Full Name:
(block capitals)
Student Number:
CM111A Calculus 1: Test 1b
CALCULATORS MAY NOT BE USED
a
b
c
d
e
1
ANSWER GRID: put a cross in
ONE BOX for the correct answer
for each question. If you change
your mind and want to correct
your answer, obliterate your incorrect answer by shading its box,
and put a new cross in the box for
the correct answer.
2
3
4
5
6
MARKS: each correct answer = +5, incorrect =
1, none (or more than one) = 0.
Do any rough working on the back of this sheet, or on a NAMED separate sheet.
P
1. Define for n 2 Z+ the sum S = nk=0 k 2 . Which of the following is NOT true?
P
Pn
Pn 1
n
2
2
2
(a) S = 14 2n
k=0 k (b) S = 6 (n+1)(2n+1) (c) S =
k=1 k (d) S =
k= 1 (k +1)
(e) none of the above (all the other options are true)
2. Let z and w be complex numbers. Which claim below is NOT true?
(a) |Im(z)|  |z| (b) |zw| = |zw| (c) |z +w|  |z|+|w| (d) |(2+3i)(3 2i)| > 13
(e) none of the above (all the other options are true)
3. log1/5 (253 ) is equal to:
(a) 10
(b) 6
(c) 53
(d) 95/3
(e) none of the above
4. The modulus of 53 10i
is:
4i
p
(a)
5
(b) 25
(c) 11/5
(d) 5
(e) none of the above
5. If z = 2ei⇡/3 and w = 3e i⇡/12 then zw equals:
p
p
p
p
(a) 3 2(1 i)
(b) 3 + 3 3i
(c) 3 3 + 3i
(d) 3 2(1 + i)
(e) none of the above
6. If z = 3+i
then z/z is equal to
3 i
(a) 7+24i
(b)
25
(e) none of the above
End of Test
3 4i
3+4i
(c)
2 i
2+i
(d) 1
Solutions
a
b
c
d
e
⇥
1
⇥
2
⇥
3
⇥
4
⇥
5
⇥
6
Note: Your answers, if correct, will not have given the above pattern, because (as
a guard against cheating) there were several versions of the question paper, with the
possible answers arranged in various orders.
Pn
2
2
2
2
1. N.B. Let SnP=
the sum to be
k=0 (k) = 1 + 2 + . . . + n is seen by
Pnexpanding
P
n
2n
2
2
2
identical to k=1 (k) . By substituting l = 2k we have k=0 (k) =
l=0 (l/2) , but
notice that the number of terms in the sum has doubled so this is incorrect. By induction
(or otherwise) Sn = n6 (n + 1)(2n + 1). [Induction proof: Basis step (n=1) gives S1 = 1
and n6 (n + 1)(2n + 1) = 16 (2)(3) = 1. The induction step: Sn+1 = Sn + (n + 1)2 =
n
(n + 1)(2n + 1) + (n + 1)2 = (n + 1)( n6 (2n + 1) + (n + 1)) = (n + 1)( n(2n+1)+6(n+1)
)=
6
6
(n+1)
(n+1)
2
(2n + n + 6n + 6) = 6 (n + 2)(2n + 3). While, by substituting k + 1 = l we see
6 P
P
that nk= 1 1 (k + 1)2 = nl=0 l2 = S. The correct answer is option (a) above.
p
2. |(2 + 3i)(3 2i)| = (22 + 32 )(32 + 22 ) = 13 hence |(2 + 3i)(3 2i)| > 13 is false.
logb (x)
logb (a)
log5 (25)
3 log (1/5) = 3 21
5
3. log1/5 (253 ) = 3 log1/5 (25). One method is to change base using loga (x) =
x = 25 = 52 , hence choose b = 5 and a = 1/5, to give 3 log1/5 (25) =
where
=
6.
in the form a+ib, i.e. z = (5(3 10i)(3+4i)
= 15+20i2530i+40 =
4i)(3+4i)
p
p
55 10i
= 11 5 2i . Now |z| = (1/5) 121 + 4 = 5. So none of the options (a) to (d) are
25
correct, hence the correct answer is (e) none of the above.
p
p
p
p
5. zw = 6ei(⇡/3 ⇡/12) = 6ei(⇡/4) = 6(cos(⇡/4)+i sin(⇡/4)) = 6(1/ 2+i1/ 2) = 3 2+3i 2.
This option is (d) as listed above.
4. One method is to rewrite z =
6. z =
3+i
3 i
5 10i
3 4i
and so one method is to write z in the form of a + ib in order to take the complex
conjugate. Now z =
1
(7 + 24i).
25
(3+i)2
10
= 15 (4 + 3i) therefore z =
1
(4
10
3i). So z/z =
4+3i
4 3i
=
(4+3i)2
25
=
P. P. Cook —9th October, 2014