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HL Paper 1

Three planes have equations:

2 x y + z = 5

x + 3 y z = 4      , where  a b R .

3 x 5 y + a z = b

Find the set of values of a and b such that the three planes have no points of intersection.

Markscheme

attempt to eliminate a variable (or attempt to find det A )       M1

( 2 1 1 1 3 1 3 5 a | 5 4 b ) ( 2 1 1 0 7 3 0 14 a + 3 | 5 3 b 12 )   (or det  A = 14 ( a 3 ) )

(or two correct equations in two variables)       A1

( 2 1 1 0 7 3 0 0 a 3 | 5 3 b 6 )   (or solving det  A = 0 )

(or attempting to reduce to one variable, e.g.  ( a 3 ) z = b 6 )       M1

a = 3 b 6        A1A1

[5 marks]

Examiners report

[N/A]



Two distinct lines, l 1 and l 2 , intersect at a point P . In addition to P , four distinct points are marked out on l 1 and three distinct points on l 2 . A mathematician decides to join some of these eight points to form polygons.

The line l 1 has vector equation r1 = ( 1 0 1 ) + λ ( 1 2 1 ) λ R  and the line l 2 has vector equation r2  = ( 1 0 2 ) + μ ( 5 6 2 ) μ R .

The point P has coordinates (4, 6, 4).

The point A has coordinates (3, 4, 3) and lies on l 1 .

The point B has coordinates (−1, 0, 2) and lies on l 2 .

Find how many sets of four points can be selected which can form the vertices of a quadrilateral.

[2]
a.i.

Find how many sets of three points can be selected which can form the vertices of a triangle.

[4]
a.ii.

Verify that P is the point of intersection of the two lines.

[3]
b.

Write down the value of λ corresponding to the point A .

[1]
c.

Write down PA and PB .

[2]
d.

Let C be the point on l 1 with coordinates (1, 0, 1) and D be the point on l 2 with parameter μ = 2 .

Find the area of the quadrilateral CDBA .

[8]
e.

Markscheme

appreciation that two points distinct from P need to be chosen from each line   M1

4 C 2 × 3 C 2

=18    A1

[2 marks]

a.i.

EITHER

consider cases for triangles including P or triangles not including P       M1

3 × 4 + 4 × 3 C 2 + 3 × 4 C 2      (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

OR

consider total number of ways to select 3 points and subtract those with 3 points on the same line      M1

8 C 3 5 C 3 4 C 3      (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

56−10−4

THEN

= 42    A1

[4 marks]

a.ii.

METHOD 1

substitution of (4, 6, 4) into both equations       (M1)

λ = 3 and  μ = 1        A1A1

(4, 6, 4)       AG

METHOD 2

attempting to solve two of the three parametric equations      M1

λ = 3 and  μ = 1        A1

check both of the above give (4, 6, 4)       M1AG

Note: If they have shown the curve intersects for all three coordinates they only need to check (4,6,4) with one of " λ " or " μ ".

[3 marks]

b.

λ = 2       A1

[1 mark]

c.

PA = ( 1 2 1 ) ,   PB = ( 5 6 2 )     A1A1

Note: Award A1A0 if both are given as coordinates.

[2 marks]

d.

METHOD 1

area triangle  ABP = 1 2 | PB × PA |     M1

( = 1 2 | ( 5 6 2 ) × ( 1 2 1 ) | ) = 1 2 | ( 2 3 4 ) |     A1

= 29 2     A1

EITHER

PC = 3 PA PD = 3 PB        (M1)

area triangle  PCD = 9 ×  area triangle ABP        (M1)A1

= 9 29 2     A1

OR

D  has coordinates (−11, −12, −2)    A1

area triangle  PCD = 1 2 | PD × PC | = 1 2 | ( 15 18 6 ) × ( 3 6 3 ) |     M1A1

Note: A1 is for the correct vectors in the correct formula.

= 9 29 2     A1

THEN

area of  CDBA = 9 29 2 29 2

= 4 29     A1

 

METHOD 2

D  has coordinates (−11, −12, −2)    A1

area  = 1 2 | CB × CA | + 1 2 | BC × BD |       M1

Note: Award M1 for use of correct formula on appropriate non-overlapping triangles.

Note: Different triangles or vectors could be used.

CB = ( 2 0 1 ) CA = ( 2 4 2 )     A1

CB × CA = ( 4 6 8 )     A1

BC = ( 2 0 1 ) BD = ( 10 12 4 )     A1

BC × BD = ( 12 18 24 )     A1

Note: Other vectors which might be used are  DA = ( 14 16 5 ) BA = ( 4 4 1 ) DC = ( 12 12 3 ) .

Note: Previous A1A1A1A1 are all dependent on the first M1.

valid attempt to find a value of  1 2 | a × b |       M1

Note: M1 independent of triangle chosen.

area  = 1 2 × 2 × 29 + 1 2 × 6 × 29

= 4 29     A1

Note: accept  1 2 116 + 1 2 1044 or equivalent.

 

[8 marks]

e.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.



Let S be the sum of the roots found in part (a).

Find the roots of  z 24 = 1 which satisfy the condition 0 < arg ( z ) < π 2 , expressing your answers in the form r e i θ , where r , θ R + .

[5]
a.

Show that Re S = Im S.

[4]
b.i.

By writing  π 12 as ( π 4 π 6 ) , find the value of cos  π 12 in the form a + b c , where a b and  c are integers to be determined.

[3]
b.ii.

Hence, or otherwise, show that S = 1 2 ( 1 + 2 ) ( 1 + 3 ) ( 1 + i ) .

[4]
b.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

( r ( cos θ + i sin θ ) ) 24 = 1 ( cos 0 + i sin 0 )

use of De Moivre’s theorem       (M1)

r 24 = 1 r = 1       (A1)

24 θ = 2 π n θ = π n 12 ( n Z )       (A1)

0 < arg ( z ) < π 2 n = 1, 2, 3, 4, 5

z = e π i 12 or  e 2 π i 12 or  e 3 π i 12 or  e 4 π i 12 or  e 5 π i 12       A2

Note: Award A1 if additional roots are given or if three correct roots are given with no incorrect (or additional) roots.

 

[5 marks]

a.

Re S =  cos π 12 + cos 2 π 12 + cos 3 π 12 + cos 4 π 12 + cos 5 π 12

Im S =  sin π 12 + sin 2 π 12 + sin 3 π 12 + sin 4 π 12 + sin 5 π 12       A1

Note: Award A1 for both parts correct.

but  sin 5 π 12 = cos π 12 ,   sin 4 π 12 = cos 2 π 12 ,   sin 3 π 12 = cos 3 π 12 ,   sin 2 π 12 = cos 4 π 12 and  sin π 12 = cos 5 π 12       M1A1

⇒ Re S = Im S       AG

Note: Accept a geometrical method.

 

[4 marks]

b.i.

cos π 12 = cos ( π 4 π 6 ) = cos π 4 cos π 6 + sin π 4 sin π 6       M1A1

= 2 2 3 2 + 2 2 1 2

= 6 + 2 4        A1

 

[3 marks]

b.ii.

 

cos 5 π 12 = cos ( π 6 + π 4 ) = cos π 6 cos π 4 sin π 6 sin π 4       (M1)

Note: Allow alternative methods eg  cos 5 π 12 = sin π 12 = sin ( π 4 π 6 ) .

= 3 2 2 2 1 2 2 2 = 6 2 4       (A1)

Re S =  cos π 12 + cos 2 π 12 + cos 3 π 12 + cos 4 π 12 + cos 5 π 12

Re S =  2 + 6 4 + 3 2 + 2 2 + 1 2 + 6 2 4       A1

= 1 2 ( 6 + 1 + 2 + 3 )       A1

= 1 2 ( 1 + 2 ) ( 1 + 3 )

S = Re(S)(1 + i) since Re S = Im S,      R1

S =  1 2 ( 1 + 2 ) ( 1 + 3 ) ( 1 + i )       AG

 

[4 marks]

b.iii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.



Let z = 1 cos 2 θ i sin 2 θ ,   z C ,   0 θ π .

Solve 2 sin ( x + 60 ) = cos ( x + 30 ) ,   0 x 180 .

[5]
a.

Show that sin 105 + cos 105 = 1 2 .

[3]
b.

Find the modulus and argument of z in terms of θ . Express each answer in its simplest form.

[9]
c.i.

Hence find the cube roots of z  in modulus-argument form.

[5]
c.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

2 sin ( x + 60 ) = cos ( x + 30 )

2 ( sin x cos 60 + cos x sin 60 ) = cos x cos 30 sin x sin 30      (M1)(A1)

2 sin x × 1 2 + 2 cos x × 3 2 = cos x × 3 2 sin x × 1 2      A1

3 2 sin x = 3 2 cos x

tan x = 1 3      M1

x = 150      A1

[5 marks]

a.

EITHER

choosing two appropriate angles, for example 60° and 45°     M1

sin 105 = sin 60 cos 45 + cos 60 sin 45 and

cos 105 = cos 60 cos 45 sin 60 sin 45      (A1)

sin 105 + cos 105 = 3 2 × 1 2 + 1 2 × 1 2 + 1 2 × 1 2 3 2 × 1 2      A1

= 1 2      AG

OR

attempt to square the expression     M1

( sin 105 + cos 105 ) 2 = sin 2 105 + 2 sin 105 cos 105 + cos 2 105

( sin 105 + cos 105 ) 2 = 1 + sin 210      A1

= 1 2      A1

sin 105 + cos 105 = 1 2   AG

 

[3 marks]

b.

EITHER

z = ( 1 cos 2 θ ) i sin 2 θ

| z | = ( 1 cos 2 θ ) 2 + ( sin 2 θ ) 2      M1

| z | = 1 2 cos 2 θ + cos 2 2 θ + sin 2 2 θ      A1

= 2 ( 1 cos 2 θ )      A1

= 2 ( 2 sin 2 θ )

= 2 sin θ      A1

let arg ( z ) = α

tan α = sin 2 θ 1 cos 2 θ      M1

= 2 sin θ cos θ 2 sin 2 θ      (A1)

= cot θ      A1

arg ( z ) = α = arctan ( tan ( π 2 θ ) )      A1

= θ π 2      A1

OR

z = ( 1 cos 2 θ ) i sin 2 θ

= 2 sin 2 θ 2 i sin θ cos θ      M1A1

= 2 sin θ ( sin θ i cos θ )      (A1)

= 2 i sin θ ( cos θ + i sin θ )      M1A1

= 2 sin θ ( cos ( θ π 2 ) + i sin ( θ π 2 ) )      M1A1

| z | = 2 sin θ      A1

arg ( z ) = θ π 2      A1

[9 marks]

c.i.

attempt to apply De Moivre’s theorem     M1

( 1 cos 2 θ i sin 2 θ ) 1 3 = 2 1 3 ( sin θ ) 1 3 [ cos ( θ π 2 + 2 n π 3 ) + i sin ( θ π 2 + 2 n π 3 ) ]      A1A1A1

 

Note:     A1 for modulus, A1 for dividing argument of z by 3 and A1 for 2 n π .

 

Hence cube roots are the above expression when n = 1 ,   0 ,   1 . Equivalent forms are acceptable.     A1

[5 marks]

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.



Consider the three planes

1: 2x-y+z=4

2: x-2y+3z=5

3:-9x+3y-2z=32

Show that the three planes do not intersect.

[4]
a.

Verify that the point P(1, -2, 0) lies on both 1 and 2.

[1]
b.i.

Find a vector equation of L, the line of intersection of 1 and 2.

[4]
b.ii.

Find the distance between L and 3.

[6]
c.

Markscheme

METHOD 1

attempt to eliminate a variable                 M1

obtain a pair of equations in two variables


EITHER

-3x+z=-3 and          A1

-3x+z=44          A1


OR

-5x+y=-7 and          A1

-5x+y=40          A1


OR

3x-z=3 and          A1

3x-z=-795          A1


THEN

the two lines are parallel (-344 or -740 or 3-795)          R1

 

Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.

 

hence the three planes do not intersect          AG

 

METHOD 2

vector product of the two normals =-1-5-3  (or equivalent)          A1

r=1-20+λ153  (or equivalent)          A1

 

Note: Award A0 if “r=” is missing. Subsequent marks may still be awarded.

 

Attempt to substitute 1+λ,-2+5λ,3λ in 3                 M1

-91+λ+3-2+5λ-23λ=32

-15=32, a contradiction          R1

hence the three planes do not intersect          AG

 

METHOD 3

attempt to eliminate a variable                M1

-3y+5z=6          A1

-3y+5z=100          A1

0=94, a contradiction           R1

 

Note: Accept other equivalent alternatives. Accept other valid methods.
To obtain the final R1, at least the initial M1 must have been awarded.

 

hence the three planes do not intersect          AG

 

[4 marks]

a.

1:2+2+0=4  and  2:1+4+0=5          A1

 

[1 mark]

b.i.

METHOD 1

attempt to find the vector product of the two normals          M1

2-11×1-23

=-1-5-3          A1

r=1-20+λ153          A1A1

 

Note: Award A1A0 if “r=” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize lack of “r=” only once.

 

METHOD 2

attempt to eliminate a variable from 1 and 2          M1

3x-z=3  OR  3y-5z=-6  OR  5x-y=7

Let x=t

substituting x=t in 3x-z=3 to obtain

z=-3+3t  and  y=5t-7 (for all three variables in parametric form)          A1

r=0-7-3+λ153          A1A1

 

Note: Award A1A0 if “r=” is missing.
Accept any multiple of the direction vector. Accept other position vectors which satisfy both the planes 1 and 2 .

 

[4 marks]

b.ii.

METHOD 1

the line connecting L and 3 is given by L1

attempt to substitute position and direction vector to form L1           (M1)

s=1-20+t-93-2          A1

substitute 1-9t,-2+3t,-2t in 3             M1

-91-9t+3-2+3t-2-2t=32

94t=47t=12          A1

attempt to find distance between 1,-2,0 and their point -72,-12,-1           (M1)

=1-20+12-93-2-1-20=12-92+32+-22

=942          A1

 

METHOD 2

unit normal vector equation of 3 is given by -932·xyz81+9+4           (M1)

=3294          A1

let 4 be the plane parallel to 3 and passing through P
then the normal vector equation of 4 is given by

-932·xyz=-932·1-20=-15             M1

 

unit normal vector equation of 4 is given by

-932·xyz81+9+4=-1594          A1

distance between the planes is 3294--1594           (M1)

=4794=942          A1

 

[6 marks]

c.

Examiners report

Part (a) was well attempted using a variety of approaches. Most candidates were able to gain marks for part (a) through attempts to eliminate a variable with many subsequently making algebraic errors. Part (b)(i) was well done. For part (b)(ii) few successful attempts were noted, many candidates failed to use an appropriate notation "r =" while giving the vector equation of a line. Part (c) proved to be challenging for most candidates with very few correct answers seen. Many candidates did not attempt part (c).

a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.



In the following diagram, OA = a, OB = b. C is the midpoint of [OA] and OF = 1 6 FB .

N17/5/MATHL/HP1/ENG/TZ0/09

It is given also that AD = λ AF and CD = μ CB , where λ ,   μ R .

Find, in terms of a and OF .

[1]
a.i.

Find, in terms of a and AF .

[2]
a.ii.

Find an expression for  OD in terms of a, b and λ ;

[2]
b.i.

Find an expression for OD in terms of a, b and μ .

[2]
b.ii.

Show that μ = 1 13 , and find the value of λ .

[4]
c.

Deduce an expression for CD in terms of a and b only.

[2]
d.

Given that area Δ OAB = k ( area  Δ CAD ) , find the value of k .

[5]
e.

Markscheme

OF = 1 7 b     A1

[1 mark]

a.i.

AF = OF OA     (M1)

= 1 7 ba     A1

[2 marks]

a.ii.

OD = a + λ ( 1 7 b a )   ( = ( 1 λ ) a + λ 7 b )     M1A1

[2 marks]

b.i.

OD = 1 2 a + μ ( 1 2 a + b )   ( = ( 1 2 μ 2 ) a + μ b )     M1A1

[2 marks]

b.ii.

equating coefficients:     M1

λ 7 = μ ,   1 λ = 1 μ 2     A1

solving simultaneously:     M1

λ = 7 13 ,   μ = 1 13     A1AG

[4 marks]

c.

CD = 1 13 CB

= 1 13 ( b 1 2 a )   ( = 1 26 a + 1 13 b )     M1A1

[2 marks]

d.

METHOD 1

area  Δ ACD = 1 2 CD × AC × sin A C ^ B     (M1)

area  Δ ACB = 1 2 CB × AC × sin A C ^ B     (M1)

ratio  area  Δ ACD area  Δ ACB = CD CB = 1 13     A1

k = area  Δ OAB area  Δ CAD = 13 area  Δ CAB × area  Δ OAB     (M1)

= 13 × 2 = 26     A1

 

METHOD 2

area  Δ OAB = 1 2 | a × b |     A1

area  Δ CAD = 1 2 | CA × CD | or 1 2 | CA × AD |     M1

= 1 2 | 1 2 a × ( 1 26 a + 1 13 b ) |

= 1 2 | 1 2 a × ( 1 26 a ) + 1 2 a × 1 13 b |     (M1)

= 1 2 × 1 2 × 1 13 | a × b |   ( = 1 52 | a × b | )     A1

area  Δ OAB = k ( area  Δ CAD )

1 2 | a × b | = k 1 52 | a × b |

k = 26     A1

[5 marks]

e.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.
[N/A]
e.



Show that ( sin x + cos x ) 2 = 1 + sin 2 x .

[2]
a.

Show that sec 2 x + tan 2 x = cos x + sin x cos x sin x .

[4]
b.

Hence or otherwise find  0 π 6 ( sec 2 x + tan 2 x ) d x  in the form  ln ( a + b ) where a b Z .

[9]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

( sin x + cos x ) 2 = si n 2 x + 2 sin x cos x + co s 2 x       M1A1

Note: Do not award the M1 for just  si n 2 x + co s 2 x .

Note: Do not award A1 if correct expression is followed by incorrect working.

= 1 + sin 2 x       AG

[2 marks]

a.

sec 2 x + tan 2 x = 1 cos 2 x + sin 2 x cos 2 x      M1

Note: M1 is for an attempt to change both terms into sine and cosine forms (with the same argument) or both terms into functions of tan x .

= 1 + sin 2 x cos 2 x

= ( sin x + cos x ) 2 co s 2 x si n 2 x          A1A1

Note: Award A1 for numerator, A1 for denominator.

= ( sin x + cos x ) 2 ( cos x sin x ) ( cos x + sin x )      M1

= cos x + sin x cos x sin x       AG

Note: Apply MS in reverse if candidates have worked from RHS to LHS.

Note: Alternative method using tan 2 x and sec 2 x in terms of tan x .

[4 marks]

b.

METHOD 1

0 π 6 ( cos x + sin x cos x sin x ) d x        A1

Note: Award A1 for correct expression with or without limits.

EITHER

= [ ln ( cos x sin x ) ] 0 π 6   or   [ ln ( cos x sin x ) ] π 6 0        (M1)A1A1

Note: Award M1 for integration by inspection or substitution, A1 for  ln ( cos x sin x ) A1 for completely correct expression including limits.

= ln ( cos π 6 sin π 6 ) + ln ( cos 0 sin 0 )        M1

Note: Award M1 for substitution of limits into their integral and subtraction.

= ln ( 3 2 1 2 )        (A1)

OR

let  u = cos x sin x        M1

d u d x = sin x cos x = ( sin x + cos x )

1 3 2 1 2 ( 1 u ) d u        A1A1

Note: Award A1 for correct limits even if seen later, A1 for integral.

= [ ln u ] 1 3 2 1 2   or   [ ln u ] 3 2 1 2 1        A1

= ln ( 3 2 1 2 ) (  + ln 1 )        M1

THEN

= ln ( 2 3 1 )

Note: Award M1 for both putting the expression over a common denominator and for correct use of law of logarithms.

= ln ( 1 + 3 )        (M1)A1

 

METHOD 2

[ 1 2 ln ( tan 2 x + sec 2 x ) 1 2 ln ( cos 2 x ) ] 0 π 6       A1A1

= 1 2 ln ( 3 + 2 ) 1 2 ln ( 1 2 ) 0        A1A1(A1)

= 1 2 ln ( 4 + 2 3 )        M1

= 1 2 ln ( ( 1 + 3 ) 2 )        M1A1

= ln ( 1 + 3 )       A1

 

 

[9 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.



The following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).

The Cartesian equation of the plane  Π 2 , passing through the points B , C and D , is  y + z = 1 .

The plane  Π 3  passes through O and is normal to the line BD.

Π 3  cuts AD and BD at the points P and Q respectively.

Find the Cartesian equation of the plane  Π 1 , passing through the points A , B and D.

[3]
a.

Find the angle between the faces ABD and BCD.

[4]
b.

Find the Cartesian equation of  Π 3 .

[3]
c.

Show that P is the midpoint of AD.

[4]
d.

Find the area of the triangle OPQ.

[5]
e.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognising normal to plane or attempting to find cross product of two vectors lying in the plane      (M1)

for example,  AB × AD = ( 0 1 0 ) × ( 1 0 1 ) = ( 1 0 1 )      (A1)

Π 1 : x + z = 1      A1

[3 marks]

a.

EITHER

( 1 0 1 ) ( 0 1 1 ) = 1 = 2 2 cos θ      M1A1

OR

| ( 1 0 1 ) × ( 0 1 1 ) | = 3 = 2 2 sin θ      M1A1

Note: M1 is for an attempt to find the scalar or vector product of the two normal vectors.

θ = 60 ( = π 3 )      A1

angle between faces is  20 ( = 2 π 3 )      A1

[4 marks]

b.

DB = ( 1 1 1 ) or  BD = ( 1 1 1 )      (A1)

Π 3 : x + y z = k      (M1)

Π 3 : x + y z = 0      A1

[3 marks]

c.

METHOD 1

line AD : (r =) ( 0 0 1 ) + λ ( 1 0 1 )      M1A1

intersects  Π 3 when  λ ( 1 λ ) = 0      M1

so  λ = 1 2      A1

hence P is the midpoint of AD      AG

 

METHOD 2

midpoint of AD is (0.5, 0, 0.5)      (M1)A1

substitute into  x + y z = 0      M1

0.5 + 0.5 − 0.5 = 0     A1

hence P is the midpoint of AD     AG

[4 marks]

d.

METHOD 1

OP = 1 2 , O P Q = 90 , O Q P = 60       A1A1A1

PQ = 1 6      A1

area  = 1 2 12 = 1 4 3 = 3 12      A1

 

METHOD 2

line BD : ( =) ( 1 1 0 ) + λ ( 1 1 1 )

λ = 2 3      (A1)

OQ = ( 1 3 1 3 2 3 )     A1

area =  1 2 | OP × OQ |      M1

OP = ( 1 2 0 1 2 )     A1

Note: This A1 is dependent on M1.

area =  3 12      A1

[5 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.



Find the value of sin π 4 + sin 3 π 4 + sin 5 π 4 + sin 7 π 4 + sin 9 π 4 .

[2]
a.

Show that 1 cos 2 x 2 sin x sin x ,   x k π  where k Z .

[2]
b.

Use the principle of mathematical induction to prove that

sin x + sin 3 x + + sin ( 2 n 1 ) x = 1 cos 2 n x 2 sin x ,   n Z + ,   x k π where k Z .

[9]
c.

Hence or otherwise solve the equation sin x + sin 3 x = cos x  in the interval 0 < x < π .

[6]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

sin π 4 + sin 3 π 4 + sin 5 π 4 + sin 7 π 4 + sin 9 π 4 = 2 2 + 2 2 2 2 2 2 + 2 2 = 2 2    (M1)A1

 

Note: Award M1 for 5 equal terms with \) + \) or signs.

 

[2 marks]

a.

1 cos 2 x 2 sin x 1 ( 1 2 sin 2 x ) 2 sin x    M1

2 sin 2 x 2 sin x    A1

sin x    AG

[2 marks]

b.

let  P ( n ) : sin x + sin 3 x + + sin ( 2 n 1 ) x 1 cos 2 n x 2 sin x

if  n = 1

P ( 1 ) : 1 cos 2 x 2 sin x sin x which is true (as proved in part (b))     R1

assume P ( k )  true, sin x + sin 3 x + + sin ( 2 k 1 ) x 1 cos 2 k x 2 sin x      M1

 

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let n = k only. Subsequent marks are independent of this M1.

 

consider P ( k + 1 ) :

P ( k + 1 ) : sin x + sin 3 x + + sin ( 2 k 1 ) x + sin ( 2 k + 1 ) x 1 cos 2 ( k + 1 ) x 2 sin x

L H S = sin x + sin 3 x + + sin ( 2 k 1 ) x + sin ( 2 k + 1 ) x    M1

1 cos 2 k x 2 sin x + sin ( 2 k + 1 ) x    A1

1 cos 2 k x + 2 sin x sin ( 2 k + 1 ) x 2 sin x

1 cos 2 k x + 2 sin x cos x sin 2 k x + 2 sin 2 x cos 2 k x 2 sin x    M1

1 ( ( 1 2 sin 2 x ) cos 2 k x sin 2 x sin 2 k x ) 2 sin x    M1

1 ( cos 2 x cos 2 k x sin 2 x sin 2 k x ) 2 sin x    A1

1 cos ( 2 k x + 2 x ) 2 sin x    A1

1 cos 2 ( k + 1 ) x 2 sin x

so if true for n = k , then also true for  n = k + 1

as true for n = 1 then true for all n Z +      R1

 

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

 

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

 

[9 marks]

c.

EITHER

sin x + sin 3 x = cos x 1 cos 4 x 2 sin x = cos x    M1

1 cos 4 x = 2 sin x cos x ,   ( sin x 0 )    A1

1 ( 1 2 sin 2 2 x ) = sin 2 x    M1

sin 2 x ( 2 sin 2 x 1 ) = 0    M1

sin 2 x = 0  or sin 2 x = 1 2      A1

2 x = π ,   2 x = π 6 and  2 x = 5 π 6

OR

sin x + sin 3 x = cos x 2 sin 2 x cos x = cos x    M1A1

( 2 sin 2 x 1 ) cos x = 0 ,   ( sin x 0 )    M1A1

sin 2 x = 1 2 of cos x = 0     A1

2 x = π 6 ,   2 x = 5 π 6 and  x = π 2

THEN

x = π 2 ,   x = π 12  and x = 5 π 12      A1

 

Note: Do not award the final A1 if extra solutions are seen.

 

[6 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.



Use the binomial theorem to expand cosθ+isinθ4. Give your answer in the form a+bi where a and b are expressed in terms of sinθ and cosθ.

[3]
a.

Use de Moivre’s theorem and the result from part (a) to show that cot4θ=cot4θ-6cot2θ+14cot3θ-4cotθ.

[5]
b.

Use the identity from part (b) to show that the quadratic equation x2-6x+1=0 has roots cot2π8 and cot23π8.

[5]
c.

Hence find the exact value of cot23π8.

[4]
d.

Deduce a quadratic equation with integer coefficients, having roots cosec2π8 and cosec23π8.

[3]
e.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

uses the binomial theorem on cosθ+isinθ4       M1

=C04cos4θ+C14cos3θisinθ+C24cos2θi2sin2θ+C34cosθi3sin3θ+C44i4sin4θ        A1

=cos4θ-6cos2θsin2θ+sin4θ+i4cos3θsinθ-4cosθsin3θ        A1

 

[3 marks]

a.

(using de Moivre’s theorem with n=4 gives) cos4θ+isin4θ        (A1)

equates both the real and imaginary parts of cos4θ+isin4θ and cos4θ-6cos2θsin2θ+sin4θ+i4cos3θsinθ-4cosθsin3θ       M1

cos4θ=cos4θ-6cos2θsin2θ+sin4θ  and  sin4θ=4cos3θsinθ-4cosθsin3θ

recognizes that cot4θ=cos4θsin4θ        (A1)

substitutes for sin4θ and cos4θ into cos4θsin4θ       M1

cot4θ=cos4θ-6cos2θsin2θ+sin4θ4cos3θsinθ-4cosθsin3θ

divides the numerator and denominator by sin4θ to obtain

cot4θ=cos4θ-6cos2θsin2θ+sin4θsin4θ4cos3θsinθ-4cosθsin3θsin4θ        A1

cot4θ=cot4θ-6cot2θ+14cot3θ-4cotθ        AG

 

[5 marks]

b.

setting cot4θ=0 and putting x=cot2θ in the numerator of cot4θ=cot4θ-6cot2θ+14cot3θ-4cotθ gives x2-6x+1=0        M1

attempts to solve cot4θ=0 for θ        M1

4θ=π2,3π2, 4θ=122n+1π,n=0,1,        (A1)

θ=π8,3π8        A1

 

Note: Do not award the final A1 if solutions other than θ=π8,3π8 are listed.

 

finding the roots of cot4θ=0 θ=π8,3π8 corresponds to finding the roots of x2-6x+1=0 where x=cot2θ        R1

so the equation x2-6x+1=0 as roots cot2π8 and cot23π8        AG

 

[5 marks]

c.

attempts to solve x2-6x+1=0 for x        M1

x=3±22        A1

since cot2π8>cot23π8, cot23π8 has the smaller value of the two roots        R1

 

Note: Award R1 for an alternative convincing valid reason.

 

so cot23π8=3-22        A1

 

[4 marks]

d.

let y=cosec2θ

uses cot2θ=cosec2θ-1 where x=cot2θ        (M1)

x2-6x+1=0y-12-6y-1+1=0        M1

y2-8y+8=0        A1

 

[3 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.



In the following diagram, the points  A B C and  D  are on the circumference of a circle with centre O and radius r [ AC ]  is a diameter of the circle.  BC = r AD = CD and  A B C = A D C = 90 .

Given that  cos 75 = q , show that  cos 105 = q .

[1]
a.

Show that B A D = 75 .

[3]
b.

By considering triangle  ABD , show that  B D 2 = 5 r 2 2 r 2 q 6 .

[4]
c.i.

By considering triangle CBD , find another expression for B D 2 in terms of r and q .

[3]
c.ii.

Use your answers to part (c) to show that  cos 75 = 1 6 + 2 .

[3]
d.

Markscheme

cos 105 = cos ( 180 75 ) = cos 75       R1

= q        AG

Note: Accept arguments using the unit circle or graphical/diagrammatical considerations.

[1 mark]

a.

AD = CD C A D = 45       A1

valid method to find  B A C         (M1)

for example:  BC = r B C A = 60

B A C = 30       A1

hence  B A D = 45 + 30 = 75       AG

[3 marks]

b.

AB = r 3 AD = ( CD ) = r 2        A1A1

applying cosine rule        (M1)

B D 2 = ( r 3 ) 2 + ( r 2 ) 2 2 ( r 3 ) ( r 2 ) cos 75        A1

= 3 r 2 + 2 r 2 2 r 2 6 cos 75

=5r22r2q6       AG

[4 marks]

c.i.

B C D = 105         (A1)

attempt to use cosine rule on  Δ BCD         (M1)

B D 2 = r 2 + ( r 2 ) 2 2 r ( r 2 ) cos 105

= 3 r 2 + 2 r 2 q 2        A1

[3 marks]

c.ii.

5 r 2 2 r 2 q 6 = 3 r 2 + 2 r 2 q 2         (M1)(A1)

2 r 2 = 2 r 2 q ( 6 + 2 )        A1

Note: Award A1 for any correct intermediate step seen using only two terms.

q = 1 6 + 2        AG

Note: Do not award the final A1 if follow through is being applied.

[3 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.



Given any two non-zero vectors, a and b, show that a×b2=a2b2-a·b2.

Markscheme

METHOD 1

use of a×b=absinθ on the LHS            (M1)

a×b2=a2b2sin2θ                   A1

=a2b21-cos2θ            M1

=a2b2-a2b2cos2θ  OR  =a2b2-abcosθ2                   A1

=a2b2-a·b2                   AG

 

METHOD 2

use of a·b=abcosθ on the RHS            (M1)

=a2b2-a2b2cos2θ                   A1

=a2b21-cos2θ            M1

=a2b2sin2θ  OR  =absinθ2                   A1

=a×b2                   AG

 

Note: If candidates attempt this question using cartesian vectors, e.g

a=a1a2a3  and  b=b1b2b3,

award full marks if fully developed solutions are seen.
Otherwise award no marks.

 

[4 marks]

Examiners report

[N/A]



The following diagram shows the graph of y=arctan2x+1+π4 for x, with asymptotes at y=-π4 and y=3π4.

Describe a sequence of transformations that transforms the graph of y=arctan x to the graph of y=arctan2x+1+π4 for x.

[3]
a.

Show that arctanp+arctanqarctanp+q1-pq where p, q>0 and pq<1.

[4]
b.

Verify that arctan 2x+1=arctan xx+1+π4 for x, x>0.

[3]
c.

Using mathematical induction and the result from part (b), prove that Σr=1narctan12r2=arctannn+1 for n+.

[9]
d.

Markscheme

EITHER
horizontal stretch/scaling with scale factor 12


Note: Do not allow ‘shrink’ or ‘compression’


followed by a horizontal translation/shift 12 units to the left           A2


Note: Do not allow ‘move’


OR

horizontal translation/shift 1 unit to the left

followed by horizontal stretch/scaling with scale factor 12     A2


THEN

vertical translation/shift up by π4 (or translation through 0π4          A1
(may be seen anywhere)

 

[3 marks]

a.

let α=arctanp and β=arctanq        M1

p=tanα and q=tanβ        (A1)

tanα+β=p+q1-pq        A1

α+β=arctanp+q1-pq        A1

so arctanp+arctanqarctanp+q1-pq where p, q>0 and pq<1.       AG

 

[4 marks]

b.

METHOD 1

π4=arctan1 (or equivalent)        A1

arctanxx+1+arctan1=arctanxx+1+11-xx+11        A1

=arctanx+x+1x+1x+1-xx+1        A1

=arctan2x+1       AG

 

METHOD 2

tanπ4=1 (or equivalent)        A1

Consider arctan2x+1-arctanxx+1=π4

tanarctan2x+1-arctanxx+1

=arctan2x+1-xx+11+x2x+1x+1        A1

=arctan2x+1x+1-xx+1+x2x+1        A1

=arctan 1       AG

 

METHOD 3

tan arctan2x+1=tanarctanxx+1+π4

tanπ4=1 (or equivalent)        A1

LHS=2x+1        A1

RHS=xx+1+11-xx+1=2x+1        A1

 

[3 marks]

c.

let Pn be the proposition that  Σr=1narctan12r2=arctannn+1 for n+

consider P1

when n=1, Σr=11arctan12r2=arctan12=RHS and so P1 is true          R1

assume Pk is true, ie. Σr=1karctan12r2=arctankk+1 k+           M1

 

Note: Award M0 for statements such as “let n=k”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

 

consider Pk+1:

Σr=1k+1arctan12r2=Σr=1karctan12r2+arctan12k+12           (M1)

=arctankk+1+arctan12k+12        A1

=arctankk+1+12k+121-kk+112k+12           M1

=arctank+12k2+2k+12k+13-k        A1

 

Note: Award A1 for correct numerator, with (k+1) factored. Denominator does not need to be simplified

 

=arctank+12k2+2k+12k3+6k2+5k+2        A1

 

Note: Award A1 for denominator correctly expanded. Numerator does not need to be simplified. These two A marks may be awarded in any order

 

=arctank+12k2+2k+1k+22k2+2k+1=arctank+1k+2        A1

 

Note: The word ‘arctan’ must be present to be able to award the last three A marks

 

Pk+1 is true whenever Pk is true and P1 is true, so

Pn is true for for n+         R1

 

Note: Award the final R1 mark provided at least four of the previous marks have been awarded.
Note: To award the final R1, the truth of Pk must be mentioned. ‘Pk implies Pk+1’ is insufficient to award the mark.

 

[9 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.



Consider the lines l 1 and l 2 defined by

l 1 :  r = ( 3 2 a ) + β ( 1 4 2 )  and l 2 : 6 x 3 = y 2 4 = 1 z where a is a constant.

Given that the lines l 1 and l 2 intersect at a point P,

find the value of a ;

[4]
a.

determine the coordinates of the point of intersection P.

[2]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

l 1 : r  = ( 3 2 a ) = β ( 1 4 2 ) { x = 3 + β y = 2 + 4 β z = a + 2 β      M1

6 ( 3 + β ) 3 = ( 2 + 4 β ) 2 4 4 = 4 β 3 β = 3    M1A1

6 ( 3 + β ) 3 = 1 ( a + 2 β ) 2 = 5 a a = 7    A1

METHOD 2

{ 3 + β = 6 3 λ 2 + 4 β = 4 λ + 2 a + 2 β = 1 λ    M1

attempt to solve     M1

λ = 2 ,   β = 3    A1

a = 1 λ 2 β = 7    A1

[4 marks]

a.

OP = ( 3 2 7 ) + 3 ( 1 4 2 )    (M1)

= ( 0 10 1 )    A1

P ( 0 ,  10,  1 )

[2 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.



Consider a triangle OAB such that O has coordinates (0, 0, 0), A has coordinates (0, 1, 2) and B has coordinates (2 b , 0, b − 1) where b < 0.

Let M be the midpoint of the line segment [OB].

Find, in terms of b , a Cartesian equation of the plane Π containing this triangle.

[5]
a.

Find, in terms of b , the equation of the line L which passes through M and is perpendicular to the plane П.

[3]
b.

Show that L does not intersect the y -axis for any negative value of b .

 

[7]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

n = ( 0 1 2 ) × ( 2 b 0 b 1 )       (M1)

= ( b 1 4 b 2 b )       (M1)A1

(0, 0, 0) on Π so  ( b 1 ) x + 4 b y 2 b z = 0       (M1)A1

 

METHOD 2

using equation of the form  p x + q y + r z = 0       (M1)

(0, 1, 2) on Π ⇒  q + 2 r = 0

(2 b , 0, b − 1) on Π ⇒  2 b p + r ( b 1 ) = 0       (M1)A1

Note: Award (M1)A1 for both equations seen.

solve for  p q and  r       (M1)

( b 1 ) x + 4 b y 2 b z = 0       A1

 

[5 marks]

a.

M has coordinates  ( b , 0 , b 1 2 )       (A1)

r ( b 0 b 1 2 ) + λ ( b 1 4 b 2 b )       M1A1

Note: Award M1A0 if r = (or equivalent) is not seen.

Note: Allow equivalent forms such as  x b b 1 = y 4 b = 2 z b + 1 4 b .

 

[3 marks]

b.

METHOD 1

x = z = 0       (M1)

Note: Award M1 for either x = 0 or z = 0 or both.

b + λ ( b 1 ) = 0 and  b 1 2 2 λ b = 0       A1

attempt to eliminate λ        M1

b b 1 = b 1 4 b       (A1)

4 b 2 = ( b 1 ) 2       A1

EITHER

consideration of the signs of LHS and RHS       (M1)

the LHS is negative and the RHS must be positive (or equivalent statement)       R1

OR

4 b 2 = b 2 2 b + 1

5 b 2 2 b + 1 = 0

Δ = ( 2 ) 2 4 × 5 × 1 = 16 ( < 0 )      M1

no real solutions       R1

THEN

so no point of intersection       AG

 

METHOD 2

x = z = 0       (M1)

Note: Award M1 for either  x = 0 or  z = 0 or both.

b + λ ( b 1 ) = 0 and  b 1 2 2 λ b = 0       A1

attempt to eliminate b       M1

λ 1 + λ = 1 1 4 λ       (A1)

4 λ 2 = 1 ( λ 2 = 1 4 )       A1

consideration of the signs of LHS and RHS       (M1)

there are no real solutions (or equivalent statement)       R1

so no point of intersection       AG

 

[7 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.



The lengths of two of the sides in a triangle are 4 cm and 5 cm. Let θ be the angle between the two given sides. The triangle has an area of 5 15 2  cm2.

Show that sin θ = 15 4 .

[1]
a.

Find the two possible values for the length of the third side.

[6]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

EITHER

5 15 2 = 1 2 × 4 × 5 sin θ       A1

OR

height of triangle is  5 15 4  if using 4 as the base or  15  if using 5 as the base      A1

THEN

sin θ = 15 4         AG

[1 mark]

a.

let the third side be x

x 2 = 4 2 + 5 2 2 × 4 × 5 × cos θ        M1

valid attempt to find  cos θ        (M1)

Note: Do not accept writing  cos ( arcsin ( 15 4 ) ) as a valid method.

cos θ = ± 1 15 16

= 1 4 , 1 4        A1A1

x 2 = 16 + 25 2 × 4 × 5 × ± 1 4

x = 31   or   51        A1A1

[6 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.



The points A and B are given by A ( 0 ,   3 ,   6 ) and B ( 6 ,   5 ,   11 ) .

The plane Π is defined by the equation 4 x 3 y + 2 z = 20 .

Find a vector equation of the line L passing through the points A and B.

[3]
a.

Find the coordinates of the point of intersection of the line L with the plane Π.

[3]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

AB = ( 6 8 17 )     (A1)

 

r = ( 0 3 6 ) + λ ( 6 8 17 ) or r = ( 6 5 11 ) + λ ( 6 8 17 )     M1A1

 

Note:     Award M1A0 if r = is not seen (or equivalent).

 

[3 marks]

a.

substitute line L in Π : 4 ( 6 λ ) 3 ( 3 8 λ ) + 2 ( 6 + 17 λ ) = 20     M1

82 λ = 41

λ = 1 2     (A1)

 

r = ( 0 3 6 ) + 1 2 ( 6 8 17 ) = ( 3 1 5 2 )

so coordinate is ( 3 ,   1 ,   5 2 )     A1

 

Note:     Accept coordinate expressed as position vector ( 3 1 5 2 ) .

 

[3 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.



The function f is defined by  f ( x ) = e x cos 2 x , where 0 ≤  x  ≤ 5. The curve  y = f ( x )  is shown on the following graph which has local maximum points at A and C and touches the x -axis at B and D.

Use integration by parts to show that e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x + c ,   c R .

[5]
a.

Hence, show that e x cos 2 x d x = e x 5 sin 2 x + e x 10 cos 2 x + e x 2 + c ,   c R .

[3]
b.

Find the x -coordinates of A and of C , giving your answers in the form  a + arctan b , where  a b R .

[6]
c.

Find the area enclosed by the curve and the x -axis between B and D, as shaded on the diagram.

[5]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

attempt at integration by parts with  u = e x ,   d v d x = cos 2 x       M1

e x cos 2 x d x = e x 2 sin 2 x d x e x 2 sin 2 x d x       A1

= e x 2 sin 2 x 1 2 ( e x 2 cos 2 x + e x 2 cos 2 x )       M1A1

e x 2 sin 2 x + e x 4 cos 2 x 1 4 e x cos 2 x d x

5 4 e x cos 2 x d x = e x 2 sin 2 x + e x 4 cos 2 x       M1

e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x ( + c )     AG

 

 

METHOD 2

attempt at integration by parts with u = cos 2 x d v d x = e x       M1

e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x d x       A1

= e x cos 2 x + 2 ( e x sin 2 x 2 e x cos 2 x d x )       M1A1

= e x cos 2 x + 2 e x sin 2 x 4 e x cos 2 x d x

5 e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x       M1

e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x ( + c )    AG

 

METHOD 3

attempt at use of table      M1

eg

      A1A1 

Note: A1 for first 2 lines correct, A1 for third line correct.

e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x 4 e x cos 2 x d x       M1

5 e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x       M1

e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x ( + c )    AG

 

[5 marks]

a.

e x co s 2 x d x = e x 2 ( cos 2 x + 1 ) d x      M1A1

= 1 2 ( 2 e x 5 sin 2 x + e x 5 cos 2 x ) + e x 2       A1

= e x 5 sin 2 x + e x 10 cos 2 x + e x 2 ( + c )       AG

Note: Do not accept solutions where the RHS is differentiated.

 

[3 marks]

b.

f ( x ) = e x co s 2 x 2 e x sin x cos x       M1A1

Note: Award M1 for an attempt at both the product rule and the chain rule.

e x cos x ( cos x 2 sin x ) = 0       (M1)

Note: Award M1 for an attempt to factorise  cos x  or divide by  cos x ( cos x 0 ) .

discount  cos x = 0  (as this would also be a zero of the function)

cos x 2 sin x = 0

tan x = 1 2       (M1)

x = arctan ( 1 2 ) (at A) and  x = π + arctan ( 1 2 )  (at C)      A1A1

Note: Award A1 for each correct answer. If extra values are seen award A1A0.

 

[6 marks]

c.

 

cos x = 0 x = π 2 or  3 π 2       A1

Note: The A1may be awarded for work seen in part (c).

π 2 3 π 2 ( e x co s 2 x ) d x = [ e x 5 sin 2 x + e x 10 cos 2 x + e x 2 ] π 2 3 π 2       M1

= ( e 3 π 2 10 + e 3 π 2 2 ) ( e π 2 10 + e π 2 2 ) ( = 2 e 3 π 2 5 2 e π 2 5 )       M1(A1)A1

Note: Award M1 for substitution of the end points and subtracting, (A1) for  sin 3 π = sin π = 0 and  cos 3 π = cos π = 1  and A1 for a completely correct answer.

 

[5 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.



The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.

It is given that  AB = DC .

The position vectors  OA OB OC and  OD are given by

a = i + 2j − 3k

b = 3ij + pk

c = qi + j + 2k

d = −i + rj − 2k

where p , q and r are constants.

The point where the diagonals of ABCD intersect is denoted by M.

The plane Π cuts the x, y and z axes at X , Y and Z respectively.

Explain why ABCD is a parallelogram.

[1]
a.i.

Using vector algebra, show that AD = BC .

[3]
a.ii.

Show that p = 1, q = 1 and r = 4.

[5]
b.

Find the area of the parallelogram ABCD.

[4]
c.

Find the vector equation of the straight line passing through M and normal to the plane Π  containing ABCD.

[4]
d.

Find the Cartesian equation of Π .

[3]
e.

Find the coordinates of X, Y and Z.

[2]
f.i.

Find YZ.

[2]
f.ii.

Markscheme

a pair of opposite sides have equal length and are parallel      R1

hence ABCD is a parallelogram      AG

[1 mark]

a.i.

attempt to rewrite the given information in vector form       M1

ba = cd      A1

rearranging d − a = c − b       M1

hence   AD = BC      AG

Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.

[3 marks]

a.ii.

EITHER

use of  AB = DC      (M1)

( 2 3 p + 3 ) = ( q + 1 1 r 4 )        A1A1

OR

use of  AD = BC       (M1)

( 2 r 2 1 ) = ( q 3 2 2 p )       A1A1

THEN

attempt to compare coefficients of i, j, and k in their equation or statement to that effect       M1

clear demonstration that the given values satisfy their equation       A1
p = 1, q = 1, r = 4       AG

[5 marks]

b.

attempt at computing AB × AD  (or equivalent)       M1

( 11 10 2 )      A1

area  = | AB × AD | ( = 225 )       (M1)

= 15       A1

[4 marks]

c.

valid attempt to find  OM = ( 1 2 ( a + c ) )       (M1)

( 1 3 2 1 2 )      A1

the equation is

r ( 1 3 2 1 2 ) + t ( 11 10 2 )  or equivalent       M1A1

Note: Award maximum M1A0 if 'r = …' (or equivalent) is not seen.

[4 marks]

d.

attempt to obtain the equation of the plane in the form ax + by + cz = d       M1

11x + 10y + 2z = 25      A1A1

Note: A1 for right hand side, A1 for left hand side.

[3 marks]

e.

putting two coordinates equal to zero       (M1)

X ( 25 11 , 0 , 0 ) , Y ( 0 , 5 2 , 0 ) , Z ( 0 , 0 , 25 2 )       A1

[2 marks]

f.i.

YZ = ( 5 2 ) 2 + ( 25 2 ) 2      M1

= 325 2 ( = 5 104 4 = 5 26 2 )      A1

[4 marks]

f.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.i.
[N/A]
f.ii.



A function f is defined by fx=1x2-2x-3, where x, x-1, x3.

A function g is defined by gx=1x2-2x-3, where x, x>3.

The inverse of g is g-1.

A function h is defined by hx=arctanx2, where x.

Sketch the curve y=f(x), clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.

[6]
a.

Show that g-1x=1+4x2+xx.

[6]
b.i.

State the domain of g-1.

[1]
b.ii.

Given that hga=π4, find the value of a.

Give your answer in the form p+q2r, where p, q, r+.

[7]
c.

Markscheme

 

y-intercept 0,-13         A1


Note:
Accept an indication of -13 on the y-axis.


vertical asymptotes x=-1 and x=3          A1

horizontal asymptote y=0          A1

uses a valid method to find the x-coordinate of the local maximum point          (M1)


Note:
For example, uses the axis of symmetry or attempts to solve f'x=0.


local maximum point 1,-14          A1


Note:
Award (M1)A0 for a local maximum point at x=1 and coordinates not given.


three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other          A1

 

[6 marks]

a.

x=1y2-2y-3           M1


Note: Award M1 for interchanging x and y (this can be done at a later stage).

 

EITHER

attempts to complete the square           M1

y2-2y-3=y-12-4          A1

x=1y-12-4

y-12-4=1xy-12=4+1x          A1

y-1=±4+1x =±4x+1x

 

OR

attempts to solve xy2-2xy-3x-1=0 for y         M1

y=--2x±-2x2+4x3x+12x         A1


Note:
Award A1 even if - (in ±) is missing


=2x±16x2+4x2x         A1

 

THEN

=1±4x2+xx         A1

y>3 and hence y=1-4x2+xx is rejected                R1 

 

Note: Award R1 for concluding that the expression for y must have the ‘+’ sign.
The R1 may be awarded earlier for using the condition x>3.

 

y=1+4x2+xx

g-1x=1+4x2+xx         AG

 

[6 marks]

b.i.

domain of g-1 is x>0         A1

 

[1 mark]

b.ii.

attempts to find hga          (M1)

hga=arctanga2   hga=arctan12a2-2a-3          (A1)

arctanga2=π4   arctan12a2-2a-3=π4

attempts to solve for ga         M1

ga=2  1a2-2a-3=2

 

EITHER

a=g-12         A1

attempts to find their g-12         M1

a=1+422+22         A1

 

Note: Award all available marks to this stage if x is used instead of a.


OR

2a2-4a-7=0         A1

attempts to solve their quadratic equation         M1

a=--4±-42+4274  =4±724         A1


Note: Award all available marks to this stage if x is used instead of a.


THEN

a=1+322  (as a>3)         A1

p=1, q=3, r=2

 

Note: Award A1 for a=1+1218  p=1, q=1, r=18

 

[7 marks]

c.

Examiners report

Part (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.

Part (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.

Part (c) was well done in general, with some algebraic errors seen in occasions.

a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.



By using the substitution u=secx or otherwise, find an expression for 0π3secnxtanxdx in terms of n, where n is a non-zero real number.

Markscheme

METHOD 1

u=secxdu=secxtanxdx         (A1)

attempts to express the integral in terms of u         M1

12un-1du         A1

=1nun12  (=1nsecnx0π3)          A1

 

Note: Condone the absence of or incorrect limits up to this point.

 

=2n-1nn         M1

=2n-1n          A1

 

Note: Award M1 for correct substitution of their limits for u into their antiderivative for u (or given limits for x into their antiderivative for x).

 

METHOD 2

secnxtanxdx=secn-1xsecxtanxdx         (A1)

applies integration by inspection         (M1)

=1nsecnx0π3          A2

 

Note: Award A2 if the limits are not stated.

 

=1nsecnπ3-secn0         M1

 

Note: Award M1 for correct substitution into their antiderivative.

 

=2n-1n          A1

  

[6 marks]

Examiners report

[N/A]



A straight line,  L θ , has vector equation r  = ( 5 0 0 ) + λ ( 5 sin θ cos θ ) λ θ R .

The plane Πp, has equation x = p p R .

Show that the angle between  L θ and Πp is independent of both  θ and  p .

Markscheme

a vector normal to Πp is  ( 1 0 0 )        (A1)

Note: Allow any scalar multiple of  ( 1 0 0 ) , including  ( p 0 0 )

attempt to find scalar product (or vector product) of direction vector of line with any scalar multiple of  ( 1 0 0 )         M1

( 1 0 0 ) ( 5 sin θ cos θ ) = 5   (or  ( 1 0 0 ) × ( 5 sin θ cos θ ) = ( 0 cos θ sin θ ) )       A1

(if  α  is the angle between the line and the normal to the plane)

cos α = 5 1 × 25 + si n 2 θ + co s 2 θ (or  sin α = 1 1 × 25 + si n 2 θ + co s 2 θ )       A1

cos α = 5 26 or  sin α = 1 26        A1

this is independent of  p and  θ , hence the angle between the line and the plane,  ( 90 α ) , is also independent of  p and  θ        R1

Note: The final R mark is independent, but is conditional on the candidate obtaining a value independent of p and  θ .

[6 marks]

Examiners report

[N/A]



The lines l1 and l2 have the following vector equations where λ,μ and m.

l1:r1=3-20+λ21m l2:r2=-1-4-2m+μ2-5-m

The plane Π has Cartesian equation x+4y-z=p where p.

 

Given that l1 and Π have no points in common, find

Show that l1 and l2 are never perpendicular to each other.

[3]
a.

the value of m.

[2]
b.i.

the condition on the value of p.

[2]
b.ii.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

attempts to calculate 21m·2-5-m        (M1)

=-1-m2        A1

since m20, -1-m2<0 for m         R1

so l1 and l2 are never perpendicular to each other        AG

 

[3 marks]

a.

(since l1 is parallel to Π, l1 is perpendicular to the normal of Π and so)

21m·14-1=0         R1

2+4-m=0

m=6        A1

 

[2 marks]

b.i.

since there are no points in common, 3,-2,0 does not lie in Π       

 

EITHER

substitutes 3,-2,0 into x+4y-zp        (M1)

 

OR

3-20·14-1p        (M1)

 

THEN

p-5        A1

 

[2 marks]

b.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.



Consider the line L1 defined by the Cartesian equation x+12=y=3-z.

Consider a second line L2 defined by the vector equation r=012+ta1-1, where t and a.

Show that the point (-1, 0, 3) lies on L1.

[1]
a.i.

Find a vector equation of L1.

[3]
a.ii.

Find the possible values of a when the acute angle between L1 and L2 is 45°.

[8]
b.

It is given that the lines L1 and L2 have a unique point of intersection, A, when ak.

Find the value of k, and find the coordinates of the point A in terms of a.

[7]
c.

Markscheme

-1+12=0=3-3          A1

the point (-1, 0, 3) lies on L1.          AG

 

[1 mark]

a.i.

attempt to set equal to a parameter or rearrange cartesian form          (M1)

x+12=y=3-z=λx=2λ-1, y=λ, z=3-λ  OR  x+12=y-01=z-3-1

correct direction vector 21-1 or equivalent seen in vector form          (A1)

r=-103+λ21-1 (or equivalent)          A1

 

Note: Award A0 if =r is omitted.

 

[3 marks]

a.ii.

attempt to use the scalar product formula          (M1)

21-1a1-1=±6a2+2cos45°          (A1)(A1)

 

Note: Award A1 for LHS and A1 for RHS

 

2a+2=±6a2+2222a+2=±3a2+2         A1A1

 

Note: Award A1 for LHS and A1 for RHS

 

4a2+8a+4=3a2+6         A1

a2+8a-2=0        M1

attempt to solve their quadratic

a=-8±64+82=-8±722=-4±32         A1

 

[8 marks]

b.

METHOD 1

attempt to equate the parametric forms of L1 and L2         (M1)

2λ-1=taλ=1+t3-λ=2-t         A1

attempt to solve equations by eliminating λ or t         (M1)

2+2t-1=ta1=ta-2  or  2λ-1=λ-1aa-1=λa-2

Solutions exist unless a-2=0

k=2         A1

 

Note: This A1 is independent of the following marks.

 

t=1a-2  or  λ=a-1a-2         A1

A has coordinates aa-2, 1+1a-2, 2-1a-2 =aa-2, a-1a-2, 2a-5a-2         A2

 

Note: Award A1 for any two correct coordinates seen or final answer in vector form.

 

 

METHOD 2

no unique point of intersection implies direction vectors of L1 and L2 parallel

k=2         A1

 

Note: This A1 is independent of the following marks.

 

attempt to equate the parametric forms of L1 and L2         (M1)

2λ-1=taλ=1+t3-λ=2-t         A1

attempt to solve equations by eliminating λ or t         (M1)

2+2t-1=ta1=ta-2  or  2λ-1=λ-1aa-1=λa-2

t=1a-2  or  λ=a-1a-2         A1

A has coordinates aa-2, 1+1a-2, 2-1a-2 =aa-2, a-1a-2, 2a-5a-2         A2

 

Note: Award A1 for any two correct coordinates seen or final answer in vector form.

 

[7 marks]

c.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.



Points A (0 , 0 , 10) , B (0 , 10 , 0) , C (10 , 0 , 0) , V ( p , p , p ) form the vertices of a tetrahedron.

Consider the case where the faces ABV and ACV are perpendicular.

The following diagram shows the graph of θ against p . The maximum point is shown by X .

Show that  AB × AV = 10 ( 10 2 p p p )  and find a similar expression for AC × AV .

[3]
a.i.

Hence, show that, if the angle between the faces  ABV and  ACV is  θ , then cos θ = p ( 3 p 20 ) 6 p 2 40 p + 100 .

[5]
a.ii.

Find the two possible coordinates of V .

[3]
b.i.

Comment on the positions of V in relation to the plane ABC .

[1]
b.ii.

At X , find the value of p and the value of θ .

[3]
c.i.

Find the equation of the horizontal asymptote of the graph.

[2]
c.ii.

Markscheme

AV = ( p p p 10 )       A1

AB × AV = ( 0 10 10 ) × ( p p p 10 ) = ( 10 ( p 10 ) + 10 p 10 p 10 p )       A1

= ( 20 p 100 10 p 10 p ) = 10 ( 10 2 p p p )       AG

AC × AV = ( 10 0 10 ) × ( p p p 10 ) = ( 10 p 100 20 p 10 p ) ( = 10 ( p 10 2 p p ) )       A1

 

[3 marks]

a.i.

attempt to find a scalar product        M1

10 ( 10 2 p p p ) 10 ( p 10 2 p p ) = 100 ( 3 p 2 20 p )

OR  ( 10 2 p p p ) ( p 10 2 p p ) = 3 p 2 20 p       A1

attempt to find magnitude of either  AB × AV   or   AC × AV         M1

| 10 ( 10 2 p p p ) | = | 10 ( p 10 2 p p ) | = 10 ( 10 2 p ) 2 + 2 p 2         A1

100 ( 3 p 2 20 p ) = 100 ( ( 10 2 p ) 2 + 2 p 2 ) 2 cos θ

cos θ = 3 p 2 20 p ( 10 2 p ) 2 + 2 p 2         A1

Note: Award A1 for any intermediate step leading to the correct answer.

= p ( 3 p 20 ) 6 p 2 40 p + 100       AG

Note: Do not allow FT marks from part (a)(i).

[8 marks]

a.ii.

p ( 3 p 20 ) = 0 p = 0   or   p = 20 3         M1A1

coordinates are (0, 0, 0) and  ( 20 3 20 3 20 3 )       A1

Note: Do not allow column vectors for the final A mark.

[3 marks]

b.i.

two points are mirror images in the plane
or opposite sides of the plane
or equidistant from the plane
or the line connecting the two Vs is perpendicular to the plane       R1

[1 mark]

b.ii.

geometrical consideration or attempt to solve  1 = p ( 3 p 20 ) 6 p 2 40 p + 100        (M1)

p = 10 3 θ = π   or   θ = 180        A1A1

[3 marks]

c.i.

p cos θ 1 2          M1

hence the asymptote has equation θ = π 3         A1

[2 marks]

c.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.



Find the coordinates of the point of intersection of the planes defined by the equations x + y + z = 3 ,   x y + z = 5 and x + y + 2 z = 6 .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

for eliminating one variable from two equations     (M1)

eg, { ( x + y + z = 3 ) 2 x + 2 z = 8 2 x + 3 z = 11      A1A1

for finding correctly one coordinate

eg, { ( x + y + z = 3 ) ( 2 x + 2 z = 8 ) z = 3      A1

for finding correctly the other two coordinates     A1

{ x = 1 y = 1 z = 3

the intersection point has coordinates  ( 1 ,   1 ,   3 )

METHOD 2

for eliminating two variables from two equations or using row reduction     (M1)

eg, { ( x + y + z = 3 ) 2 = 2 z = 3  or  ( 1 1 1 0 2 0 0 0 1 | 3 2 3 )      A1A1

for finding correctly the other coordinates     A1A1

{ x = 1 y = 1 ( z = 3 )  or  ( 1 0 0 0 1 0 0 0 1 | 1 1 3 )

the intersection point has coordinates  ( 1 ,   1 ,   3 )

METHOD 3

| 1 1 1 1 1 1 1 1 2 | = 2    (A1)

attempt to use Cramer’s rule     M1

x = | 3 1 1 5 1 1 6 1 2 | 2 = 2 2 = 1    A1

y = | 1 3 1 1 5 1 1 6 2 | 2 = 2 2 = 1    A1

z = | 1 1 3 1 1 5 1 1 6 | 2 = 6 2 = 3    A1

 

Note:     Award M1 only if candidate attempts to determine at least one of the variables using this method.

 

[5 marks]

Examiners report

[N/A]



Consider the function  g ( x ) = 4 cos x + 1 a x π 2 where  a < π 2 .

For  a = π 2 , sketch the graph of  y = g ( x ) . Indicate clearly the maximum and minimum values of the function.

[3]
a.

Write down the least value of a such that g has an inverse.

[1]
b.

For the value of a found in part (b), write down the domain of g 1 .

[1]
c.i.

For the value of a found in part (b), find an expression for g 1 ( x ) .

[2]
c.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

concave down and symmetrical over correct domain       A1

indication of maximum and minimum values of the function (correct range)       A1A1

 

[3 marks]

a.

a = 0      A1

Note: Award A1 for a = 0 only if consistent with their graph.

 

[1 mark]

b.

1 x 5      A1

Note: Allow FT from their graph.

 

[1 mark]

c.i.

y = 4 cos x + 1

x = 4 cos y + 1

x 1 4 = cos y       (M1)

y = arccos ( x 1 4 )

g 1 ( x ) = arccos ( x 1 4 )       A1

 

[2 marks]

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.



Consider the vectors a =  i   3 j    2 k, b  =   3 j  +   2 k.

Find a  ×  b.

[2]
a.

Hence find the Cartesian equation of the plane containing the vectors a and b, and passing through the point ( 1 ,   0 ,   1 ) .

[3]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

a  ×  b  = 12 i    2 j    3 k     (M1)A1

[2 marks]

a.

METHOD 1

12 x 2 y 3 z = d    M1

12 × 1 2 × 0 3 ( 1 ) = d    (M1)

d = 9    A1

12 x 2 y 3 z = 9   ( or  12 x + 2 y + 3 z = 9 )

METHOD 2

( x y z ) ( 12 2 3 ) = ( 1 0 1 ) ( 12 2 3 )    M1A1

12 x 2 y 3 z = 9   ( or  12 x + 2 y + 3 z = 9 )    A1

[3 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.



A and B  are acute angles such that  cos A = 2 3 and  sin B = 1 3 .

Show that cos ( 2 A + B ) = 2 2 27 4 5 27 .

Markscheme

attempt to use  cos ( 2 A + B ) = cos 2 A cos B sin 2 A sin B (may be seen later)       M1

attempt to use any double angle formulae (seen anywhere)       M1

attempt to find either sin A or cos B (seen anywhere)       M1

cos A = 2 3 sin A ( = 1 4 9 ) = 5 3        (A1)

sin B = 1 3 cos B ( = 1 1 9 = 8 3 ) = 2 2 3        A1

cos 2 A ( = 2 co s 2 A 1 ) = 1 9        A1

sin 2 A ( = 2 sin A cos A ) = 4 5 9        A1

So   cos ( 2 A + B ) = ( 1 9 ) ( 2 2 3 ) ( 4 5 9 ) ( 1 3 )

= 2 2 27 4 5 27        AG

[7 marks]

Examiners report

[N/A]



The acute angle between the vectors 3i − 4j − 5k and 5i − 4j + 3k is denoted by θ.

Find cos θ.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

cos θ =  ( 3 i 4 j 5 k ) ( 5 i 4 j + 3 k ) | 3 i 4 j 5 k | | 5 i 4 j + 3 k |       (M1)

= 16 50 50      A1A1

Note: A1 for correct numerator and A1 for correct denominator.

= 8 25 ( = 16 50 = 0.32 )       A1

[4 marks]

 

Examiners report

[N/A]



ABCD is a parallelogram, where AB = –i + 2j + 3k and AD = 4ij – 2k.

Find the area of the parallelogram ABCD.

[3]
a.

By using a suitable scalar product of two vectors, determine whether A B ^ C is acute or obtuse.

[4]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

AB × AD = i + 10 j – 7k     M1A1

area = | AB × AD |  =  1 2 + 10 2 + 7 2

  = 5 6 ( 150 )     A1

[3 marks]

a.

METHOD 1

AB AD = 4 2 6      M1A1

= 12

considering the sign of the answer

AB AD < 0 , therefore angle D A ^ B is obtuse     M1

(as it is a parallelogram), A B ^ C is acute     A1

[4 marks]

METHOD 2

BA BC = + 4 + 2 + 6      M1A1

= 12 considering the sign of the answer     M1

BA BC > 0 A B ^ C is acute     A1

[4 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.



It is given that cosecθ=32, where π2<θ<3π2. Find the exact value of cotθ.

Markscheme

METHOD 1

attempt to use a right angled triangle        M1

correct placement of all three values and θ seen in the triangle        (A1)

cotθ<0 (since cosecθ>0 puts θ in the second quadrant)        R1

cotθ=-52        A1

Note: Award M1A1R0A0 for cotθ=52 seen as the final answer
         The R1 should be awarded independently for a negative value only given as a final answer.

 

METHOD 2

Attempt to use 1+cot2θ=cosec2θ        M1

1+cot2θ=94

cot2θ=54        (A1)

cotθ=±52

cotθ<0 (since cosecθ>0 puts θ in the second quadrant)        R1

cotθ=-52        A1

Note: Award M1A1R0A0 for cotθ=52 seen as the final answer
         The R1 should be awarded independently for a negative value only given as a final answer.

 

METHOD 3

sinθ=23

attempt to use sin2θ+cos2θ=1        M1

49+cos2θ=1

cos2θ=59        (A1)

cosθ=±53

cosθ<0 (since cosecθ>0 puts θ in the second quadrant)        R1

cosθ=-53

cotθ=-52        A1

 

Note: Award M1A1R0A0 for cotθ=52 seen as the final answer
         The R1 should be awarded independently for a negative value only given as a final answer.

 

[4 marks]

Examiners report

[N/A]



Let a =  ( 2 k 1 ) and b = ( 3 k + 2 k ) , k R .

Given that a and b are perpendicular, find the possible values of k .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

a b =  ( 2 k 1 ) ( 3 k + 2 k )

= 6 + k ( k + 2 ) k       A1

a b = 0        (M1)

k 2 + k 6 = 0

attempt at solving their quadratic equation        (M1)

( k + 3 ) ( k 2 ) = 0

k = 3 , 2       A1

Note: Attempt at solving using |a||b| = |a × b| will be M1A0A0A0 if neither answer found M1(A1)A1A0
for one correct answer and M1(A1)A1A1 for two correct answers.

[4 marks]

Examiners report

[N/A]



Three points in three-dimensional space have coordinates A(0, 0, 2), B(0, 2, 0) and C(3, 1, 0).

Find the vector  AB .

[1]
a.i.

Find the vector  AC .

[1]
a.ii.

Hence or otherwise, find the area of the triangle ABC.

[4]
b.

Markscheme

AB = ( 0 2 2 )       A1

Note: Accept row vectors or equivalent.

[1 mark]

a.i.

AC = ( 3 1 2 )       A1

Note: Accept row vectors or equivalent.

[1 mark]

a.ii.

METHOD 1

attempt at vector product using  AB and  AC .      (M1)

±(2i + 6j +6k)      A1

attempt to use area  = 1 2 | AB × AC |        M1

= 76 2 ( = 19 )       A1

 

METHOD 2

attempt to use  AB AC = | AB | | AC | cos θ        M1

( 0 2 2 ) ( 3 1 2 ) = 0 2 + 2 2 + ( 2 ) 2 3 2 + 1 2 + ( 2 ) 2 cos θ

6 = 8 14 cos θ       A1

cos θ = 6 8 14 = 6 112

attempt to use area  = 1 2 | AB × AC | sin θ        M1

= 1 2 8 14 1 36 112 ( = 1 2 8 14 76 112 )

= 76 2 ( = 19 )       A1

[4 marks]

b.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.



Solve the equation sec 2 x + 2 tan x = 0 ,   0 x 2 π .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

use of sec 2 x = tan 2 x + 1     M1

tan 2 x + 2 tan x + 1 = 0

( tan x + 1 ) 2 = 0     (M1)

tan x = 1     A1

x = 3 π 4 ,   7 π 4     A1A1

METHOD 2

1 cos 2 x + 2 sin x cos x = 0     M1

1 + 2 sin x cos x = 0

sin 2 x = 1     M1A1

2 x = 3 π 2 ,   7 π 2

x = 3 π 4 ,   7 π 4     A1A1

 

Note:     Award A1A0 if extra solutions given or if solutions given in degrees (or both).

 

[5 marks]

Examiners report

[N/A]



The lines l1 and l2 have the following vector equations where λ, μ.

l1:r1=32-1+λ2-22

l2:r2=204+μ1-11

Show that l1 and l2 do not intersect.

[3]
a.

Find the minimum distance between l1 and l2.

[5]
b.

Markscheme

METHOD 1

setting at least two components of l1 and l2 equal           M1

3+2λ=2+μ      1

2-2λ=-μ        2

-1+2λ=4+μ  3

attempt to solve two of the equations eg. adding 1 and 2           M1

gives a contradiction (no solution), eg 5=2           R1

so l1 and l2 do not intersect                   AG

 

Note: For an error within the equations award M0M1R0.
Note: The contradiction must be correct to award the R1.

 

METHOD 2

l1 and l2 are parallel, so l1 and l2 are either identical or distinct.           R1

Attempt to subtract two position vectors from each line,

e.g. 32-1-204=12-5           M1

32-1k1-11               A1

 

[3 marks]

a.

METHOD 1

l1 and l2 are parallel (as 2-22 is a multiple of 1-11)

let A be 3,2,-1 on l1 and let B be 2,0,4 on l2

Attempt to find vector AB=-1-25            (M1)

Distance required is v×ABv              M1

=131-11×-1-25            (A1)

=13363                    A1

minimum distance is 18=32                    A1

 

METHOD 2

l1 and l2 are parallel (as 2-22 is a multiple of 1-11)

let A be a fixed point on l1 eg 3,2,-1 and let B be a general point on l2 2+μ,-μ,4+μ

attempt to find vector AB            (M1)

AB=-1-25+μ1-11 μ                   A1

AB=-1+μ2+-2-μ2+5+μ2 =3μ2+12μ+30              M1


EITHER

null                   A1


OR

AB=3μ+22+18 to obtain μ=-2                   A1


THEN

minimum distance is 18=32                   A1

 

METHOD 3

let A be 3,2,-1 on l1 and let B be 2+μ,-μ,4+μ on l2              (M1)

(or let A be 2,0,4 on l2 and let B be 3+2λ,2-2λ,-1+2λ on l1)

AB=-1-25+μ1-11 μ  (or AB=2λ+1-2λ+22λ-5)                 A1

μ-1-μ-2μ+5·1-11=0  (or 2λ+1-2λ+22λ-5·1-11=0)              M1

μ=-2  or  λ=1                   A1

minimum distance is 18=32                   A1

 

[5 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.



Let  a = sin b , 0 < b < π 2 .

Find, in terms of b, the solutions of sin 2 x = a , 0 x π .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

sin 2 x = sin b

EITHER

sin 2 x = sin ( b ) or  sin 2 x = sin ( π + b ) or  sin 2 x = sin ( 2 π b ) …      (M1)(A1)

Note: Award M1 for any one of the above, A1 for having final two.

OR

     (M1)(A1)

Note: Award M1 for one of the angles shown with b clearly labelled, A1 for both angles shown. Do not award A1 if an angle is shown in the second quadrant and subsequent A1 marks not awarded.

THEN

2 x = π + b or  2 x = 2 π b      (A1)(A1)

x = π 2 + b 2 , x = π b 2      A1

[5 marks]

Examiners report

[N/A]



Consider the functions f and g defined on the domain  0 < x < 2 π by  f ( x ) = 3 cos 2 x and  g ( x ) = 4 11 cos x .

The following diagram shows the graphs of  y = f ( x ) and  y = g ( x )

Find the x -coordinates of the points of intersection of the two graphs.

[6]
a.

Find the exact area of the shaded region, giving your answer in the form  p π + q 3 , where p q Q .

[5]
b.

At the points A and B on the diagram, the gradients of the two graphs are equal.

Determine the y -coordinate of A on the graph of g .

[6]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

3 cos 2 x = 4 11 cos x

attempt to form a quadratic in  cos x      M1

3 ( 2 co s 2 x 1 ) = 4 11 cos x      A1

( 6 co s 2 x + 11 cos x 7 = 0 )

valid attempt to solve their quadratic     M1

( 3 cos x + 7 ) ( 2 cos x 1 ) = 0

cos x = 1 2      A1

x = π 3 , 5 π 3      A1A1

Note: Ignore any “extra” solutions.

[6 marks]

a.

consider (±)  π 3 5 π 3 ( 4 11 cos x 3 cos 2 x ) d x      M1

= ( ± ) [ 4 x 11 sin x 3 2 sin 2 x ] π 3 5 π 3      A1

Note: Ignore lack of or incorrect limits at this stage.

attempt to substitute their limits into their integral     M1

= 20 π 3 11 sin 5 π 3 3 2 sin 10 π 3 ( 4 π 3 11 sin π 3 3 2 sin 2 π 3 )

= 16 π 3 + 11 3 2 + 3 3 4 + 11 3 2 + 3 3 4

= 16 π 3 + 25 3 2      A1A1

[5 marks]

b.

attempt to differentiate both functions and equate     M1

6 sin 2 x = 11 sin x      A1

attempt to solve for x      M1

11 sin x + 12 sin x cos x = 0

sin x ( 11 + 12 cos x ) = 0

cos x = 11 12 ( or sin x = 0 )      A1

y = 4 11 ( 11 12 )      M1

y = 169 12 ( = 14 1 12 )      A1

[6 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.



Consider quadrilateral PQRS where PQ is parallel to SR.

In PQRS, PQ=x, SR=y, RS^P=α and QR^S=β.

Find an expression for PS in terms of x,y,sinβ and sinα+β.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

METHOD 1

from vertex P, draws a line parallel to QR that meets SR at a point X        (M1)

uses the sine rule in ΔPSX        M1

PSsinβ=y-xsin180°-α-β        A1

sin180°-α-β=sinα+β        (A1)

PS=y-xsinβsinα+β        A1

 

METHOD 2

let the height of quadrilateral PQRS be h

h=PSsinα        A1

attempts to find a second expression for h        M1

h=y-x-PScosαtanβ

PSsinα=y-x-PScosαtanβ

writes tanβ as sinβcosβ, multiplies through by cosβ and expands the RHS        M1

PSsinαcosβ=y-xsinβ-PScosαsinβ

PS=y-xsinβsinαcosβ+cosαsinβ        A1

PS=y-xsinβsinα+β        A1

 

[5 marks]

Examiners report

[N/A]



Consider the complex numbers z1=1+bi and z2=1-b2-2bi, where b, b0.

Find an expression for z1z2 in terms of b.

[3]
a.

Hence, given that argz1z2=π4, find the value of b.

[3]
b.

Markscheme

z1z2=1+bi1-b2-2bi

=1-b2-2i2b2+i-2b+b-b3             M1

=1+b2+i-b-b3            A1A1


Note: Award A1 for 1+b2 and A1 for -bi-b3i.

 

[3 marks]

a.

argz1z2=arctan-b-b31+b2=π4            (M1)


EITHER
arctan-b=π4 (since 1+b20, for b)            A1


OR

-b-b3=1+b2  (or equivalent)            A1


THEN

b=-1            A1

 

[3 marks]

b.

Examiners report

Part (a) was generally well done with many completely correct answers seen. Part (b) proved to be challenging with many candidates incorrectly equating the ratio of their imaginary and real parts to π4 instead of tanπ4. Stronger candidates realized that when θ=π4, it forms an isosceles right-angled triangle and equated the real and imaginary parts to obtain the value of b .

a.
[N/A]
b.



The vectors a and b are defined by ( 1 1 t ) b  ( 0 t 4 t ) , where t R .

Find and simplify an expression for a • b in terms of t .

[2]
a.

Hence or otherwise, find the values of t for which the angle between a and b is obtuse .

[4]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

a • b =  ( 1 × 0 ) + ( 1 × t ) + ( t × 4 t )       (M1)

t + 4 t 2       A1

 

[2 marks]

a.

recognition that  a • b = |a||b|cos θ      (M1)

a • b < 0 or  t + 4 t 2 < 0 or cos θ < 0      R1

Note: Allow ≤ for R1.

 

attempt to solve using sketch or sign diagram      (M1)

0 < t < 1 4       A1

 

[4 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.



Let fx=1+x for x>-1.

Show that f''x=-141+x3.

[3]
a.

Use mathematical induction to prove that fnx=-14n-12n-3!n-2!1+x12-n for n, n2.

[9]
b.

Let gx=emx, m.

Consider the function h defined by hx=fx×gx for x>-1.

It is given that the x2 term in the Maclaurin series for h(x) has a coefficient of 74.

Find the possible values of m.

[8]
c.

Markscheme

attempt to use the chain rule            M1

f'x=121+x-12         A1

f''x=-141+x-32         A1

=-141+x3         AG

 

Note: Award M1A0A0 for f'x=11+x or equivalent seen

  

[3 marks]

a.

let n=2

f''x=-141+x3=-1411!0!1+x12-2         R1

 

Note: Award R0 for not starting at n=2. Award subsequent marks as appropriate.

 

assume true for n=k, (so fkx=-14k-12k-3!k-2!1+x12-k)       M1

 

Note: Do not award M1 for statements such as “let n=k” or “n=k is true”. Subsequent marks can still be awarded.

 

consider n=k+1

LHS=fk+1x=dfkxdx            M1

=-14k-12k-3!k-2!12-k1+x12-k-1 (or equivalent)         A1

 

EITHER

RHS=fk+1x=-14k2k-1!k-1!1+x12-k-1 (or equivalent)         A1

=-14k2k-12k-22k-3!k-1k-2!1+x12-k-1        A1

 

Note: Award A1 for 2k-1!k-1!=2k-12k-22k-3!k-1k-2!=22k-12k-3!k-2!

 

=-14-14k-12k-12k-22k-3!k-1k-2!1+x12-k-1        A1

=-12-14k-12k-12k-3!k-2!1+x12-k-1

 

Note: Award A1 for leading coefficient of -14.

 

=12-k-14k-12k-3!k-2!1+x12-k-1        A1

 

OR

Note: The following A marks can be awarded in any order.

 

=-14k-12k-3!k-2!1-2k21+x12-k-1

=-12-14k-12k-12k-3!k-2!1+x12-k-1        A1

 

Note: Award A1 for isolating (2k1) correctly.

 

=-12-14k-12k-1!2k-2k-2!1+x12-k-1        A1

 

Note: Award A1 for multiplying top and bottom by (k1) or 2(k1).

 

=-14-14k-12k-1!k-1k-2!1+x12-k-1        A1

 

Note: Award A1 for leading coefficient of -14.

 

=-14k2k-1!k-1!1+x12-k-1        A1

 

=-14k+1-12k+1-3!k+1-2!1+x12-k+1=RHS

 

THEN

since true for n=2, and true for n=k+1 if true for n=k, the statement is true for all, n, n2  by mathematical induction           R1

 

Note: To obtain the final R1, at least four of the previous marks must have been awarded.

 

[9 marks]

b.

METHOD 1

hx=1+xemx

using product rule to find h'x        (M1)

h'x=1+xmemx+121+xemx         A1

h''x=m1+xmemx+121+xemx+121+xmemx-141+x3emx         A1

substituting x=0 into h''x       M1

h''0=m2+12m+12m-14=m2+m-14         A1

hx=h0+xh'0+x22!h''0+

equating x2 coefficient to 74       M1

h''02!=74 h''0=72

4m2+4m-15=0         A1

2m+52m-3=0

m=-52  or  m=32         A1

 

METHOD 2

EITHER

attempt to find f0, f'0, f''0        (M1)

fx=1+x12                    f0=1

f'x=121+x-12            f'0=12

f''x=-141+x-32      f''0=-14

fx=1+12x-18x2+         A1

 

OR

attempt to apply binomial theorem for rational exponents        (M1)

fx=1+x12=1+12x+12-122!x2

fx=1+12x-18x2+         A1

 

THEN

gx=1+mx+m22x2+        (A1)

hx=1+12x-18x2+1+mx+m22x2+        (M1)

coefficient of x2 is m22+m2-18         A1

attempt to set equal to 74 and solve             M1

m22+m2-18=74

4m2+4m-15=0          A1

2m+52m-3=0

m=-52  or  m=32         A1

 

METHOD 3

g'x=memx and g''x=m2emx        (A1)

hx=h0+xh'0+x22!h''0+

equating x2 coefficient to 74       M1

h''02!=74 h''0=72

using product rule to find h'x and h''x        (M1)

h'x=fxg'x+f'xgx

h''x=fxg''x+2f'xg'x+f''xgx         A1

substituting x=0 into h''x       M1

h''0=f0g''0+2g'0f'0+g0f''0

=1×m2+2m×12+1×-14  =m2+m-14         A1

4m2+4m-15=0          A1

2m+52m-3=0

m=-52  or  m=32         A1

 

[8 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.



Solve the equation 2cos2x+5sinx=4, 0x2π.

Markscheme

attempt to use cos2x=1-sin2x              M1

2sin2x-5sinx+2=0                   A1

 

EITHER

attempting to factorise              M1

(2sinx1)(sinx2)                   A1

 

OR

attempting to use the quadratic formula            M1

sinx=5±52-4×2×24=5±34         A1

 

THEN

sinx=12           (A1)

x=π6,5π6                  A1A1

 

[7 marks]

Examiners report

[N/A]



A sector of a circle with radius r  cm , where r > 0, is shown on the following diagram.
The sector has an angle of 1 radian at the centre.

Let the area of the sector be A  cm2 and the perimeter be P  cm. Given that A = P , find the value of r .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

A = P

use of the correct formula for area and arc length       (M1)

perimeter is r θ + 2 r        (A1)

Note: A1 independent of previous M1.

1 2 r 2 ( 1 ) = r ( 1 ) + 2 r       A1

r 2 6 r = 0

r = 6   (as  r > 0)        A1

Note: Do not award final A1 if r = 0 is included.

[4 marks]

Examiners report

[N/A]



The plane П has the Cartesian equation  2 x + y + 2 z = 3

The line L has the vector equation r = ( 3 5 1 ) + μ ( 1 2 p ) , μ , p R . The acute angle between the line L and the plane П is 30°.

Find the possible values of p .

Markscheme

recognition that the angle between the normal and the line is 60° (seen anywhere)       R1

attempt to use the formula for the scalar product       M1

cos 60° = | ( 2 1 2 ) ( 1 2 p ) | 9 × 1 + 4 + p 2      A1

1 2 = | 2 p | 3 5 + p 2       A1

3 5 + p 2 = 4 | p |

attempt to square both sides         M1

9 ( 5 + p 2 ) = 16 p 2 7 p 2 = 45

p = ± 3 5 7  (or equivalent)       A1A1

[7 marks]

Examiners report

[N/A]