ESAT Mock Maths 2: Questions & Worked Solutions

Question 1

1 mark

For

x > 0

, let

y

be defined by the expression:

y = \frac{(8x^6)^{\frac{1}{3}} \times (4x^2)^{\frac{3}{2}}}{2x^4}

Which one of the following is equal to

y

?

A. $2x$
B. $4x$
C. $8x$
D. $16x$
E. $32x$

Answer: C

Worked solution

To simplify the expression, we apply the laws of indices to each part of the numerator:
1.

(8x^6)^{\frac{1}{3}} = 8^{\frac{1}{3}} \times (x^6)^{\frac{1}{3}} = 2x^2

.
2.

(4x^2)^{\frac{3}{2}} = (4^{\frac{1}{2}})^3 \times (x^2)^{\frac{3}{2}} = 2^3 \times x^3 = 8x^3

.

Multiplying these parts together gives the numerator:

2x^2 \times 8x^3 = 16x^5

.

Finally, we divide by the denominator:

y = \frac{16x^5}{2x^4} = 8x^{5-4} = 8x

.

Therefore, the correct option is C.

Question 2

1 mark

For

x > 0

, which one of the following is equivalent to the expression below?

\frac{x\sqrt{x} + 4\sqrt{x}}{x^{-\frac{1}{2}}}

A. $x + 4$
B. $x^2 + 4$
C. $x + 4x$
D. $x^2 + 4x$
E. $x^2 + 4x^2$

Answer: D

Worked solution

First, we rewrite the terms in the numerator and denominator using rational exponents:
-

x\sqrt{x} = x^1 \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}}

-

4\sqrt{x} = 4x^{\frac{1}{2}}

The expression is:

\frac{x^{\frac{3}{2}} + 4x^{\frac{1}{2}}}{x^{-\frac{1}{2}}}

Dividing by

x^{-\frac{1}{2}}

is equivalent to multiplying by

(x^{-\frac{1}{2}})^{-1} = x^{\frac{1}{2}}

:

(x^{\frac{3}{2}} + 4x^{\frac{1}{2}}) \times x^{\frac{1}{2}}

Distributing the

x^{\frac{1}{2}}

:

x^{\frac{3}{2}} \cdot x^{\frac{1}{2}} + 4x^{\frac{1}{2}} \cdot x^{\frac{1}{2}}

= x^{\frac{3}{2} + \frac{1}{2}} + 4x^{\frac{1}{2} + \frac{1}{2}}

= x^2 + 4x^1 = x^2 + 4x

This matches option D.

Question 3

1 mark

The vertex of the parabola

y = x^2 + bx + c

lies on the line

y = x

. If the quadratic equation

x^2 + bx + c = 0

has two distinct real roots, which one of the following must be true?

A. $b > 0$
B. $b < 0$
C. $c > 0$
D. $c < 0$
E. $b^2 + 4c = 0$

Answer: A

Worked solution

The x-coordinate of the vertex of the parabola

y = x^2 + bx + c

is given by

x_v = -\frac{b}{2a} = -\frac{b}{2}

. The y-coordinate is

y_v = f(-\frac{b}{2}) = (-\frac{b}{2})^2 + b(-\frac{b}{2}) + c = \frac{b^2}{4} - \frac{b^2}{2} + c = c - \frac{b^2}{4}

. Since the vertex lies on the line

y = x

, we have

c - \frac{b^2}{4} = -\frac{b}{2}

. Multiplying by

-4

gives

b^2 - 4c = 2b

. The discriminant of the quadratic equation

x^2 + bx + c = 0

is

D = b^2 - 4c

. For the equation to have two distinct real roots, we require

D > 0

. Substituting the previous identity, we get

2b > 0

, which implies

b > 0

. Thus, option A is the only statement that must be true.

Question 4

1 mark

The quadratic function

f(x) = (c-1)x^2 + 4x + (c+2)

takes only positive values for all real values of

x

. What is the complete range of possible values for the constant

c

?

A. $c > 1$
B. $c > 2$
C. $c < -3$ or $c > 2$
D. $1 < c < 2$
E. $c > -3$

Answer: B

Worked solution

For a quadratic

f(x) = ax^2 + bx + c

to be positive for all real

x

, it must have no real roots (the graph does not touch or cross the x-axis) and it must open upwards. This implies two conditions: (1)

a > 0

and (2) the discriminant

D < 0

. Condition 1:

c - 1 > 0

, so

c > 1

. Condition 2:

D = 4^2 - 4(c-1)(c+2) < 0

. Dividing by 4:

4 - (c^2 + c - 2) < 0

, which simplifies to

4 - c^2 - c + 2 < 0

, or

c^2 + c - 6 > 0

. Factorising the inequality gives

(c+3)(c-2) > 0

. This inequality is satisfied when

c < -3

or

c > 2

. Combining this with the first condition (

c > 1

), the only valid range for

c

is

c > 2

.

Question 5

1 mark

The polynomial

f(x) = x^4 + kx^3 - x^2 + mx + 6

is exactly divisible by

x^2 + 2x - 3

, where

k

and

m

are constants.

What is the remainder when

f(x)

is divided by

(x + 1)

?

A. $-6$
B. $0$
C. $6$
D. $12$
E. $18$

Answer: D

Worked solution

If

f(x)

is divisible by

x^2 + 2x - 3

, it must be divisible by its factors.
Factorising

x^2 + 2x - 3

gives

(x + 3)(x - 1)

.
Using the Factor Theorem:
1)

f(1) = 0 \implies 1^4 + k(1)^3 - 1^2 + m(1) + 6 = 0 \implies 1 + k - 1 + m + 6 = 0 \implies k + m = -6

.
2)

f(-3) = 0 \implies (-3)^4 + k(-3)^3 - (-3)^2 + m(-3) + 6 = 0 \implies 81 - 27k - 9 - 3m + 6 = 0 \implies 78 - 27k - 3m = 0

.

Divide the second equation by 3:

26 - 9k - m = 0 \implies 9k + m = 26

.

Now solve the system:

k + m = -6

9k + m = 26

Subtract the first from the second:

(9k - k) = 26 - (-6) \implies 8k = 32 \implies k = 4

.
Substitute

k = 4

into

k + m = -6

:

4 + m = -6 \implies m = -10

.

The polynomial is

f(x) = x^4 + 4x^3 - x^2 - 10x + 6

.
To find the remainder when

f(x)

is divided by

(x + 1)

, we evaluate

f(-1)

:

f(-1) = (-1)^4 + 4(-1)^3 - (-1)^2 - 10(-1) + 6

f(-1) = 1 - 4 - 1 + 10 + 6 = 12

.

The remainder is 12.

Question 6

1 mark

Which of the following functions

f(x)

, defined for all real numbers

x

, is a one-to-one mapping?

A. $f(x) = (x - 2)^2$
B. $f(x) = |x + 5|$
C. $f(x) = 1 - x^3$
D. $f(x) = \cos x$
E. $f(x) = x^2 + 4$

Answer: C

Worked solution

A function is one-to-one if every value in the range is mapped to by exactly one value in the domain. We check each option:
A:

f(x) = (x - 2)^2

is a parabola with vertex at

(2, 0)

. It is many-to-one because, for example,

f(1) = 1

and

f(3) = 1

.
B:

f(x) = |x + 5|

is a V-shaped graph with a vertex at

x = -5

. It is many-to-one because, for example,

f(-4) = 1

and

f(-6) = 1

.
C:

f(x) = 1 - x^3

is a strictly decreasing function for all

x

. Since it is strictly monotonic, it never takes the same value twice, so it is one-to-one.
D:

f(x) = \cos x

is periodic. It is many-to-one because, for example,

\cos(0) = 1

and

\cos(2\pi) = 1

.
E:

f(x) = x^2 + 4

is a parabola. It is many-to-one because

f(1) = 5

and

f(-1) = 5

.
Only C is one-to-one.

Question 7

1 mark

The sequence

a_n

is defined by the formula

a_n = \frac{n}{(n+1)!}

for

n \ge 1

. Find the value of the sum

\sum_{k=1}^{10} a_k

.

A. $1 - \frac{1}{10!}$
B. $1 - \frac{1}{11!}$
C. $\frac{10}{11!}$
D. $1 + \frac{1}{11!}$
E. $1 - \frac{11}{12!}$

Answer: B

Worked solution

We can rewrite the expression for the

n^{th}

term to reveal a telescoping sum:

a_n = \frac{n}{(n+1)!} = \frac{(n+1) - 1}{(n+1)!} = \frac{n+1}{(n+1)!} - \frac{1}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}

.
Now we write out the summation:

\sum_{k=1}^{10} a_k = \left(\frac{1}{1!} - \frac{1}{2!}\right) + \left(\frac{1}{2!} - \frac{1}{3!}\right) + \dots + \left(\frac{1}{10!} - \frac{1}{11!}\right)

.
This is a telescoping series where almost all terms cancel out:

\sum_{k=1}^{10} a_k = \frac{1}{1!} - \frac{1}{11!} = 1 - \frac{1}{11!}

.

Question 8

1 mark

A sequence

x_n

is defined by

x_1 = 2.5

and the recurrence relation

x_{n+1} = x_n^2 - 2

for

n \ge 1

. What is the value of

x_6

?

A. $2^{16} + 2^{-16}$
B. $2^{32} + 2^{-32}$
C. $2^{32} - 2^{-32}$
D. $2^{64} + 2^{-64}$
E. $2.5^{32}$

Answer: B

Worked solution

Let's calculate the first few terms of the sequence:

x_1 = 2.5 = 2 + \frac{1}{2} = 2^1 + 2^{-1}

x_2 = (2 + 2^{-1})^2 - 2 = (2^2 + 2(2)(2^{-1}) + 2^{-2}) - 2 = (2^2 + 2 + 2^{-2}) - 2 = 2^2 + 2^{-2}

x_3 = (2^2 + 2^{-2})^2 - 2 = (2^4 + 2(2^2)(2^{-2}) + 2^{-4}) - 2 = (2^4 + 2 + 2^{-4}) - 2 = 2^4 + 2^{-4}

We can see a pattern emerging:

x_n = 2^{2^{n-1}} + 2^{-(2^{n-1})}

.
To verify for

n=4

:

x_4 = (2^4 + 2^{-4})^2 - 2 = 2^8 + 2 + 2^{-8} - 2 = 2^8 + 2^{-8}

.
Following this pattern, for

n=6

:

x_6 = 2^{2^{6-1}} + 2^{-(2^{6-1})} = 2^{2^5} + 2^{-(2^5)} = 2^{32} + 2^{-32}

.

Question 9

1 mark

The sum of the first

n

positive integers is given by

S_n = \sum_{i=1}^n i

. For a given positive integer

m

, let

n = 2m + 1

. If

S_n = k S_m

, which of the following is an expression for

k

in terms of

m

?

A. $k = \frac{2m+1}{m}$
B. $k = \frac{4m+2}{m}$
C. $k = \frac{4m+4}{m}$
D. $k = \frac{2m+2}{m}$
E. $k = \frac{m+1}{m}$

Answer: B

Worked solution

We are given the sum of the first

n

natural numbers as

S_n = \frac{n(n+1)}{2}

.
First, express

S_n

in terms of

m

using

n = 2m+1

:

S_n = S_{2m+1} = \frac{(2m+1)(2m+1+1)}{2} = \frac{(2m+1)(2m+2)}{2} = (2m+1)(m+1)

.
Next, express

S_m

:

S_m = \frac{m(m+1)}{2}

.
We are told

S_n = k S_m

, so we solve for

k

:

k = \frac{S_n}{S_m} = \frac{(2m+1)(m+1)}{\frac{m(m+1)}{2}}

.
Cancelling

(m+1)

(since

m

is a positive integer,

m+1 \neq 0

):

k = \frac{2(2m+1)}{m} = \frac{4m+2}{m}

.
This matches option B.

Question 10

1 mark

The first three terms in the expansion of

(1 + ax)^n

in ascending powers of

x

are

1

,

24x

, and

264x^2

, where

a

is a constant and

n

is a positive integer. Find the value of

n - a

.

A.8
B.10
C.12
D.14
E.22

Answer: B

Worked solution

The expansion of

(1 + ax)^n

is

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

We are given the coefficients of

x

and

x^2

:
1)

na = 24 \implies a = \frac{24}{n}

2)

\frac{n(n-1)}{2} a^2 = 264

Substitute

a = \frac{24}{n}

into the second equation:

\frac{n(n-1)}{2} \left(\frac{24}{n}\right)^2 = 264

\frac{n(n-1)}{2} \cdot \frac{576}{n^2} = 264

\frac{n-1}{n} \cdot 288 = 264

Divide both sides by 24:

\frac{n-1}{n} \cdot 12 = 11

12n - 12 = 11n \implies n = 12

Now find

a

using

na = 24

:

12a = 24 \implies a = 2

The value required is

n - a = 12 - 2 = 10

.

Question 11

1 mark

The line

L

has the equation

(k+2)x + (2k-1)y = 5k+5

, where

k

is a real constant. It can be shown that all such lines pass through a fixed point

P

. Find the equation of the straight line that passes through

P

and is perpendicular to the line

2x + 5y + 7 = 0

.

A. $5x - 2y - 13 = 0$
B. $5x - 2y + 17 = 0$
C. $2x + 5y - 11 = 0$
D. $2x - 5y - 1 = 0$
E. $5x + 2y - 17 = 0$

Answer: A

Worked solution

First, we find the coordinates of the fixed point

P

. We can expand the equation of

L

and group the terms involving

k

:

(k+2)x + (2k-1)y = 5k+5

kx + 2x + 2ky - y = 5k + 5

k(x + 2y - 5) + (2x - y - 5) = 0

For this to hold for all

k

, we must have

x + 2y - 5 = 0

and

2x - y - 5 = 0

. Solving these simultaneously:
From the second equation,

y = 2x - 5

. Substituting into the first:

x + 2(2x - 5) - 5 = 0 \implies x + 4x - 10 - 5 = 0 \implies 5x = 15 \implies x = 3

.
Then

y = 2(3) - 5 = 1

. Thus,

P

is

(3, 1)

.
The given line

2x + 5y + 7 = 0

can be rewritten as

5y = -2x - 7

, which has a gradient of

m_1 = -\frac{2}{5}

.
The gradient of a line perpendicular to this is

m_2 = -\frac{1}{m_1} = \frac{5}{2}

.
The equation of the line through

P(3, 1)

with gradient

\frac{5}{2}

is:

y - 1 = \frac{5}{2}(x - 3)

2(y - 1) = 5(x - 3)

2y - 2 = 5x - 15

5x - 2y - 13 = 0

.
This matches option A.

Question 12

1 mark

A rectangle has two of its sides on the parallel lines

3x - 4y + 6 = 0

and

3x - 4y - 9 = 0

. A third side of the rectangle passes through the point

(1, 2)

. If the area of the rectangle is

15

square units, which of the following could be the equation of the fourth side?

A. $4x + 3y - 35 = 0$
B. $4x + 3y - 15 = 0$
C. $3x - 4y - 10 = 0$
D. $4x + 3y + 25 = 0$
E. $3x + 4y - 25 = 0$

Answer: A

Worked solution

First, find the distance

h

between the two parallel sides

3x - 4y + 6 = 0

and

3x - 4y - 9 = 0

.
The distance between

ax + by + c_1 = 0

and

ax + by + c_2 = 0

is

d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}

.

h = \frac{|6 - (-9)|}{\sqrt{3^2 + (-4)^2}} = \frac{15}{5} = 3

.
Since the area is

15

, the width

w

of the rectangle must be

w = \frac{15}{h} = \frac{15}{3} = 5

.
The third and fourth sides must be perpendicular to the first two sides. The gradient of the original lines is

m_1 = \frac{3}{4}

, so the perpendicular gradient is

m_2 = -\frac{4}{3}

.
The equation of the third side passing through

(1, 2)

is:

y - 2 = -\frac{4}{3}(x - 1) \implies 3y - 6 = -4x + 4 \implies 4x + 3y - 10 = 0

.
The fourth side is parallel to the third side, so its equation is

4x + 3y + C = 0

.
The distance between these two sides must be

w = 5

:

\frac{|C - (-10)|}{\sqrt{4^2 + 3^2}} = 5 \implies \frac{|C + 10|}{5} = 5 \implies |C + 10| = 25

.
This gives

C + 10 = 25 \implies C = 15

, or

C + 10 = -25 \implies C = -35

.
Thus, the possible equations are

4x + 3y + 15 = 0

and

4x + 3y - 35 = 0

.
Comparing with the options, only A is a match.

Question 13

1 mark

The points

P(1, 2)

and

Q(5, 6)

are the endpoints of a diameter of a circle. What is the area of this circle?

A. $4\pi$
B. $8\pi$
C. $16\pi$
D. $32\pi$
E. $64\pi$

Answer: B

Worked solution

First, we find the length of the diameter

PQ

using the distance formula:

d = \sqrt{(5 - 1)^2 + (6 - 2)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}

.
The radius

r

of the circle is half of the diameter:

r = \frac{4\sqrt{2}}{2} = 2\sqrt{2}

.
The area of the circle is given by

A = \pi r^2

. Substituting our value for

r

:

A = \pi (2\sqrt{2})^2 = \pi (4 \times 2) = 8\pi

.
Therefore, the correct option is B.

Question 14

1 mark

A circle with centre

(3, -4)

is tangent to the

x

-axis. Which of the following is the equation of this circle?

A. $x^2 + y^2 - 6x + 8y + 9 = 0$
B. $x^2 + y^2 - 6x + 8y + 16 = 0$
C. $x^2 + y^2 + 6x - 8y + 9 = 0$
D. $x^2 + y^2 - 6x + 8y + 25 = 0$
E. $x^2 + y^2 - 6x + 8y = 0$

Answer: A

Worked solution

The circle is tangent to the

x

-axis, which means the distance from the centre

(3, -4)

to the line

y = 0

is the radius. The distance is

|-4| = 4

, so

r = 4

and

r^2 = 16

.
The standard equation of a circle with centre

(a, b)

and radius

r

is

(x - a)^2 + (y - b)^2 = r^2

. Substituting the values:

(x - 3)^2 + (y - (-4))^2 = 16

(x - 3)^2 + (y + 4)^2 = 16

Expanding the squares:

x^2 - 6x + 9 + y^2 + 8y + 16 = 16

x^2 + y^2 - 6x + 8y + 9 = 0

.
This matches option A.

Question 15

1 mark

The circle

C

has the equation

x^2 + y^2 - 4x + 6y - 12 = 0

. What is the equation of the tangent to

C

at the point

(5, 1)

?

A. $4x - 3y = 17$
B. $3x - 4y = 11$
C. $3x + 4y = 19$
D. $4x + 3y = 23$
E. $3x + 4y = 25$

Answer: C

Worked solution

First, find the centre of the circle by completing the square for

x

and

y

:

x^2 - 4x = (x - 2)^2 - 4

and

y^2 + 6y = (y + 3)^2 - 9

. Substituting these into the circle equation gives

(x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0

, so

(x - 2)^2 + (y + 3)^2 = 25

. The centre is at

(2, -3)

. At the point of tangency

(5, 1)

, the radius from the centre has a gradient of

m_{\text{radius}} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}

. Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent is

m_{\text{tangent}} = -\frac{3}{4}

. Using the point-slope form

y - y_1 = m(x - x_1)

with

(5, 1)

, we get

y - 1 = -\frac{3}{4}(x - 5)

. Multiplying by 4 gives

4y - 4 = -3x + 15

, which simplifies to

3x + 4y = 19

.

Question 16

1 mark

A circle has the equation

(x - 1)^2 + (y + 2)^2 = 100

. A horizontal chord of this circle lies on the line

y = 6

. What is the length of this chord?

A.12
B.16
C.6
D.8
E.20

Answer: A

Worked solution

The circle has centre

(1, -2)

and radius

r = 10

. The chord lies on the line

y = 6

. The distance

d

from the centre to the line

y = 6

is

|6 - (-2)| = 8

. The perpendicular from the centre to the chord bisects the chord. This forms a right-angled triangle where the radius (10) is the hypotenuse, the distance from the centre to the line (8) is one leg, and half the chord length (

x

) is the other leg. By Pythagoras:

x^2 + 8^2 = 10^2 \implies x^2 + 64 = 100 \implies x^2 = 36

, so

x = 6

. The total length of the chord is

2x = 12

.

Question 17

1 mark

A pyramid has a square base

PQRS

in the

xy

-plane with vertices at

P(0,0,0)

,

Q(4,0,0)

,

R(4,4,0)

, and

S(0,4,0)

. The vertex of the pyramid is at

V(2,2,6)

. Let

\theta

be the angle

\angle VQP

. Find

\cos \theta

.

0. $\frac{2\sqrt{11}}{11}$
A. $\frac{1}{11}$
B. $\frac{\sqrt{11}}{11}$
C. $\frac{\sqrt{11}}{22}$
E. $\frac{1}{2}$

Answer: B

Worked solution

We consider the triangle

VQP

. To find

\cos \theta

where

\theta

is the angle at vertex

Q

, we need the lengths of the three sides

VQ

,

QP

, and

VP

.

1. The length

PQ

is the distance between

(0,0,0)

and

(4,0,0)

, which is

4

.
2. The length

VQ

is the distance between

(4,0,0)

and

(2,2,6)

:

VQ = \sqrt{(2-4)^2 + (2-0)^2 + (6-0)^2} = \sqrt{(-2)^2 + 2^2 + 6^2} = \sqrt{4 + 4 + 36} = \sqrt{44} = 2\sqrt{11}

.
3. The length

VP

is the distance between

(0,0,0)

and

(2,2,6)

:

VP = \sqrt{(2-0)^2 + (2-0)^2 + (6-0)^2} = \sqrt{2^2 + 2^2 + 6^2} = \sqrt{4 + 4 + 36} = \sqrt{44} = 2\sqrt{11}

.

In triangle

VQP

, we have side lengths

VQ = 2\sqrt{11}

,

PQ = 4

, and

VP = 2\sqrt{11}

. We use the cosine rule for the angle at

Q

:

VP^2 = VQ^2 + PQ^2 - 2(VQ)(PQ)\cos \theta

44 = 44 + 4^2 - 2(2\sqrt{11})(4)\cos \theta

44 = 44 + 16 - 16\sqrt{11}\cos \theta

0 = 16 - 16\sqrt{11}\cos \theta

16\sqrt{11}\cos \theta = 16

\cos \theta = \frac{16}{16\sqrt{11}} = \frac{1}{\sqrt{11}} = \frac{\sqrt{11}}{11}

.

Question 18

1 mark

A triangle

ABC

has side length

AB = \sqrt{2}

and angle

\angle ABC = 135^{\circ}

. Given that the area of the triangle is

Z

, which of the following is an expression for the length of side

AC

?

A. $\sqrt{2Z^2 + 2Z + 2}$
B. $\sqrt{4Z^2 + 2}$
C. $\sqrt{4Z^2 - 4Z + 2}$
D. $\sqrt{4Z^2 + 4Z + 2}$
E. $2Z + \sqrt{2}$

Answer: D

Worked solution

Let the side length

BC = x

.
The area of triangle

ABC

is given by

Z = \frac{1}{2}(AB)(BC)\sin(\angle ABC)

.

Z = \frac{1}{2}(\sqrt{2})(x)\sin(135^{\circ})

.
Since

\sin(135^{\circ}) = \sin(45^{\circ}) = \frac{\sqrt{2}}{2}

:

Z = \frac{1}{2} \cdot \sqrt{2} \cdot x \cdot \frac{\sqrt{2}}{2} = \frac{2}{4}x = \frac{1}{2}x

.
Thus,

x = 2Z

.

Now, we use the cosine rule to find side

AC

(the side opposite the

135^{\circ}

angle):

AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)

AC^2 = (\sqrt{2})^2 + (2Z)^2 - 2(\sqrt{2})(2Z)\cos(135^{\circ})

Since

\cos(135^{\circ}) = -\cos(45^{\circ}) = -\frac{\sqrt{2}}{2}

:

AC^2 = 2 + 4Z^2 - 4\sqrt{2}Z \left(-\frac{\sqrt{2}}{2}\right)

AC^2 = 2 + 4Z^2 + \frac{4 \cdot 2}{2}Z

AC^2 = 4Z^2 + 4Z + 2

AC = \sqrt{4Z^2 + 4Z + 2}

.

Question 19

1 mark

A sector of a circle is formed using a piece of wire of length

L

. The area of the sector is

A = rac{3L^2}{50}

. Which one of the following is a possible value for the angle of the sector in radians?

A.2
B.2.5
C.3
D.3.5
E.4

Answer: C

Worked solution

Let the radius of the sector be

r

and the angle subtended at the centre be

heta

radians. The perimeter of the sector is given by the length of the wire

L

, so

L = 2r + r heta = r(2 + heta)

. The area of the sector is given by

A = rac{1}{2}r^2 heta

. Substituting the given relationship

A = rac{3L^2}{50}

, we have:

\frac{1}{2}r^2 heta = \frac{3}{50}(r(2+ heta))^2

\frac{1}{2}r^2 heta = \frac{3}{50}r^2(2+ heta)^2

Assuming

r \neq 0

, we can divide by

r^2

:

\frac{1}{2}\theta = \frac{3}{50}(4 + 4\theta + \theta^2)

Multiplying by 50 to clear the fraction:

25\theta = 3(4 + 4\theta + \theta^2)

25\theta = 12 + 12\theta + 3\theta^2

Rearranging into a quadratic equation:

3\theta^2 - 13\theta + 12 = 0

Factoring the quadratic:

(3\theta - 4)(\theta - 3) = 0

The possible values for

heta

are

\theta = \frac{4}{3}

and

\theta = 3

. Among the given options, only 3 is present.

Question 20

1 mark

Two sectors,

S_1

and

S_2

, are defined within the same circle of radius

r

. The angle subtended at the centre by

S_1

is

\alpha

radians and the angle subtended by

S_2

is

\beta

radians. Given that the area of

S_1

is three times the area of

S_2

, and the perimeter of

S_1

is exactly twice the perimeter of

S_2

, what is the value of

\alpha

?

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

Answer: E

Worked solution

Let

A_1, P_1

be the area and perimeter of

S_1

, and

A_2, P_2

be the area and perimeter of

S_2

. Using the formulas for area and perimeter in radians:

A_1 = \frac{1}{2}r^2\alpha

and

A_2 = \frac{1}{2}r^2\beta

P_1 = 2r + r\alpha

and

P_2 = 2r + r\beta

From

A_1 = 3A_2

, we have:

\frac{1}{2}r^2\alpha = 3 \left( \frac{1}{2}r^2\beta \right) \implies \alpha = 3\beta

From

P_1 = 2P_2

, we have:

2r + r\alpha = 2(2r + r\beta)

Dividing by

r

(since

r > 0

):

2 + \alpha = 4 + 2\beta

Substitute

\alpha = 3\beta

into this equation:

2 + 3\beta = 4 + 2\beta

\beta = 2

Now, calculate

\alpha

:

\alpha = 3(2) = 6

The value of

\alpha

is 6 radians.

Question 21

1 mark

The graph of

y = 3^x

is reflected in the line

y = 9

to produce the graph of

y = f(x)

. The graph of

y = f(x)

intersects the graph of

y = 3^{x+2} - 9

at the point

P(p, q)

. What is the value of

3^p

?

A.0.9
B.1.8
C.2.7
D.4.5
E.9.0

Answer: C

Worked solution

To reflect a function

y = g(x)

in a horizontal line

y = L

, we use the transformation

y' = 2L - g(x)

. Here,

g(x) = 3^x

and

L = 9

, so the equation for the reflected graph is

f(x) = 2(9) - 3^x = 18 - 3^x

.

We find the intersection of

y = 18 - 3^x

and

y = 3^{x+2} - 9

by setting them equal:

18 - 3^x = 3^{x+2} - 9

18 - 3^x = 3^2 \cdot 3^x - 9

18 - 3^x = 9 \cdot 3^x - 9

Rearranging to group the

3^x

terms:

18 + 9 = 9 \cdot 3^x + 3^x

27 = 10 \cdot 3^x

3^x = \frac{27}{10} = 2.7

Since

P(p, q)

is the point of intersection,

p

is the

x

-coordinate, so

3^p = 2.7

.

Question 22

1 mark

Consider the equation

4^x - (k+1)2^x + k = 0

, where

k

is a real constant. For which set of values of

k

does this equation have exactly one real solution for

x

?

A. $k = 1$ only
B. $k \le 0$ only
C. $k \le 1$
D. $k \le 0$ or $k = 1$
E. $k \le 0$ or $k \ge 1$

Answer: D

Worked solution

Let

u = 2^x

. Since

x

is real,

u

must be a positive real number (

u > 0

). Substituting this into the equation gives a quadratic in

u

:

u^2 - (k+1)u + k = 0

This quadratic factorises as

(u - k)(u - 1) = 0

. The roots are

u = 1

and

u = k

.

For each root

u

, there is a corresponding real solution for

x

if and only if

u > 0

:
1. The root

u = 1

gives

2^x = 1

, so

x = 0

. This is always one real solution regardless of

k

.
2. The root

u = k

gives a real solution for

x

if and only if

k > 0

, in which case

x = \log_2 k

.

To have exactly one real solution for

x

, we consider the cases:
- If

k = 1

, the roots are identical (

u = 1

), resulting in only one value for

x

(

x = 0

).
- If

k \le 0

, the root

u = k

does not produce a real value for

x

(since

2^x

cannot be zero or negative), so the only solution is

x = 0

from the first root.
- If

k > 0

and

k \neq 1

, there are two distinct positive roots for

u

, leading to two distinct real solutions for

x

.

Therefore, there is exactly one solution when

k = 1

or

k \le 0

.

Question 23

1 mark

What is the complete set of real values of

x

that satisfy the equation

2\log_2 x - \log_2(x+4) = 1

?

A. $x = 4$ only
B. $x = 2$ only
C. $x = 4$ or $x = -2$
D. $x = 2$ or $x = -4$
E. $x = 4$ or $x = 2$

Answer: A

Worked solution

We start by using the power law and the quotient law of logarithms. The equation is:

2\log_2 x - \log_2(x+4) = 1

Using

k\log_a x = \log_a(x^k)

, we have:

\log_2(x^2) - \log_2(x+4) = 1

Using

\log_a A - \log_a B = \log_a(\frac{A}{B})

, we have:

\log_2\left(\frac{x^2}{x+4}\right) = 1

From the definition

b = \log_a c \Leftrightarrow a^b = c

, we can write:

\frac{x^2}{x+4} = 2^1 = 2

Rearranging the equation gives a quadratic:

x^2 = 2(x + 4)

x^2 - 2x - 8 = 0

Factorising the quadratic gives:

(x - 4)(x + 2) = 0

This yields potential solutions

x = 4

and

x = -2

. However, we must check the domain of the original logarithmic expressions. In the expression

2\log_2 x

, the argument

x

must be strictly positive (

x > 0

). In the expression

\log_2(x+4)

, we require

x+4 > 0

, or

x > -4

. The intersection of these conditions is

x > 0

. Thus,

x = -2

is an invalid solution. The only valid solution is

x = 4

.

Question 24

1 mark

The sum of the real roots of the equation

3^{2x+1} + 3 = 10 \cdot 3^x

is

A.-1
B.0
C.1
D. $\frac{10}{3}$
E. $\log_3 10 - 1$

Answer: B

Worked solution

We can rewrite the equation as a quadratic in terms of

3^x

. Let

u = 3^x

. Then

3^{2x+1} = 3^1 \cdot (3^x)^2 = 3u^2

.
The equation becomes:

3u^2 + 3 = 10u

3u^2 - 10u + 3 = 0

Factorising the quadratic:

(3u - 1)(u - 3) = 0

So the solutions for

u

are

u = \frac{1}{3}

and

u = 3

.
Now we solve for

x

:
Case 1:

3^x = \frac{1}{3} = 3^{-1} \implies x = -1

Case 2:

3^x = 3 = 3^1 \implies x = 1

Both solutions are real numbers. The sum of these roots is:

x_1 + x_2 = -1 + 1 = 0

Alternatively, using properties of roots for the quadratic in

u

, the product of the roots

u_1 u_2 = \frac{c}{a} = \frac{3}{3} = 1

. Since

u_1 u_2 = 3^{x_1} \cdot 3^{x_2} = 3^{x_1+x_2}

, we have

3^{x_1+x_2} = 1

, which implies

x_1 + x_2 = 0

.

Question 25

1 mark

The function

f

is defined for

x \neq 0

by

f(x) = \frac{(x - k)^2}{x}

where

k

is a non-zero constant. The tangent to the graph

y = f(x)

at

x = 2

is parallel to the line

4y - 3x = 12

. What is the value of

f''(2)

?

A. $\frac{1}{16}$
B. $\frac{1}{8}$
C. $\frac{1}{4}$
D. $\frac{1}{2}$
E. $1$

Answer: C

Worked solution

First, we rewrite

f(x)

for easier differentiation:

f(x) = \frac{x^2 - 2kx + k^2}{x} = x - 2k + k^2x^{-1}

Then, we find the first derivative:

f'(x) = 1 - k^2x^{-2} = 1 - \frac{k^2}{x^2}

The line

4y - 3x = 12

can be written as

y = \frac{3}{4}x + 3

, which has a gradient of

\frac{3}{4}

.
Since the tangent at

x = 2

is parallel to this line,

f'(2) = \frac{3}{4}

:

1 - \frac{k^2}{2^2} = \frac{3}{4} \Rightarrow 1 - \frac{k^2}{4} = \frac{3}{4} \Rightarrow \frac{k^2}{4} = \frac{1}{4} \Rightarrow k^2 = 1

Now we find the second derivative:

f''(x) = \frac{d}{dx}(1 - k^2x^{-2}) = 2k^2x^{-3} = \frac{2k^2}{x^3}

Substituting

x = 2

and

k^2 = 1

:

f''(2) = \frac{2(1)}{2^3} = \frac{2}{8} = \frac{1}{4}

.

Question 26

1 mark

A continuous function

f(x)

is defined on the interval

1 \le x \le 5

. Let

A = \int_1^5 |f(x)| \,dx

represent the total area enclosed between the curve

y = f(x)

and the

x

-axis, and let

I = \int_1^5 f(x) \,dx

represent the definite integral over the same interval.

Consider the following three statements:
I.

A \ge |I|

II. If

f(x) = 0

for at least one value

x = c

where

1 < c < 5

, then

A > I

.
III. If

f(1) < 0

and

f(5) > 0

, then

A > I

.

Which of the above statements must be true?

A.I only
B.III only
C.1 and 2 only
D.1 and 3 only
E.1, 2 and 3

Answer: D

Worked solution

Statement I:

A = \int_1^5 |f(x)| dx

and

I = \int_1^5 f(x) dx

. By the integral triangle inequality,

\int |f| \ge |\int f|

, so

A \ge |I|

is always true for any integrable function. Thus, Statement I must be true.

Statement II: This is not necessarily true. A function can have a root without crossing the

x

-axis. For example, if

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

, the function has a root at

x=3

, but

f(x) \ge 0

for all

x

. In this case,

|f(x)| = f(x)

, so

A = I

. Therefore,

A > I

is not guaranteed.

Statement III: If

f(1) < 0

and

f(5) > 0

, then because

f(x)

is continuous, the Intermediate Value Theorem guarantees that there is at least one point where the curve crosses the

x

-axis. This implies there exists an interval within

(1, 5)

where

f(x) < 0

. In any region where

f(x) < 0

, the integrand for

A

is

|f(x)| = -f(x)

, which is strictly greater than the integrand for

I

, which is

f(x)

. Since

|f(x)| \ge f(x)

everywhere and

|f(x)| > f(x)

on at least one sub-interval, the integral

A

must be strictly greater than

I

. Thus, Statement III must be true.

The correct statements are I and III only.

Question 27

1 mark

The function

f

is continuous for all

x

and satisfies the equation:

\int_{2}^{x} f(t) \, dt = x^2 + ax + 2

where

a

is a constant. What is the value of

f(5)

?

A.4
B.7
C.10
D.12
E.13

Answer: B

Worked solution

To find the value of the constant

a

, we first evaluate the given equation at the lower limit of the integral,

x = 2

. Since the integral of any continuous function from

c

to

c

is zero (

\int_{c}^{c} f(t) \, dt = 0

), we can write:

0 = 2^2 + a(2) + 2

0 = 4 + 2a + 2

2a = -6 \implies a = -3

Now, according to the Fundamental Theorem of Calculus,

\frac{d}{dx} \int_{c}^{x} f(t) \, dt = f(x)

. By differentiating both sides of the original equation with respect to

x

, we obtain an expression for

f(x)

:

f(x) = \frac{d}{dx}(x^2 + ax + 2)

f(x) = 2x + a

Substituting the value

a = -3

into this derivative:

f(x) = 2x - 3

Finally, we evaluate the function at

x = 5

:

f(5) = 2(5) - 3 = 10 - 3 = 7

Thus, the correct option is B.