New test - February 23, 2016 [276 marks]

Texto completo

(1)

New test - February 23, 2016

[276 marks]

[2 marks] 1a.

The first three terms of a infinite geometric sequence are , where .

Write down an expression for the common ratio, .

Markscheme

correct expression for A1 N1 eg

[2 marks]

Examiners report

[N/A]

m− 1, 6, m+ 4 m∈Z

r

r r=m6−1, m+4

6

[2 marks] 1b. Hence, show that satisfies the equation .

Markscheme

correct equation A1 eg

correct working (A1) eg

correct working A1 eg

AG N0 [2 marks]

Examiners report

[N/A]

m m2+ 3m− 40 = 0

= , =

6

m−1

m+4 6

6

m+4

m−1 6

(m+ 4)(m− 1) = 36

m+ 4m− 4 = 36, + 3m− 4 = 36

m2 m2

+ 3m− 40 = 0 m2

[3 marks] 1c. Find the two possible values of .

Markscheme

valid attempt to solve (M1) eg

A1A1 N3 [3 marks]

Examiners report

[N/A]

m

(m+ 8)(m− 5) = 0, m=−3± 9+4×40√2 m= −8, m= 5

(2)

Markscheme

attempt to substitute any value of to find (M1) eg

A1A1 N3 [3 marks]

Examiners report

[N/A]

m r

,

6 −8−1

5+4 6

r= , r32 = −23

1e. The sequence has a finite sum. [3 marks]

State which value of leads to this sum and justify your answer.

Markscheme

(may be seen in justification) A1

valid reason R1 N0 eg

Notes: Award R1 for only if A1 awarded.

[2 marks]

Examiners report

[N/A]

r

r= −23

|r| < 1, − 1 <−2< 1

3

|r| < 1

1f. The sequence has a finite sum. [3 marks]

Calculate the sum of the sequence.

Markscheme

finding the first term of the sequence which has (A1)

eg

(may be seen in formula) (A1)

correct substitution of and their into , as long as A1

eg

A1 N3 [4 marks]

Examiners report

[N/A]

|r| < 1 −8 − 1, 6 ÷−2

3

= −9 u1

u1 r 1−u1r |r| < 1

= ,

S∞ −9 1−(−2)

3

−9

5 3

= − (= −5.4) S∞ 275

2a. Expand as the sum of four terms. [1 mark] r=4

7

(3)

Markscheme

(accept ) A1 N1

[1 mark]

Examiners report

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

= + + +

r=4 7

2r 24 25 26 27 16 + 32 + 64 + 128

2b. (i) Find the value of . [6 marks]

(ii) Explain why cannot be evaluated.

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 report

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

r=4 30 2rr=4 ∞ 2r =

u1 24 r= 2 n= 27

= S27 2( −1)

4227

2−1 = 2147483632 S27 = − ∑ r=4 30 ∑ r=1 30 ∑ r=1 3 = 2

u1 r= 2 n= 30

= S30 2(2 −1)

30

2−1

= 214783646

= 2147483646 − (2 + 4 + 8)

r=4 30

2r

= 2147483632

r≥ 1 r> 1

n 27 30

[1 mark] 3a.

The n term of an arithmetic sequence is given by .

Write down the common difference.

(4)

Markscheme

A1 N1 [1 mark]

Examiners report

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

d= 2

th

3b. (i) Given that the n term of this sequence is 115, find the value of n . [5 marks]

(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 report

Part (b) demonstrated that candidates were not familiar with expression, "n term". Many stated that the first term was 5 and then decided 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

= 7 u1

= (7 + 115)

S55 552 S55= (2(7) + 54(2))552 ∑(5 + 2k)

k=1 55

= 3355

S55 3360

th

[3 marks] 4a.

The diagram shows a circle of radius metres. The points ABCD lie on the circumference of the circle.

BC = m, CD = m, AD = m, , and .

Find AC.

8

(5)

Markscheme

evidence of choosing cosine rule (M1)

eg ,

correct substitution A1

eg ,

AC (m) A1 N2 [3 marks]

Examiners report

There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible

combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus.

= + − 2abcosC

c2 a2 b2 CD2+ AD2− 2 × CD × ADcosD

+ − 2 × 11.5 × 8 cos104

11.52 82 196.25 − 184 cos104

= 15.5

4b. (i) Find . [5 marks]

(ii) Hence, find .

Markscheme

(i) METHOD 1

evidence of choosing sine rule (M1)

eg ,

correct substitution A1

eg

A1 N2

METHOD 2

evidence of choosing cosine rule (M1) eg

correct substitution A1

e.g.

A1 N2

(ii) subtracting their from (M1) eg ,

A1 N2

[5 marks] A DC^

A BC^

=

sinA a

sinB

b =

sin A DCˆ AD

sinD

AC

=

sin A DC^ 8

sin 104 15.516…

A D =C^ 30.0∘

= + − 2abcosC c2 a2 b2

= + 15.516 − 2(11.5)(15.516…)cosC

82 11.52 2

A D =C^ 30.0∘

A DC^ 73

73 − A DC^ 70 − 30.017…

(6)

Examiners report

There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible

combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus.

[2 marks] 4c. Find the area of triangle ADC.

Markscheme

correct substitution (A1) eg area

area (m ) A1 N2 [2 marks]

Examiners report

There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible

combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and most could find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrong mode or were rounding values prematurely while some misinformed candidates treated ADC as a right-angled triangle.

ΔADC = (8)(11.5)sin 1041 2

= 44.6 2

4d. (c) Find the area of triangle ADC. [6 marks]

(d) Hence or otherwise, find the total area of the shaded regions.

Markscheme

(c) correct substitution (A1) eg area

area (m ) A1 N2 [2 marks]

(d) attempt to subtract (M1)

eg ,

area (A1)

correct working A1

eg ,

shaded area is (m ) A1 N3 [4 marks]

Total [6 marks]

ΔADC = (8)(11.5)sin 1041 2

= 44.6 2

circle − ABCD πr2− ΔADC − ΔACB ΔACB = (15.516…)(14)sin 42.981

2

π(8 − 44.6336… − (15.516…)(14)sin 42.98)2 1

2 64π− 44.6 − 74.1

(7)

Examiners report

There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible

combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and most could find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrong mode or were rounding values prematurely while some misinformed candidates treated ADC as a right-angled triangle. In part (d), most candidates recognized what to do and often obtained follow through marks from errors made in previous parts.

[4 marks] 4e. Hence or otherwise, find the total area of the shaded regions.

Markscheme

attempt to subtract (M1)

eg ,

area (A1)

correct working A1

eg ,

shaded area is (m ) A1 N3 [4 marks]

Total [6 marks]

Examiners report

There was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer the question, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possible

combinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended. Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due to the fact that cyclic quadrilaterals is not part of the syllabus.

In part (d), most candidates recognized what to do and often obtained follow through marks from errors made in previous parts.

circle − ABCD πr2− ΔADC − ΔACB

ΔACB = (15.516…)(14)sin 42.981 2

π(8 − 44.6336… − (15.516…)(14)sin 42.98)2 1

2 64π− 44.6 − 74.1

(8)

5a. [6 marks]

The diagram below shows part of the graph of a function .

The graph has a maximum at A( , ) and a minimum at B( , ) .

The function can be written in the form . Find the value of

(a) (b) (c) .

Markscheme

(a) valid approach to find (M1) eg amplitude ,

A1 N2 [2 marks]

(b) valid approach to find (M1) eg period = 4 ,

A1 N2 [2 marks]

(c) valid approach to find (M1) eg axis = , sketch of horizontal axis,

A1 N2 [2 marks]

Total [6 marks]

Examiners report

Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters , and . The successful candidates were able to find the answers using the given points and their understanding of the different transformations.

Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between and the period of the function. There were also some candidates who showed working such as without explaining what represented.

f

1 5 3 −1

f f(x) =psin(qx) +r

p

q

r

p

=max−min

2 p= 6

p= 3

q

q=period2π

q=π2

r

max+min

2 f(0)

r= 2

p q r

q

2π

(9)

[2 marks] 5b.

Markscheme

valid approach to find (M1) eg amplitude ,

A1 N2 [2 marks]

Examiners report

Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters , and . The successful candidates were able to find the answers using the given points and their understanding of the different transformations.

Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between and the period of the function. There were also some candidates who showed working such as without explaining what represented.

p

p

=max−min

2 p= 6

p= 3

p q r

q

2π

b b

[2 marks] 5c.

Markscheme

valid approach to find (M1) eg period = 4 ,

A1 N2 [2 marks]

Examiners report

Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters , and . The successful candidates were able to find the answers using the given points and their understanding of the different transformations.

Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between and the period of the function. There were also some candidates who showed working such as without explaining what represented.

q

q

q=period2π

q=π2

p q r

q

2π

b b

[2 marks] 5d. .

Markscheme

valid approach to find (M1)

eg axis = , sketch of horizontal axis, A1 N2

[2 marks]

Total [6 marks] r

r

max+min

2 f(0)

r= 2

p q r

q

2π

(10)

[2 marks] 6a.

The following diagram shows a right-angled triangle, , where .

Show that .

Markscheme

METHOD 1

approach involving Pythagoras’ theorem (M1) eg , labelling correct sides on triangle finding third side is 12 (may be seen on diagram) A1

AG N0

METHOD 2

approach involving (M1) eg

correct working A1 eg

AG N0 [2 marks]

Examiners report

[N/A]

ABC sin A = 5

13

cosA=12 13

+ =

52 x2 132

cosA=12 13

θ+ θ= 1 sin2 cos2

+ θ= 1, + = 1

( )135 2 cos2 x2 25 169

θ= cos2 144

169

cosA=12 13

[3 marks] 6b. Find .

Markscheme

correct substitution into (A1) eg

correct working (A1) eg

A1 N2 [3 marks]

Examiners report

[N/A] cos2A

cos2θ

1 − 2( )135 2, 2( )12 − 1, −

13 2

( )12 13

2 ( )135 2

1 − 50, − 1, −

169 288 169 144169

25 169

(11)

[2 marks] 7a.

The following diagram shows a circle with centre and radius .

The points , and lie on the circumference of the circle, and radians.

Find the length of the arc .

Markscheme

correct substitution into arc length formula (A1) eg

arc length (cm) A1 N2 [2 marks]

Examiners report

[N/A]

O 5 cm

A rmB rmC A C = 0.7O^

ABC

0.7 × 5 = 3.5

[2 marks] 7b. Find the perimeter of the shaded sector.

Markscheme

valid approach (M1) eg

perimeter (cm) A1 N2 [2 marks]

Examiners report

[N/A]

3.5 + 5 + 5, arc + 2r = 13.5

[2 marks] 7c. Find the area of the shaded sector.

Markscheme

correct substitution into area formula (A1) eg

A1 N2 [2 marks]

Examiners report

[N/A] (0.7)(5

1

2 )2

(12)

[2 marks] 8a.

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

The graph has a maximum at and a minimum at .

Write down the value of .

Markscheme

A2 N2

Note: Award A1for .

[2 marks]

Examiners report

[N/A]

f(x) =pcos(q(x+r)) + 10 0⩽x⩽20 f

(4,18) (16,2)

r

r= −4

r= 4

[2 marks] 8b. Find .

Markscheme

evidence of valid approach (M1) eg , distance from

A1 N2 [2 marks]

Examiners report

[N/A] p

maxy value -- y value

2 y= 10

p= 8

[2 marks] 8c. Find .

Markscheme

valid approach (M1)

eg period is , , substitute a point into their

, (do not accept degrees) A1 N2 [2 marks]

Examiners report

[N/A] q

24 360

24 f(x)

q=2π( , exact) 24

π

(13)

[2 marks] 8d. Solve .

Markscheme

valid approach (M1) eg line on graph at

(accept ) A1 N2 [2 marks]

Note: Do not award the final A1 if additional values are given. If an incorrect value of leads to multiple solutions, award the final

A1 only if all solutions within the domain are given.

Examiners report

[N/A] f(x) = 7

y= 7, 8 cos(2π(x− 4))+ 10 = 7

24

x= 11.46828

x= 11.5 (11.5,7)

q

[3 marks] 9a.

Let . The graph of f passes through the point .

Find the value of .

Markscheme

METHOD 1

attempt to substitute both coordinates (in any order) into (M1) eg

correct working (A1) eg

A1 N2 [3 marks]

METHOD 2

recognizing shift of left means maximum at R1)

recognizing is difference of maximum and amplitude (A1) eg

A1 N2 [3 marks]

Examiners report

[N/A]

f(x) = sin(x+π)+k

4 ( , 6)

π

4

k

f

f( )π4 = 6, = sinπ4 (6 +π4)+k

sin = 1, 1 +π2 k= 6

k= 5

π

4 6

k 6 − 1 k= 5

[2 marks] 9b. Find the minimum value of .

Markscheme

evidence of appropriate approach (M1) eg minimum value of is

minimum value is A1 N2 [2 marks]

Examiners report

[N/A]

f(x)

sinx −1, − 1 +k, (x) = 0, f′ (54π, 4)

(14)

9c. Let . The graph of g is translated to the graph of by the vector . [2 marks] Write down the value of and of .

Markscheme

A1A1 N2

[2 marks]

Examiners report

[N/A]

g(x) = sinx f ( )p

q

p q

p= − , qπ = 5 (accept ( ))

4

π

4

5

[3 marks] 10a.

Consider a circle with centre and radius cm. Triangle is drawn such that its vertices are on the circumference of the circle.

cm, cm and radians.

Find .

Markscheme

Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in

subsequent parts. Accept answers that are consistent with their working.

Candidates may have their GDCs in degree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate.

Ignore missing or incorrect units.

evidence of choosing sine rule (M1) eg

correct substitution (A1) eg

A1 N2 [3 marks]

Examiners report

[N/A]

O 7 ABC

AB = 12.2 BC = 10.4 A B = 1.058C^

B CA^

=

sinA^ a

sinB^ b

=

sinA^

10.4

sin 1.058 12.2

B C = 0.837A^

(15)

Markscheme

Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in

subsequent parts. Accept answers that are consistent with their working.

Candidates may have their GDCs in degree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate.

Ignore missing or incorrect units.

METHOD 1

evidence of subtracting angles from (M1) eg

correct angle (seen anywhere) A1

attempt to substitute into cosine or sine rule (M1)

correct substitution (A1) eg

A1 N3

METHOD 2

evidence of choosing cosine rule M1 eg

correct substitution (A2) eg

A2 N3 [5 marks]

Examiners report

[N/A]

π A C =B^ πAC

A C =B^ π− 1.058 − 0.837, 1.246, 71.4∘

+ − 2 × 12.2 × 10.4cos71.4, =

12.22 10.42 AC

sin 1.246 sin 1.05812.2

AC = 13.3 (cm)

= + − 2bccosA a2 b2 c2

= + − 2 × 10.4bcos1.058 12.22 10.42 b2

AC = 13.3 (cm)

(16)

Markscheme

Notes: In this question, there may be slight differences in answers, depending on which values candidates carry through in

subsequent parts. Accept answers that are consistent with their working.

Candidates may have their GDCs in degree mode, leading to incorrect answers. If working shown, award marks in line with the markscheme, with FT as appropriate.

Ignore missing or incorrect units.

METHOD 1

valid approach (M1)

eg , correct working (A1)

eg

(A1)

EITHER

correct substitution for arc length (seen anywhere) A1 eg

subtracting arc from circumference (M1) eg

OR

attempt to find reflex (M1) eg

correct substitution for arc length (seen anywhere) A1 eg

THEN

A1 N4

METHOD 2

valid approach to find or (M1) eg choosing cos rule, twice angle at circumference correct working for finding one value, or (A1) eg ,

two correct calculations for arc lengths

eg (A1)(A1)

adding their arc lengths (seen anywhere)

eg M1

A1 N4

Note: Candidates may work with other interior triangles using a similar method. Check calculations carefully and award marks in line

with markscheme.

[6 marks]

Examiners report

[N/A]

cosA C =O^ OA2+OC2−AC2

2×OA×OC A C = 2 × A CO^ B^

13. =32 72+72− 2 × 7 × 7 cosA C, OO^ = 2 × 1.246 A C = 2.492 (O^ 142.8∘)

2.492 = , ll = 17.4, 14π×

7

142.8 360

2πr−l, 14π= 17.4

A CO^

2π− 2.492, 3.79, 360 − 142.8

l= 7 × 3.79, 14π×217.2360

arc ABC = 26.5

A BO^ B CO^

A BO^ B CO^ cosA B =O^ 72+ −72 12.22

2×7×7 A B = 2.116,B C = 1.6745O^ O^

AB = 7 × 2 × 1.058 (= 14.8135), 7 × 1.6745 (= 11.7216)

rA B +O^ rB C, 14.8135 + 11.7216, 7(2.116 + 1.6745)O^ arc ABC = 26.5 (cm)

[3 marks]

(17)

Markscheme

Note: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not award

the final A1FT mark.

METHOD 1

valid approach involving Pythagoras (M1)

e.g. , labelled diagram

correct working (may be on diagram) (A1)

e.g. ,

A1 N2 [3 marks]

METHOD 2

valid approach involving tan identity (M1) e.g.

correct working (A1) e.g.

A1 N2 [3 marks]

Examiners report

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because , then . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

x+ x= 1 sin2 cos2

+ (cos100 = 1

m2 )2 1 −−−−−m−2

cos100 = − 1 −−−−−−m−2

tan =sin cos

cos100 = sin 100 tan 100

cos100 = m

tan 100

= 1 −m 1 −m2

−−−−−− √

sin100∘=m sin200= 2m

[1 mark]

11b.Let . Find an expression for in terms of m.

Markscheme

METHOD 1

(accept ) A1 N1

[1 mark]

METHOD 2

A1 N1 [1 mark]

sin

100

=

m

tan 100

tan 100 = − m

1−m2

m

− 1− m2

tan 100 = m

(18)

Examiners report

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because , then . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

= 1 −m 1 −m2

−−−−−− √

sin100∘=m sin200= 2m

[2 marks] 11c.Let . Find an expression for in terms of m.

Markscheme

METHOD 1

valid approach involving double angle formula (M1) e.g.

(accept ) A1 N2

Note: If candidates find , award full FT in parts (b) and (c), even though the values may not have appropriate

signs for the angles.

[2 marks]

METHOD 2

valid approach involving double angle formula (M1)

e.g. ,

A1 N2 [2 marks]

Examiners report

While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because , then . In addition, some candidates did not seem to understand what writing an expression "in terms of m" meant.

sin100∘=m sin 200∘

sin 2θ= 2 sinθcosθ

sin 200 = −2m1 −−−−−−m−2 2m(1 −−−−−−m−2)

cos100 = 1 −−−−−−m−2

sin 2θ= 2 sinθcosθ 2m×tan 100m

sin 200 = 2m2 (= 2mcos100)

tan 100

= 1 −m 1 −m2

−−−−−− √

(19)

12a. [2 marks]

Let . The diagram below shows part of the graph of f , for .

The graph has a local maximum at P(3, 5) , a local minimum at Q(7, − 5) , and crosses the x-axis at R.

Write down the value of (i) ;

(ii) .

Markscheme

(i) (accept ) A1 N1

(ii) (accept , if ) A1 N1

Note: Accept other correct values of c, such as 11, , etc.

[2 marks]

Examiners report

Part (a) (i) was well answered in general. There were more difficulties in finding the correct value of the parameter c.

f(x) =acos(b(x−c)) 0 ≤x≤ 10

a

c

a= 5 −5

c= 3 c= 7 a= −5

−5

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

Markscheme

attempt to find period (M1)

e.g. 8 ,

(exact), , 0.785 [ ] (do not accept 45) A1 N2 [2 marks]

Examiners report

Finding the correct value of b in part (b) also proved difficult as many did not realize the period was equal to 8.

b=period2π

0.785398…

b=2π

8

π

4 0.785, 0.786

(20)

Markscheme

valid approach (M1)

e.g. , symmetry of curve (accept A1 N2 [2 marks]

Examiners report

Most candidates could handle part (c) without difficulties using their GDC or working with the symmetry of the curve although follow through from errors in part (b) was often not awarded because candidates failed to show any working by writing down the equations they entered into their GDC.

f(x) = 0

x= 5 (5 ,0))

[3 marks] 13a.

The following diagram shows a circular play area for children.

The circle has centre O and a radius of 20 m, and the points A, B, C and D lie on the circle. Angle AOB is 1.5 radians.

(21)

Markscheme

Note: In this question, do not penalise for missing or incorrect units. They are not included in the markscheme, to avoid complex answer lines.

METHOD 1

choosing cosine rule (must have cos in it) (M1)

e.g.

correct substitution (into rhs) A1

e.g. ,

A1 N2 [3 marks]

METHOD 2

choosing sine rule (M1)

e.g. ,

correct substitution A1

e.g.

A1 N2 [3 marks]

Examiners report

Candidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a right-angled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors in accuracy.

= + − 2abcosC c2 a2 b2

+ − 2(20)(20)cos1.5

202 202 AB = 800 − 800 cos1.5−−−−−−−−−−−−−

AB = 27.26555…

AB = 27.3 [27.2, 27.3]

=

sinA a

sinB

b =

AB sinO

AO sinB

=

AB sin 1.5

20 sin(0.5(π−1.5))

AB = 27.26555… AB = 27.3 [27.2, 27.3]

[2 marks] 13b.Find the area of triangle AOB.

Markscheme

correct substitution into area formula A1

e.g. ,

(accept , from using 27.3) A1 N1

[2 marks]

Examiners report

Candidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a right-angled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors in accuracy.

(20)(20)sin 1.5

1

2 12(20)(27.2655504…)sin(0.5(π− 1.5))

area = 199.498997… 199.75106 = 200 area = 199 [199, 200]

13c.Angle BOC is 2.4 radians. [3 marks]

(22)

Markscheme

appropriate method to find angle AOC (M1)

e.g.

correct substitution into arc length formula (A1)

e.g. ,

(i.e. do not accept ) A1 N2

Notes: Candidates may misread the question and use . If working shown, award M0 then A0MRA1 for the answer 48.

Do not then penalize in part (d) which, if used, leads to the answer

However, if they use the prematurely rounded value of 2.4 for , penalise 1 mark for premature rounding for the answer 48 in

(c). Do not then penalize for this in (d).

[3 marks]

Examiners report

In part (c), some candidates misread the question and used 2.4 as the size of angle AOC while others rounded prematurely leading to the inaccurate answer of 48. In either case, marks were lost.

2π− 1.5 − 2.4

(2π− 3.9) × 20 2.3831853… × 20 arc length = 47.6637…

arc length = 47.7 (47.6, 47.7] 47.6

A C = 2.4Oˆ

A COˆ 679.498…

A COˆ

13d.Angle BOC is 2.4 radians. [3 marks]

Find the area of the shaded region.

Markscheme

calculating sector area using their angle AOC (A1)

e.g. , ,

shaded area = their area of triangle AOB + their area of sector (M1)

e.g. ,

(accept from using 199) A1 N2

[3 marks]

Examiners report

Part (d) proved to be straightforward and candidates were able to obtain full FT marks from errors made in previous parts.

(2.38…)( )

1

2 20

2 200(2.38…) 476.6370614…

199.4989973… + 476.6370614… 199 + 476.637 shaded area = 676.136… 675.637… = 676

shaded area = 676 [676, 677]

13e.Angle BOC is 2.4 radians. [4 marks]

The shaded region is to be painted red. Red paint is sold in cans which cost each. One can covers . How much does it cost to buy the paint?

(23)

Markscheme

dividing to find number of cans (M1)

e.g. ,

5 cans must be purchased (A1)

multiplying to find cost of cans (M1)

e.g. ,

cost is 160 (dollars) A1 N3 [4 marks]

Examiners report

Most candidates had a suitable strategy for part (e) and knew to work with a whole number of cans of paint.

676

140 4.82857…

5(32) 676× 32

140

14a. [3 marks]

The following diagram shows the graph of , for .

There is a minimum point at P(2, − 3) and a maximum point at Q(4, 3) .

(i) Write down the value of a . (ii) Find the value of b .

Markscheme

(i) A1 N1

(ii) METHOD 1

attempt to find period (M1)

e.g. 4 , ,

A1 N2 [3 marks]

METHOD 2

attempt to substitute coordinates (M1)

e.g. , A1 N2 [3 marks]

f(x) =acos(bx) 0 ≤x≤ 4

a= 3

b= 4 2π b

b=2π(= ) 4

π

2

3 cos(2b) = −3 3 cos(4b) = 3

b=2π(= ) 4

π

(24)

Examiners report

In part (a), many candidates were able to successfully write down the value of a as instructed by inspecting the graph and seeing the amplitude of the function is 3. Many also used a formulaic approach to reach the correct answer. When finding the value of b, there were many candidates who thought b was the period of the function, rather than 2π .

period

[1 mark] 14b.Write down the gradient of the curve at P.

Markscheme

0 A1 N1 [1 mark]

Examiners report

In part (b), the directions asked candidates to write down the gradient of the curve at the local minimum point P. However, many candidates spent a good deal of time finding the derivative of the function and finding the value of the derivative for the given value of x, rather than simply stating that the gradient of a curve at a minimum point is zero.

[2 marks] 14c.Write down the equation of the normal to the curve at P.

Markscheme

recognizing that normal is perpendicular to tangent (M1)

e.g. , , sketch of vertical line on diagram (do not accept 2 or ) A1 N2

[2 marks]

Examiners report

For part (c), finding the equation of the normal to the curve, many candidates tried to work with algebraic equations involving negative reciprocal gradients, rather than recognizing that the equation of the vertical line was . There were also candidates who had trouble expressing the correct equation of a line parallel to the y-axis.

× = −1

m1 m2 m= −10

x= 2 y= 2

(25)

[2 marks] 15a.

The diagram below shows part of the graph of , where .

The point is a maximum point and the point is a minimum point.

Find the value of a .

Markscheme

evidence of valid approach (M1)

e.g. , distance from A1 N2

[2 marks]

Examiners report

A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph.

f(x) =acos(b(xc)) − 1 a> 0

P(π,2)

4 Q( ,−4)

3π

4

max y value−min y value

2 y= −1

a= 3

15b.(i) Show that the period of f is . [4 marks]

(ii) Hence, find the value of b .

Markscheme

(i) evidence of valid approach (M1)

e.g. finding difference in x-coordinates, evidence of doubling A1

e.g.

AG N0

(ii) evidence of valid approach (M1)

e.g.

A1 N2 [4 marks]

π

π

2

2 ×( )π

2

period =π

b=2π π

(26)

Examiners report

A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble showing that the period was , either incorrectly adding the given and or using the value of b that they found first for part (b)(ii).

π π/4 3π/4

[1 mark] 15c.Given that , write down the value of c .

Markscheme

A1 N1 [1 mark]

Examiners report

A pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had trouble showing that the period was , either incorrectly adding the given and or using the value of b that they found first for part (b)(ii).

0 <c<π

c=π

4

π π/4 π/3

[2 marks] 16a.

Let .

Show that can be expressed as .

Markscheme

attempt to expand (M1)

e.g. ; at least 3 terms correct expansion A1

e.g.

AG N0 [2 marks]

Examiners report

Simplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidates were certainly helped by the fact that it was a "show that" question.

f(x) = (sinx+ cosx)2

f(x) 1 + sin 2x

(sinx+ cosx)(sinx+ cosx)

x+ 2 sinxcosx+ x

sin2 cos2

(27)

16b.The graph of f is shown below for . [2 marks]

Let . On the same set of axes, sketch the graph of g for .

Markscheme

A1A1 N2

Note: Award A1 for correct sinusoidal shape with period and range , A1 for minimum in circle.

Examiners report

More candidates had difficulty with part (b) with many assuming the first graph was and hence sketching a horizontal translation of for the graph of g; some attempts were not even sinusoidal. While some candidates found the stretch factor p

correctly or from follow-through on their own graph, very few successfully found the value and direction for the translation.

0 ≤x≤ 2π

g(x) = 1 + cosx 0 ≤x≤ 2π

2π [0, 2]

1 + sin(x) π/2

16c.The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the [2 marks]

vector .

Write down the value of p and a possible value of k .

Markscheme

, A1A1 N2 [2 marks]

( )k 0

(28)

Examiners report

Part (c) certainly served as a discriminator between the grade 6 and 7 candidates.

[2 marks] 17a.

A Ferris wheel with diameter metres rotates clockwise at a constant speed. The wheel completes rotations every hour. The bottom of the wheel is metres above the ground.

A seat starts at the bottom of the wheel.

Find the maximum height above the ground of the seat.

Markscheme

valid approach (M1)

eg ,

maximum height (m) A1 N2 [2 marks]

Examiners report

Most candidates were successful with part (a).

122 2.4

13

13 + diameter 13 + 122

= 135

17b. [2 marks]

After tminutes, the height metres above the ground of the seat is given by

(i) Show that the period of is minutes. (ii) Write down the exact value of .

h

h= 74 +acosbt.

h 25

b

= 60 2.4

= 25

b=2π

(29)

Examiners report

A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at thereby making the value of a negative. Incorrect values of were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges.

25

b

t

= 0

61

17c.(b) (i) Show that the period of is minutes. [9 marks]

(ii) Write down the exact value of . (c) Find the value of .

(d) Sketch the graph of , for .

h 25

b

a

h 0 ≤t≤ 50

= 60 2.4

= 25

b=225π (= 0.08π)

max − 74 |a| = 135−13

2 74 − 13

|a| = 61 a= 61

a= −61

135 = 74 +acos(2π×12.5) 25

135 = 74 +acos(π) 13 = 74 +a

a= −61

(30)

Examiners report

A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at thereby making the value of a negative. Incorrect values of were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges.

25

b

t

= 0

61

[3 marks] 17d.Find the value of .

Markscheme

METHOD 1

valid approach (M1)

eg , ,

(accept ) (A1)

A1 N2

METHOD 2

attempt to substitute valid point into equation for h (M1) eg

correct equation (A1)

eg ,

A1 N2 [3 marks]

Examiners report

A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at thereby making the value of a negative. Incorrect values of were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges.

a

max − 74 |a| = 135−13

2 74 − 13

|a| = 61 a= 61

a= −61

135 = 74 +acos(2π×12.5) 25

135 = 74 +acos(π) 13 = 74 +a

a= −61

25

b

t

= 0

61

(31)

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

A1 for approximately correct sinusoidal shape with cycles.

Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]

Examiners report

A surprising number had difficulty producing enough work to show that the period was ; writing down the exact value of also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on the Ferris wheel is a minimum at thereby making the value of a negative. Incorrect values of were often seen with correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show key features in approximately correct locations and candidates lost marks for incorrect domains and ranges.

2

25

b

t

= 0

61

[5 marks] 17f.In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground.

Markscheme

setting up inequality (accept equation) (M1)

eg , , sketch of graph with line any two correct values for t (seen anywhere) A1A1

eg , , ,

valid approach M1

eg , , , A1 N2

[5 marks]

Examiners report

Part (e) was very poorly done for those who attempted the question and most did not make the connection between height, time and probability. The idea of linking probability with a real-life scenario proved beyond most candidates. That said, there were a few novel approaches from the strongest of candidates using circles and angles to solve this part of question 10.

105

h> 105 105 = 74 +acosbt y= 105

t= 8.371… t= 16.628… t= 33.371… t= 41.628…

16.628−8.371 25

t1 t2

25

2×8.257 50

2(12.5−8.371) 25

p= 0.330

[3 marks] 18a.

Let , where .

Find .

sinθ= 2

13

√ <θ<π

π

2

(32)

Markscheme

METHOD 1

evidence of choosing (M1)

correct working (A1)

e.g. , ,

A1 N2

Note: If no working shown, award N1 for .

METHOD 2

approach involving Pythagoras’ theorem (M1)

e.g. ,

finding third side equals 3 (A1)

A1 N2

Note: If no working shown, award N1 for .

[3 marks]

Examiners report

While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a negative value.

θ+ θ= 1 sin2 cos2

θ= cos2 9

13 cosθ= ± 3 13

√ cosθ= 9 13

−− √

cosθ= − 3 13 √

3 13 √

+ = 13 22 x2

cosθ= −313

3 13 √

[5 marks] 18b.Find .

Markscheme

correct substitution into (seen anywhere) (A1)

e.g.

correct substitution into (seen anywhere) (A1)

e.g. , ,

valid attempt to find (M1)

e.g. ,

correct working A1

e.g. , ,

A1 N4

Note: If students find answers for which are not in the range , award full FT in (b) for correct FT working shown.

[5 marks] tan 2θ

sin 2θ

2( 2 ) (− )

13 √ 3 13 √ cos2θ − (− 3 )

13 √

2

( 2 )

13 √

2

2(− 3 ) − 1

13 √

2

1 − 2( 2 )

13 √

2

tan 2θ

2( 2 )(− )

13 √ 3 13 √ − (− 3 )

13 √

2

( 2 )

13 √

2

2(−2)

3

1−(−2)

3 2 (2)(2)(−3) 13 − 9 13 4 13 − 12 (√13)2

−1 18 13 −12 13 5 13 tan 2θ= −12

5

(33)

Examiners report

In part (b), many candidates incorrectly tried to calculate tan 2θ as 2 × tanθ , rather than using the double-angle identities.

19a. [3 marks]

The following diagram shows the graph of , for .

There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of (i) a ;

(ii) c ; (iii) d .

Markscheme

(i) A1 N1

(ii) A1 N1

(iii) A1 N1 [3 marks]

Examiners report

Part (a) of this question proved challenging for most candidates.

f(x) =asin(b(xc)) +d 2 ≤x≤ 10

a= 8

c= 2

d= 4

(34)

Markscheme

METHOD 1

recognizing that period (A1)

correct working A1

e.g. , AG N0

METHOD 2

attempt to substitute M1

e.g.

correct working A1

e.g.

AG N0 [2 marks]

Examiners report

Although a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize that this period could be found using the given coordinates of the maximum and minimum points.

= 8

8 =2π b b=

2π

8

b=π4

12 = 8 sin(b(4 − 2)) + 4

sin 2b= 1

b=π

4

[3 marks] 19c.Find .

Markscheme

evidence of attempt to differentiate or choosing chain rule (M1)

e.g. ,

(accept ) A2 N3 [3 marks]

Examiners report

In part (c), not many candidates found the correct derivative using the chain rule.

(x) f

cos (xπ − 2)

4 × 8

π

4

(x) = 2πcos( (x− 2))

fπ4 2πcos (xπ − 2)

4

(35)

Markscheme

recognizing that gradient is (M1)

e.g.

correct equation A1

e.g. , correct working (A1)

e.g.

using (seen anywhere) (A1)

e.g.

simplifying (A1)

e.g.

A1 N4 [6 marks]

Examiners report

For part (d), a good number of candidates correctly set their expression equal to , but errors in their previous values kept most from correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.

(x) f

(x) =m f

−2π= 2πcos(π(x− 2))

4 −1 = cos( (x− 2))

π

4

(−1) = (x− 2)

cos−1 π

4

(−1) =π cos−1

π= (xπ4 − 2)

4 = (x− 2)

x= 6

−2π

20a. [2 marks]

The following diagram represents a large Ferris wheel, with a diameter of 100 metres.

Let P be a point on the wheel. The wheel starts with P at the lowest point, at ground level. The wheel rotates at a constant rate, in an anticlockwise (counter-clockwise) direction. One revolution takes 20 minutes.

Write down the height of P above ground level after (i) 10 minutes;

(ii) 15 minutes.

Markscheme

(i) 100 (metres) A1 N1

(ii) 50 (metres) A1 N1 [2 marks]

Examiners report

Nearly all candidates answered part (a) correctly, finding the height of the wheel at and of a revolution.1 2

(36)

20b. [4 marks]

Let metres be the height of P above ground level after t minutes. Some values of are given in the table below.

(i) Show that . (ii) Find .

Markscheme

(i) identifying symmetry with (M1)

subtraction A1

e.g. , AG N0

(ii) recognizing period (M1)

e.g.

A1 N2 [4 marks]

Examiners report

While many candidates were successful in part (b), there were many who tried to use right-angled triangles or find a function for height, rather than recognizing the symmetry of the wheel in its different positions and using the values given in the table.

h(t) h(t)

h(8) = 90.5 h(21)

h(2) = 9.5

100 −h(2) 100 − 9.5

h(8) = 90.5

h(21) =h(1) h(21) = 2.4

[3 marks] 20c.Sketch the graph of h , for 0 ≤t≤ 40 .

(37)

Examiners report

In part (c), most candidates were able to sketch a somewhat accurate representation of the height of the wheel over two full cycles. However, it seems that many candidates are not familiar with the shape of a sinusoidal wave, as many of the candidates' graphs were constructed of line segments, rather than a curve.

[5 marks] 20d.Given that h can be expressed in the form , find a , b and c .

Markscheme

evidence of a quotient involving 20, or to find b (M1)

e.g. ,

(accept if working in degrees) A1 N2

, A2A1 N3 [5 marks]

Examiners report

For part (d), candidates were less successful in finding the parameters of the cosine function. Even candidates who drew accurate sketches were not always able to relate their sketch to the function. These candidates understood the context of the problem, that the position on the wheel goes up and down, but they did not relate this to a trigonometric function. Only a small number of candidates recognized that the value of a would be negative. Candidates should be aware that while working in degrees may be acceptable, the expectation is that radians will be used in these types of questions.

h(t) =acosbt+c

2π 360∘

= 20

2π

b b=

360 20

b=2π

20 (= )

π

10 b= 18

a= −50 c= 50

[2 marks] 21a.Show that .

Markscheme

attempt to substitute for (M1)

correct substitution A1

e.g.

AG N0 [2 marks]

Examiners report

In part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidates who worked backward, starting with the required answer and manipulating the equation in various ways. As this was a "show that" question, working backward from the given answer is not a valid method.

4 − cos2θ+ 5 sinθ= 2sin2θ+ 5 sinθ+ 3

1 − 2sin2θ cos2θ

4 − (1 − 2sin2θ) + 5 sinθ

4 − cos2θ+ 5 sinθ= 2sin2θ+ 5 sinθ+ 3

[5 marks]

(38)

Markscheme

evidence of appropriate approach to solve (M1)

e.g. factorizing, quadratic formula correct working A1

e.g. , ,

correct solution (do not penalise for including (A1)

A2 N3 [5 marks]

Examiners report

In part (b), many candidates seemed to realize what was required by the word “hence”, though some had trouble factoring the quadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solution to the equation , and there were a few who did not realize that the equation has no solution.

(2 sinθ+ 3)(sinθ+ 1) (2x+ 3)(x+ 1) = 0 sinx=−5± 14

sinθ= −1 sinθ= −3

2

θ=32π

sinθ= −1 sinθ= −3

2

[5 marks] 22a.

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 .

Markscheme

evidence 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 report

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

2 m θ 0 <θ<π2

y= 4 sinθ+ 2 sin 2θ

sinθ=h

2 2 sinθ

cosθ=b

2 2 cosθ

θ

2(12× 2 cosθ× 2 sinθ)+ 2 × 2 sinθ 12(2 sinθ)(2 + 2 + 4 cosθ)

2 sinθcosθ sin 2θ 4 sinθ+ 2(2 sinθcosθ)

(39)

[4 marks] 22b.Zoe wants a window to have an area of . Find the two possible values of .

Markscheme

correct equation A1

e.g. ,

evidence of attempt to solve (M1)

e.g. a sketch,

, A1A1 N3 [4 marks]

Examiners report

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

5 m2 θ

y= 5 4 sinθ+ 2 sin 2θ= 5

4 sinθ+ 2 sinθ− 5 = 0

θ= 0.856 (49.0∘) θ= 1.25 (71.4)

22c.John wants two windows which have the same area A but different values of . [7 marks]

Find all possible values for A .

Markscheme

recognition 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 report

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

θ

θ=π

2

θ=π

2

4 sin( )π + 2 sin(2 × ) 2

π

2

A= 4

A= 5.19615…

4 <A< 5.20 4 <A< 5.19

[3 marks] 23a.

Let , . Let .

Find an expression for .

Markscheme

attempt to form any composition (even if order is reversed) (M1)

correct composition (A1)

A1 N3

[3 marks]

f(x) =3x+ 1

2 g(x) = 4 cos( )− 1

x

3 h(x) = (gf)(x)

h(x)

h(x) =g(3x+ 1) 2

h(x) = 4 cos(3+1)− 1

x

2

3 (4 cos(12x+13)− 1,4 cos( )− 1) 3x+2

(40)

Examiners report

The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period and range, but this approach was not the most efficient.

[1 mark] 23b.Write down the period of .

Markscheme

period is A1 N1 [1 mark]

Examiners report

The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period and range, but this approach was not the most efficient.

h

4π(12.6)

[2 marks] 23c.Write down the range of .

Markscheme

range is A1A1 N2 [2 marks]

Examiners report

The majority of candidates handled the composition of the two given functions well. However, a large number of candidates had difficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidates knew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the period and range, but this approach was not the most efficient.

h

−5 ≤h(x) ≤ 3 ([−5,3])

[3 marks] 24a.

Let , for .

Sketch the graph of f .

(41)

Markscheme

A1A1A1 N3

Note: Award A1 for approximately sinusoidal shape, A1 for end points approximately correct , A1 for

approximately correct position of graph, (y-intercept , maximum to right of y-axis).

[3 marks]

Examiners report

Some graphs in part (a) were almost too detailed for just a sketch but more often, the important features were far from clear. Some graphs lacked scales on the axes.

(−2π, 4) (2π, 4) (0, 4)

24b.Write down [3 marks]

(i) the amplitude; (ii) the period;

(iii) the x-intercept that lies between and 0.

Markscheme

(i) 5 A1 N1

(ii) (6.28) A1 N1

(iii) A1 N1 [3 marks]

Examiners report

A number of candidates had difficulty finding the period in part (b)(ii).

π

2

−0.927

[3 marks] 24c.Hence write in the form .

Markscheme

(accept , , ) A1A1A1 N3 [3 marks]

Examiners report

A number of candidates had difficulty writing the correct value of q in part (c).

f(x) psin(qx+r)

(42)

[2 marks] 24d.Write down one value of x such that .

Markscheme

evidence of correct approach (M1)

e.g. max/min, sketch of indicating roots

one 3 s.f. value which rounds to one of , , , A1 N2

[2 marks]

Examiners report

The most common approach in part (d) was to differentiate and set . Fewer students found the values of x given by the maximum or minimum values on their graphs.

(x) = 0 f

(x) f

−5.6 −2.5 0.64 3.8

(x) = 0 f

[2 marks] 24e.Write down the two values of k for which the equation has exactly two solutions.

Markscheme

, A1A1 N2 [2 marks]

Examiners report

Part (e) proved challenging for many candidates, although if candidates answered this part, they generally did so correctly.

f(x) =k

k= −5 k= 5

24f.Let , for . There is a value of x, between and , for which the gradient of f is equal to the [5 marks]

gradient of g(x) = ln(xg. Find this value of + 1) 0 ≤ x.

(43)

Markscheme

METHOD 1

graphical approach (but must involve derivative functions) M1

e.g.

each curve A1A1

A2 N2

METHOD 2

A1

A1

evidence of attempt to solve M1

A2 N2 [5 marks]

Examiners report

In part (f), many candidates were able to get as far as equating the two derivatives but fewer used their GDC to solve the resulting equation. Again, many had trouble demonstrating their method of solution.

x= 0.511

(x) = gx+11

(x) = 3 cosx− 4 sinx

f(5 cos(x+ 0.927))

(x) = (x) gf

x= 0.511

25a. [6 marks]

The graph of , for , is shown below.

There is a minimum point at (0, −3) and a maximum point at (4, 7) .

Find the value of (i) p ; (ii) q ; (iii) r.

Figure

Actualización...

Referencias

Actualización...