2017 Singapore- CAMBRIDGE A Level H2 Math P1 Suggested

2017 Singapore-

CAMBRIDGE

A Level

H2 Math P1

Suggested Answer

Key (9758)

Written and Prepared by Mr Mitch Peh

Preface

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2017 A Level H2 Math P1 Answers Done by Mr Mitch Peh

3

Overall Remarks for 2017 A Level H2 Math Paper 1

Overall, this is a manageable paper with tedious manipulations for some questions where we have to be careful

so that we avoid making unnecessary careless mistakes and lose precious marks.

There is quite a heavy emphasis on differentiation, graphing skills and vectors in this paper so you must

definitely be strong with your Calculus and familiar with the content knowledge in the above topics in order to

perform well. In particular, always remember to check that you have applied chain rule and product rule

correctly during differentiation manipulations. Also, significant amount of marks are allocated to the topics of

integration, complex numbers and sequence and series so you need to be familiar with these topics as well.

o Differentiation: Q3, Q4(i), 5(ii), Q10(iii)

o Graphing Skills: Q2, Q4(ii) and (iii)

o Vectors: Q6, Q10

o Integration: Q7,Q11

o Complex Numbers: Q8

o Sequence and Series: Q9

Students also have to be very comfortable manipulating expressions and equations with variables rather than

concrete values in this paper. This applies particularly to question 6 on the topic of vectors, question 7 on

integration with the use of trigonometric factor formula and question 9(c) on sequence and series. To do so,

strong content knowledge in these topics is definitely a must so that you have the confidence to work through

the intimidating and complicated looking expressions.

In this paper, getting the correct results for the earlier part of the questions are also important. Making careless

mistakes will likely cause us to lose significant amount of marks in the later part of the questions since we will

not have the correct expressions for manipulation. This applies especially to questions 2, 3, 8b, 9, 11.

In terms of O level knowledge, being familiar with techniques like completing the square, using the discriminant

method, solving simultaneous equations are important to answer questions 5, 8 and 9.

In this Paper, the last 2 questions are contextual based questions largely based on the topics of vectors and

differential equations.

Due to the focus of the new H2 Math syllabus on the ability to formulate and solve real world problems

mathematically, you can expect the last 2 questions of future A Level H2 Math Paper 1 to continue to be

contextual questions likely based on topics such as differential equations, vectors, APGP, maximisation and

minimisation problems.

For question 10 testing on 3 dimensional vector geometry, I would strongly encourage students to draw

diagrams in order to visualise the information provided in the question.

For question 11 testing on differential equations, it is important to know how to interpret the information

provided correctly and form the differential equation accordingly as well as knowing how to solve differential

equations with direct integration and method of separable variables.

We should also pay attention to the topics that have not been tested so that we can better prepare ourselves for

Paper 2. A typical A Level H2 Math Paper 2 will have 4 questions on the Pure Math section.

o Topics not tested: APGP, Parametric Equations, Functions, Integration to find area and volume


4

2017 A Level 9758 H2 Math P1 Suggested Solution

1. Topic: Maclaurin series expansion using standard series

Using the standard series expansion for ex and ln(1+x), we have:

2 322 4

ln 1 1 2 ... ...2 2 3

xax axx

e ax x ax

2 2 3 3

21 2 2 ... ...2 3

a x a xx x ax

2 2 3 32 2 3 3

22 3

2 2 ...2 3

2 1 24 3

a x a xax ax a x ax

a aax a x a a x

Hence, the value of a for which there is no term in x2 is a =0 or a=4.

Comments

Just be careful when modifying the standard series expansion to suit the context here.

It is alright if you did not factorise a out from the coefficients of x2 and x3.

Do not miss out that a can be 0 as well for which there is no term in x2.

[4]

2. (i) Topic: Graph Sketching

Comments

Always remember that for graph sketching, the stationary points, intercepts with the axes and

asymptotes must always be indicated.

For the intercepts with the axes, we should give in terms of the coordinate form.

Graphical transformation skills is also required here where we should know the general shape of

1y

x and y x graphs. Then translate and scale accordingly to obtain the graphs for this

question.

[2]


5

(ii) Topic: Solving inequalities by graphical methods

We need to find the x-coordinate of the intersection point between the 2 graphs.

1

b x ax a

2 1

0x ab

1 10x a x a

b b

1 1 or = x a x a

b b

Based on the graph above, we are interested in the intersection point where x>a.

Hence, it occurs when 1

x ab

.

Therefore, for1

b x ax a

, x < a or1

x ab

.

Comments

From the graphs sketched, we should look for the sections where the graph of 1

yx a

is below

the graph of y b x a in order to find the conditions for 1

b x ax a

to hold.

[4]

3 (i) Topic: Implicit Differentiation

Given2 22 5 10 0.....(1)y xy x , differentiate with respect to x,

2 22 5 10 0

2 2 2 10 0

2 2 2 10

y xy x

dy dyy y x x

dx dx

dyy x y x

dx

2 10 5

2 2

dy y x y x

dx y x y x

For stationary points, y=5x

Subst y=5x in (1), 2 2 2

2

25 10 5 10 0

20 10

1

2

x x x

x

x

Hence, the exact x-coordinates of the stationary points of C are 1

2and

1

2 .

[4]


6

Comments

Always remember to apply chain rule and product rule during the differentiation

The correct x coordinate is needed to obtain the correct answer in (ii).

(ii) Topic: Differentiation on determining whether a stationary point is a maximum or minimum

Using second derivative test,

Given 5dy

y x y xdx

, differentiate with respect to x,

2

21 5.....(2)

dy dy d y dyy x

dx dx dx dx

When x=1

2,

55

2y x , 0

dy

dx

Hence, substituting these values in (2), we have: 2

2

5 15

2 2

d y

dx

2

2

5 20

4

d y

dx

Hence, for the stationary point with x > 0, it is a maximum point.

Comments

It is alright to leave the answer for our second derivative in the unrationalised form 2

2

5

2 2

d y

dx . We will still get the same result that the stationary point is a maximum point.

We should not be using the first derivative test as the use of GC is not allowed here. This is

because we will need to find the value of 1

2x in 3 significant figures with our calculator in

order to know the smaller value and larger value of x to be used in the table.

[3]

4. (i) Topic: Differentiation

Given4 9

2

xy

x

, 2x , multiplying both sides by x+2, we have:

2 4 9xy y x

Differentiating with respect to x,

2 4

2 4

4.....(1)

2

dy dyy x

dx dx

dyx y

dx

dy y

dx x

Substituting 4 9

in (1), 2

xy

x

[3]


7

2

2

4 94 4 2 4 92

2 2

1

2

xx xdy x

dx x x

x

Since 2

2 0x for , 2x x ,

0dy

dx for , 2x x .

Hence, the gradient of C is negative for all points on C.

Comments

The mathematical manipulations can be slightly tedious here, so just be careful during the

process.

We can also use quotient rule for the differentiation instead of implicit differentiation here.

(ii) Topic: Graphing Techniques

4 2 14 9

2 2

14

2

xxy

x x

x

Hence, the equations of the asymptotes of C are x = -2 and y = 4

Comments

You can also use long division instead of factorisation to obtain the equation of y in similar form,

just more tedious.

I would consider this to be a relatively easy question so you should be getting the correct answer

here.

[3]

(iii) Topic: Graphical Transformation

To transform1

42

yx

on to the graph of 1

yx

,

Replacement in equation to perform Action to perform

Replace x with x-2

14y

x

Translate in the positive x

direction by 2 units

Replace y with y+4

1y

x

Translate in the negative y

direction by 4 units

Hence, the pair of transformations are:

1. Translate in the positive x direction by 2 units

2. Translate in the negative y direction by 4 units

[2]


8

Comments

You will need to be familiar with topic of graphical transformations to know the replacements in

equation to perform and actions to perform.

Do not misinterpret the question as transforming the graph of 1

yx

on C.

5. (i) Find the values of a, b and c.

Topic: System of linear equations

For3 2x ax bx c ,

When 1,x 1 8a b c

7.....(1)a b c

When 2,x 4 2 8 12a b c

4 2 4.....(2)a b c

When 3,x 9 3 27 25a b c

9 3 2.....(3)a b c

Using GC to solve (1), (2) and (3), we have:

3

2a ,

3

2b and 7c

Comments

We will need to be familiar with the topic of remainder and factor theorem learnt in secondary

school to substitute the relevant values and form the 3 equations accordingly before solving them.

To solve the system of linear equations, we just use our GC, go to “apps”→ “2: Simult equation

solver” and key in the equations.

[4]

(ii) Topic: Differentiation and solving equation

Substituting in the values of a, b and c obtained in (i),

3 23 3( ) 7

2 2f x x x x

2

2

2

3'( ) 3 3

2

33

2

1 3 33

2 2 4

f x x x

x x

x

21 3

32 4

x

Since

21

02

x

for all real values of x,

21 3

'( ) 3 02 4

f x x

for all real values of x.

Hence, the gradient is always positive.

[3]


9

The equation ( ) 0f x has only one real root because ( )y f x is strictly increasing so it will only

intersect the x axis exactly once.

3 23 3( ) 7 0

2 2f x x x x

Using GC, 1.33(3 . .)x s f

Comments

To show the gradient of the curve is always positive, besides using complete the square method as

shown above, we can also use the discriminant method, showing that the discriminant is less than

0 and the coefficient of x2 is positive. Hence, '( )f x is always greater than 0 for all real values of

x.

As the question did not require us to solve for the exact value of x for which ( ) 0f x , it is alright

to just use our GC and solve the cubic equation and express our answer to 3 significant figures.

(iii) Topic: Inferring the gradient of line value is m given y mx c

For tangent to the curve to be parallel to the line '( ) 2f x ,

2

2

3'( ) 3 3 2

2

13 3 0

2

f x x x

x x

Using GC, 0.145x or 1.15x (3s.f.)

Comments

Easy question which tests secondary school concept

Note that since the question did not require us to find the exact x-coordinates of the points, it is

alright to use our GC again to solve the quadratic equation and express our answers to 3 significant

figures.

[3]

6. (i) Topic: Vectors on geometrical interpretation of vector equation

t r a b is the equation of the line where a is the position vector of a point lying on the line and

is parallel to the vector b

Comments

We should not just say a is a position vector of the line as it may not be clear what that means.

Similarly, we should not simply say that b is the direction vector of the line as it may not be clear

what that means.

[2]

(ii) Topic: Vectors on geometrical interpretation of vector equation

d r n is the equation of the plane which is normal/perpendicular to the constant unit vector n

and d is the perpendicular displacement from origin to the plane.

Comments

Note that we cannot say d is the shortest distance from origin to the plane as d can be a negative

value. |d| is then the shortest distance from origin to the plane since n is a unit vector.

[3]


10

(iii) Topic: Vectors on finding intersection point between line and plane

Given that 0 b n , this means that the line and plane are not parallel to one another and thus the

line would intersect the plane at exactly one point.

Substituting t r a b in d r n ,

t d a b n

t d a n b n

dt

a n

b n

Substituting d

t

a n

b nin t r a b , we have:

d

a nr a b

b n

This solution is the position vector of the point of intersection between the line t r a b and the

plane d r n .

Comments

The challenge to solving for the intersection point between the line and the plane is that actual

position vector and normal vector are not provided.

We would need to know that we should solve for the value of t and substitute into the line

equation.

[3]

7. (i) Topic: Integration with the use of trigonometric factor formula

Using MF26, we know that 1 1

cos cos 2sin sin2 2

P Q P Q P Q . Hence,

1

sin 2 sin 2 cos 2 2 cos 2 22

mx nxdx mx nx mx nx dx

1 1

cos 2 2 cos 2 22 2

mx nx dx mx nx dx

sin 2 2 sin 2 2

2 2 2 2 2 2

mx nx mx nxC

m n m n

sin 2 2 sin 2 2

4 4

mx nx mx nxC

m n m n

Comments

A common issue that many students face is not knowing how to make use of the trigonometric

factor formula, transforming the expression in RHS to the expression in LHS.

We should notice that if we add the terms inside the sine functions together,

1 1

2 2P Q P Q P

We should also notice that if we subtract the terms inside the sine functions,

1 1

2 2P Q P Q Q

Hence, we can simply add/minus the terms 2mx and 2nx to transform the trigonometric

expression sin2 sin2mx nx in order to perform the integration.

[3]


11

Getting the correct result here is important as we need to apply it in (ii) of the question.

(ii) Topic: Integration

2 2 2

0 0sin 2 sin 2 sin 2 2sin 2 sin 2 sin 2mx nx dx mx mx nx nx dx

0

1 cos4 1 cos42sin 2 sin 2

2 2

mx nxmx nxdx

0 0 0

1 11 cos4 1 cos4 2 sin 2 sin 2

2 2mxdx nxdx mx nxdx

0 0 0

sin 2 2 sin 2 21 sin 4 1 sin 42

2 4 2 4 4 4

mx nx mx nxmx nxx x

m n m n m n

1 1

0 0 0 0 0 0 2 0 0 0 02 2

Comments

Before the integration is even done, we need to be careful and watch out especially for the

addition sign in ( ) sin 2 sin 2f x mx nx .

It is not ( ) sin 2 sin 2f x mx nx which is the function in the integral sign for (i).

The manipulation during the integration can be rather complicated due to the presence of many

variables, so we have to be careful and work out slowly to avoid careless mistakes.

We need to be familiar with the use of double angle formulae so that the expressions can be

transformed for suitable integrations to be done.

We also need to be familiar with the sine function i.e. sin 0k where k is an integer in order

to simplify the expressions after substituting 0x and x into them.

[5]

8. (a) Find the roots of the equation 2 1 2 5 5 0z i z i , giving your answers in cartesian form

a+ib.

Topic: Complex Numbers on the use of the quadratic formula to solve for roots of equation

2 1 2 5 5 0z i z i

22 2 4 1 5 5

2 1

i iz

i

2 4 20(2) 2 6

2 1 2 1

iz

i i

1 3 1

1 1

i iz

i i

or

1 3 1

1 1

i iz

i i

1

1 3 32

1 2

i i

i

11 3 3

2

2

i i

i

[3]


12

Comments

We should be using the quadratic formula to solve the quadratic equation here. Other methods

like comparing coefficients is not appropriate as we would have too many unknowns, making

the solving complicated.

We can check that our rationalised values of z are correct with the GC by keying in 1 3

1

i

i

and

1 3

1

i

i

.

(b)

(i)

Topic: Complex Numbers

Given 1 i

22 1 1 2 1 2i i i

23 1 1 2 1

2 2

i i i i

i

2 2 24 1 1 2

4

i i i

Given that4 3 239 58 0p q ,

Substituting in the values of , 2 ,

3 and 4 ,

4 2 2 39 2 1 58 0p i i q i

4 2 2 39 2 1 58 0

4 2 58 2 78 0

p i i q i

p q p q i

Comparing real parts,

2 54.....(1)p q

Comparing imaginary parts,

2 78......(2)p q

(1)+(2):

4 24 6p p

Subst 6p in (1),

2( 6) 54 66q q

Hence, 6p , 66q

Comments

Even though this question does not allow us to use the calculator, we can still use our GC to

verify the answers we obtained manually in Cartesian form for 2 ,

3 and 4 are correct.

We should be familiar with the method of comparing real parts and imaginary parts to solve for

the values of p and q.

Note that we need to solve for the values of p and q manually but can still check with the help of

our calculator.

We will need to get the values of p and q correct in order to obtain the correct answer for (ii) as

well.

[4]


13

(ii) Topic: Complex Numbers on expressing polynomial equation as the product of two quadratic

factors

Substituting in the values of p and q obtained in (b)(i),we have: 4 3 26 39 66 58 0

We also know in (b)(i) that 1 i is one of the roots of the polynomial equation.

Since all the coefficients of the polynomial equation are real, by Conjugate Root theorem,

1 i is another root of the polynomial equation.

Hence, we can express the polynomial equation in the following form:

4 3 2 2

2 2

2 2

6 39 66 58 1 1

1 1

2 2

i i a b

a b

a b

By comparing the coefficients of constant,

58 2 29b b

By comparing the coefficients of ,

66 2 2 29 4a a

4 3 2

2 2

6 39 66 58

2 2 4 29

Comments

We need to be able to tell that 1 i is one of the roots of the equation already.

We can use the formula you have learnt in secondary school a2-b2 = (a-b)(a+b) to expand

1 1i i easily.

You can also use long division to obtain the other quadratic factor of 2 4 29 .

[3]

9 (a)

(i)

Topic: Sequence and Series

22

1

2 2

1 1

2

2

n n nu S S An Bn A n B n

An Bn An An A Bn B

An A B

Comments

We have to recall this relationship 1n n nu S S in order to solve this question.

Be careful especially with the expansion of 2

1A n .

Note that we need to get this part of the question right in order to obtain the correct answer for

(ii).

[3]

(ii) Topic: Solving simultaneous equations

10 2 10

19 48.....(1)

u A A B

A B

17 2 17

33 90.....(2)

u A A B

A B

[2]


14

(2)-(1), 14 42A

3A

Subst A=3 in (1),

19 3 48

9

B

B

Comments

Simple application of solving simultaneous equations that you can do with secondary school

knowledge.

You can also use the GC to solve the simultaneous equations.

(b) Topic: Sequence and series on method of differences

2 22 2 2 2 2 21 1 2 1 2 1r r r r r r r r r r

4 3 2 4 3 22 2r r r r r r 34r (Shown)

2 23 2 2

1 1

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

2 22 2

11 1

4

1 2 0 1

2 3 1 2

3 4 2 31

4 ...

1 2 1

1 1

n n

r r

r r r r r

n n n n

n n n n

221

14

n n

Comments

A standard question where we have to be familiar with the method of difference to simplify the

series.

Do not forget about the 1

4 that is outside of the bracket after performing the method of

difference.

[4]

(c) Topic: Sequence and series

Let !

r

r

xa

r .

1

1 n!

1 ! 1

n

n

n

n

a x x

a n x n

As n→ꚙ, 01

x

n

[4]


15

1lim 0 1n

xn

a

a

Thus, by D’Alembert’s ratio test, the series 0 !

r

r

x

r

converges for all real values of x.

From MF26,

0 !

rx

r

xe

r

Comments

This question can appear intimidating for many students due to the sheer amount of information

provided. You would also have to react on the spot on how to make use of the information

provided to show that the series converges.

Besides that, you may also not know how to deduce the sum to infinity of the series.

Note that the sum to infinity is not 0 as 0 is for 1lim n

xn

a

a

instead.

We would have to use MF26 standard series expansion to know the sum to infinity of the series.

10. (i) Topic: Vectors on finding the intersection point between 2 lines

Let the new cable be D.

For the existing cable C,

0 3

: 0 1 ,

0 2

Cl

r

To form the equation of cable D,

5 1 4

7 2 5

1 1

PQ OQ OP

a a

1 4

: 2 5 ,

1 1

Dl

a

r

Equating the 2 lines together,

0 3 1 4

0 1 2 5

0 2 1 1a

[4]


16

Hence, we can obtain 3 equations:

3 1 4 .....(1)

2 5 ......(2)

2 1 1 .....(3)a

(2) x 3: 3 6 15 ......(4)

Equating (1) and (4) together,

51 4 6 15

11

Subst 5

11 in (2),

5 32 5

11 11

Then, we substitute the values of λ and µ in (3).

3 5

2 1 111 11

6 5 51

11 11 11

5 22

11 11

a

a

a

22 24

5 5a

Comments

The difficulty of this question lies in having the ability to interpret the information provided and

form the required equations of the cables.

We should also be familiar on how to solve a system of linear equations in order to resolve for the

values of λ, µ and a.

(ii) Topic: Vectors on the dot product value of 2 vectors = 0 when the angle between the 2 vectors is

900

0 3

: 0 1 ,

0 2

Cl

r

[4]


17

For the angle PRQ to be 900, 0RP RQ

1 3 1 3

2 2

1 2 1 2

RP OP OR

5 3 5 3

7 7

3 2 3 2

RQ OQ OR

1 3 5 3

2 7

1 2 3 2

RP RQ

2 2 2

2

5 15 3 9 14 7 2 3 6 2 4

14 35 22

Discriminant = 2

35 4 14 22 7 0

Hence, 214 35 22 is always positive for all real values of x.

Thus, there is no real values of λ for which 0RP RQ

As a result, angle PRQ cannot be 900.

Comments

Drawing a 3D diagram can be helpful in interpreting the information provided in the question.

We can just see PQ as being above existing cable C and point R can be any point on the existing

cable C.

To show that214 35 22 cannot be 0 here, the mathematical manipulation is easier when

we use the discriminant method rather than completing the square method.

(iii) Topic: Minimisation of length problem

Method 1: Using differentiation

From (ii), we have found

3 1

2

2 1

PR

Length of PR = 2 2 2

3 1 2 2 1PR

2 2 29 6 1 4 4 4 4 1

214 14 6

To minimise the length of PR, we can ensure that 2

PR is a minimum.

2

28 14 0d PR

d

Hence, 1

2

[5]


18

22

228 0

d PR

d

Hence, 1

2 will give us the minimum length of PR.

3 0.5 1.5

0.5 0.5

12 0.5

OR

Thus, the coordinates of R is (1.5,0.5,-1)

Minimum length PR

2 5 10

14 0.5 14 0.5 62 2

units

Method 2: Using Vector Properties

Let the position of the main switching site which lies on cable C be O

For the length of PR to be the shortest, we need 0OR PR

From (ii),

3 1

2

2 1

PR

and

3

2

OR

2 2 2

3 3 1

2 9 3 2 4 2

2 2 1

OR PR

214 7 0

2 1 0

10 or

2

Length of PR = 2 2 2

3 1 2 2 1PR

When 0 , 6PR

When 1

2 ,

10

2PR


19

Since 10

6,2

for the length of PR to be the shortest, 1

2

.

Hence, the coordinate of R is (1.5,0.5,-1).

Comments

It is good to understand the difference between the question asked in (ii) and (iii). For (ii), the

question wants to minimise the total length of PR+RQ while for (iii), the question only wants to

minimise the length of PR.

11. (i)

(a)

Topic: Differential equation on forming equation based on information provided

dvc

dt

Comments

Simple interpretation of the statement: The rate of change of velocity, v ms-1, with respect to

time, t seconds, is a constant, c.

[1]

(b) Topic: Solving differential equation

1 dv

c dv c dtdt

v ct d where d is another constant.

Substituting t=0 and v=4, we have d=4

Substituting t=2.5, v=29, we have:

29 2.5 4

10

c

c

Hence, 10 4v t

Comments

Very standard and relatively simple manipulations to solve for the particular solution of a

differential equation

[3]

(ii) Topic: Differential equation on forming equation based on information provided

10dv

kvdt

Dividing by 10 kv throughout,

11

10

dv

kv dt

Hence, 1

110

dv dtkv

11

10

kdv dt

k kv

1

ln 10 kv t Ak

where A is a constant.

ln 10 kv k t A

[5]


20

10 kt Ak ktkv e Be where AkB e

Substituting t=0 and v=0,

10B

Hence, 10 10 ktkv e

10 10

10 10

kt

kt

kv e

v ek k

Comments

The manipulations here is a little bit more complicated and tedious to obtain the particular

solution of the differential solution as it involves the ln function.

Requires the correct interpretation of the statement: The new rate of change of v is modelled as

the difference between the value of c found in part (i)(b)and an amount proportional to the

velocity, v, with a constant of proportionality, k.

Some students may not know how to proceed to solve 10dv

kvdt

. Incorrect approaches

include adding kv on both sides of the equation, dividing by v throughout.

We can use round brackets in the ln function after integration because 10-kv>0 for all the

possible values that v can be.

For the manipulation to be simpler, the substitution of t=0 and v=0 should be done after getting

rid of the ln function.

We should double check our final expression for v, making sure that it is in terms of k and t only.

(iii) Topic: Forming and solving equations

10 10 ktv ek k

Terminal velocity can be found by letting t→ꚙ

As t→ꚙ, 0kte

Hence, 10

vk

so the terminal velocity is 10

kms-1.

Since the question states that terminal velocity is 40ms-1,

10 140

4k

k

0.2540 40 tv e

To find the time for object to reach 90% of terminal velocity,

0.250.9 40 40 40 te

Dividing by 40 throughout, 0.250.9 1 te

0.25 1

10

te

1 1ln

4 10t

[4]


21

14ln ln10000 9.21

10t s

(3s.f.)

Hence, the time it takes s for the object to reach 90% of terminal velocity is 9.21s.

Comments

We will need the correct equation of v from (ii) in order to perform the manipulations and

obtain the correct time for the object to reach 90% of its terminal velocity.

In the final answer, t should not just be expressed in terms of k as the value of k can be solved

here.

Even though we have the exact value of t which is ln10000, it is better to leave our answers as

3 significant figures since 9.21s would have more meaning in the real world context.

End of Solutions

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2017 Singapore- CAMBRIDGE A Level H2 Math P1 Suggested