Question 1

1 mark

The first three terms of a geometric sequence are

\sqrt{2}

,

2

,

2\sqrt{2}

. What is the seventh term?

A. $8$
B. $8\sqrt{2}$
C. $16$
D. $16\sqrt{2}$
E. $32$

Answer: B

Worked solution

The common ratio is

r=\frac{2}{\sqrt{2}}=\sqrt{2}

and the first term is

\sqrt{2}

, so the

n

th term is

(\sqrt{2})^{n}

. The seventh term is

(\sqrt{2})^{7}=2^{7/2}=2^{3}\cdot 2^{1/2}=8\sqrt{2}

. Option A (

8

) comes from stopping at

(\sqrt{2})^{6}

, an off-by-one in the term count.

Question 2

1 mark

Evaluate

\displaystyle\int_{1}^{4}\frac{3x-2}{\sqrt{x}}\,dx

.

A. $6$
B. $8$
C. $10$
D. $12$
E. $14$

Answer: C

Worked solution

Split first:

\frac{3x-2}{\sqrt{x}}=3x^{1/2}-2x^{-1/2}

, which integrates to

2x^{3/2}-4x^{1/2}

. At

x=4

this is

2(8)-4(2)=8

; at

x=1

it is

2-4=-2

. The value is

8-(-2)=10

. Option B (

8

) forgets the lower-limit contribution.

Question 3

1 mark

The line

y=2x+c

does not intersect the curve

y=x^{2}-4x+7

. Find the set of values of

c

.

A. $c>-2$
B. $c<2$
C. $c>2$
D. $c<-2$
E. $-2<c<2$

Answer: D

Worked solution

Setting the two equal gives

x^{2}-6x+(7-c)=0

. No intersection means the discriminant is negative:

36-4(7-c)<0

, i.e.

8+4c<0

, so

c<-2

. Option A flips the inequality; the strict boundary is what separates

c<-2

from

c\le -2

.

Question 4

1 mark

Find the sum of all real solutions of

x-5\sqrt{x}+6=0

.

A. $5$
B. $13$
C. $6$
D. $25$
E. $90$

Answer: B

Worked solution

Let

t=\sqrt{x}\ge 0

, so

t^{2}-5t+6=0

, giving

t=2

or

t=3

; both are non-negative and valid, so

x=4

or

x=9

and the sum is

13

. Option A (

5

) sums the values of

t

rather than of

x

.

Question 5

1 mark

Find

\displaystyle\sum_{n=0}^{\infty}\frac{\cos(n\pi)}{3^{n}}

.

A. $\frac{3}{2}$
B. $\frac{2}{3}$
C. $\frac{3}{4}$
D. $\frac{1}{2}$
E. $\frac{9}{4}$

Answer: C

Worked solution

Since

\cos(n\pi)=(-1)^{n}

, the sum is geometric with first term

1

and ratio

-\frac{1}{3}

:

\frac{1}{1-(-1/3)}=\frac{1}{4/3}=\frac{3}{4}

. Option A (

\frac{3}{2}

) uses ratio

+\frac{1}{3}

, missing the alternating sign.

Question 6

1 mark

The circle

C

has equation

x^{2}+y^{2}-6x+8y+21=0

. Find the length of the tangent to

C

drawn from the origin.

A. $\sqrt{13}$
B. $\sqrt{5}$
C. $21$
D. $\sqrt{29}$
E. $\sqrt{21}$

Answer: E

Worked solution

Completing the square gives

(x-3)^{2}+(y+4)^{2}=4

, so the centre is

(3,-4)

and the radius is

2

. The distance from the origin to the centre is

\sqrt{9+16}=5

, so the tangent length is

\sqrt{5^{2}-2^{2}}=\sqrt{21}

. Option D (

\sqrt{29}

) adds the squares instead of subtracting.

Question 7

1 mark

Evaluate

\displaystyle\prod_{n=2}^{99}\left(1-\frac{1}{n^{2}}\right)

.

A. $\frac{50}{99}$
B. $\frac{1}{99}$
C. $\frac{100}{99}$
D. $\frac{1}{2}$
E. $\frac{25}{99}$

Answer: A

Worked solution

Write

1-\frac{1}{n^{2}}=\frac{(n-1)(n+1)}{n^{2}}

, so the product splits into

\left(\prod\frac{n-1}{n}\right)\left(\prod\frac{n+1}{n}\right)

. The first telescopes to

\frac{1}{99}

and the second to

\frac{100}{2}=50

, giving

\frac{50}{99}

. Option B keeps only the first telescoping product.

Question 8

1 mark

Find the area of the region enclosed between the graphs of

y=|x-1|

and

y=3-|x-1|

.

A. $9$
B. $2.25$
C. $4.5$
D. $3$
E. $6$

Answer: C

Worked solution

The graphs meet where

|x-1|=\frac{3}{2}

, at

x=-\tfrac12

and

x=\tfrac52

. The region is a kite whose diagonals are the vertical segment from

(1,0)

to

(1,3)

(length

3

) and the horizontal segment between the intersection points (length

3

). Its area is

\frac{1}{2}(3)(3)=4.5

. Option A (

9

) omits the factor

\frac{1}{2}

.

Question 9

1 mark

Find the maximum value of

f(x,y)=-2x^{2}-y^{2}+8x+6y-5

.

A. $8$
B. $12$
C. $9$
D. $3$
E. $20$

Answer: B

Worked solution

Complete the square in each variable:

-2(x-2)^{2}+8

and

-(y-3)^{2}+9

, so

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

. The maximum is

12

, attained at

(2,3)

. Options A and C are the separate partial contributions before combining.

Question 10

1 mark

Find the coefficient of

x^{5}

in the expansion of

(1+x)^{3}(1+2x)^{4}

.

A. $168$
B. $144$
C. $96$
D. $208$
E. $120$

Answer: A

Worked solution

The

x^{5}

term collects contributions

x^{i}

from

(1+x)^{3}

and

x^{j}

from

(1+2x)^{4}

with

i+j=5

:

\binom{3}{1}\binom{4}{4}2^{4}+\binom{3}{2}\binom{4}{3}2^{3}+\binom{3}{3}\binom{4}{2}2^{2}=48+96+24=168

. Dropping the

2^{j}

powers of

2

is the usual slip.

Question 11

1 mark

For how many integer values of

k

does the equation

x^{3}-3x^{2}=k

have three distinct real solutions?

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

Answer: D

Worked solution

Let

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

, so

f'(x)=3x(x-2)

with a local maximum

f(0)=0

and a local minimum

f(2)=-4

. Three distinct roots occur exactly when

-4<k<0

, giving integer values

k=-3,-2,-1

: three of them. Option B (

5

) wrongly includes the endpoints

k=-4

and

k=0

, where a root is repeated.

Question 12

1 mark

A quadratic curve has a minimum point at

(2,-3)

and passes through

(0,5)

. What is the coefficient of

x^{2}

?

A. $2$
B. $\frac{1}{2}$
C. $4$
D. $8$
E. $-2$

Answer: A

Worked solution

Write the curve in vertex form

y=a(x-2)^{2}-3

. Substituting

(0,5)

gives

4a-3=5

, so

a=2

. Option D (

8

) is the value of

4a

before dividing.

Question 13

1 mark

Given that

\log_{a}b=3

and

\log_{b}c=4

, find

\log_{a}c

.

A. $7$
B. $\frac{4}{3}$
C. $12$
D. $64$
E. $81$

Answer: C

Worked solution

By the change-of-base chain,

\log_{a}c=\log_{a}b\cdot\log_{b}c=3\times 4=12

. Option A (

7

) adds the logarithms instead of multiplying.

Question 14

1 mark

How many solutions does

2\sin^{2}x=\sin x+1

have in the interval

0\le x<2\pi

?

A. $4$
B. $2$
C. $5$
D. $3$
E. $1$

Answer: D

Worked solution

Rearrange to

2\sin^{2}x-\sin x-1=0

, which factors as

(2\sin x+1)(\sin x-1)=0

. Then

\sin x=1

gives one solution

x=\frac{\pi}{2}

, and

\sin x=-\frac{1}{2}

gives two,

x=\frac{7\pi}{6},\frac{11\pi}{6}

: three in total. Counting

\sin x=-\frac{1}{2}

as one solution gives the distractor

2

.

Question 15

1 mark

Let

F(n)=\displaystyle\int_{0}^{n}(2x-1)\,dx

for positive integers

n

. What is the largest

n

for which

F(n)<90

?

A. $9$
B. $10$
C. $8$
D. $11$
E. $45$

Answer: A

Worked solution

Here

F(n)=\big[x^{2}-x\big]_{0}^{n}=n^{2}-n

. Solving

n^{2}-n<90

gives

(n-10)(n+9)<0

, i.e.

-9<n<10

, so the largest positive integer is

9

. Option B (

10

) treats the inequality as non-strict, but

F(10)=90

is not less than

90

.

Question 16

1 mark

How many real solutions does

9^{x}-10\cdot 3^{x}+9=0

have?

A. $1$
B. $0$
C. $2$
D. $3$
E. $4$

Answer: C

Worked solution

Let

t=3^{x}>0

, so

t^{2}-10t+9=0

, giving

t=1

or

t=9

. Both are positive, so

x=0

and

x=2

: two solutions. Neither root is rejected, so the answer is not

1

.

Question 17

1 mark

The parabola

y=x^{2}

is rotated

180^{\circ}

about the origin and then translated by the vector

\binom{2}{0}

. What is the equation of the resulting curve?

A. $y=-x^{2}+4x-4$
B. $y=-x^{2}-4x-4$
C. $y=x^{2}-4x+4$
D. $y=-x^{2}+4x+4$
E. $y=-(x+2)^{2}$

Answer: A

Worked solution

A

180^{\circ}

rotation about the origin sends

(x,y)

to

(-x,-y)

, turning

y=x^{2}

into

y=-x^{2}

. Translating by

\binom{2}{0}

replaces

x

with

x-2

, giving

y=-(x-2)^{2}=-x^{2}+4x-4

. Option B translates the wrong way; option C forgets the rotation.

Question 18

1 mark

Evaluate

\cos^{2}1^{\circ}+\cos^{2}2^{\circ}+\cos^{2}3^{\circ}+\cdots+\cos^{2}89^{\circ}

.

A. $45$
B. $44.5$
C. $44$
D. $89$
E. $88.5$

Answer: B

Worked solution

Pairing the term for

k^{\circ}

with that for

(90-k)^{\circ}

gives

\cos^{2}k^{\circ}+\sin^{2}k^{\circ}=1

. The

89

terms form

44

such pairs plus the middle term

\cos^{2}45^{\circ}=\frac{1}{2}

, so the sum is

44+\frac{1}{2}=44.5

. Option A (

45

) forgets that the central term is only

\frac{1}{2}

.

Question 19

1 mark

The region between the curves

y=x

and

y=x^{3}

for

0\le x\le 1

is divided by the vertical line

x=a

into two parts of equal area. Which equation does

a

satisfy?

A. $2a^{4}-4a^{2}+1=0$
B. $2a^{4}-4a^{2}-1=0$
C. $a^{4}-2a^{2}+1=0$
D. $2a^{4}-4a^{2}+2=0$
E. $4a^{4}-2a^{2}+1=0$

Answer: A

Worked solution

On

[0,1]

the total area is

\int_{0}^{1}(x-x^{3})\,dx=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}

, so each part has area

\frac{1}{8}

. The left part is

\int_{0}^{a}(x-x^{3})\,dx=\frac{a^{2}}{2}-\frac{a^{4}}{4}=\frac{1}{8}

; multiplying by

8

gives

4a^{2}-2a^{4}=1

, i.e.

2a^{4}-4a^{2}+1=0

.

Question 20

1 mark

The line

L

is a tangent to the curve

y=x^{4}-4x^{3}+8x

at two distinct points. Find the

y

-intercept of

L

.

A. $4$
B. $8$
C. $-4$
D. $-8$
E. $-2$

Answer: C

Worked solution

If

L\colon y=mx+k

touches the quartic at two points, then

x^{4}-4x^{3}+8x-(mx+k)

must be a perfect square

(x^{2}+bx+c)^{2}

. Matching coefficients gives

2b=-4

so

b=-2

; then

b^{2}+2c=0

so

c=-2

; then

2bc=8-m

gives

m=0

; and

c^{2}=-k

gives

k=-4

. So

L

is

y=-4

, and indeed

x^{4}-4x^{3}+8x+4=(x^{2}-2x-2)^{2}

, touching at

x=1\pm\sqrt{3}

.

TMUA Mock Paper 1

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