Question 1

1 mark

It is given that the expansion of

(ax + b)^3

is

8x^3 – px^2 + 18x – 3\sqrt{3}

, where

a

,

b

and

p

are real constants.

What is the value of

p

?

A. $-12\sqrt{3}$
B. $-6\sqrt{3}$
C. $-4\sqrt{3}$
D. $-\sqrt{3}$
E. $\sqrt{3}$
F. $4\sqrt{3}$
G. $6\sqrt{3}$
H. $12\sqrt{3}$

Answer: H

Worked solution

We expand

(ax + b)^3

and compare with the given expansion.

(ax + b)^3 = a^3x^3 + 3a^2bx^2 + 3ab^2x + b^3 = 8x^3 - px^2 + 18x - 3\sqrt{3}.

So we must have:

a^3 = 8

3a^2b = -p

3ab^2 = 18

b^3 = -3\sqrt{3}

The final equation can be rewritten as

b^3 = -3^{\frac{1}{2}}

, so

b = -3^{\frac{1}{2}} = -\sqrt{3}.

Then the third equation becomes

3a \times 3 = 18

, so

a = 2.

Finally, the second equation gives

3 \times 2^2 \times (-\sqrt{3}) = -p

, so

p = 12\sqrt{3}

and the answer is H.

Question 2

1 mark

The expression

3x^3 + 13x^2 + 8x + a

, where

a

is a constant, has

(x + 2)

as a factor.

Which one of the following is a complete factorisation of the expression?

A. $(x + 2)(x − 1)(3x – 2)$
B. $(x + 2)(x + 1)(3x - 2)$
C. $(x + 2)(x + 1)(3x + 2)$
D. $(x + 2)(x − 3)(3x + 2)$
E. $(x + 2)(x + 3)(3x - 2)$
F. $(x + 2)(x + 3)(3x + 2)$

Answer: E

Worked solution

We can use the factor theorem to find the value of

a

; alternatively we can just divide by

(x+2)

and deduce

a

that way, as we will need to divide by

(x+2)

anyway to complete the factorisation. We show both methods.

The factor theorem states that in this case,

f(-2) = 0

, where

f(x)

is the polynomial. So

3(-2)^3 + 13(-2)^2 + 8(-2) + a = 0

, giving

-24 + 52 - 16 + a = 0

, so

a = -12

. We can now divide the polynomial by

(x + 2)

.

Alternatively, dividing the given polynomial by

(x + 2)

by considering the

x^3, x^2

and

x

terms, but ignoring the constant term, gives:

3x^3 + 13x^2 + 8x + a = (x + 2)(3x^2 + 7x - 6)

and so

a = -12

, though that is actually not needed to answer the question.

Finally, we can factorise

3x^2 + 7x - 6

to determine the answer. Alternatively, we can use the suggested options: only options D and E give a constant term in the quadratic of

-6

, so it must be one of those. D expands to

3x^2 - 7x - 6

and E expands to

3x^2 + 7x - 6

, so the answer is E.

Question 3

1 mark

A line is drawn normal to the curve

y = \frac{2}{x^2}

at the point on the curve where

x = 1

.

This line cuts the x-axis at

P

and the y-axis at

Q

.

The length of

PQ

is

A. $\frac{3\sqrt{5}}{2}$
B. $\frac{3\sqrt{17}}{4}$
C. $\frac{7\sqrt{17}}{4}$
D. $\frac{35}{4}$
E. $\frac{35\sqrt{5}}{2}$
F. $\frac{3\sqrt{17}}{2}$

Answer: C

Worked solution

We need to find the gradient of the curve at

(1,2)

. We have

y = 2x^{-2}

and so

\frac{dy}{dx} = -4x^{-3}.

At

x=1

, this gives

\frac{dy}{dx} = -4

and so the normal has gradient

\frac{1}{4}.

As it passes through

(1,2)

, it has equation

y - 2 = \frac{1}{4}(x - 1).

It cuts the x-axis

(y = 0)

at

P

, so at

P

, we have

-2 = \frac{1}{4}(x - 1)

, giving

x= -7

, and so

P

has coordinates

(-7,0)

.
It cuts the y-axis

(x = 0)

at

Q

, so at

Q

, we have

y - 2 = \frac{1}{4}(-1)

, giving

y = \frac{7}{4}

, and so

Q

has coordinates

(0,\frac{7}{4}).

Thus

PQ

has length

\sqrt{7^2 + (\frac{7}{4})^2} = \frac{7}{4}\sqrt{4^2 + 1} = \frac{7\sqrt{17}}{4}

(where we have written

7 = \frac{7}{4} \times 4

for the first equality).
Hence the answer is C.

Question 4

1 mark

The sequence

a_n

is defined by the rule:

a_n = (-1)^n – (-1)^{n-1} + (-1)^{n+2}

for

n \geq 1

.

Find the value of

\sum_{n=1}^{39} a_n

A. $-39$
B. $-3$
C. $-1$
D. $0$
E. $1$
F. $3$
G. $39$

Answer: B

Worked solution

Commentary: This looks quite scary! It is unlikely that you have ever seen a sequence looking like this, so a sensible thing to do is to work out the first few values and look for any patterns.
We calculate the first few terms of the sequence:

a_1 = (-1)^1 - (-1)^0 + (-1)^3 = (-1) - 1 + (-1) = -3

a_2 = (-1)^2 - (-1)^1 + (-1)^4 = 1 - (-1) + 1 = 3

a_3 = (-1)^3 - (-1)^2 + (-1)^5 = (-1) - 1 + (-1) = -3

a_4 = (-1)^4 - (-1)^3 + (-1)^6 = 1 - (-1) + 1 = 3

The pattern is now clear (and we could prove it if we wished to): the sequence goes

-3, 3, -3, 3, \dots -3, 3

and so on. So the sum of each pair of terms is zero:

a_1 + a_2 = 0

,

a_3 + a_4 = 0

,

\dots

,

a_{37} + a_{38} = 0.

Thus

\sum_{n=1}^{39} a_n = (a_1 + a_2) + (a_3 + a_4) + \dots + (a_{37} + a_{38}) + a_{39} = 0+0+\dots+0+(-3) = -3

and the answer is B.

Question 5

1 mark

What is the total area enclosed between the curve

y = x^2 – 1

, the x-axis and
the lines

x = -2

and

x = 2

?

A. $\frac{4}{3}$
B. $\frac{8}{3}$
C. $4$
D. $\frac{16}{3}$
E. $12$
F. $16$

Answer: C

Worked solution

To answer this question, it is worth drawing a sketch.
The graph is of

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

, so the parabola intersects the x-axis at

(1,0)

and

(-1,0)

:

To find the area enclosed, we integrate over the three separate regions, from

-2

to

-1

, from

-1

to

1

, and from

1

to

2

:

\int_{-2}^{-1} x^2 - 1 dx = [\frac{1}{3}x^3-x]_{-2}^{-1} = (-\frac{1}{3}+1) - (-\frac{8}{3}+2) = \frac{4}{3}

\int_{-1}^{1} x^2 - 1 dx = [\frac{1}{3}x^3-x]_{-1}^{1} = (\frac{1}{3}-1) - (-\frac{1}{3}+1) = -\frac{4}{3}

\int_{1}^{2} x^2 - 1 dx = [\frac{1}{3}x^3-x]_{1}^{2} = (\frac{8}{3}-2) - (\frac{1}{3}-1) = \frac{4}{3}

Thus the three areas are each

\frac{4}{3}

, and the total area is

4

, so the answer is C.

Question 6

1 mark

P, Q, and R are each mixtures of red and white paint.
The percentage by volume of red paint in P is 30%.
The percentage by volume of red paint in Q is 20%.
The mixtures P, Q, and R are combined in the proportion 12 : 5 : 3 respectively.
If the resulting mixture contains 25% by volume of red paint, what percentage by volume of mixture R is red paint?

A. $25\%$
B. $23\%$
C. $13\frac{1}{2}\%$
D. $19\frac{1}{2}\%$
E. $9\frac{3}{4}\%$
F.It is impossible to achieve this result.

Answer: C

Worked solution

We can either work in fractions or percentages. This solution works in fractions.
The fraction of red paint in P is

\frac{3}{10}

and in Q is

\frac{1}{5}.

Let the fraction of red paint in R be

x

.
The combination is in the ratio

12 : 5 : 3

, with

\frac{1}{4}

red paint, so the composition is as follows:

We can now sum the individual red amounts, giving:

12\times\frac{3}{10}+5\times\frac{1}{5}+ 3x = 20 \times \frac{1}{4}

so

\frac{23}{10} + 3x = 5

, giving

3x = \frac{7}{10}

, so

x = 13 \frac{1}{3}\%.

(Note: The arithmetic in the original solution for

23/10

and

7/10

is inconsistent with

12 \times 3/10 + 5 \times 1/5 = 4.6

. However, the final percentage

13 \frac{1}{3}\%

is correct for

x = 2/15

, which satisfies

4.6 + 3x = 5

.)
The correct answer is C.

Question 7

1 mark

60% of a sports club's members are women and the remainder are men.

This sports club offers the opportunity to play tennis or cricket. Every member plays
exactly one of the two sports.

\frac{2}{5}

of the male members of the club play cricket;

\frac{2}{3}

of the cricketing members of the club are women.

What is the probability that a member of the club, chosen at random, is a woman who
plays tennis?

A. $\frac{1}{5}$
B. $\frac{7}{25}$
C. $\frac{1}{3}$
D. $\frac{11}{25}$
E. $\frac{3}{5}$

Answer: B

Worked solution

There are two factors for each member: whether the member is a man or woman, and whether the member plays tennis or cricket. We can either display this situation as a tree diagram or as a two-way table. Since the information given has both the fraction of male members who play cricket and the fraction of the cricketing members who are women, it will be simpler to use a two-way table. (In a tree diagram, we could easily display one of the two, but the other will be hard.)

We could also work with fractions or with numbers of people. In this solution, we will work with numbers. The

60\%

suggests that we work with a multiple of 100 people, and the presence of a fraction

\frac{1}{3}

, suggests that we work with a multiple of

3

. So we start with 300 people. This is the information we obtain from the

60\%

of members are women and

\frac{1}{3}

of the resulting 120 male members play cricket:

We are also told that

\frac{1}{3}

of the cricketing members are women, so the 48 cricketing men are

\frac{2}{3}

of the cricketers, hence 96 cricketers are women, leaving 84 women to play tennis:

(The parenthesised figures are not needed to finish the question.)
Therefore the probability that a member of the club, chosen at random, is a woman who plays tennis is

\frac{84}{300} = \frac{28}{100} = \frac{7}{25}

, and the answer is B.

Question 8

1 mark

Find the maximum angle

x

in the range

0^\circ \leq x \leq 360^\circ

which satisfies the equation

\cos^2(2x) + \sqrt{3}\sin(2x) - \frac{7}{4} = 0

A. $30^\circ$
B. $60^\circ$
C. $120^\circ$
D. $150^\circ$
E. $210^\circ$
F. $240^\circ$
G. $300^\circ$
H. $330^\circ$

Answer: F

Worked solution

We note that the only angle involved is

2x

, and

0^\circ \le x \le 360^\circ

gives

0^\circ \le 2x < 720^\circ

.
We start by writing everything in terms of

\sin 2x

, giving:

(1 - \sin^2 2x) + \sqrt{3} \sin 2x - \frac{7}{4} = 0

which rearranges to give

\sin^2 2x - \sqrt{3}\sin 2x + \frac{3}{4} = 0.

We can apply the quadratic formula to this to obtain

\sin 2x = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4(1)(\frac{3}{4})}}{2} = \frac{\sqrt{3} \pm \sqrt{3-3}}{2} = \frac{\sqrt{3}}{2}

and hence the possible values of

2x

in the range are

2x = 60^\circ, 120^\circ, 420^\circ \text{ and } 480^\circ

. Thus the largest possible value of

x

in the range

0^\circ \le x \le 360^\circ

is

240^\circ

, and the answer is F.

Question 9

1 mark

The line segment joining the points

(3, 3)

and

(7, 5)

is a diameter of a circle.

This circle is translated by 3 units in the negative x-direction, then reflected in the x-axis,
and then enlarged by a scale factor of 4 about the centre of the resulting circle.

The equation of the final circle is

A. $(x - 2)^2 + (y – 4)^2 = 320$
B. $(x – 2)^2 + (y + 4)^2 = 320$
C. $(x - 2)^2 + (y – 4)^2 = 80$
D. $(x – 2)^2 + (y + 4)^2 = 80$
E. $(x – 2)^2 + (y – 4)^2 = 20$
F. $(x - 2)^2 + (y + 4)^2 = 20$

Answer: D

Worked solution

The initial circle looks like this:

The centre of the circle is at the midpoint of this diameter, which is

(\frac{3+7}{2}, \frac{3+5}{2}) = (5,4)

,
and the radius is therefore

\sqrt{(5-3)^2 + (4-3)^2} = \sqrt{5}.

The transformations do the following to the centre and radius of the circle:

Therefore the final circle has equation

(x - 2)^2 + (y + 4)^2 = (4\sqrt{5})^2 = 80

and the answer is D.

Question 10

1 mark

How many solutions does the equation

x \tan x = 1

have in the interval

-2\pi \leq x \leq 2\pi

?

A. $0$
B. $1$
C. $2$
D. $3$
E. $4$
F. $5$
G. $6$

Answer: E

Worked solution

We can rearrange this equation as

\tan x = \frac{1}{x}

. (We note that

x = 0

is not a solution of the original equation

x \tan x = 1

, so we can divide by

x

without losing any solutions.)

It is impossible to solve this equation exactly, but we are only asked for the number of solutions. Therefore we sketch the graphs of

y = \tan x

and

y = \frac{1}{x}

on the same axes, and count the number of intersections.

There are four points of intersection with

-2\pi \le x \le 2\pi

, so the answer is 4, which is option E.

Question 11

1 mark

The real roots of the equation

4^{2x} + 12 = 2^{2x+3}

are

p

and

q

, where

p > q

.

The value of

p – q

can be expressed as

A. $\frac{3}{4}$
B. $1$
C. $4$
D. $-\frac{1}{2} + \log_{10} \frac{3}{2}$
E. $\frac{\log_{10} 3}{\log_{10} 4}$
F. $\frac{\log_{10} 3}{\log_{10} 2}$

Answer: E

Worked solution

This is a quadratic in

2^{2x}

, so we write

y = 2^{2x}

and solve for

y

. The quadratic becomes

y^2 + 12 = 8y

as

4^{2x} = (2^2)^{2x} = 2^{4x} = (2^{2x})^2

and

2^{2x+3} = 2^{2x} \times 2^3

.
This rearranges to

y^2 - 8y + 12 = 0

, which factorises as

(y - 6)(y - 2) = 0

, so

y = 6

or

y = 2

.
Thus

2^{2x} = 6

or

2^{2x} = 2

, with the value of

x

being larger in the first than in the second, so

2^{2p} = 6

and

2^{2q} = 2

. The options are given in terms of log to base 10, so we take

\log_{10}

of the first equation. Thus

2p\log_{10} 2 = \log_{10} 6

and

2q = 1

, and hence

p=\frac{\log_{10} 6}{2\log_{10} 2}

and

q = \frac{1}{2}

Therefore

p-q=\frac{\log_{10} 6}{2\log_{10} 2} - \frac{1}{2}

= \frac{\log_{10} 6 - \log_{10} 2}{2 \log_{10} 2}

= \frac{\log_{10} 3}{\log_{10} 4}

so the answer is E.

Question 12

1 mark

A right circular cylinder is contained within a sphere of radius 5 cm in such a way that the
whole of the circumferences of both ends of the cylinder are in contact with the sphere.

The diagram shows a planar cross section through the centre of the sphere and cylinder.

[diagram not to scale]

Find, in cubic centimetres, the maximum possible volume of the cylinder.

A. $250\pi$
B. $500\pi$
C. $1000\pi$
D. $\frac{250\sqrt{3}}{3}\pi$
E. $\frac{500\sqrt{3}}{9}\pi$
F. $\frac{1000\sqrt{3}}{9}\pi$

Answer: E

Worked solution

Let the radius of the cylinder be

r

. Then the diagram is as follows (with all measurements in cm), where we have drawn in two extra lines:

The 5cm is the radius of the sphere and hence of the circle shown (as the cross section is through the centre of the sphere).

Pythagoras's theorem gives

h^2 + r^2 = 5^2

, and the volume of the cylinder is

V = \pi r^2(2h) = \pi(5^2 - h^2)(2h) = 2\pi(25h - h^3).

We can maximise this by differentiating with respect to

h

. (We could write everything in terms of

r

instead, but that would involve square roots.)

We have

\frac{dV}{dh} = 2\pi(25 - 3h^2)

, which is zero when

25 = 3h^2

, so

h = \frac{5}{\sqrt{3}}

. Substituting this into the formula for

V

gives the largest possible

V

as

V = 2\pi(5^2 - h^2)h = 2\pi \left( 25 - \frac{25}{3} \right) \frac{5}{\sqrt{3}} = 2\pi \left( \frac{50}{3} \right) \frac{5}{\sqrt{3}} = \frac{500\pi}{3\sqrt{3}} = \frac{500\sqrt{3}}{9}\pi

hence the answer is E.

Question 13

1 mark

How many real roots does the equation

3x^5 – 10x^3 – 120x + 30 = 0

have?

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

Answer: C

Worked solution

Commentary: This is not something that you will have learnt how to do, and in general, this is a very difficult problem indeed. But we can use the tools at our disposal to do something helpful. We don't know how to solve a quintic (

x^5

) equation, but we can at least try to find the turning (stationary) points, and see where that gets us.

We attempt to sketch the graph of

y = 3x^5 - 10x^3 - 120x + 30

, or at least think about what a sketch might look like.

We calculate

\frac{dy}{dx} = 15x^4 - 30x^2 - 120 = 15(x^4 - 2x^2 - 8) = 15(x^2 - 4)(x^2 + 2)

so there are stationary points when

\frac{dy}{dx} = 0

, that is, when

x = 2

and when

x = -2

. We can now calculate the y-coordinates: when

x = -2

,

y = -36 + 80 +240 + 30 > 0

(we do not need to calculate the exact value), and when

x = 2

,

y = 36 - 80 - 240 + 30 < 0

.
So the graph must look something like the following:

There are clearly 3 points of intersection of this graph with

y = 0

, so the original equation has 3 real roots, and the answer is C.

Question 14

1 mark

The terms of an infinite series

S

are formed by adding together the corresponding terms in
two infinite geometric series,

T

and

U

.

The first term of

T

and the first term of

U

are each 4.

In order, the first three terms of the combined series

S

are

8

,

3

, and

\frac{5}{4}

.

What is the sum to infinity of

S

?

A. $\frac{32}{5}$
B. $\frac{20}{3}$
C. $\frac{64}{5}$
D. $\frac{40}{3}$
E. $16$
F. $32$

Answer: D

Worked solution

Let us write out the first few terms of the three sequences explicitly. Let the common ratio of T be

R

and the common ratio of U be

r

.

We are given the first three terms of S, so

8=8

4R+4r=3

4R^2 + 4r^2 = \frac{5}{4}

Therefore

R + r = \frac{3}{4}

and

R^2 + r^2 = \frac{5}{16}.

We can work out the values of

R

and

r

by guessing, or we can work them out by solving these equations. We show the second approach here.

The first equation gives

R = \frac{3}{4} - r

, which we can substitute into the second equation to give

(\frac{3}{4}-r)^2 + r^2 = \frac{5}{16}

so

2r^2 - \frac{3}{2}r + \frac{1}{4} = 0

, hence

8r^2 - 6r + 1 = 0

, which factorises as

(4r - 1)(2r - 1) = 0

, giving

r=\frac{1}{4}

or

r = \frac{1}{2}

and

R = \frac{1}{2}

or

R = \frac{1}{4}

respectively.
We can therefore work out the sums to infinity of T and U, taking

R = \frac{1}{2}

and

r = \frac{1}{4}

:

\text{sum to infinity of T} = \frac{4}{1 - \frac{1}{2}} = 8

\text{sum to infinity of U} = \frac{4}{1 - \frac{1}{4}} = \frac{16}{3}

giving the sum to infinity of

S

as

\frac{16}{3} + 8 = \frac{40}{3}

, since S is the sum of T and U. If we took the values of R and r to be the other way round, we would have the series T and U swapped, but their sum would be unchanged.
So the answer is D.

Question 15

1 mark

The least possible value of the gradient of the curve

y = (2x + a)(x – 2a)^2

at the point
where

x = 1

, as

a

varies, is

A. $-\frac{49}{4}$
B. $-8$
C. $-\frac{25}{4}$
D. $-\frac{7}{4}$
E. $-\frac{47}{16}$

Answer: C

Worked solution

We start by expanding the brackets to get

y = (2x + a)(x^2 - 4ax + 4a^2) = 2x^3 - 7ax^2 + 4a^2x + 8a^3

so the derivative is given by

\frac{dy}{dx} = 6x^2 - 14ax + 4a^2.

Therefore at

x = 1

, the gradient is

6 - 14a + 4a^2

. To find the least possible value of this as

a

varies, we can either differentiate this expression with respect to

a

or complete the square.
The latter approach gives

4a^2 - 14a + 6 = 4(a^2 - \frac{7}{2}a) + 6 = 4\left(\left(a-\frac{7}{4}\right)^2 - \left(\frac{7}{4}\right)^2\right) + 6 = 4\left(a-\frac{7}{4}\right)^2 - \frac{49}{4} + 6 = 4\left(a-\frac{7}{4}\right)^2 - \frac{25}{4}

so the minimum value is

-\frac{25}{4}

and the answer is C.

Question 16

1 mark

Given the simultaneous equations

\log_{10} 2 + \log_{10}(y - 1) = 2 \log_{10} x

\log_{10}(y + 3 - 3x) = 0

the values of

y

are

A. $\frac{5}{2} \pm \frac{3\sqrt{5}}{2}$
B. $3\pm\sqrt{3}$
C. $7\pm 3\sqrt{3}$
D. $3,9$
E. $1, 13$

Answer: C

Worked solution

The first equation involves the sum of two logarithms, so we can rewrite it as:

\log_{10}(2(y - 1)) = \log_{10} x^2

and so

2(y - 1) = x^2

.
The second equation becomes

y + 3 - 3x = 1

, as

\log_a 1 = 0

for any base

a

.
We therefore have two simultaneous equations:

2y - 2 = x^2

y - 3x + 2 = 0.

We want to find the values of

y

, so we rewrite the second equation to eliminate

x

:

x = \frac{y+2}{3}

and so the first equation becomes

2y - 2 = \frac{(y+2)^2}{9}

or

18y - 18 = y^2 + 4y + 4

which rearranges to give the quadratic

y^2 - 14y + 22 = 0

. The solutions to this quadratic are

\frac{14 \pm \sqrt{14^2 - 4 \times 22}}{2} = 7 \pm \sqrt{7^2 - 22} = 7 \pm \sqrt{27}

and so the correct answer is C.
It would be good to check whether both of these solutions are valid (though the question does not require us to do so). To be valid, we require both

x > 0

and

y > 1

. Since

\sqrt{27} < 6

, we see that

y > 1

. Also, as

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

and

y > 0

, it follows that

x > 0

. So there are, indeed, two valid solutions.

Question 17

1 mark

It is given that

y = (1 + 2 \cos x) \cos 2x

for

0 < x < \pi

The complete set of values of

x

for which

y

is negative is

A. $0<x<\frac{2\pi}{3}$ or $\frac{3\pi}{4}<x<\pi$
B. $0<x<\frac{3\pi}{4}$ or $\frac{3\pi}{4}<x<\pi$
C. $0<x<\frac{2\pi}{3}$ or $\frac{3\pi}{4}<x<\pi$
D. $\frac{\pi}{4}<x<\frac{2\pi}{3}$ or $\frac{3\pi}{4}<x<\pi$
E. $\frac{\pi}{4}<x<\frac{2\pi}{3}$
F. $\frac{\pi}{4}<x<\frac{3\pi}{4}$

Answer: D

Worked solution

The formula for

y

is the product of two factors, namely

1 + 2 \cos x

and

\cos 2x.

The whole expression is negative when one of the two factors is positive and the other is negative. The factors change sign when they cross a point where the factor is zero, and we can find these points:

•

1+2 \cos x = 0

when

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

, which is when

x = \frac{2\pi}{3}

(within the range

0 < x < \pi

).$

\n•

\cos 2x = 0

when

x=\frac{\pi}{4}

and when

x = \frac{3\pi}{4}.

We can now make a table showing the signs of the two factors in the different parts of the interval

0 < x < \pi

. (This is a useful technique in general.)

Therefore

y

is negative when

\frac{\pi}{4} < x < \frac{2\pi}{3}

and when

\frac{3\pi}{4} < x < \pi

, and so the answer is D.

Question 18

1 mark

The function

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

is defined for all

x \neq 0

.

The complete set of values of

x

for which the function is decreasing is

A. $x \leq −2, x > 0$
B. $-2 < x < 0$
C. $x \leq 1, x\neq 0$
D. $x \geq 1$
E. $-2 \leq x \leq 1, x \neq 0$
F. $x \leq -2, x \geq 1$

Answer: A

Worked solution

This requires us to first differentiate the function. We therefore write

y = \frac{1-x}{\sqrt[3]{x^2}} = x^{-\frac{2}{3}} - x^{\frac{1}{3}}

which we can differentiate to get

\frac{dy}{dx} = -\frac{2}{3}x^{-\frac{5}{3}} - \frac{1}{3}x^{-\frac{2}{3}}

This is zero when

-\frac{2}{3}x^{-\frac{5}{3}} - \frac{1}{3}x^{-\frac{2}{3}} = 0

,
so multiplying by

3x^{\frac{5}{3}}

to clear fractions in both the coefficients and in the powers gives

-2-x=0

so

x = -2

. (We could also have obtained this by writing

-\frac{2}{3}x^{-\frac{5}{3}} = \frac{1}{3}x^{-\frac{2}{3}}

and dividing one side by the other.)
We next need to determine the sign of

\frac{dy}{dx}

in each region. We note that the function is not defined at

x = 0

, so we have to deal with

x < 0

and

x > 0

separately. It is also not clear how to find the sign of

\frac{dy}{dx}

directly from its given form, so we first factorise it, giving

\frac{dy}{dx} = -\frac{1}{3}x^{-\frac{5}{3}}(2 + x).

(Incidentally, this gives yet another way to see that the derivative is zero at

x = -2

.) We can now work out the signs of the two factors, and hence of the derivative, in the various ranges:

It is therefore increasing in the region

x < -2

and

x > 0

. (It it not increasing at

x = -2

, but rather it is stationary at that point.) The correct answer is therefore A (though the question mistakenly says

x \le -2

).

Question 19

1 mark

The coefficient of

x^3

in the expansion of

(1 + 2x + 3x^2)^6

is equal to twice the coefficient
of

x^4

in the expansion of

(1 – ax^2)^5

.

Find all possible values of the constant

a

.

A. $\pm 2\sqrt{2}$
B. $\pm \sqrt{17}$
C. $\pm \sqrt{34}$
D. $\pm 2\sqrt{17}$
E.There are no possible values of $a$ .

Answer: B

Worked solution

We expand the first expression up to powers of

x^3

to find its coefficient. We can think of the first expression as

(1 + (2x + 3x^2))^6

and use the binomial theorem to expand it. We note that the first few binomial coefficients are

\binom{6}{0} = 1; \binom{6}{1} = 6; \binom{6}{2} = \frac{6 \times 5}{2!} = 15; \binom{6}{3} = \frac{6 \times 5 \times 4}{3!} = 20.

We thus have

(1 + (2x + 3x^2))^6 = 1 + \binom{6}{1}(2x + 3x^2) + \binom{6}{2}(2x + 3x^2)^2 + \binom{6}{3}(2x + 3x^2)^3 + \dots

= 1 + 6(2x + 3x^2) + 15((2x)^2 + 2(2x)(3x^2) + \dots) + 20((2x)^3 + \dots) + \dots

where we have stopped when the powers reach 3. We can read off the coefficient of

x^3

from this; it is

15 \times 2 \times 2 \times 3 + 20 \times 8 = 340.

(Note: This is interpreted as

15 \times (2 \times (2 \times 3)) + 20 \times 8 = 15 \times 12 + 160 = 180 + 160 = 340

)

We can expand

(1 - ax^2)^5

similarly, and obtain

1+ \binom{5}{1} (-ax^2) + \binom{5}{2} (-ax^2)^2 + \dots = 1 - 5ax^2 + 10a^2x^4 + \dots

so the coefficient of

x^4

is

10a^2

.
We are told how these two coefficients relate to each other: we have

340 = 2 \times 10a^2

so

a^2 = 17

and

a = \pm\sqrt{17}

, giving the answer as option B.

Question 20

1 mark

The diagram shows a square-based pyramid with base

PQRS

and vertex

O

. All the edges
of the pyramid are of length 20 metres.

[diagram not to scale]

Find the shortest distance, in metres, along the outer surface of the pyramid from

P

to the
midpoint of

OR

.

A. $10\sqrt{5}-2\sqrt{3}$
B. $10\sqrt{3}$
C. $10\sqrt{5}$
D. $10\sqrt{7}$
E. $10\sqrt{5}+ 2\sqrt{3}$

Answer: D

Worked solution

Let us label the midpoint of

OR

as

T

, so that we can refer to it.
We first observe that since all the edges of the pyramid are of length 20m, the triangles are all equilateral.
There are two ways to get from

P

to the midpoint of

OR

in a short distance: one can either travel along the base to

RS

and then along the triangular face to

T

, or one could travel along the triangular face

OPS

to the edge

OS

and from there to

T

. It is not obvious which route is shorter, so we will calculate both. (There are also routes which are mirror-reflections of these going via the edge

QR

or

OQ

.)
To find these distances, we can "unfold" the tetrahedron, that is, make a net for the tetrahedron.
For the first route, we draw just the faces

PQRS

and

ORS

:

The height of the triangle OSR is

20 \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}

, so

T

lies

5\sqrt{3}

above the line

SR

;

T

also lies

\frac{3}{4} \times 20 = 15

to the right of

S

(as it is at the midpoint of OR). We can therefore find the length

PT

using Pythagoras's theorem:

PT^2 = 15^2 + (20+5\sqrt{3})^2 = 225 + 400 + 200\sqrt{3} + 75 = 700 + 200\sqrt{3}

We now look at the other route, which we can do by drawing the triangles

OPS

and

OSR

:

Using the height of the triangle as calculated above,

P

lies

10\sqrt{3}

to the left of

OS

and

T

lies

5\sqrt{3}

to its right, while

T

lies

5

above

P

(being a quarter of OS). Pythagoras's theorem therefore gives

PT^2 = (15\sqrt{3})^2 + 5^2 = 225 \times 3 + 25 = 700

This is therefore the shorter of the two routes, and its length is

\sqrt{700} = 10\sqrt{7}

, hence the solution is option D.

TMUA 2016

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