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Diagnostic test solutions
(Unofficial) Solutions (updated 1st October, the other one had errors):
Q1. “Raise e (the natural base) to the power e. Raise the result to the power
e. Take the (natural) logarithm and raise the result to the power e. Take the
logarithm and raise 2 to the power of the result. Square the result and take
the logarithm. Divide by e and take e to the power of the result. (Write the
answer in simplest form.”
Answer: 16.
Step by Step:
“Raise e (the natural base) to the power e.”
ee
“Raise the result to the power e.”
ee e = e2
“Take the (natural) logarithm...”
log(e2 ) = 2e
“... and raise the result to the power of e”
e2e
“Take the logarithm..”
log(e2e ) = 2e
“and raise 2 to the power of the result.”
22e = 4e
“Square the result..”
42e = 16e
“... and take the logarithm”
log(16e ) = e log(16)
“Divide by e..”
e log(16)/e = log(16)
“... and take e to the power of the result”
elog(16) = 16
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Diagnostic test solutions
Q2. What’s (1 + i)20 .
Answer: (a) = −1024.

√ ! 20
2
2 
+
i
2
2
√

(1 + i)20
2
= √
2
√
1/2 20
= (2
)
×
√ !20
2
2
+
i
2
2
= (21/2 )20 × (eiπ/4 )20
= 1024 × (eiπ )5
= 1024 × −15
= −1024
Q3: What’s the minimum value of 4 cos4 θ − 2 cos 2θ + sin2 2θ + 5?
Answer: 7.
2 cos 2θ = 2 cos2 θ − 2 sin2 θ
= 2 cos2 θ − 2 + 2 cos2 θ
= 4 cos2 θ − 2
Also
sin2 2θ = 1 − cos2 2θ
= 1 − (cos2 θ − sin2 θ)2
= 1 − (2 cos2 θ − 1)2
= 1 − 4 cos4 θ + 1 − 4 cos2 θ
So the expression simplifies to
4 cos4 θ − 4 cos2 θ + 2 + 1 − 4 cos4 θ + 1 − 4 cos2 θ + 5 = 7.
So 7 is the minimum value.
Q4: For what value(s) of α does
x + αy = −1
αx + 4y = 2,
have more than one solution?
Answer: (e) α = −2.
Substituting x = −1 − αy, we find that
y=
2−α
α2 − 4
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Diagnostic test solutions
So we only get one solution when α2 6= 4, i.e. there might be more than one
solution if α2 = 4 =⇒ α = 2 or α = −2.
It suffices to check that when α = 2, y = 0, x = 1, which is inconsistent so no
solution. But for α = −2, x = 2y − 1 =⇒ infinite solutions.
If you knew some linear algebra you could also note that the system has a
solution if
α 4
det
= α2 − 4 6= 0
1 α
And then just check the cases α = 2, α = −2.
Q5: How many of the statements are true. x denotes a real number:
(a) 2 > 1 if x3 < 3.
(b) x = 3 only if x2 = 9.
(c) If x + sin(x2 ) > 2 then ex > 0.
(d) x2 < 3 if x2 < −1.
Answer: All 4 are true. This was the answered by far the worst iirc - you
might want to look at Tim Gower’s blog posts on basic logic (the posts come
up in reverse order).
For (a), 2 > 1 is true, so the statement is Anything =⇒ True, which is
always true. For (b), it’s x = 3 =⇒ x2 = 9 which is true (the reverse is false
though). For (c), ex > 0, so again like (a) it’s anything =⇒ True, which is
true. For (d), If x2 < −1 then obviously x2 < 3.
Q6: If f (x)/1/x, g(x) = 1 + x, h(x) = x − 1, then what is the sum of
h(f (g(f (x)))) and f (g(x))?
Answer: (a) 0.
Calculate
f (g(x)) =
1
1+x
h(f (g(f (x)))) = h(f (g(1/x)))
= h(f (1 + 1/x))
1
= h(
)
1 + 1/x
1
=
−1
1 + 1/x
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Diagnostic test solutions
So adding these
1
1
1
x
+
−1=
+
−1
1 + x 1 + 1/x
1+x 1+x
1+x
=
−1
1+x
= 0.
Q7: Not until your second year will you learn how to evaluate
Z ∞
cos2 x
dx
2
−∞ 1 + x
But one of these are correct:
(a) π, (b) 21 π(1 + 1/e2 ), (c) 0, (d) − 41 log(2), (e) infinity.
Answer: (b) 12 π(1 + 1/e2 ).
You should know that
Z
∞
−∞
i∞
1
−1
=
tan
(x)
1 + x2
−∞
π
π
= −−
2
2
=π
2
1
cos x
Now comparing the graphs of 1+x
2 and 1+x2 , it’s obvious it has to be (b),
because the integral is positive, not infinite and the area is less than the area
1
2
of 1+x
2 = π since cos x ranges from 0 to 1.
Q8 If ⊕ is associative : (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c) for all a, b, c, which of these
are associative.
Answer: (b) and (c) are associative. Just plug in x, y, z and check:
For (a): x ⊕ y = 2x + y. So (x ⊕ y) ⊕ z = (2x + y) ⊕ z = 4x + 2y + z. While
x ⊕ (y ⊕ z) = x ⊕ (2y + z) = 2x + 2y + z. So it’s not associative.
For (b): x ⊕ y = x. So (x ⊕ y) ⊕ z = x ⊕ z = x. Also x ⊕ (y ⊕ z) = x ⊕ y = x,
they’re equal so it’s associative.



if x ≥ 0
x
Consider (c): x ⊕ y = y
if x < 0 and y ≥ 0


x + y if x < 0 and y < 0
Case by case:
(i) If x ≥ 0, then (x ⊕ y) ⊕ z = x, while x ⊕ (y ⊕ z) = x ⊕ (something) = x.
(ii) If x < 0, y ≥ 0, or x < 0, y < 0, z ≥ 0, we get a similar thing to (i).
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Diagnostic test solutions
(iii) If x < 0, y < 0, z < 0 this is just addition, which better be associative!
Q9:If McCoy first switches with p or doesn’t switch with probability 1 − p,
what is the probability that the week 4 question is easy given the week 1
question was hard.
Answer: (a) p3 + 3p(1 − p)2 .
P (4H|1E) = P ( 3 switches) + P (1 switch, 2 same)
= p3 + p(1 − p)2 × 3
Q10: Label graphs of sin−1 (x), sin(x−1 ) and (sin x)−1 .
Answer: Blue - sin−1 x, Red - sin(x−1 ), Green - (sin x)−1 .
Lots of ways to get the result, e.g. (sin x)−1 → ∞ as x → 0, so the green is
(sin x)−1 . at x = 0, sin−1 (x) = 0, so the blue one is sin−1 (x).
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