Question 1

1 mark

Exactly one of the following statements is true. Which one?

A. $\sqrt{144}=\pm 12$ .
B.For every real number $x$ , $\sqrt{x^{2}}=x$ .
C. $0.\overline{9}=1$ .
D. $3^{0}=0$ .
E.The product of any two irrational numbers is irrational.

Answer: C

Worked solution

The radical sign denotes the non-negative root, so

\sqrt{144}=12

(A false) and

\sqrt{x^{2}}=|x|

, which fails for

x<0

(B false). Also

3^{0}=1

(D false), and

\sqrt{2}\cdot\sqrt{2}=2

is rational (E false). The recurring decimal

0.\overline{9}

equals

1

exactly, so C is the true statement.

Question 2

1 mark

Let

n

be a positive integer. Consider statement P: '

n

is prime', and statement Q: '

n

is of the form

6k\pm 1

for some integer

k

'. Then Q is:

A.necessary but not sufficient for P.
B.sufficient but not necessary for P.
C.both necessary and sufficient for P.
D.neither necessary nor sufficient for P.

Answer: D

Worked solution

Q is not necessary for P, since

2

and

3

are prime yet not of the form

6k\pm 1

. Q is not sufficient for P either, since

25=6(4)+1

is of the form but is not prime. So Q is neither necessary nor sufficient. The tempting error is to recall 'primes are

6k\pm 1

' while forgetting it holds only for primes greater than

3

.

Question 3

1 mark

Consider the claim: 'For every integer

n\ge 2

, the value

n^{2}-n+11

is prime.' Which of the following is a counterexample?

A. $n=5$
B. $n=10$
C. $n=11$
D. $n=2$
E.there is no counterexample

Answer: C

Worked solution

A counterexample needs

n^{2}-n+11

composite. At

n=11

,

n^{2}-n+11=121-11+11=121=11^{2}

, which is composite. The others give primes (

n=5\to 31

,

n=10\to 101

,

n=2\to 13

), and the expression is prime for every

n

from

2

to

10

, which makes option E tempting.

Question 4

1 mark

Let

x

and

y

be real numbers. Consider statement P: '

x+y>0

', and statement Q: '

x^{3}+y^{3}>0

'. Then Q is:

A.necessary but not sufficient for P.
B.sufficient but not necessary for P.
C.necessary and sufficient for P.
D.neither necessary nor sufficient for P.

Answer: C

Worked solution

Factor

x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})

. The second factor

x^{2}-xy+y^{2}=\left(x-\frac{y}{2}\right)^{2}+\frac{3y^{2}}{4}

is positive except at

x=y=0

, where the whole product is

0

. Hence

x^{3}+y^{3}

has the same sign as

x+y

, so

x+y>0\iff x^{3}+y^{3}>0

: the conditions are equivalent.

Question 5

1 mark

f

is a function defined for all real numbers, with

f(1)=3

and

f(4)=-2

. Consider the statements: I:

f(x)=0

has a solution with

1<x<4

; II:

f(x)=0

has an odd number of solutions with

1<x<4

; III:

f

takes the value

1

somewhere in

1<x<4

. Which of these must be true?

A.none of them
B.I only
C.I and II only
D.I and III only
E.I, II and III

Answer: A

Worked solution

The intermediate value theorem requires continuity, but

f

is only stated to be defined for all real

x

. Without continuity

f

can jump straight from positive to negative values, taking neither

0

nor

1

in between, so none of the statements must hold. The trap is to apply the intermediate value theorem reflexively and conclude I (or II).

Question 6

1 mark

f

is a differentiable function. Consider statement P: '

x=2

is a turning point of

f

', and statement Q: '

f'(2)=0

'. Then Q is:

A.necessary but not sufficient for P.
B.sufficient but not necessary for P.
C.both necessary and sufficient for P.
D.neither necessary nor sufficient for P.

Answer: A

Worked solution

A turning point forces

f'(2)=0

, so Q is necessary for P. But Q is not sufficient:

f(x)=(x-2)^{3}

has

f'(2)=0

with a stationary point of inflection, not a turning point. So Q is necessary but not sufficient.

Question 7

1 mark

The following argument claims to show that every real number equals

0

. Line 1: Let

a=b

. Line 2: Then

a^{2}=ab

. Line 3: So

a^{2}-b^{2}=ab-b^{2}

. Line 4: Factorising,

(a-b)(a+b)=b(a-b)

. Line 5: Dividing by

a-b

,

a+b=b

. Line 6: Hence

a=0

. At which line does the first error occur?

A.Line 2
B.Line 3
C.Line 4
D.Line 5
E.Line 6

Answer: D

Worked solution

Lines 1 to 4 are all valid. The first error is at Line 5, which divides by

a-b

; but

a=b

makes

a-b=0

, and division by zero is not permitted. Every step after inherits this invalid move.

Question 8

1 mark

Consider the statement: 'If a function

g

is defined on

[0,1]

, then

\int_{0}^{1}g(x)\,dx

equals the area enclosed between the graph of

g

and the

x

-axis over

[0,1]

.' Which of the following is a counterexample?

A. $g(x)=x^{2}$
B. $g(x)=x-1$
C. $g(x)=\frac{1}{x}$
D. $g(x)=1$
E.there is no counterexample

Answer: B

Worked solution

A counterexample must satisfy the hypothesis (be defined on

[0,1]

) yet make the conclusion fail. For

g(x)=x-1

, which is

\le 0

on

[0,1]

, the integral is

-\frac{1}{2}

while the enclosed area is

+\frac{1}{2}

. Option C is not valid:

\frac{1}{x}

is undefined at

x=0

, so it fails the hypothesis and cannot serve as a counterexample.

Question 9

1 mark

Which of the following statements are true? I: There exists a real

y

such that for all real

x

,

x^{2}\ge y

. II: For all real

x

, there exists a real

y

such that

y>x^{2}

. III: There exists a real

y

such that for all real

x

,

y>x^{2}

.

A.none of them
B.I only
C.II only
D.III only
E.I and II only
F.I and III only
G.II and III only
H.I, II and III

Answer: E

Worked solution

For I, take

y=0

:

x^{2}\ge 0

for all

x

, true. For II, given any

x

take

y=x^{2}+1

, true. For III we would need a single

y

exceeding every

x^{2}

, but

x^{2}

is unbounded, so III is false. Only I and II hold; the order of the quantifiers is what separates II (true) from III (false).

Question 10

1 mark

How many real solutions with

x>0

does the equation

(\ln x)^{2}-3\ln x+1=0

have?

A. $0$
B. $1$
C. $2$
D. $3$
E.infinitely many

Answer: C

Worked solution

Substituting

t=\ln x

, which is a one-to-one map from

x>0

onto all real

t

, gives

t^{2}-3t+1=0

. Its discriminant is

9-4=5>0

, so there are two real values of

t

, each corresponding to exactly one

x>0

. Hence there are two solutions.

Question 11

1 mark

For a positive integer

n

, let

R_{n}

be the repunit consisting of

n

ones (so

R_{1}=1

,

R_{2}=11

,

R_{3}=111

, and so on). The statement '

3

divides

R_{n}

' is true if and only if:

A. $n$ is even
B. $3$ divides $n$
C. $9$ divides $n$
D. $n$ is odd
E. $n>3$

Answer: B

Worked solution

The digit sum of

R_{n}

is

n

, and an integer is divisible by

3

exactly when its digit sum is. So

3\mid R_{n}

if and only if

3\mid n

. Option C is too strong (divisibility by

3

, not

9

, is what the digit-sum test gives).

Question 12

1 mark

What is the maximum value of

f(x)=2x^{3}-9x^{2}+12x+1

on the closed interval

0\le x\le 3

?

A. $6$
B. $10$
C. $5$
D. $1$
E. $12$

Answer: B

Worked solution

Here

f'(x)=6(x-1)(x-2)

, giving a local maximum at

x=1

and a local minimum at

x=2

. Comparing all candidates,

f(0)=1

,

f(1)=6

,

f(2)=5

and

f(3)=10

, the greatest value on

[0,3]

is

f(3)=10

, at the endpoint. Option A (

6

) is the local maximum, which is not the global one here.

Question 13

1 mark

S(x)

is a property of a real number

x

, and exactly one of the following four claims about

S

is true. Which one? A:

S(x)

holds whenever

x^{2}<4

. B:

S(x)

holds precisely when

x^{2}<4

. C:

S(x)

holds whenever

x^{2}<9

. D:

S(x)

holds precisely when

x^{2}<9

.

A.A
B.B
C.C
D.D

Answer: A

Worked solution

Each of B, C and D forces A to hold as well. B (an 'if and only if') implies the corresponding 'if', which is A. And C says

x^{2}<9\Rightarrow

holds; since

x^{2}<4

implies

x^{2}<9

, C also gives

x^{2}<4\Rightarrow

holds, which is A; D implies C, hence A. So if exactly one is true it cannot be B, C or D (each would make A true too), leaving only A able to stand alone.

Question 14

1 mark

b

is a real number, and

x=b

is a root of the equation

bx^{2}-(b^{2}+1)x+b=0

. Which of the following describes the complete set of possible values of

b

?

A. $b=0$ or $b=1$
B. $b=\pm 1$
C.all real numbers
D. $b\ne 0$
E.there is no such $b$

Answer: C

Worked solution

Substituting

x=b

gives

b\cdot b^{2}-(b^{2}+1)b+b=b^{3}-b^{3}-b+b=0

, which holds identically. So

x=b

is a root for every real

b

, and the answer is all real numbers. The trap is to divide through by

b

, which silently discards the case

b=0

and yields a wrongly restricted set.

Question 15

1 mark

What is the maximum value of

x+2y

subject to

x^{2}+y^{2}\le 5

?

A. $5$
B. $\sqrt{5}$
C. $2\sqrt{5}$
D. $3$
E. $10$

Answer: A

Worked solution

On the disc of radius

\sqrt{5}

, the largest value of

x+2y

is

\sqrt{5}\cdot\sqrt{1^{2}+2^{2}}=\sqrt{5}\cdot\sqrt{5}=5

, attained where

(x,y)

points along

(1,2)

. Testing lattice points on the boundary instead gives smaller values and misses the maximum.

Question 16

1 mark

Consider the statement about an integer

n

: 'if

n^{2}

is even then

n

is even'. Which of the following is its contrapositive?

A.If $n$ is even then $n^{2}$ is even.
B.If $n$ is odd then $n^{2}$ is odd.
C.If $n^{2}$ is odd then $n$ is odd.
D.If $n$ is odd then $n^{2}$ is even.
E.If $n^{2}$ is even then $n$ is odd.

Answer: B

Worked solution

The contrapositive of 'if P then Q' is 'if not Q then not P', with both parts negated and the order reversed. Here P is '

n^{2}

is even' and Q is '

n

is even', so the contrapositive is 'if

n

is odd then

n^{2}

is odd'. Option C ('if

n^{2}

is odd then

n

is odd') is the inverse, not the contrapositive.

Question 17

1 mark

Which of the following statements are true? I: no integer of the form

4k+3

is the sum of two perfect squares. II: the product of two integers, each of the form

4k+1

, is again of the form

4k+1

. III: every prime number is of the form

4k+1

or

4k+3

.

A.none of them
B.I only
C.II only
D.III only
E.I and II only
F.I and III only
G.II and III only
H.I, II and III

Answer: E

Worked solution

A perfect square is

0

or

1

modulo

4

, so a sum of two squares is

0

,

1

or

2

modulo

4

, never

3

: I is true. Also

(4a+1)(4b+1)=4(4ab+a+b)+1

, so II is true. But III fails:

2

is prime and is

4(0)+2

, of neither form. Hence I and II only.

Question 18

1 mark

f

is a continuous function and

a<b

. Consider statement P: '

f(a)

and

f(b)

have opposite signs', and statement Q: '

f

has a root in the open interval

(a,b)

'. Then P is:

A.necessary but not sufficient for Q.
B.sufficient but not necessary for Q.
C.both necessary and sufficient for Q.
D.neither necessary nor sufficient for Q.

Answer: B

Worked solution

By the intermediate value theorem, opposite signs at the endpoints of a continuous function force a root, so P is sufficient for Q. But P is not necessary:

f(x)=(x-\frac{a+b}{2})^{2}

style curves, or a function crossing zero twice, can have a root in

(a,b)

while

f(a)

and

f(b)

share a sign. So P is sufficient but not necessary.

Question 19

1 mark

For which of these functions is it true that

f'(x)>0

for all real

x

? I:

f(x)=x^{3}

. II:

f(x)=e^{x}

. III:

f(x)=x+\sin x

.

A.none of them
B.I only
C.II only
D.III only
E.I and II only
F.II and III only
G.I, II and III
H.I and III only

Answer: C

Worked solution

For I,

f'(x)=3x^{2}

is zero at

x=0

, so not positive for all

x

. For III,

f'(x)=1+\cos x

is zero when

\cos x=-1

(for example

x=\pi

), so again not strictly positive everywhere. Only II has

f'(x)=e^{x}>0

for all real

x

. Both I and III are increasing overall, but their derivatives touch zero, which the strict inequality forbids.

Question 20

1 mark

A sequence of integers satisfies

u_{n+1}=3u_{n}-4

, with

u_{1}

an integer. Which of the following statements are true? I:

u_{n}-2

is a multiple of

3^{n-1}

for every

n

. II: if

u_{1}

is even then every term is even. III:

u_{10}

is divisible by

3

.

A.none of them
B.I only
C.II only
D.III only
E.I and II only
F.I and III only
G.II and III only
H.I, II and III

Answer: E

Worked solution

The fixed point is

2

, and

u_{n}-2=3^{n-1}(u_{1}-2)

, so I holds. If

u_{1}

is even then

u_{2}=3u_{1}-4

is even, and by induction every term is even, so II holds. For III,

u_{10}=2+3^{9}(u_{1}-2)\equiv 2\pmod 3

, so

u_{10}

is not divisible by

3

: III is false. Hence I and II only.

TMUA Mock Paper 2

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20