New test - October 20, 2014 - David Delgado...New test - October 20, 2014 [389 marks]1a.The first...

New test - October 20, 2014 [389 marks]

1a. [2 marks]The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .Find the common difference.

Markschemevalid method (M1)

e.g. subtracting terms, using sequence formula

A1 N2

[2 marks]

Examiners reportMost candidates performed well on this question. A few were confused between the term number and the value of a term.

d = 1.7

1b. [2 marks]The first three terms of an arithmetic sequence are 5 , 6.7 , 8.4 .

Find the 28 term of the sequence.

Markschemecorrect substitution into term formula (A1)

e.g.

28 term is 50.9 (exact) A1 N2

[2 marks]


th

5 + 27(1.7)

th

1c. [2 marks]The first three terms of an arithmetic sequence are , , .Find the sum of the first 28 terms.

Markschemecorrect substitution into sum formula (A1)

e.g. ,

(exact) [ , ] A1 N2

[2 marks]


5 6.7 8.4

= (2(5) + 27(1.7))S28282

(5 + 50.9)282

= 782.6S28 782 783

(2x + p 6 60 4

2. [7 marks]The third term in the expansion of is . Find the possible values of p .

Markschemeattempt to expand binomial (M1)

e.g. ,

one correct calculation for term in in the expansion for power 6 (A1)

e.g. 15 ,

correct expression for term in (A1)

e.g. ,

Notes: Accept sloppy notation e.g. omission of brackets around .

Accept absence of in middle factor.

correct term (A1)

e.g. (accept absence of )

setting up equation with their coefficient equal to 60 M1

e.g. , ,

A1A1 N3

[7 marks]

Examiners reportThis question proved challenging for many students. Most candidates recognized the need to expand a binomial but many executedthis task incorrectly by selecting the wrong term, omitting brackets, or ignoring the binomial coefficient. Other candidates did notrecognize that there were two values for p when solving their quadratic equation.

(2x + p)6 60x4

(2x + ( ) (2x (p + …)6p0 61

)5 )1 ( ) (2x (pn

r)r )n−r

x4

16x4

x4

( ) (2x (p62

)4 )2 15.24p2

2x

x

240p2x4 x4

( ) (2 (p = 6062

)4 )2 240 = 60p2x4 x4 =p2 60240

p = ± (p = ±0.5)12

3. [7 marks]Given that , find the value of n .

Markschemeattempt to expand (M1)

e.g. Pascal's triangle,

correct first two terms of (seen anywhere) (A1)

e.g.

correct first two terms of quadratic (seen anywhere) (A1)

e.g. 9 , ,

correct calculation for the x-term A2

e.g. , ,

correct equation A1

e.g. ,

A1 N1

[7 marks]

(3 + nx = 9 + 84 + …(1 + x)23

n

)2

(1 + x)23

n

= 1 + nx + …(1 + x)23

n23

(1 + x)23

n

1 + nx23

6nx (9 + 6nx + )n2x2

nx × 9 + 6nx23

6n + 6n 12n

6n + 6n = 84 12nx = 84x

n = 7

Examiners reportThis question proved quite challenging for the majority of candidates, although there were a small number who were able to find thecorrect value of n using algebraic and investigative methods. While most candidates recognized the need to apply the binomialtheorem, the majority seemed to have no idea how to do so when the exponent was a variable, n, rather than a known integer. Mostcandidates who attempted this question did expand the quadratic correctly, but many went no further, or simply set the x-term of thequadratic equal to , ignoring the expansion of the first binomial altogether.84x

4a. [4 marks]The first term of a geometric sequence is 200 and the sum of the first four terms is 324.8.

Find the common ratio.

Markschemecorrect substitution into sum of a geometric sequence (A1)

e.g. ,

attempt to set up an equation involving a sum and 324.8 M1

e.g. ,

(exact) A2 N3

[4 marks]

Examiners reportIn part (a), although most candidates substituted correctly into the formula for the sum of a geometric series and set it equal to 324.8,some used the formula for the sum to infinity and a few the formula for the sum of an arithmetic series. The overwhelming error madewas in attempting to solve the equation algebraically and getting nowhere, or getting a wrong answer. The great majority did notrecognize the need to use the GDC to find the value of r.

200 ( )1−r4

1−r200 + 200r + 200 + 200r2 r3

200 ( ) = 324.81−r4

1−r200 + 200r + 200 + 200 = 324.8r2 r3

r = 0.4

4b. [2 marks]The first term of a geometric sequence is 200 and the sum of the first four terms is 324.8.

Find the tenth term.

Markschemecorrect substitution into formula A1

e.g.

(exact), A1 N1

[2 marks]

Examiners reportIn part (b) many did not obtain any marks since they weren't able to find an answer to part (a). Those who were able to get a value forr in part (a) generally went on to gain full marks in (b). However, this was one of the most common places for rounding errors to bemade.

= 200 ×u10 0.49

= 0.0524288u10 0.0524

5a. [3 marks]

The first three terms of an arithmetic sequence are 36, 40, 44,….

(i) Write down the value of d .

(ii) Find .u8

Markscheme(i) A1 N1

(ii) evidence of valid approach (M1)

e.g. , repeated addition of d from 36

A1 N2

[3 marks]

Examiners reportThe majority of candidates were successful with this question. Most had little difficulty with part (a).

d = 4

= 36 + 7(4)u8

= 64u8

5b. [3 marks](i) Show that .

(ii) Hence, write down the value of .

Markscheme(i) correct substitution into sum formula A1

e.g. ,

evidence of simplifying

e.g. A1

AG N0

(ii) A1 N1

[3 marks]

Examiners reportSome candidates were unable to show the required result in part (b), often substituting values for n rather than working with theformula for the sum of an arithmetic series.

= 2 + 34nSn n2

S14

= {2 (36) + (n − 1)(4)}Snn

2{72 + 4n − 4}n

2

{4n + 68}n

2

= 2 + 34nSn n2

868

6a. [3 marks]

Consider the expansion of .

Find b.

Markschemevalid attempt to find term in (M1)

e.g. ,

correct equation A1

e.g.

A1 N2

[3 marks]

= 256 + 3072 + … + k + …(2 + )x3 b

x

8x24 x20 x0

x20

( ) ( )(b)81

27 (2 ( ) = 3072x3)7 b

x

( ) ( )(b) = 307281

27

b = 3

Examiners reportAn unfamiliar presentation confused a number of candidates who attempted to set up an equation with the wrong term in part (a).Time and again, candidates omitted the binomial coefficient in their set up leading to an incorrect result. In part (b) it was common tosee the constant term treated as the last term of the expansion rather than the 7th term.

6b. [3 marks]Find k.

Markschemeevidence of choosing correct term (M1)

e.g. 7th term,

correct expression A1

e.g.

(accept ) A1 N2

[3 marks]

Examiners reportAn unfamiliar presentation confused a number of candidates who attempted to set up an equation with the wrong term in part (a).Time and again, candidates omitted the binomial coefficient in their set up leading to an incorrect result. In part (b) it was common tosee the constant term treated as the last term of the expansion rather than the 7th term.

r = 6

( ) (286

x3)2( )3x

6

k = 81648 81600

7a. [5 marks]

The following diagram shows two ships A and B. At noon, ship A was 15 km due north of ship B. Ship A was moving south at 15 km hand ship B was moving east at 11 km h .

Find the distance between the ships

(i) at 13:00;

(ii) at 14:00.

–1

–1

Markscheme(i) evidence of valid approach (M1)

e.g. ship A where B was, B away

A1 N2

(ii) evidence of valid approach (M1)

e.g. new diagram, Pythagoras, vectors

(A1)

A1 N2

Note: Award M0A0A0 for using the formula given in part (b).

[5 marks]

Examiners reportPart (a) was generally well done although some candidates incorrectly used the function given in part (b) to find the required values.There was evidence that some candidates are not comfortable with a 24-hour clock.

11 km

distance = 11

s = +152 222− −−−−−−√

= 26.62705709−−−√

s = 26.6

7b. [6 marks]Let be the distance between the ships t hours after noon, for .

Show that .

Markschemeevidence of valid approach (M1)

e.g. a table, diagram, formula

distance ship A travels t hours after noon is (A2)

distance ship B travels in t hours after noon is (A1)

evidence of valid approach M1

e.g.

correct simplification A1

e.g.

AG N0

[6 marks]

Examiners reportCandidates had difficulty generalizing the problem and therefore, were unable to show how the function was obtained in part (b).

s(t) 0 ≤ t ≤ 4

s(t) = 346 − 450t + 225t2− −−−−−−−−−−−−−√

d = r × t

15(t − 1)

11t

s(t) = +[15(t − 1)]2 (11t)2− −−−−−−−−−−−−−−√

225( − 2t + 1) + 121t2 t2− −−−−−−−−−−−−−−−−−√

s(t) = 346 − 450t + 225t2− −−−−−−−−−−−−−√

s(t)

7c. [3 marks]Sketch the graph of .s(t)

Markscheme

A1A1A1 N3

Note: Award A1 for shape, A1 for minimum at approximately , A1 for domain.

[3 marks]

Examiners reportSurprisingly, the graph in part (c) was not well done. Candidates often ignored the given domain, provided no indication of scale, anddrew "V" shapes or parabolas.

(0.7, 9)

7d. [3 marks]Due to poor weather, the captain of ship A can only see another ship if they are less than 8 km apart. Explain why the captaincannot see ship B between noon and 16:00.

Markschemeevidence of valid approach (M1)

e.g. , find minimum of , graph, reference to "more than 8 km"

(accept 2 or more sf) A1

since , captain cannot see ship B R1 N0

[3 marks]

Examiners reportIn part (d), candidates simply regurgitated the question without providing any significant evidence for their statements that the twoships must have been more than 8 km apart.

(t) = 0s′ s(t)

min = 8.870455…

> 8smin

8a. [4 marks]

Let . The line L is the tangent to the curve of f at (4, 6) .

Find the equation of L .

f(x) = + 214

x2

Markschemefinding A1

attempt to find (M1)

correct value A1

correct equation in any form A1 N2

e.g. ,

[4 marks]

Examiners reportWhile most candidates answered part (a) correctly, finding the equation of the tangent, there were some who did not consider thevalue of their derivative when .

(x) = xf ′ 12

(4)f ′

(4) = 2f ′

y − 6 = 2(x − 4) y = 2x − 2

x = 4

8b. [6 marks]

Let , for . The following diagram shows the graph of g .

Find the area of the region enclosed by the curve of g , the x-axis, and the lines and . Give your answer in theform , where .

Markscheme

correct integral A1A1

e.g.

substituting limits and subtracting (M1)

e.g. ,

correct working (A1)

e.g.

correct application of (A1)

e.g.

A1 N4

[6 marks]

g(x) = 903x+4

2 ≤ x ≤ 12

x = 2 x = 12a lnb a,b ∈ Z

area = dx∫ 122

903x+4

30 ln(3x + 4)

30 ln(3 × 12 + 4) − 30 ln(3 × 2 + 4) 30 ln40 − 30 ln10

30(ln40 − ln10)

lnb − lna

30 ln 4010

area = 30 ln4

Examiners reportIn part (b), most candidates knew that they needed to integrate to find the area, but errors in integration, and misapplication of therules of logarithms kept many from finding the correct area.

9a. [4 marks]

Consider the triangle ABC, where AB =10 , BC = 7 and = .

Find the two possible values of .

MarkschemeNote: accept answers given in degrees, and minutes.

evidence of choosing sine rule (M1)

e.g.

correct substitution A1

e.g. ,

, A1A1 N1N1

Note: If candidates only find the acute angle in part (a), award no marks for (b).

[4 marks]

Examiners reportMost candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating alack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse anglegenerally had no difficulty with part (b).

C BA 30∘

A BC

=sin A

a

sin B

b

=sin θ

10sin 30∘

7sin θ = 5

7

A B =C 45.6∘ A B =C 134∘

9b. [2 marks]Hence, find , given that it is acute.

Markschemeattempt to substitute their larger value into angle sum of triangle (M1)

e.g.

A1 N2

[2 marks]

Examiners reportMost candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating alack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse anglegenerally had no difficulty with part (b).

A CB

− (134.415 + )180∘ …∘ 30∘

A C =B 15.6∘

10a. [1 mark]

Consider the expansion of .

Write down the number of terms in the expansion.

Markscheme10 terms A1 N1

[1 mark]

(3 + 2x2 )9

Examiners reportMany candidates were familiar with the binomial expansion, although some expanded entirely which at times led to careless errors.Others attempted to use Pascal's Triangle. Common errors included misidentifying the binomial coefficient corresponding to this termand not squaring the 3 in ( ) . 3x2

10b. [5 marks]Find the term in .

Markschemeevidence of binomial expansion (M1)

e.g. , , Pascal’s triangle

evidence of correct term (A1)

e.g. 8th term, , ,

correct expression of complete term (A1)

e.g. , ,

(accept ) A1 N2

[4 marks]

Examiners reportMany candidates were familiar with the binomial expansion, although some expanded entirely which at times led to careless errors.Others attempted to use Pascal's triangle. Common errors included misidentifying the binomial coefficient corresponding to this termand not squaring the 3 in .

x4

+ ( ) b + ( ) + …a9b0 91

a8 92

a7b2 ( ) (a (b9r

)n−r )r

r = 7 ( )97

(3x2)227

( ) (3 (297

x2)2 )7 C(3 (292 x2)2 )7 36 × 9 × 128

41472x4 41500x4

(3 )x2

11a. [4 marks]Consider an infinite geometric sequence with and .

(i) Find .

(ii) Find the sum of the infinite sequence.

Markscheme(i) correct approach (A1)

e.g. , listing terms

A1 N2

(ii) correct substitution into formula for infinite sum (A1)

e.g. ,

A1 N2

[4 marks]

Examiners reportMost candidates found part (a) straightforward, although a common error in (a)(ii) was to calculate 40 divided by as 20.

= 40u1 r = 12

u4

= (40)u412

(4−1)

= 5u4

=S∞40

1−0.5=S∞

400.5

= 80S∞

12

11b. [5 marks]Consider an arithmetic sequence with n terms, with first term ( ) and eighth term ( ) .

(i) Find the common difference.

(ii) Show that .

Markscheme(i) attempt to set up expression for (M1)

e.g.

correct working A1

e.g. ,

A1 N2

(ii) correct substitution into formula for sum (A1)

e.g.

correct working A1

e.g. ,

AG N0

[5 marks]

Examiners reportIn part (b), some candidates had difficulty with the "show that" and worked backwards from the answer given.

−36 −8

= 2 − 38nSn n2

u8

−36 + (8 − 1)d

−8 = −36 + (8 − 1)d−8−(−36)

7

d = 4

= (2(−36) + (n − 1)4)Snn

2

= (4n − 76)Snn

2−36n + 2 − 2nn2

= 2 − 38nSn n2

11c. [5 marks]The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n .

Markschememultiplying (AP) by 2 or dividing S (infinite GP) by 2 (M1)

e.g. , , 40

evidence of substituting into A1

e.g. , ( )

attempt to solve their quadratic (equation) (M1)

e.g. intersection of graphs, formula

A2 N3

[5 marks]

Examiners reportMost candidates obtained the correct equation in part (c), although some did not reject the negative value of n as impossible in thiscontext.

Sn

2SnS∞

2

2 =Sn S∞

2 − 38n = 40n2 4 − 76n − 80n2 = 0

n = 20

12a. [3 marks]Find the value of .40 − 5log2 log2

Markschemeevidence of correct formula (M1)

eg , ,

Note: Ignore missing or incorrect base.


eg ,

A1 N2

[3 marks]

Examiners reportMany candidates readily earned marks in part (a). Some interpreted to mean , an error which led to no further

marks. Others left the answer as where an integer answer is expected.

loga − logb = log a

blog( )40

5log8 + log5 − log5

8log2 = 823

40 − 5 = 3log2 log2

40 − 5log2 log240log2

5log2

5log2

12b. [4 marks]Find the value of .

Markschemeattempt to write as a power of (seen anywhere) (M1)

eg , ,

multiplying powers (M1)

eg ,


eg , ,

A1 N3

[4 marks]

Examiners reportPart (b) proved challenging for most candidates, with few recognizing that changing to base is a helpful move. Some made it asfar as yet could not make that final leap to an integer.

8 5log2

8 2

(23) 5log2 = 823 2a

23 5log2 a 5log2

2 125log2 log253 ( )2 5log23

= 1258 5log2

8 223 5log2

13. [7 marks]

An arithmetic sequence is given by , , , ….

(a) Write down the value of .

(b) Find

(i) ;

(ii) .

(c) Given that , find the value of .

5 8 11

d

u100

S100

= 1502un n

Markscheme(a) A1 N1

[1 mark]

(b) (i) correct substitution into term formula (A1)

e.g. ,

A1 N2

(ii) correct substitution into sum formula (A1)

eg ,

A1 N2

[4 marks]

(c) correct substitution into term formula (A1)

eg ,

A1 N2

[2 marks]

Total [7 marks]

Examiners reportThe majority of candidates had little difficulty with this question. If errors were made, they were normally made out of carelessness. Avery few candidates mistakenly used the formulas for geometric sequences and series.

d = 3

= 5 + 3(99)u100 5 + 3(100 − 1)

= 302u100

= (2(5) + 99(3))S100100

2= (5 + 302)S100

1002

= 15350S100

1502 = 5 + 3(n − 1) 1502 = 3n + 2

n = 500

14. [5 marks]

In the expansion of , the term in can be expressed as .

(a) Write down the value of , of and of .

(b) Find the coefficient of the term in .

Markscheme(a) , , (accept ) A1A1A1 N3

[3 marks]

(b) correct working (A1)

eg , , , ,

coefficient of term in is A1 N2

Note: Do not award the final A1 for an answer that contains .

[2 marks]

Total [5 marks]

(3x − 2)12 x5 ( ) × (3x × (−212r

)p )q

p q r

x5

p = 5 q = 7 r = 7 r = 5

( ) × (3x × (−2127

)5 )7 792 243 −27 24634368

x5 −24634368

x

Examiners reportCandidates frequently made reasonable attempts at both parts of the question. Those who correctly stated the values in (a) weregenerally successful in part (b). Many candidates offered the whole term rather than the coefficient in part (b) and lost the final mark.Some candidates appeared to have misread the order of the variables, stating that (instead of ), (instead of )and or (instead of or ). A large number of candidates did not make the connection between parts (a) and (b).

p = 7 r = 7 q = 5 p = 5r = 5 7 q = 5 7

15. [7 marks]

Let and .

(a) Find .

(b) Find .

(c) Find .

p = 6log3 q = 7log3

log3p2

( )log3p

q

(9p)log3

Markscheme(a) METHOD 1

evidence of correct formula (M1)

eg ,

A1 N2

METHOD 2

valid method using (M1)

eg , ,

A1 N2

[2 marks]

(b) METHOD 1


eg ,

A1 N2

METHOD 2

valid method using and (M1)

eg , ,

A1 N2

[2 marks]

(c) METHOD 1


eg ,

(may be seen in expression) A1

eg

A1 N2

METHOD 2

valid method using (M1)

eg ,

correct working A1

eg ,

A1 N2

[3 marks]

Total [7 marks]

Examiners reportThis question proved to be surprisingly challenging for many candidates. A common misunderstanding was to set equal to and equal to . A large number of candidates had trouble applying the rules of logarithms, and made multiple errors in each part of the

question. Common types of errors included incorrect working such as in part (a), or

in part (b), and in part (c).

log = n loguun 2 plog3

( ) = 12log3 p2

p = 36

(log3 36)2 log312 12 3log3

( ) = 12log3 p2

log( ) = logp − logqp

q6 − 7

( ) = −1log3p

q

p = 36 q = 37

( )log336

37log3−1 − 3log3

( ) = −1log3p

q

uv = u + vlog3 log3 log3 log9 + logp

9 = 2log3

2 + logp

(9p) = 8log3

p = 36

(9 × )log3 36 ( × )log3 32 36

9 +log3 log336 log338

(9p) = 8log3

p 6 q

7

= 36log3p2 ( ) =log3p

q

6log3

7log3

( ) = 6 − 7log3p

qlog3 log3 (9p) = 54log3

16. [6 marks]The sum of the first three terms of a geometric sequence is , and the sum of the infinite sequence is . Find thecommon ratio.

Markschemecorrect substitution into sum of a geometric sequence A1

eg ,

correct substitution into sum to infinity A1

eg

attempt to eliminate one variable (M1)

eg substituting

correct equation in one variable (A1)

eg ,

evidence of attempting to solve the equation in a single variable (M1)

eg sketch, setting equation equal to zero,

A1 N4

[6 marks]

Examiners reportMany candidates were able to successfully obtain two equations in two variables, but far fewer were able to correctly solve for thevalue of . Some candidates had misread errors for either or , with some candidates taking the French and Spanish examsmistaking the decimal comma for a thousands comma. Many candidates who attempted to solve algebraically did not cancel the from both sides and ended up with a 4 degree equation that they could not solve. Some of these candidates obtained the extraneousanswer of as well. Some candidates used a minimum of algebra to eliminate the first term and then quickly solved the resultingequation on their GDC.

62.755 440

62.755 = ( )u11−r3

1−r+ r + = 62.755u1 u1 u1r2

= 440u1

1−r

= 440(1 − r)u1

62.755 = 440(1 − r)( )1−r3

1−r440(1 − r)(1 + r + ) = 62.755r2

62.755 = 440(1 − )r3

r = −0.95 = 1920

r 440 62.7551 − r

th

r − 1

17. [7 marks]The constant term in the expansion of , where is . Find .

Markschemeevidence of binomial expansion (M1)

eg selecting correct term,

evidence of identifying constant term in expansion for power (A1)

eg , 4 term

evidence of correct term (may be seen in equation) A2

eg ,

attempt to set up their equation (M1)

eg ,

correct equation in one variable (A1)

eg ,

A1 N4

[7 marks]

( + )x

a

a2

x

6a ∈ R 1280 a

+ ( ) + …( )x

a

6( )a2

x

0 61

( )x

a

5( )a2

x

1

6

r = 3 th

20 a6

a3( )6

3( )x

a

3( )a2

x

3

( ) = 128063

( )x

a

3( )a2

x

3= 1280a3

a

20 = 1280a3 = 64a3

a = 4

Examiners reportMany candidates struggled with this question. Some had difficulty with the binomial expansion, while others did not understand thatthe constant term had no , while still others were unable to simplify a ratio of exponentials with a common base. Some candidatesfound using algebraic methods while others found it by writing out the first several terms. In some cases, candidates just set theentire expansion equal to 1280.

x

r = 3

18. [1 mark]

The following diagram shows a circle with centre O and radius cm.

Points A and B are on the circumference of the circle and radians .

The point C is on [OA] such that radians .

Show that .

Markschemeuse right triangle trigonometry A1

eg

AG N0

[1 mark]

Examiners reportAs to be expected, candidates found this problem challenging. In part (a), many were able to use right angle trigonometry to find thelength of OC.

r

A B = 1.4O

B O =C π

2

OC = rcos1.4

cos1.4 = OCr

OC = rcos1.4

19a. [1 mark]

The first three terms of an infinite geometric sequence are 32, 16 and 8.

Write down the value of r .

Markscheme A1 N1

[1 mark]

Examiners reportThis question was very well done by the majority of candidates. There were some who used a value of r greater than one, with themost common error being .

r = (= )1632

12

r = 2

19b. [2 marks]Find .

Markschemecorrect calculation or listing terms (A1)

e.g. , , 32, 4, 2, 1

A1 N2

[2 marks]

Examiners reportThis question was very well done by the majority of candidates. There were some who used a value of r greater than one, with themost common error being .

u6

32 × ( )12

6−18 × ( )1

2

3…

= 1u6

r = 2

19c. [2 marks]Find the sum to infinity of this sequence.

Markschemeevidence of correct substitution in A1

e.g. ,

A1 N1

[2 marks]

Examiners reportA handful of candidates struggled with the basic computation involved in part (c).

S∞

32

1− 12

3212

= 64S∞

20a. [5 marks]

The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, −1) .

(i) Show that .

(ii) Find .

(iii) Show that is perpendicular to .

Markscheme(i) correct approach A1

e.g. ,

AG N0

(ii) appropriate approach (M1)

e.g. , , move 3 to the right and 6 down

A1 N2

(iii) finding the scalar product A1

e.g. ,

valid reasoning R1

e.g. , scalar product is zero

is perpendicular to AG N0

[5 marks]

Examiners reportThe majority of candidates were successful on part (a), finding vectors between two points and using the scalar product to show twovectors to be perpendicular.

= ( )AC−→− 4

2

BD−→

AC−→−

BD−→

−OC−→−

OA−→− ( ) − ( )5

210

= ( )AC−→− 4

2

D − B ( ) − ( )4−1

15

= ( )BD−→ 3

−6

4(3) + 2(−6) 12 − 12

4(3) + 2(−6) = 0

AC−→−

BD−→

20b. [4 marks]The line (AC) has equation .

(i) Write down vector u and vector v .

(ii) Find a vector equation for the line (BD).

Markscheme(i) correct “position” vector for u; “direction” vector for v A1A1 N2

e.g. , ; ,

accept in equation e.g.

(ii) any correct equation in the form , where

, A2 N2

[4 marks]

Examiners reportAlthough a large number of candidates answered part (b) correctly, there were many who had trouble with the vector equation of aline. Most notably, there were those who confused the position vector with the direction vector, and those who wrote their equation inan incorrect form.

r = u + sv

u = ( )52

u = ( )10

v = ( )42

v = ( )−2−1

( ) + t ( )52

−4−2

r = a + tb b = BD−→

r = ( ) + t ( )15

3−6

( ) = ( ) + t ( )x

y

4−1

−12

20c. [3 marks]The lines (AC) and (BD) intersect at the point .

Show that .

MarkschemeMETHOD 1

substitute (3, k) into equation for (AC) or (BD) (M1)

e.g. ,

value of t or s A1

e.g. , , ,

substituting A1

e.g.

AG N0

METHOD 2

setting up two equations (M1)

e.g. , ; setting vector equations of lines equal

value of t or s A1

e.g. , , ,

substituting A1

e.g.

AG N0

[3 marks]

P(3, k)

k = 1

3 = 1 + 4s 3 = 1 + 3t

s = 12

− 12

t = 23

− 13

k = 0 + (2)12

k = 1

1 + 4s = 4 + 3t 2s = −1 − 6t

s = 12

− 12

t = 23

− 13

r = ( ) − ( )4−1

13

3−6

k = 1

Examiners reportIn part (c), most candidates seemed to know what was required, though there were many who made algebraic errors when solving forthe parameters. A few candidates worked backward, using , which is not allowed on a "show that" question.k = 1

20d. [5 marks]The lines (AC) and (BD) intersect at the point .

Hence find the area of triangle ACD.

Markscheme (A1)

(A1)

(A1)

area ( ) M1

A1 N4

[5 marks]

Examiners reportIn part (d), candidates attempted many different geometric and vector methods to find the area of the triangle. As the question said"hence", it was required that candidates should use answers from their previous working - i.e. and . Somegeometric approaches, while leading to the correct answer, did not use "hence" or lacked the required justification.

P(3, k)

= ( )PD−→ 1

−2

| | =PD−→

+22 12− −−−−√ (= )5√

| | =AC−→−

+42 22− −−−−√ (= )20−−√

= × | | × | |12

AC−→−

PD−→

= × ×12

20−−√ 5√

= 5

AC⊥BD P(3, 1)

f(x) = 3

21a. [7 marks]

Let . The following diagram shows part of the graph of f .

The point , where , lies on the graph of f . The tangent at P crosses the x-axis at the point . This tangent intersects

the graph of f at the point R(−2, −8) .

(i) Show that the gradient of [PQ] is .

(ii) Find .

(iii) Hence show that .

f(x) = x3

P(a,f(a)) a > 0 Q( ,0)23

a3

a− 23

(a)f ′

a = 1

Markscheme(i) substitute into gradient (M1)

e.g.

substituting

e.g. A1

gradient AG N0

(ii) correct answer A1 N1

e.g. , ,

(iii) METHOD 1

evidence of approach (M1)

e.g. ,

simplify A1

e.g.

rearrange A1

e.g.

evidence of solving A1

e.g.

AG N0

METHOD 2

gradient RQ A1

simplify A1

e.g.

evidence of approach (M1)

e.g. , ,

simplify A1

e.g. ,

AG N0

[7 marks]

Examiners reportPart (a) seemed to be well-understood by many candidates, and most were able to earn at least partial marks here. Part (ai) was a"show that" question, and unfortunately there were some candidates who did not show how they arrived at the given expression.

= −y1 y2

−x1 x2

f(a)−0

a− 23

f(a) = a3

−0a3

a− 23

a3

a− 23

3a2 (a) = 3f ′ (a) =f ′ a3

a− 23

(a) = gradientf ′ 3 =a2 a3

a− 23

3 (a − ) =a2 23

a3

3 − 2 =a3 a2 a3

2 − 2 = 2 (a − 1) = 0a3 a2 a2

a = 1

= −8

−2− 23

,3−8

− 83

(a) = gradientf ′ 3 =a2 −8

−2− 23

= 3a3

a− 23

3 = 3a2 = 1a2

a = 1

21b. [9 marks]

The equation of the tangent at P is . Let T be the region enclosed by the graph of f , the tangent [PR] and the line , between and where . This is shown in the diagram below.

Given that the area of T is , show that k satisfies the equation .

Markschemeapproach to find area of T involving subtraction and integrals (M1)

e.g. , ,

correct integration with correct signs A1A1A1

e.g. ,

correct limits and k (seen anywhere) A1

e.g. ,

attempt to substitute k and (M1)

correct substitution into their integral if 2 or more terms A1

e.g.

setting their integral expression equal to (seen anywhere) (M1)

simplifying A1

e.g.

AG N0

[9 marks]

Examiners reportIn part (b), the concept seemed to be well-understood. Most candidates saw the necessity of using definite integrals and subtractingthe two functions, and the integration was generally done correctly. However, there were a number of algebraic and arithmetic errorswhich prevented candidates from correctly showing the desired final result.

y = 3x − 2 x = k

x = −2 x = k −2 < k < 1

2k + 4 − 6 + 8 = 0k4 k2

∫ f − (3x − 2)dx (3x − 2) −∫ k

−2 ∫ k

−2 x3 ∫ ( − 3x + 2)x3

− + 2x14

x4 32

x2 − 2x −32

x2 14

x4

−2

( − 3x + 2)dx∫ k

−2 x3 [ − + 2x]14

x4 32

x2k

−2

−2

( − + 2k) − (4 − 6 − 4)14

k4 32

k2

2k + 4

− + 2 = 014

k4 32

k2

− 6 + 8 = 0k4 k2

22a. [1 mark]

The n term of an arithmetic sequence is given by .

Write down the common difference.

th = 5 + 2nun

Markscheme A1 N1

[1 mark]

Examiners reportThe majority of candidates could either recognize the common difference in the formula for the n term or could find it by writing outthe first few terms of the sequence.

d = 2

th

22b. [5 marks](i) Given that the n term of this sequence is 115, find the value of n .

(ii) For this value of n , find the sum of the sequence.

Markscheme(i) (A1)

A1 N2

(ii) (may be seen in above) (A1)

correct substitution into formula for sum of arithmetic series (A1)

e.g. , ,

(accept ) A1 N3

[5 marks]

Examiners reportPart (b) demonstrated that candidates were not familiar with expression, "n term". Many stated that the first term was 5 and thendecided to use their own version of the nth term formula leading to a great many errors in (b) (ii).

th

5 + 2n = 115

n = 55

= 7u1

= (7 + 115)S55552

= (2(7) + 54(2))S55552

(5 + 2k)∑k=1

55

= 3355S55 3360

th

23a. [2 marks]

The following diagram represents a large Ferris wheel at an amusement park.

The points P, Q and R represent different positions of a seat on the wheel.

The wheel has a radius of 50 metres and rotates clockwise at a rate of one revolution every 30 minutes.

A seat starts at the lowest point P, when its height is one metre above the ground.

Find the height of a seat above the ground after 15 minutes.

Markschemevalid approach (M1)

e.g. 15 mins is half way, top of the wheel,

height (metres) A1 N2

[2 marks]

Examiners reportPart (a) was well done with most candidates obtaining the correct answer.

d + 1

= 101

23b. [5 marks]After six minutes, the seat is at point Q. Find its height above the ground at Q.

Markschemeevidence of identifying rotation angle after 6 minutes A1

e.g. , of a rotation,

evidence of appropriate approach (M1)

e.g. drawing a right triangle and using cosine ratio

correct working (seen anywhere) A1

e.g. ,

evidence of appropriate method M1

e.g. height

height (metres) (accept 35.6) A1 N2

[5 marks]

Examiners reportPart (b) however was problematic with most errors resulting from incorrect, missing or poorly drawn diagrams. Many did notrecognize this as a triangle trigonometric problem while others used the law of cosines to find the chord length rather than the verticalheight, but this was only valid if they then used this to complete the problem. Many candidates misinterpreted the question as one thatwas testing arc length and area of a sector and made little to no progress in part (b).

Still, others recognized that 6 minutes represented of a rotation, but the majority then thought the height after 6 minutes should be of the maximum height, treating the situation as linear. There were even a few candidates who used information given later in thequestion to answer part (b). Full marks are not usually awarded for this approach.

2π

515

72∘

cos =2π

5x

5015.4(508 …)

= radius + 1 − 15.45…

= 35.5

15

15

23c. [6 marks]The height of the seat above ground after t minutes can be modelled by the function .

Find the value of b and of c .

h(t) = 50 sin(b(t − c)) + 51

MarkschemeMETHOD 1

evidence of substituting into (M1)

correct substitution

e.g. period = 30 minutes, A1

A1 N2

substituting into (M1)

e.g. ,


A1 N2

METHOD 2

evidence of setting up a system of equations (M1)

two correct equations

e.g. , A1A1

attempt to solve simultaneously (M1)

e.g. evidence of combining two equations

, A1A1 N2N2

[6 marks]

Examiners reportPart (c) was not well done. It was expected that candidates simply use the formula to find the value of b and then substitute backinto the equation to find the value of c. However, candidates often preferred to set up a pair of equations and attempt to solve themanalytically, some successful, some not. No attempts were made to solve this system on the GDC indicating that candidates do not getexposed to many “systems” that are not linear. Confusing radians and degrees here did nothing to improve the lack of success.

b = 2π

period

b = 2π

30

b = 0.209 ( )π

15

h(t)

h(0) = 1 h(15) = 101

1 = 50 sin (− c) + 51π

15

c = 7.5

1 = 50 sin b(0 − c) + 51 101 = 50 sin b(15 − c) + 51

b = 0.209 ( )π

15c = 7.5

2π

period

23d. [3 marks]The height of the seat above ground after t minutes can be modelled by the function .

Hence find the value of t the first time the seat is above the ground.

Markschemeevidence of solving (M1)

e.g. equation, graph

(minutes) A2 N3

[3 marks]

Examiners reportIn part (d), candidates were clear on what was required and set their equation equal to 96. Yet again however, solving this equationgraphically using a GDC proved too daunting a task for most.

h(t) = 50 sin(b(t − c)) + 51

96 m

h(t) = 96

t = 12.8

24a. [3 marks]

In an arithmetic sequence , and .

Find the value of the common difference.

= 7u1 = 64u20 = 3709un

Markschemeevidence of choosing the formula for 20th term (M1)

e.g.

correct equation A1

e.g. ,

A1 N2

[3 marks]

Examiners reportThe majority of candidates gained full marks in this question. However, the presentation of work was often disappointing, and therewere a few incorrect trial and error approaches in part (a) resulting in division by 20 rather than 19.

= + 19du20 u1

64 = 7 + 19d d = 64−719

d = 3

24b. [2 marks]Find the value of n .

Markschemecorrect substitution into formula for A1

e.g. ,

A1 N1

[2 marks]

Examiners reportThe majority of candidates gained full marks in this question. However, the presentation of work was often disappointing, and therewere a few incorrect trial and error approaches in part (a) resulting in division by 20 rather than 19.

un

3709 = 7 + 3(n − 1) 3709 = 3n + 4

n = 1235

25a. [2 marks]

Let , for .

Show that .

Markschemecombining 2 terms (A1)

e.g. ,

expression which clearly leads to answer given A1

e.g. ,

AG N0

[2 marks]

Examiners reportFew candidates had difficulty with part (a) although it was often communicated using some very sloppy applications of the rules of

logarithm, writing instead of .

f(x) = + 16 − 4log3x

2log3 log3 x > 0

f(x) = 2xlog3

8x − 4log3 log3 x + 4log312

log3

log38x

4log3

4x

2

f(x) = 2xlog3

log 16log 4

log( )164

25b. [3 marks]Find the value of and of .f(0.5) f(4.5)

Markschemeattempt to substitute either value into f (M1)

e.g. ,

, A1A1 N3

[3 marks]

Examiners reportPart (b) was generally done well.

1log3 9log3

f(0.5) = 0 f(4.5) = 2

25c. [6 marks]The function f can also be written in the form .

(i) Write down the value of a and of b .

(ii) Hence on graph paper, sketch the graph of f , for , , using a scale of 1 cm to 1 unit on each axis.

(iii) Write down the equation of the asymptote.

Markscheme(i) , A1A1 N1N1

(ii)

A1A1A1 N3

Note: Award A1 for sketch approximately through , A1 for approximately correct shape, A1 for sketchasymptotic to the y-axis.

(iii) (must be an equation) A1 N1

[6 marks]

Examiners reportPart (c) (i) was generally done well; candidates seemed quite comfortable changing bases. There were some very good sketches in (c)(ii), but there were also some very poor ones with candidates only considering shape and not the location of the x-intercept or theasymptote. A surprising number of candidates did not use the scale required by the question and/or did not use graph paper to sketchthe graph. In some cases, it was evident that students simply transposed their graphs from their GDC without any analyticalconsideration.

f(x) = ln ax

ln b

−5 ≤ x ≤ 5 −5 ≤ y ≤ 5

a = 2 b = 3

(0.5 ± 0.1, 0 ± 0.1)

x = 0

(0)−1

25d. [1 mark]Write down the value of .

Markscheme A1 N1

[1 mark]

Examiners reportPart (d) was poorly done as candidates did not consider the command term, “write down” and often proceeded to find the inversefunction before making the appropriate substitution.

(0)f−1

(0) = 0.5f−1

25e. [4 marks]The point A lies on the graph of f . At A, .

On your diagram, sketch the graph of , noting clearly the image of point A.

Markscheme

A1A1A1A1 N4

Note: Award A1 for sketch approximately through , A1 for approximately correct shape of the graph reflectedover , A1 for sketch asymptotic to x-axis, A1 for point clearly marked and on curve.

[4 marks]

Examiners reportPart (e) eluded a great many candidates as most preferred to attempt to find the inverse analytically rather than simply reflecting thegraph of f in the line . This graph also suffered from the same sort of problems as the graph in (c) (ii). Some students did nothave their curve passing through nor did they clearly indicate its position as instructed. This point was often mislabelled onthe graph of f. The efforts in this question demonstrated that students often work tenuously from one question to the next, withoutconsidering the "big picture", thereby failing to make important links with earlier parts of the question.

x = 4.5

f−1

(0 ± 0.1, 0.5 ± 0.1)y = x (2 ± 0.1, 4.5 ± 0.1)

y = x

(2, 4.5)

26a. [2 marks]

In an arithmetic sequence, and .

Find d .

= 2u1 = 8u3

Markschemeattempt to find d (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportThis question was answered correctly by the large majority of candidates. The few mistakes seen were due to either incorrectsubstitution into the formula or simple arithmetic errors. Even where candidates made mistakes, they were usually able to earn fullfollow-through marks in the subsequent parts of the question.

−u3 u1

28 = 2 + 2d

d = 3


Markschemecorrect substitution (A1)

e.g. ,

A1 N2

[2 marks]


u20

= 2 + (20 − 1)3u20 = 3 × 20 − 1u20

= 59u20

26c. [2 marks]Find .


e.g. ,

A1 N2

[2 marks]


S20

= (2 + 59)S20202

= (2 × 2 + 19 × 3)S20202

= 610S20

27a. [4 marks]

Let and .

Express in the form , where .

f(x) = 3 lnx g(x) = ln5x3

g(x) f(x) + lna a ∈ Z+

Markschemeattempt to apply rules of logarithms (M1)

e.g. ,

correct application of (seen anywhere) A1

e.g.

correct application of (seen anywhere) A1

e.g.

so

(accept ) A1 N1

[4 marks]

Examiners reportThis question was very poorly done by the majority of candidates. While candidates seemed to have a vague idea of how to apply therules of logarithms in part (a), very few did so successfully. The most common error in part (a) was to begin incorrectly with

. This error was often followed by other errors.

ln = b lnaab lnab = lna + lnb

ln = b lnaab

3 lnx = lnx3

lnab = lna + lnb

ln5 = ln5 + lnx3 x3

ln5 = ln5 + 3 lnxx3

g(x) = f(x) + ln5 g(x) = 3 lnx + ln5

ln5 = 3 ln5xx3

27b. [3 marks]The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.

Markschemetransformation with correct name, direction, and value A3

e.g. translation by , shift up by , vertical translation of

[3 marks]

Examiners reportIn part (b), very few candidates were able to describe the transformation as a vertical translation (or shift). Many candidates attemptedto describe numerous incorrect transformations, and some left part (b) entirely blank.

( )0ln5

ln5 ln5

28a. [5 marks]

The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are long. The angle between the sloping sides of the window and the base is , where .

Show that the area of the window is given by .

2 m θ 0 < θ < π

2

y = 4 sin θ + 2 sin 2θ

Markschemeevidence of finding height, h (A1)

e.g. ,

evidence of finding base of triangle, b (A1)

e.g. ,

attempt to substitute valid values into a formula for the area of the window (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of ) A1

e.g. ,

attempt to replace by M1

e.g.

AG N0

[5 marks]

Examiners reportAs the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) wasgenerally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in aclear and organized manner. Some tried a "working backwards" approach, earning no marks.

sin θ = h

22 sin θ

cosθ = b

22 cosθ

θ

2 ( × 2 cosθ × 2 sin θ) + 2 × 2 sin θ12

(2 sin θ)(2 + 2 + 4 cosθ)12

2 sin θ cosθ sin 2θ

4 sin θ + 2(2 sin θ cosθ)

y = 4 sin θ + 2 sin 2θ

28b. [4 marks]Zoe wants a window to have an area of . Find the two possible values of .

Markschemecorrect equation A1

e.g. ,

evidence of attempt to solve (M1)

e.g. a sketch,

, A1A1 N3

[4 marks]

Examiners reportIn part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and insteadattempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a secondsolution outside the domain.

5 m2 θ

y = 5 4 sin θ + 2 sin 2θ = 5

4 sin θ + 2 sin θ − 5 = 0

θ = 0.856 ( )49.0∘ θ = 1.25 ( )71.4∘

28c. [7 marks]John wants two windows which have the same area A but different values of .

Find all possible values for A .

θ

Markschemerecognition that lower area value occurs at (M1)

finding value of area at (M1)

e.g. , draw square

(A1)

recognition that maximum value of y is needed (M1)

(A1)

(accept ) A2 N5

[7 marks]

Examiners reportA pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or themaximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident fromcandidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken intoconsideration for future paper writing.

θ = π

2

θ = π

2

4 sin ( ) + 2 sin (2 × )π

2π

2

A = 4

A = 5.19615…

4 < A < 5.20 4 < A < 5.19

29a. [3 marks]Expand and simplify your result.

Markschemeevidence of expanding M1

e.g. ,

A2 N2

[3 marks]

Examiners reportSurprisingly few candidates employed the binomial theorem, choosing instead to expand by repeated use of the distributive property.This earned full marks if done correctly, but often proved prone to error.

(2 + x)4

+ 4( )x + 6( ) + 4(2) +24 23 22 x2 x3 x4 (4 + 4x + )(4 + 4x + )x2 x2

(2 + x = 16 + 32x + 24 + 8 +)4 x2 x3 x4

29b. [3 marks]Hence, find the term in in .

Markschemefinding coefficients 24 and 1 (A1)(A1)

term is A1 N3

[3 marks]

Examiners reportCandidates often expanded the entire expression in part (b). Few recognized that only two distributions are required to answer thequestion. Some gave the coefficient as the final answer.

x2 (2 + x (1 + ))4 1x2

25x2

30a. [1 mark]

The straight line with equation makes an acute angle with the x-axis.

Write down the value of .

y = x34

θ

tanθ

Markscheme (do not accept ) A1 N1

[1 mark]

Examiners reportMany candidates drew a diagram to correctly find , although few recognized that a line through the origin can be expressed as

, with gradient , which is explicit in the syllabus.

tanθ = 34

x34

tanθ

y = x tanθ tanθ

30b. [6 marks]Find the value of

(i) ;

(ii) .

Markscheme(i) , (A1)(A1)


e.g.

A1 N3

(ii) correct substitution A1

e.g. ,

A1 N1

[6 marks]

Examiners reportA surprising number were unable to find the ratios for and from . It was not uncommon for candidates to useunreasonable values, such as and , or to write nonsense such as .

sin 2θ

cos2θ

sin θ = 35

cosθ = 45

sin 2θ = 2 ( ) ( )35

45

sin 2θ = 2425

cos2θ = 1 − 2( )35

2−( )4

5

2 ( )35

2

cos2θ = 725

sin θ cosθ tanθ

sin θ = 3 cosθ = 4 2 sin cos35

45

31a. [2 marks]

Let , for .

Show that .

Markschemeinterchanging x and y (seen anywhere) (M1)

e.g. (accept any base)

evidence of correct manipulation A1

e.g. , , ,

AG N0

[2 marks]

Examiners reportCandidates were generally skilled at finding the inverse of a logarithmic function.

f(x) = log3 x√ x > 0

(x) =f−1 32x

x = log y√

=3x y√ =3y x12 x = y1

2log3 2y = xlog3

(x) =f−1 32x

31b. [1 mark]Write down the range of .

Markscheme , A1 N1

[1 mark]

Examiners reportFew correctly gave the range of this function, often stating “all real numbers” or “ ”, missing the idea that the range of aninverse is the domain of the original function.

f−1

y > 0 (x) > 0f−1

y ≥ 0

31c. [4 marks]Let , for .

Find the value of , giving your answer as an integer.

MarkschemeMETHOD 1

finding (seen anywhere) A1

attempt to substitute (M1)

e.g.

evidence of using log or index rule (A1)

e.g. ,

A1 N1

METHOD 2

attempt to form composite (in any order) (M1)

e.g.

evidence of using log or index rule (A1)

e.g. ,

A1

A1 N1

[4 marks]

Examiners reportSome candidates answered part (c) correctly, although many did not get beyond . Some attempted to form the composite in theincorrect order. Others interpreted as multiplication by 2.

g(x) = xlog3 x > 0

( ∘ g)(2)f−1

g(2) = lo 2g3

( ∘ g)(2) =f−1 32 log 23

( ∘ g)(2) =f−1 3log 43 3log322

( ∘ g)(2) = 4f−1

( ∘ g)(x) =f−1 32 xlog3

( ∘ g)(x) =f−1 3log3x23log3x 2

( ∘ g)(x) =f−1 x2

( ∘ g)(2) = 4f−1

32 2log3

( ∘ g)(2)f−1

32. [6 marks]Let . The point lies on the curve of f . At P, the normal to the curve is parallel to . Find the valueof k.

f(x) = kx4 P(1, k) y = − x18

Markschemegradient of tangent (seen anywhere) (A1)

(seen anywhere) A1

recognizing the gradient of the tangent is the derivative (M1)

setting the derivative equal to 8 (A1)

e.g. ,

substituting (seen anywhere) (M1)

A1 N4

[6 marks]

Examiners reportCandidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient ofthe normal line were not bothered by the lack of structure in the question, solving clearly with only a few steps, earning full marks.Those who were unclear often either gained a few marks for finding the derivative and substituting , or no marks for workingthat did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting thederivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among thecandidates who demonstrated greater understanding, more used the gradient of the normal (the equation ) than thegradient of the tangent ( ) ; this led to more algebraic errors in obtaining the final answer of . A number of unsuccessfulcandidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.

= 8

(x) = 4kf ′ x3

4k = 8x3 k = 2x3

x = 1

k = 2

x = 1

− k = −14

18

4k = 8 k = 2

33. [7 marks]Solve , for .

Markschemerecognizing (seen anywhere) (A1)

e.g. ,

recognizing (A1)

e.g.

correct simplification A1

e.g. ,

evidence of correct approach to solve (M1)

e.g. factorizing, quadratic formula

correct working A1

e.g. ,

A2 N3

[7 marks]

Examiners reportCandidates secure in their understanding of logarithm properties usually had success with this problem, solving the resulting quadraticeither by factoring or using the quadratic formula. The majority of successful candidates correctly rejected the solution that was not inthe domain. A number of candidates, however, were unclear on logarithm properties. Some unsuccessful candidates were able todemonstrate understanding of one property but without both were not able to make much progress. A few candidates employed a“guess and check” strategy, but this did not earn full marks.

x + (x − 2) = 3log2 log2 x > 2

loga + logb = logab

(x(x − 2))log2 − 2xx2

b = x ⇔ = bloga ax

= 823

x(x − 2) = 23 − 2x − 8x2

(x − 4)(x + 2) 2± 36√2

x = 4

34a. [3 marks]

Let .

Given that , find the value of .

f(x) = k xlog2

(1) = 8f−1 k

MarkschemeMETHOD 1

recognizing that (M1)

e.g.

recognizing that (A1)

e.g.

A1 N2

METHOD 2

attempt to find the inverse of (M1)

e.g. ,

substituting 1 and 8 (M1)

e.g. ,

A1 N2

[3 marks]

Examiners reportA very poorly done question. Most candidates attempted to find the inverse function for and used that to answer parts (a) and (b).Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able toactually get the correct inverse. Almost none recognized that if , then . Many thought that the letters "log" couldbe simply "cancelled out", leaving the and the .

f(8) = 1

1 = k 8log2

8 = 3log2

1 = 3k

k = 13

f(x) = k xlog2

x = k ylog2 y = 2x

k

1 = k 8log2 = 821k

k = 18log2

(k = )13

f

(1) = 8f−1 f(8) = 12 8


MarkschemeMETHOD 1

recognizing that (M1)

e.g.

(A1)

(accept ) A2 N3

METHOD 2

attempt to find inverse of (M1)

e.g. interchanging x and y , substituting into

correct inverse (A1)

e.g. ,

A2 N3

[4 marks]

( )f−1 23

f(x) = 23

= x23

13

log2

x = 2log2

( ) = 4f−1 23

x = 4

f(x) = x13

log2

k = 13

y = 2x

k

(x) =f−1 23x 23x

( ) = 4f−1 23

Examiners reportA very poorly done question. Most candidates attempted to find the inverse function for and used that to answer parts (a) and (b).Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able toactually get the correct inverse. Almost none recognized that if , then . Many thought that the letters "log" couldbe simply "cancelled out", leaving the and the .

f

(1) = 8f−1 f(8) = 12 8

35. [3 marks]

In a class of 100 boys, 55 boys play football and 75 boys play rugby. Each boy must play at least one sport from football and rugby.

(i) Find the number of boys who play both sports.

(ii) Write down the number of boys who play only rugby.

Markscheme(i) evidence of substituting into (M1)

e.g. , Venn diagram

30 A1 N2

(ii) 45 A1 N1

[3 marks]

Examiners reportOverall, this question was very well done. There were some problems with the calculation of conditional probability, where aconsiderable amount of candidates tried to use a formula instead of using its concept and analysing the problem. It is the kind ofquestion where it can be seen if the concept is not clear to candidates.

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

75 + 55 − 100

36. [6 marks]In an arithmetic sequence, and . Find the value of and of d .= 1900S40 = 106u40 u1

MarkschemeMETHOD 1

substituting into formula for (M1)


e.g.

A1 N2

substituting into formula for or (M1)


e.g. ,

A1 N2

METHOD 2



e.g.



e.g.

, A1A1 N2N2

[6 marks]

Examiners reportMost candidates answered this question correctly. Those who chose to solve with a system of equations often did so algebraically,using a fair bit of time doing so and sometimes making a careless error in the process. Few candidates took advantage of the systemsolving features of the GDC.

S40

1900 = 40( +106)u1

2

= −11u1

u40 S40

106 = −11 + 39d 1900 = 20(−22 + 39d)

d = 3

S40

20(2 + 39d) = 1900u1

u40

106 = + 39du1

= −11u1 d = 3

37. [4 marks]

Let .

Solve the equation .

Markschemecollecting like terms; using laws of logs (A1)(A1)

e.g. , , ,

simplify (A1)

e.g. ,

A1 N2

[4 marks]

Examiners reportWhere students attempted to solve the equation in (b), most treated as and created an incorrect equation from theoutset. The few who applied laws of logarithms often carried the algebra through to completion.

f(x) = ex+3

(x) = lnf−1 1x

lnx − ln( ) = 31x

lnx + lnx = 3 ln( ) = 3x1x

ln = 3x2

lnx = 32

=x2 e3

x = (= )e32 e3−−√

lnx − 3 ln(x − 3)

38. [3 marks]

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.

The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is radians, where .

Let the area of the rectangle be A.

Show that .

Markschemefinding area (M1)

e.g. ,

substituting A1

e.g. ,

A1

AG N0

[3 marks]

Examiners reportThose with answers from (a) could begin part (b), but many worked backwards and thus earned no marks. In a "show that" question,a solution cannot begin with the answer given. The area of the rectangle could be found by using , or by using the eightsmall triangles, but it was essential that the substitution of the double-angle formula was shown before writing the given answer.

θ 0 ≤ θ ≤ π

2

A = 18 sin 2θ

A = 2x × 2y A = 8 × bh12

A = 4 × 3 sin θ × 3 cosθ 8 × × 3 cosθ × 3 sin θ12

A = 18(2 sin θ cosθ)

A = 18 sin 2θ

2x × 2y

39a. [3 marks]

In an arithmetic series, the first term is –7 and the sum of the first 20 terms is 620.

Find the common difference.

Markschemeattempt to substitute into sum formula for AP (accept term formula) (M1)

e.g. ,

setting up correct equation using sum formula A1

e.g.

A1 N2

[3 marks]

Examiners reportThis question was generally well answered. Many candidates who got part (a) wrong, recovered and received full follow throughmarks in part (b). There were a few candidates who confused the term and sum formula in part (a).

= {2(−7) + 19 d}S20202

(or (−7 + ))202

u20

{2(−7) + 19d} = 620202

d = 4

39b. [2 marks]Find the value of the 78th term.


= 301 A1 N2

[2 marks]

Examiners reportThis question was generally well answered. Many candidates who got part (a) wrong, recovered and received full follow throughmarks in part (b). There were a few candidates who confused the term and sum formula in part (a).

= −7 + 77(4)u78

40a. [1 mark]

The fifth term in the expansion of the binomial is given by .

Write down the value of .

Markscheme A1 N1

[1 mark]

Examiners report[N/A]

(a + b)n ( ) (2q104

p6 )4

n

n = 10

40b. [2 marks]Write down a and b, in terms of p and/or q.

Markscheme , (or , ) A1A1 N1N1

[2 marks]

a = p b = 2q a = 2q b = p


40c. [3 marks]Write down an expression for the sixth term in the expansion.

Markscheme A1A1A1 N3

[3 marks]

Examiners reportThe most common error was in (c) where many candidates interpreted the "sixth" term as using , with accompanying powers

of 4 and 6 in the expression.

( ) (2q105

p5 )5

( )106

41a. [1 mark]Find .

Markscheme5 A1 N1

[1 mark]

Examiners reportPart (a) proved very accessible.

32log2

41b. [4 marks]Given that can be written as , find the value of p and of q.

MarkschemeMETHOD 1

(A1)

(A1)

(A1)

, (accept ) A1 N3

METHOD 2

(A1)

(A1)

(A1)

, (accept ) A1 N3

[4 marks]

( )log232x

8y px + qy

( ) = −log232x

8y log232x log28y

= x 32 − y 8log2 log2

8 = 3log2

p = 5 q = −3 5x − 3y

=32x

8y

( )25 x

( )23 y

= 25x

23y

= 25x−3y

( ) = 5x − 3ylog2 25x−3y

p = 5 q = −3 5x − 3y

Examiners reportAlthough many found part (b) accessible as well, a good number of candidates could not complete their way to a final result. Manygave q as a positive value.

42a. [1 mark]

Consider the arithmetic sequence 3, 9, 15, , 1353 .

Write down the common difference.

Markschemecommon difference is 6 A1 N1

[1 mark]

Examiners reportMost candidates did well on this question. Any errors were usually arithmetic in nature but candidates were able to obtain follow-through marks on errors made in earlier parts.

…

42b. [3 marks]Find the number of terms in the sequence.

Markschemeevidence of appropriate approach (M1)

e.g.

correct working A1

e.g. ,

A1 N2

[3 marks]


= 1353un

1353 = 3 + (n − 1)6 1353+36

n = 226

42c. [2 marks]Find the sum of the sequence.

Markschemeevidence of correct substitution A1

e.g. ,

(accept 153000) A1 N1

[2 marks]


=S226226(3+1353)

2(2 × 3 + 225 × 6)226

2

= 153228S226

43a. [2 marks]

An arithmetic sequence, has and .

Find .

Markschemeevidence of equation for M1

e.g. , ,

A1 N1

[2 marks]

Examiners reportThis problem was done well by the vast majority of candidates. Most students set out their working very neatly and logically andgained full marks.

, , … ,u1 u2 u3 d = 11 = 263u27

u1

u27

263 = + 26 × 11u1 = + (n − 1) × 11u27 u1 263 − (11 × 26)

= −23u1

43b. [4 marks](i) Given that , find the value of n .

(ii) For this value of n , find .

Markscheme(i) correct equation A1

e.g. ,

A1 N1

(ii) correct substitution into sum formula A1

e.g. ,

(accept 12300) A1 N1

[4 marks]

Examiners reportThis problem was done well by the vast majority of candidates. Most students set out their working very neatly and logically andgained full marks.

= 516un

Sn

516 = −23 + (n − 1) × 11 539 = (n − 1) × 11

n = 50

=S5050(−23+516)

2=S50

50(2×(−23)+49×11)2

= 12325S50

44. [6 marks]Find the term in in the expansion of .x4 (3 − )x2 2x

5

Markschemeevidence of substituting into binomial expansion (M1)

e.g.

identifying correct term for (M1)

evidence of calculating the factors, in any order A1A1A1

e.g. ;

Note: Award A1 for each correct factor.

A1 N2

Note: Award M1M1A1A1A1A0 for 1080 with working shown.

[6 marks]

Examiners reportAlthough a great number of students recognized they could use the binomial theorem, fewer were successful in finding the term in .

Candidates showed various difficulties when trying to solve this problem:

choosing the incorrect term

attempting to expand by hand

finding only the coefficient of the termnot being able to determine which term would yield an errors in the calculations of the coefficient.

+ ( ) b + ( ) + …a5 51

a4 52

a3b2

x4

( ) ,27 ,52

x6 4x2

10(3x2)3( )−2x

2

term = 1080x4

x4

(3 − )x2 2x

5

x4

45a. [1 mark]Consider the infinite geometric sequence .

Write down the 10th term of the sequence. Do not simplify your answer.

Markscheme A1 N1

[1 mark]

Examiners reportThis question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying

to 30 , and there were a few candidates who used the formula for the finite sum unsuccessfully.

3, 3(0.9), 3(0.9 , 3(0.9 , …)2 )3

= 3(0.9u10 )9

30.1

45b. [4 marks]Consider the infinite geometric sequence .

Find the sum of the infinite sequence.

3, 3(0.9), 3(0.9 , 3(0.9 , …)2 )3

Markschemerecognizing (A1)


e.g.

(A1)

A1 N3

[4 marks]

Examiners reportThis question was well done by most candidates. There were a surprising number of candidates who lost a mark for not simplifying

to 30, and there were a few candidates who used the formula for the finite sum unsuccessfully.

r = 0.9

S = 31−0.9

S = 30.1

S = 30

30.1

46a. [3 marks]Expand and simplify your result.

Markschemeevidence of expanding M1

e.g. A2 N2

[3 marks]

Examiners reportWhere candidates recognized the binomial nature of the expression, many completed the expansion successfully, although someomitted the negative signs.

(x − 2)4

(x − 2 = + 4 (−2) + 6 (−2 + 4x(−2 + (−2)4 x4 x3 x2 )2 )3 )4

(x − 2 = − 8 + 24 − 32x + 16)4 x4 x3 x2

46b. [3 marks]Find the term in in .

Markschemefinding coefficients, , (A1)(A1)

term is A1 N3

[3 marks]

Examiners reportWhere candidates recognized the binomial nature of the expression, many completed the expansion successfully, although someomitted the negative signs. Few recognized that only the multiplications that achieve an index of 3 are required in part (b) anddistributed over the entire expression. Others did not recognize that two terms in the expansion must be combined.

x3 (3x + 4)(x − 2)4

3 × 24(= 72) 4 × (−8)(= −32)

40x3

47a. [3 marks]Consider the arithmetic sequence .

Find .

2, 5, 8, 11,…

u101

Markscheme (A1)

evidence of substitution into (M1)

e.g.

A1 N3

[3 marks]

Examiners reportCandidates probably had the most success with this question with many good solutions which were written with the working clearlyshown. Many used the alternate approach of .

d = 3

= a + (n − 1)dun

= 2 + 100 × 3u101

= 302u101

= 3n − 1un

47b. [3 marks]Consider the arithmetic sequence .

Find the value of n so that .

Markschemecorrect approach (M1)

e.g.

correct simplification (A1)

e.g. , ,

A1 N2

[3 marks]

Examiners reportCandidates probably had the most success with this question with many good solutions which were written with the working clearlyshown. Many used the alternate approach of .

2, 5, 8, 11,…

= 152un

152 = 2 + (n − 1) × 3

150 = (n − 1) × 3 50 = n − 1 152 = −1 + 3n

n = 51

= 3n − 1un

48. [2 marks]

Let .

Show that .

Markscheme A1

A1

AG N0

[2 marks]

Examiners reportThis problem was generally well done. The “show that” question in part (a) was done correctly by most candidates, with a fewattempting to show it by working backwards, which earned no marks.

f(x) = 3(x + 1 − 12)2

f(x) = 3 + 6x − 9x2

f(x) = 3( + 2x + 1) − 12x2

= 3 + 6x + 3 − 12x2

= 3 + 6x − 9x2

49. [3 marks]

Let S be the total area of the two segments shaded in the diagram below.

Show that .

Markschemearea semicircle A1

A1

M1

AG N0

[3 marks]

Examiners reportMany candidates seemed to see why but the arguments presented for showing why this result was true were notvery convincing in many cases. Explicit evidence of why the area of the semicircle was was often missing as was an explanationfor and for subtraction.

S = 2(π − 2 sin θ)

= × π(212

)2 (= 2π)

area ΔAPB = 2 sin θ + 2 sin θ (= 4 sin θ)

S = area of semicircle − area ΔAPB (= 2π − 4 sin θ)

S = 2(π − 2 sin θ)

S = 2(π − 2 sin θ)2π

2(2 sin θ)

50a. [2 marks]

Consider the infinite geometric sequence .

Find the common ratio.

Markschemeevidence of dividing two terms (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportThis question was generally well done by most candidates.

3000, − 1800, 1080, − 648,…

− 18003000

− 18001080

r = −0.6

50b. [2 marks]Find the 10th term.

Markschemeevidence of substituting into the formula for the 10th term (M1)

e.g.

(accept the exact value ) A1 N2

[2 marks]

= 3000(−0.6u10 )9

= 30.2u10 −30.233088

Examiners reportThis question was generally well done by most candidates, although quite a few showed difficulty answering part (b) exactly or tothree significant figures.

50c. [2 marks]Find the exact sum of the infinite sequence.

Markschemeevidence of substituting into the formula for the infinite sum (M1)

e.g.

A1 N2

[2 marks]

Examiners reportThis question was generally well done by most candidates, although quite a few showed difficulty answering part (b) exactly or tothree significant figures. Some candidates reversed the division of terms to obtain a ratio of . Of these, most did not recognize thisratio as an inappropriate value when finding the sum in part (c).

S = 30001.6

S = 1875

− 53

51. [5 marks]Find the term in the expansion of .

Markschemeevidence of using binomial expansion (M1)

e.g. selecting correct term,

evidence of calculating the factors, in any order A1A1A1

e.g. 56 , , ,

(accept = to 3 s.f.) A1 N2

[5 marks]

Examiners reportCandidates produced mixed results in this question. Many showed a binomial expansion in some form, although simply writing rowsof Pascal’s triangle is insufficient evidence. A common error was to answer with the coefficient of the term, and many neglected theuse of brackets when showing working. Although sloppy notation was not penalized if candidates achieved a correct result, for somethe missing brackets led to a wrong answer.

x3 ( x − 3)23

8

+ ( ) b + ( ) + …a8b0 81

a7 82

a6b2

22

33−35 ( ) (−3

85

( x)23

3)5

−4032x3 −4030x3

52. [6 marks]

A city is concerned about pollution, and decides to look at the number of people using taxis. At the end of the year 2000, there were 280 taxisin the city. After n years the number of taxis, T, in the city is given by

(i) Find the number of taxis in the city at the end of 2005.

(ii) Find the year in which the number of taxis is double the number of taxis there were at the end of 2000.

T = 280 × .1.12n

Markscheme(i) (A1)

A1 N2

(ii) evidence of doubling (A1)

e.g. 560

setting up equation A1

e.g. ,

(A1)

in the year 2007 A1 N3

[6 marks]

Examiners reportA number of candidates found this question very accessible. In part (a), many correctly solved for n, but often incorrectly answeredwith the year 2006, thus misinterpreting that 6.12 years after the end of 2000 is in the year 2007.

n = 5

T = 280 × 1.125

T = 493

280 × = 5601.12n = 21.12n

n = 6.116…

53a. [3 marks]

In a geometric series, and .

Find the value of .

Markschemeevidence of substituting into formula for th term of GP (M1)

e.g.

setting up correct equation A1

A1 N2

[3 marks]

Examiners reportPart (a) was well done.

=u11

81=u4

13

r

n

=u41

81r3

=181

r3 13

r = 3

53b. [4 marks]Find the smallest value of n for which .> 40Sn

MarkschemeMETHOD 1

setting up an inequality (accept an equation) M1

e.g. , ,

evidence of solving M1

e.g. graph, taking logs

(A1)

A1 N2

METHOD 2

if , sum ; if , sum A2

(is the smallest value) A2 N2

[4 marks]

Examiners reportIn part (b) a good number of candidates did not realize that they could use logs to solve the problem, nor did they make good use oftheir GDCs. Some students did use a trial and error approach to check various values however, in many cases, they only checked oneof the "crossover" values. Most candidates had difficulty with notation, opting to set up an equation rather than an inequality.

> 40( −1)1

813n

2> 40

(1− )181

3n

−2> 64813n

n > 7.9888…

∴ n = 8

n = 7 = 13.49… n = 8 = 40.49…

n = 8

54a. [1 mark]Expand as the sum of four terms.

Markscheme (accept ) A1 N1

[1 mark]

Examiners reportThis question proved difficult for many candidates. A number of students seemed unfamiliar with sigma notation. Many weresuccessful with part (a), although some listed terms or found an overall sum with no working.

∑r=4

72r

= + + +∑r=4

72r 24 25 26 27 16 + 32 + 64 + 128

54b. [6 marks](i) Find the value of .

(ii) Explain why cannot be evaluated.

∑r=4

302r

∑r=4

∞2r

Markscheme(i) METHOD 1

recognizing a GP (M1)

, , (A1)

correct substitution into formula for sum (A1)

e.g.

A1 N4

METHOD 2

recognizing (M1)

recognizing GP with , , (A1)

correct substitution into formula for sum

(A1)

A1 N4

(ii) valid reason (e.g. infinite GP, diverging series), and (accept ) R1R1 N2

[6 marks]

Examiners reportThe results for part (b) were much more varied. Many candidates did not realize that was and used instead. Very fewcandidates gave a complete explanation why the infinite series could not be evaluated; candidates often claimed that the value couldnot be found because there were an infinite number of terms.

=u1 24 r = 2 n = 27

=S27( −1)24 227

2−1

= 2147483632S27

= −∑r=4

30∑r=1

30∑r=1

3

= 2u1 r = 2 n = 30

=S302( −1)230

2−1

= 214783646

= 2147483646 − (2 + 4 + 8)∑r=4

302r

= 2147483632

r ≥ 1 r > 1

n 27 30

55a. [3 marks]

In an arithmetic series, the first term is −7 and the sum of the first 20 terms is 620.

Find the common difference.

Markschemeattempt to substitute into sum formula for AP M1

e.g. ,

setting up correct equation using sum formula A1

e.g.

A1 N2

[3 marks]


= (2(−7) + 19d)S20202

(−7 + )202

u20

(2(−7) + 19d = 620202

d = 4

55b. [2 marks]Find the value of the 78 term.th

Printed for sek-catalunya

© International Baccalaureate Organization 2014 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemecorrect substitution A1

A1 N2

[2 marks]


−7 + 77(4)

= 301u78

New test - October 20, 2014 - David Delgado...New test - October 20, 2014 [389 marks]1a.The first...

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