is zero) A1
EITHER
calculating
and
M1A1
OR
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sketching an appropriate (discrete) graph of
against
M1A1
OR
finding
and stating that
M1A1
OR
using
where
M1A1
[3 marks]
4d. [2 marks]
The department store introduces a new policy to improve customer service. The number of complaints
received per day now follows a Poisson distribution with mean .
On a randomly chosen day, the probability that there are no complaints is now
.
Find the value of .
Markscheme
(A1)
A1
[2 marks]
Total [10 marks]
5. [17 marks]
The function f is defined by
(a) Write down the value of the constant term in the Maclaurin series for
(b) Find the first three derivatives of
including the
term is
.
and hence show that the Maclaurin series for
.
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up to and
(c) Use this series to find an approximate value for ln 2 .
(d) Use the Lagrange form of the remainder to find an upper bound for the error in this approximation.
(e) How good is this upper bound as an estimate for the actual error?
Markscheme
(a) Constant term = 0 A1
[1 mark]
(b)
A1
A1
A1
A1
Note: Allow FT on their derivatives.
M1A1
AG
[6 marks]
(c)
(A1)
M1
A1
[3 marks]
(d) Lagrange error
(M1)
A1
10
A2
giving an upper bound of 0.25. A1
[5 marks]
(e) Actual error
A1
The upper bound calculated is much larger that the actual error therefore cannot be considered a good
estimate. R1
[2 marks]
Total [17 marks]
6a. [1 mark]
A random variable
Sketch the graph of
has probability density function
.
Markscheme
11
A1
Note: Ignore open / closed endpoints and vertical lines.
Note: Award A1 for a correct graph with scales on both axes and a clear indication of the relevant
values.
[1 mark]
6b. [5 marks]
Find the cumulative distribution function for
.
Markscheme
considering the areas in their sketch or using integration (M1)
12
A1
A1
A1A1
Note: Accept
for
in all places and also
for
first A1.
[5 marks]
6c. [3 marks]
Find the interquartile range for
.
Markscheme
A1A1
A1
[3 marks]
Total [9 marks]
7. [9 marks]
The general term of a sequence
is given by the formula
(a) Determine whether the sequence
(b) Show that the sequence
(c) Find the smallest value of
.
is decreasing or increasing.
is convergent and find the limit L.
such that
, for all
.
Markscheme
(a)
M1A1
the sequence is decreasing (as terms are positive) A1
Note: Accept reference to the sum of a constant and a decreasing geometric sequence.
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Note: Accept use of derivative of
(and condone use of n) and graphical methods (graph
of the sequence or graph of corresponding function
or graph of its derivative
).
Accept a list of consecutive terms of the sequence clearly decreasing (eg
).
[3 marks]
(b)
M1A1
[2 marks]
(c)
M1
EITHER
(A1)
(A1)
OR
(A1)(A1)
Note: A1 for correct inequality; A1 for correct value.
THEN
therefore
A1
[4 marks]
8a. [3 marks]
Two species of plant,
and
leaves from a plant of species
is
, are identical in appearance though it is known that the mean length of
is
cm, whereas the mean length of leaves from a plant of species
cm. Both lengths can be modelled by normal distributions with standard deviation
In order to test whether a particular plant is from species
or species
,
cm.
leaves are collected at
random from the plant. The length, , of each leaf is measured and the mean length evaluated. A one14
tailed test of the sample mean,
and
, is then performed at the
level, with the hypotheses:
.
Find the critical region for this test.
Markscheme
(M1)
critical value is
(A1)
critical region is
A1
Note: Allow follow through for the final A1 from their critical value.
Note: Follow through previous values in (b), (c) and (d).
[3 marks]
8b. [2 marks]
It is now known that in the area in which the plant was found
are of species
of all the plants are of species
and
.
Find the probability that
will fall within the critical region of the test.
Markscheme
M1A1
Note: Award M1 for a weighted average of probabilities with weights
.
[2 marks]
8c. [3 marks]
If, having done the test, the sample mean is found to lie within the critical region, find the probability
that the leaves came from a plant of species
.
Markscheme
attempt to use conditional probability formula M1
15
(A1)
A1
[3 marks]
Total [10 marks]
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