Nyjc_my2007 p1 Solution

8/8/2019 Nyjc_my2007 p1 Solution

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2007 Mid Year Paper 1 Solution

1

2

6 11 2

2 5 3 3

6 11 2

( 3)(2 1) 3

by partial fractions, we have1 4 1

13 2 1 3

4 11 ,

2 1 2

3 20

2 1

1 3or

2 2

x x

x x x

x x

x x x

x x x

xx

x

x

x x

e

e

e

e {

e

u

To solve for 211 12 2 2

8 10 3 2 3

x x

x x x

" , replacex with 2x,

1 3 3 12

2 2 4 4x x e e

Markers comments for Q1: Inequalities

Majority of the students were able to multiply the square of the denominator to get rid of the denominator.However, a few of them did cross multiplication. As for the second part, many of them wrote Letx = 2x. A

minority of the students were able to write Lety = 2x or Letx = 1/2y but solved forx instead.

2 (i) Number of ways = 7! = 5040(ii)Number of ways = 3! 5! = 720(iii)Number of ways = 3! 4! = 144(iv)Number of ways = 5 3 3! 4!C v v = 1440

Markers comments for Q2: Permutations and Combinations

Most are able to answer part (i) and (ii), majority got part (iii) right and generally, part (iv) is not well

done.

For those who attempted part (iv), many counted the cases, which many cases are left out/not

considered, causing loss of marks.

3i) The maximum number of employees is 150000 600

250=

ii) 10 2( 1)n

T n= + -

600 10 2( 1)

296

n

n

= + -

=

iii) Total number of weeks in 6 years = 6 52 312=


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( ) ( )( ) ( )( )( )

296 296

7

Total bill for 1st 6 years =250 250 (312 29 6)

296250 2 10 2 296 1 312 296 250 600

2

2.497 10

S T+ -

= + - + -

=

Markers comments for Q3: AP & GP

This is an AP/GP Question. There are 3 parts to the questions. For part(ii), the weaker students could not

interpret thatn

T should be used. The weaker students tend to usen

S . In answering part (iii), the students

need to realize that there is limit to the number of employees the manufacturer can employ and hence to

calculate total wages, they need to know if the limit has been reached. Some students had no idea they

need to do this.

Generally, the question is not well-answered.

4

Markers comments for Q4: Maclaurins Series

y Generally, many could do this question and know the standard procedures in doing Maclaurins series.y The most common mistake made by students is 1cos(3cos ) 3x x ! .y Many knew that they have to use 21cos 1

2x x} for the last part of the question but they did not

know that they must show1 2cos sin(3cos )x x ax } . They claimed that the result is true and find

the value ofa which is incorrect.


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5

Markers comments for Q5: Transformation on Graphs

In general students did reasonably well for this question. In part (a), many students were unable totransformation 2 ( )fx- correctly. Some candidates could not shift the curve correctly. Other mistakes

included making out of the intercepts of they-axis as 1. For part(b), a common mistake is to stretch the original

curve by 2 along thex-axis instead of compressing by 2 along thex-axis. Inverting the curve proved to be

challenge for some students.

6 2 2 2

1 1

ax ay ax a

x x

! !

(i) 1y ax a and x! !

(ii)

2

2

2 2

( 1)

dy ax ax

dx x

!

y

x

a/2

1/2

y

x

y = 1/f(2x)

(b)

y2 = |2 f(x)|

a

-1/2

(a)


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2

2

0 2 2 0

(2 ) 4 ( 2) 0 0

4 ( 2) 0

2 0

dyax ax

dx

a a or a

a a

a

!

e

Note that when a = 0, the equation ofC is2

1y

x!

, which has a negative gradient. Thus a = 0 is

included in the range as well.

(iii) When 1a ! then 2 21

xy

x

!

2

2 2

2

1 ( 1)

1 21

1 1

21

1

x x x

x

x

x x xe e x e

x x xe

x x

xe

x

! !

! !

!

From the graph, there are two real roots.

Markers comments for Q6: Graphs of Rational Functions

Most are able to write out the asmyptotes and find the derivative, many failed to use D < 0, resulting

in part (ii) poorly done.Most students couldn't simplify the form to the form given earlier, resulting subsequent loss of marks

curve-sketching and conclusion. Marks are not awarded for conclusions made based on curves

wrongly sketched.


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Markers comments for Q7: Differentiation and Its Applications

Students omitted the portion of the graph where t is negative. Direct transfer of graph from GC to answer

paper results in the gap atx = a.

A handful of students made algebraic mistakes, and hence, unable to obtain the correct answer.

y

xO

2a

a

2a

24 ( )y a a x!


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Markers comments for Q8: Mathematical Induction

The induction part was generally well done. The common mistake is poor management of sign.

For part (ii), students wrongly assume that1 1

3 3

r r

!

, which results in the wrong answer.

9

Markers comments for Q9: Integration and its Applications

Part (a) was generally well done. Students were able to identify the use of integration by parts technique tosolve the problem. A handful of students made the wrong choice foru and dv which cost them. Algebraicmanipulation was rather weak for a group of students when cancellation of terms were made: eg

2

3

24 1

8 3

x

x x!

Most students were able to complete the integral in part (b) to a certain extent. However, a handful of


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U

A(-2,2)

Re(z)

Im(z)

2

4

O

B(-2,-2)

students did not show the relationship

2 2

cos( )a x

xU

! clearly, assuming the result is true straightaway.

As this is a show question, every step must be accounted for.

The part on finding the area was surprisingly done badly. Most students were able to identify the correctregion, but were unable to give the correct integral representation. Some students just use the fact that the

area required is one eighth of a circle of radius 3, which score 0 marks.

10(a) |z| = 2, arg(z) = 5

6T

*

1 5arg arg( )

6

k

kk z

z

T ! !

.

For*

1k

z

to be real,*

1arg

k

z

need to be integral multiple ofT.

Thus k = 6.

(b) 1 2sin ( )4 6

T

! !

least arg( 2 2 )2 6 3

z iT T T

! !

(c) z6+32+323i = 0

z6 = -32 - 323i

|-32 - 323i| = |-32||1 + 3i| = 64

arg(-32 - 323i) = -( -tan-13) = -2 /3

z6

= 64e-(2Ti/3)

=(2e-T/9

)6

Letx = z/2e-T/9, thusx

6 = 1.

Hencex6

= 1, thereforex = e-kTi/3

, k= 0, 1, . . . , 5

Accordingly, z = 2e(3k-1)Ti/9

, k = 0, 1, . . . , 5

Otherwise

z6

= -32 - 323i

|-32 - 323i| = |-32||1 + 3i| = 64

arg(-32 - 323i) = -(T-tan-13) = -2T/3

z6 = 64e -(2Ti/3)

2( 2 )

6 364k i

z e

T

T

!

( )9 32

ki

z eT T

!k = 0, 1, . . . , 5


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Markers comments for Q10: Complex Numbers

For part (a), some students still cannot use GC to compute arg(z). Thus later part of the question could not

be solved.

For part (b), students did not shade the required region and cannot identify the smallest argument.

For part (c), students found this part, solve6 32 32 3x i! , difficult .

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Markers comments for Q11:

Poor use of notations for vectors: for example, position vector should be OEuuur

rather than E.

For (ii), students mistaken that direction vectors of the plane satisfies the equation of the plane, which is

clearly wrong. As such, students will unable to verify the statement.

For (iv), majority of the students failed to see that the shortest distance is DE, and solved this problem byworking through the routine way. Students need to look at the marks allocated for the problem and choose

an appropriate strategy to solve the problem.

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Nyjc_my2007 p1 Solution