Edexcel Gce Core 4 Mathematics C4 6666 Advanced Subsidiary Jan 2007 Mark Scheme

Mark Scheme (Results) January 2007 GCE

GCE Mathematics Core Mathematics C4 (6666)

Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH

January 2007 6666 Core Mathematics C4 Mark Scheme Question Number

Scheme

Marks

** represents a constant 1.

−2

f(x) = (2 − 5x)

= ( 2)

−2

⎛ 5x ⎞ ⎜1 − 2 ⎟ ⎝ ⎠

−2

1 ⎛ 5x ⎞ = ⎜1 − 4⎝ 2 ⎟⎠

Takes 2 outside the bracket to give any of B1 (2)-2 or 41 .

−2

Expands (1 + * * x )−2 to give an M1 unsimplified 1 + ( −2)(* * x) ; ⎫

⎧ ( −2)( −3) ( −2)( −3)( −4) = 41 ⎨1 + ( −2)(* * x); + (* * x)2 + (* * x)3 + ... 2! 3! ⎭ ⎩

A correct unsimplified {..........} expansion A1 with candidate’s

(* * x )

⎧ ⎫ ( −2)( −3) − 5x 2 ( −2)( −3)( −4) − 5x 3 ( 2 ) + ( 2 ) + ...⎬ = 41 ⎨1 + ( −2)( − 25x ); + 2! 3! ⎭ ⎩

⎧ ⎫ 75x 2 125x 3 = 41 ⎨1 + 5x; + + + ...⎬ 4 2 ⎩ ⎭

Anything that cancels to 1 + 5x ; A1;

1 5x 75x 2 125x 3 = + + + ... ;+ 4 4 16 8

=

4

Simplified

75x 2 16

+

4

125x 3 8

A1

1 1 11 2 5 + 1 x; + 4 x + 15 x 3 + ... 4 4 16 8

[5] 5 marks

1

Question Number Aliter 1. Way 2

Scheme

Marks

f(x) = (2 − 5x)−2

or (2)−2 B1 Expands (2 − 5x)−2 to give an M1 unsimplifed −2 (2) + ( −2)(2)−3 (* * x) ; 1 4

⎧ −2 ⎫ ( −2)( −3) −4 −3 (2) (* * x)2 ⎪ ⎪(2) + ( −2)(2) (* * x); + ⎪ ⎪ 2! =⎨ ⎬ ( 2)( 3)( 4) − − − ⎪ ⎪ (2)−5 (* * x)3 + ... + ⎪⎩ ⎪⎭ 3!

A correct unsimplified {..........} expansion A1 with candidate’s

(* * x )

( −2)( −3) −4 ⎧ −2 ⎫ −3 (2) ( −5x)2 ⎪ ⎪⎪(2) + ( −2)(2) ( −5x); + ⎪ 2! =⎨ ⎬ ( 2)( 3)( 4) − − − ⎪ ⎪ (2)−5 ( −5x)3 + ... + 3! ⎩⎪ ⎭⎪ ⎧⎪ 41 + ( −2)( 81 )( −5x); + (3)( 161 )(25x 2 )⎫⎪ =⎨ ⎬ + ( −4)( 161 )( −125x 3 ) + ... ⎪⎭ ⎪⎩

Anything that cancels to 1 + 5x ; A1;

1 5x 75x 2 125x 3 ;+ = + + + ... 4 4 16 8

=

4

Simplified

75x 2 16

+

4

125x 3 8

A1

1 1 11 2 5 + 1 x; + 4 x + 15 x 3 + ... 4 4 16 8

[5] 5 marks Attempts using Maclaurin expansions need to be referred to your team leader.

2

Question Number

Scheme 1 2

2. (a)

Marks 1 2

2

⎛ ⎞ π 1 1 Volume = π ⎜⎜ dx ⎟⎟ dx = 2 3 1 2x 9 + ( ) 1 2x + ( ) ⎠ − 41 ⎝ − 41

∫

⎛π⎞

=⎜ ⎟ ⎝9⎠

∫

−2

B1

Can be implied. Ignore limits. Moving their power to the top. (Do not allow power of -1.) Can be implied. M1 Ignore limits and 9π

1 2

∫ (1 + 2x )

∫

Use of V = π y 2 dx .

dx

− 41

1

−1 2 ⎛ π ⎞ ⎡ (1 + 2x) ⎤ = ⎜ ⎟⎢ ⎥ ⎝ 9 ⎠ ⎣⎢ ( −1)(2) ⎦⎥ − 1

Integrating to give ±p(1 + 2x)−1

M1

− 21 (1 + 2x)−1

A1

4

1 2 ⎛π⎞ = ⎜ ⎟ ⎡ − 21 (1 + 2x)−1 ⎤ 1 ⎦ −4 ⎝9⎠⎣

⎛ π ⎞ ⎡⎛ −1 ⎞ ⎛ −1 ⎞ ⎤ = ⎜ ⎟ ⎢⎜ ⎟⎥ ⎟−⎜ ⎝ 9 ⎠ ⎣⎢⎝ 2(2) ⎠ ⎝ 2( 21 ) ⎠ ⎦⎥

⎛π⎞ = ⎜ ⎟ ⎡⎣ − 41 − ( −1)⎤⎦ ⎝9⎠

=

(b)

From Fig.1, AB = As

3 4

Use of limits to give exact values of A1 aef π 3π or 36 or 224π or aef 12

π 12 1 2

− (−

1 4

)=

[5] 3 4

units

units ≡ 3cm

then scale factor k =

3

( 34 )

= 4. ⎛ π ⎞

Hence Volume of paperweight = ( 4 ) ⎜ ⎟ ⎝ 12 ⎠ 3

V=

3 ( 4 ) × (their answer to part (a)) M1

16π cm3 = 16.75516... cm3 3

16 π 3

or awrt 16.8

or 6412π or aef

A1 [2] 7 marks

Note:

π 9

(or implied) is not needed for the middle three marks of question 2(a).

3

Question Number Aliter

Scheme 1 2

2. (a)

Marks 1 2

2

⎛ ⎞ 1 1 Volume = π ⎜⎜ dx ⎟⎟ dx = π 2 + 3 1 2x ( ) + 3 6x ( ) ⎠ − 41 ⎝ − 41

∫

∫

∫

Use of V = π y 2 dx .

B1

Can be implied. Ignore limits.

Way 2 = ( π)

Moving their power to the top. (Do not allow power of -1.) M1 Can be implied. Ignore limits and π

1 2

∫ (3 + 6x )

−2

dx

− 41

Integrating to give ±p(3 + 6x)−1

1

⎡ (3 + 6x)−1 ⎤ 2 = ( π) ⎢ ⎥ ⎣ ( −1)(6) ⎦ − 1

− 61 (3 + 6x)

4

−1

M1 A1

1

2 = ( π ) ⎡ − 61 (3 + 6x)−1 ⎤ 1 ⎣ ⎦ −4

⎡⎛ −1 ⎞ ⎛ −1 ⎞ ⎤ = ( π ) ⎢⎜ ⎟ − ⎜ 3 ⎟⎥ ⎣⎢⎝ 6(6) ⎠ ⎝ 6( 2 ) ⎠ ⎦⎥

= ( π ) ⎡⎣ − 361 − ( − 91 )⎤⎦

=

Use of limits to give exact values of A1 aef π 3π or 36 or 224π or aef 12

π 12

[5] Note: π is not needed for the middle three marks of question 2(a).

4

Question Number 3. (a)

Scheme

Marks

x = 7 cos t − cos 7t , y = 7 sin t − sin7t ,

dx dy = − 7 sin t + 7 sin 7t , = 7 cos t − 7 cos 7t dt dt

Attempt to differentiate x and y with respect to t to give dx in the form ± A sin t ± B sin7t M1 dt dy dt

in the form ±C cos t ± D cos 7t Correct

∴

dy 7 cos t − 7 cos 7t = dx −7 sin t + 7 sin 7t

dx dt

and

Candidate’s

dy dt

A1

dy dt dx dt

B1 [3]

(b)

π 6

When t = , m(T) =

=

7 3 2

(

− − 7 23

)

dy 7 cos 6π − 7 cos 76π ; = dx −7 sin 6π + 7 sin 76π

7 3 = = − 3 = awrt − 1.73 −7

− 72 − 72

Hence m(N) =

−1

1

or

− 3

When t = 6π , x = 7 cos 6π − cos 76π = y = 7 sin − sin π 6

N: y − 4 =

N: y =

or 4 =

1 3

1 3

1 3

7π 6

=

7 3 2 7 2

( )=

− −

− (−

1 2

3 2

)=

8 2

8 3 2

=4 3

=4

y=

or

(4 3 ) + c

Hence N: y =

1 3

x

or 30o into their dy dx

expression;

M1

to give any of the four underlined expressions oe A1 cso (must be correct solution only)

(

The point 4 3, 4

)

or ( awrt 6.9, 4 )

A1

oe.

B1

Finding an equation of a normal with their point and their normal M1 gradient or finds c by using y = (their gradient)x + " c " .

(x − 4 3 )

x

π 6

Uses m(T) to ‘correctly’ find m(N). Can be ft from “their tangent gradient”.

= awrt 0.58

3

Substitutes t =

3 3

Correct simplified EXACT equation of normal. A1 oe This is dependent on candidate using correct 4 3 , 4

x or 3y = 3x

(

)

⇒ c=4−4 = 0

or y =

3 3

x or 3y = 3x

[6] 9 marks 5

Question Number Aliter 3. (a) Way 2

Scheme

Marks

x = 7 cos t − cos 7t , y = 7 sin t − sin7t ,

dx dy = − 7 sin t + 7 sin 7t , = 7 cos t − 7 cos 7t dt dt

Attempt to differentiate x and y in the with respect to t to give dx dt form ± A sin t ± B sin7t M1 dy in theform ±C cos t ± D cos 7t dt A1 Correct dx and dy dt dt

dy 7cos t − 7 cos7t −7( −2 sin 4t sin3t) = = = tan 4t dx −7 sin t + 7 sin7t −7(2cos 4t sin3t)

Candidate’s

dy dt dx dt

B1 [3]

(b)

Substitutes t =

π dy When t = , m(T) = = tan 46π ; dx 6

=

2

( ) (1) 3 2

2 ( − 21 ) (1)

Hence m(N) =

1

or

− 3

3

When t = 6π , x = 7 cos 6π − cos 76π =

y = 7 sin 6π − sin 76π =

N: y − 4 =

N: y =

or 4 =

1 3

1 3

1 3

7 3 2

7 2

3 2

− ( − 21 ) =

8 2

8 3 2

Uses m(T) to ‘correctly’ find m(N). Can be ft from “their tangent gradient”.

=4 3

=4

y=

or

(4 3 ) + c

Hence N: y =

1 3

x

(

The point 4 3, 4

)

or ( awrt 6.9, 4 )

A1

oe.

B1

Finding an equation of a normal with their point and their normal M1 gradient or finds c by using y = (their gradient)x + " c " .

(x − 4 3 )

x

expression; M1

to give any of the three underlined expressions oe A1 cso (must be correct solution only)

= awrt 0.58

( )=

− −

or 30o into their dy dx

= − 3 = awrt − 1.73

−1

π 6

3 3

Correct simplified EXACT equation of normal. A1 oe This is dependent on candidate using correct 4 3 , 4

x or 3y = 3x

(

)

⇒ c=4−4 = 0

or y =

3 3

x or 3y = 3x

[6] 9 marks 6

Beware: A candidate finding an m(T) = 0 can obtain A1ft for m(N) → ∞ , but obtains M0 if they write y − 4 = ∞(x − 4 3 ) . If they write, however, N: x = 4 3 , then they can score M1. Beware: A candidate finding an m(T) = ∞ can obtain A1ft for m(N) = 0, and also obtains M1 if they write y − 4 = 0(x − 4 3 ) or y = 4.

7

Question Number 4. (a)

Scheme

Marks

2x − 1 A B ≡ + (x − 1)(2x − 3) (x − 1) (2x − 3)

Forming this identity. NB: A & B are not assigned in M1 this question

2x − 1 ≡ A(2x − 3) + B(x − 1)

Let x = 32 ,

2 = B ( 21 )

⇒ B=4

Let x = 1,

1 = A ( −1)

⇒ A = −1

either one of A = − 1 or B = 4 . A1 both correct for their A, B. A1

−1 4 + (x − 1) (2x − 3)

giving

[3] (b) & (c)

∫

∫

dy = y

(2x − 1) dx (2x − 3)(x − 1) −1

Separates variables as shown B1 Can be implied

∫ (x − 1) + (2x − 3) dx

Replaces RHS with their partial fraction to be integrated. M1

∴ ln y = − ln(x − 1) + 2ln(2x − 3) + c

At least two terms in ln’s M1 At least two ln terms correct A1 All three terms correct and ‘+ c’ A1

=

4

[5] c = ln10

y = 10, x = 2 gives c = ln10

B1

∴ ln y = − ln(x − 1) + 2ln(2x − 3) + ln10 ln y = − ln(x − 1) + ln(2x − 3)2 + ln10

Using the power law for M1 logarithms

⎛ (2x − 3)2 ⎞ ln y = ln ⎜ ⎟ + ln10 or ⎝ (x − 1) ⎠ ⎛ 10(2x − 3)2 ⎞ ln y = ln ⎜ ⎟ ⎝ (x − 1) ⎠

y=

Using the product and/or quotient laws for logarithms to obtain a M1 single RHS logarithmic term with/without constant c.

10(2x − 3)2 (x − 1)

y=

10(2x − 3)2 or aef. isw A1 aef (x − 1)

[4] 12 marks

8

Question Number Aliter 4. (b) & (c) Way 2

Scheme

∫y

dy

Marks

(2x − 1)

∫ (2x − 3)(x − 1) dx

=

=

∫

Separates variables as shown B1 Can be implied

−1 4 + dx (x − 1) (2x − 3)

Replaces RHS with their partial M1 fraction to be integrated. At least two terms in ln’s M1 At least two ln terms correct A1 All three terms correct and ‘+ c’ A1

∴ ln y = − ln(x − 1) + 2ln(2x − 3) + c

decide to award B1 here!! B1

See below for the award of B1

Using the power law for M1 logarithms

ln y = − ln(x − 1) + ln(2x − 3)2 + c

Using the product and/or quotient laws for logarithms to obtain a M1 single RHS logarithmic term with/without constant c.

⎛ (2x − 3)2 ⎞ ln y = ln ⎜ ⎟+c ⎝ x −1 ⎠ ⎛ A(2x − 3)2 ⎞ ln y = ln ⎜ ⎟ x −1 ⎠ ⎝

or e

ln y

y =

=e

⎛ (2x − 3)2 ⎞ +c ln⎜ ⎜ x − 1 ⎟⎟ ⎝ ⎠

=e

where c = ln A ⎛ (2x − 3)2 ⎞ ln⎜ ⎜ x − 1 ⎟⎟ ⎝ ⎠

ec

A(2x − 3)2 (x − 1)

y = 10, x = 2 gives A = 10

y=

A = 10 for B1

10(2x − 3)2 (x − 1)

y=

award above

10(2x − 3)2 or aef & isw A1 aef (x − 1)

[5] & [4]

Note: The B1 mark (part (c)) should be awarded in the same place on ePEN as in the Way 1 approach.

9

Question Number Aliter (b) & (c)

Scheme

∫y

dy

(2x − 1)

∫ (2x − 3)(x − 1) dx

=

Way 3 =

Marks

∫

Separates variables as shown Can B1 be implied

−1 2 + dx (x − 1) (x − 32 )

Replaces RHS with their partial M1 fraction to be integrated. At least two terms in ln’s M1 At least two ln terms correct A1 All three terms correct and ‘+ c’ A1

∴ ln y = − ln(x − 1) + 2ln(x − 32 ) + c

[5] y = 10, x = 2 gives c = ln10 − 2ln ( 21 ) = ln 40

c = ln10 − 2ln ( 21 ) or c = ln 40

B1 oe

∴ ln y = − ln(x − 1) + 2ln(x − 32 ) + ln 40

ln y = − ln(x − 1) + ln(x − 32 )2 + ln10

Using the power law for M1 logarithms

⎛ (x − 32 )2 ⎞ ln y = ln ⎜ ⎟ + ln 40 or ⎝ (x − 1) ⎠ ⎛ 40(x − 32 )2 ⎞ ln y = ln ⎜ ⎟ ⎝ (x − 1) ⎠ y=

Using the product and/or quotient laws for logarithms to obtain a M1 single RHS logarithmic term with/without constant c.

40(x − 32 )2 (x − 1)

y=

40(x − 32 )2 or aef. isw A1 aef (x − 1)

[4]

Note: Please mark parts (b) and (c) together for any of the three ways.

10

Question Number

Scheme sin x + cos y = 0.5

5. (a) ⎧ dy ⎫ =⎬ ⎨ ⎩ dx ⎭

dy cos x − sin y =0 dx

Marks ( eqn ∗ ) Differentiates implicitly to include ( eqn # )

± sin y

dy . (Ignore dx

dy cos x = dx sin y

(

dy dx

)

= .)

cos x sin y

M1

A1 cso [2]

(b)

Candidate realises that they need to solve ‘their numerator’ = 0 …or candidate sets ddyx = 0 in their M1 (eqn #) and attempts to solve the resulting equation.

dy cos x =0 ⇒ = 0 ⇒ cos x = 0 dx sin y

giving x = − 2π or x =

both x = − 2π ,

π 2

When x = , sin (

π 2

) + cos y = 0.5

x=

Substitutes either their or x = − 2π into eqn ∗ M1 2π 3

or − 23π or 120o

or −120° or awrt -2.09 or awrt 2.09

( 2π , 23π ) and ( 2π , − 23π )

A1

π 2

Only one of y =

⇒ cos y = 1.5 ⇒ y has no solutions ⇒ cos y = − 0.5 ⇒ y = 23π or − 23π

In specified range ( x, y ) =

or x = ±90o or

awrt x = ± 1.57 required here

When x = − 2π , sin ( − 2π ) + cos y = 0.5 π 2

π 2

Only exact coordinates of

( 2π , 23π ) and ( 2π , − 23π )

A1

A1

Do not award this mark if candidate states other coordinates inside the required range. [5] 7 marks

11

Question Number

Scheme

6.

y = 2x = e x ln 2

(a)

dy = ln 2.e x ln2 dx

Marks

dy = ln 2.e x ln2 dx

M1

Way 1 dy = ln 2.(2x ) = 2x ln 2 dx

Hence

2x ln 2 AG A1 cso

AG

[2] Aliter (a)

( )

ln y = ln 2x

Takes logs of both sides, then uses the power law of logarithms…

leads to ln y = x ln 2

Way 2

M1 … and differentiates implicitly to give 1y dy = ln 2 dx

1 dy = ln 2 y dx dy = y ln 2 = 2x ln 2 dx

Hence

2x ln 2 AG A1 cso

AG

[2]

(b)

y=2

2

(x )

Ax 2 ( x

dy ⇒ = 2x. 2 ( x ).ln 2 dx 2

2

)

M1

2

2x. 2 ( x ).ln 2

or 2x. y.ln 2 if y is defined dy dx ( x2 ) which is of the form ± k 2

A1

Substitutes x = 2 into their When x = 2,

dy = 2(2) 2 4 ln 2 dx

or Ax 2 ( x dy = 64 ln 2 = 44.3614... dx

2

M1

)

64ln 2 or awrt 44.4 A1

[4] 6 marks

12

Question Number Aliter 6. (b)

Scheme

( )

ln y = ln 2x

2

Marks

leads to ln y = x 2 ln 2

Way 2 1 y 1 y

1 dy = 2x.ln 2 y dx

dy = Ax.ln 2 dx dy = 2x.ln 2 dx

dy dx ( x2 ) which is of the form ± k 2

M1 A1

Substitutes x = 2 into their When x = 2,

dy = 2(2) 2 4 ln 2 dx

or Ax 2 ( x dy = 64 ln 2 = 44.3614... dx

2

M1

)

64ln 2 or awrt 44.4 A1

[4]

13

Question Number 7.

(a)

Scheme

Marks

uuur uuur a = OA = 2i + 2 j + k ⇒ OA = 3 uuur uuur b = OB = i + j − 4 k ⇒ OB = 18 uuur uuur BC = ± ( 2i + 2 j + k ) ⇒ BC = 3 uuur uuur AC = ± ( i + j − 4 k ) ⇒ AC = 18 uuur c = OC = 3i + 3 j − 3k

3i + 3 j − 3k

B1 cao [1]

(b)

⎛ 2⎞ ⎛ 1 ⎞ uuur uuur ⎜ ⎟ ⎜ ⎟ OA • OB = ⎜ 2 ⎟ • ⎜ 1 ⎟ = 2 + 2 − 4 = 0 ⎜ 1 ⎟ ⎜ −4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −1⎞ ⎛ 2 ⎞ uuur uuur ⎜ ⎟ ⎜ ⎟ BO • BC = ⎜ −1⎟ • ⎜ 2 ⎟ = −2 − 2 + 4 = 0 ⎜ 4 ⎟ ⎜ 1⎟ ⎝ ⎠ ⎝ ⎠

or…

or…

⎛ 1 ⎞ ⎛ 2⎞ uuur uuur ⎜ ⎟ ⎜ ⎟ AC • BC = ⎜ 1 ⎟ • ⎜ 2 ⎟ = 2 + 2 − 4 = 0 or… ⎜ −4 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ −2 ⎞ ⎛ 1 ⎞ uuur uuur ⎜ ⎟ ⎜ ⎟ AO • AC = ⎜ −2 ⎟ • ⎜ 1 ⎟ = −2 − 2 + 4 = 0 ⎜ −1⎟ ⎜ −4 ⎟ ⎝ ⎠ ⎝ ⎠

and therefore OA is perpendicular to OB and hence OACB is a rectangle.

An attempt to take the dot product uuur uuur between either OA and OB uuur uuur uuur uuur M1 OA and AC , AC and BC uuur uuur or OB and BC Showing the result is equal to zero. A1

perpendicular and A1 cso OACB is a rectangle Using distance formula to find M1 either the correct height or width. Multiplying the rectangle’s M1 height by its width. exact value of A1

Area = 3 × 18 = 3 18 = 9 2

3 18 , 9 2 , 162 or aef

[6] (c)

uuur OD = d =

1 2

( 3i + 3 j − 3k )

1 2

( 3i + 3 j − 3k ) B1 [1]

14

Question Number

Scheme using dot product formula

(d)

uuur DA = ± ( 21 i + 21 j + 52 k ) & uuur or BA = ± (i + j + 5k ) &

Marks

uuur DC = ± ( 32 i + 32 j − 32 k ) uuur OC = ± ( 3i + 3 j − 3k )

Identifies a set of two M1 relevant vectors Correct vectors ± A1

Way 1 ⎛ 0.5 ⎞ ⎛ 1.5 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0.5 ⎟ • ⎜ 1.5 ⎟ 3 3 15 + − ⎜ 2.5 ⎟ ⎜ −1.5 ⎟ ⎝ ⎠ ⎝ ⎠ 4 4 4 = (±) 1 cos D = ( ± ) = (±) 27 3 27 27 . 4 2 2

Attempts to find the correct angle D ddM1 rather than 180° − D .

⎛ 1⎞ D = cos−1 ⎜ − ⎟ ⎝ 3⎠

109.5° or A1 awrt 109° or 1.91c

D = 109.47122...o

Aliter (d)

Applies dot product formula on multiples of these vectors. Correct ft. application of dot product formula

using dot product formula and direction vectors uuur

d BA = ± (i + j + 5k )

&

uuur

d OC = ± (i + j − k )

[6] Identifies a set of two M1 direction vectors Correct vectors ± A1

Way 2 ⎛ 1 ⎞ ⎛ 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ • ⎜ 1⎟ ⎜ −1⎟ ⎜ 5 ⎟ 1+ 1− 5 1 cos D = ( ± ) ⎝ ⎠ ⎝ ⎠ = ( ± ) = (±) 3 3 . 27 3 . 27

Applies dot product formula on multiples dM1 of these vectors. Correct ft. application of dot A1 product formula. Attempts to find the correct angle D ddM1 rather than 180° − D .

⎛ 1⎞ D = cos−1 ⎜ − ⎟ ⎝ 3⎠

109.5° or A1 awrt 109° or 1.91c

D = 109.47122...o

[6]

15

Question Scheme Number using dot product formula and similar triangles Aliter (d)

uuur dOA = ( 2i + 2 j + k )

&

uuur

d OC = (i + j − k )

Marks Identifies a set of two M1 direction vectors Correct vectors A1

Way 3 ⎛ 2⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2⎟ •⎜ 1 ⎟ ⎜ 1 ⎟ ⎜ −1⎟ 2+ 2−1 = cos ( 21 D ) = ⎝ ⎠ ⎝ ⎠ = 9. 3 9. 3

Applies dot product formula on multiples dM1 of these vectors. Correct ft. application of dot A1 product formula.

1 3

⎛ 1 ⎞ D = 2 cos−1 ⎜ ⎟ ⎝ 3⎠

Attempts to find the correct angle D by doubling their angle ddM1 for 21 D .

D = 109.47122...o

109.5° or A1 awrt 109° or 1.91c

[6] Aliter (d) Way 4

using uuur cosine rule DA = 21 i +

uuur DA =

1 2

uuur j + 52 k , DC = 32 i +

27 , 2

uuur DC =

3 2

uuur j − 32 k , AC = i + j − 4 k

uuur AC = 18

27 , 2

Attempts to find all the lengths of all M1 three edges of ∆ ADC

All Correct A1 2

2

⎛ 27 ⎞ ⎛ 27 ⎞ ⎜⎜ ⎟ +⎜ ⎟ − 18 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎝ cos D = ⎛ 27 ⎞ ⎛ 27 ⎞ 2 ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎝ 2 ⎠⎝ 2 ⎠

(

)

2

=−

1 3

Using the cosine rule formula with correct dM1 ‘subtraction’. Correct ft application of the cosine rule A1 formula Attempts to find the correct angle D ddM1 rather than 180° − D .

⎛ 1⎞ D = cos ⎜ − ⎟ ⎝ 3⎠ −1

109.5° or A1 awrt 109 or 1.91c

D = 109.47122...o

°

[6]

16

Question Scheme Number using trigonometry on auuur right angled triangle Aliter uuur uuur 5 1 1 DA = 2 i + 2 j + 2 k OA = 2i + 2 j + k AC = i + j − 4 k (d) Way 5

Attempts to find two out of the three M1 lengths in ∆ ADX

Let X be the midpoint of AC uuur DA =

27 , 2

uuur DX =

(hypotenuse),

sin( 21 D) =

18 2 27 2

1 2

uuur 3 , OA = 2

(adjacent)

cos( 21 D) =

,

⎛ eg. D = 2 tan−1 ⎜ ⎜ ⎝

18 2 3 2

uuur AX =

,

3 2 27 2

1 2

uuur AC =

1 2

18

(opposite)

tan( 21 D) =

or

Marks

Any two correct A1 Uses correct dM1 sohcahtoa to find 21 D Correct ft application A1 of sohcahtoa

18 2 3 2

Attempts to find the correct angle D by doubling their angle ddM1 for 21 D .

⎞ ⎟ ⎟ ⎠

109.5° or A1 awrt 109° or 1.91c

D = 109.47122...o

[6] Aliter (d) Way 6

using trigonometry on auuur right angled similar triangle OAC uuur uuur OC = 3i + 3 j − 3k

uuur OC =

OA = 2i + 2 j + k

uuur OA = 3 ,

27 ,

(hypotenuse), (adjacent),

uuur AC =

AC = i + j − 4 k

Attempts to find two out of the three M1 lengths in ∆ OAC

18

(opposite) Any two correct A1

sin( 21 D) =

18 27

,

cos( 21 D) =

3 27

tan( 21 D) =

or

18 3

Uses correct dM1 sohcahtoa to find 21 D Correct ft application of sohcahtoa A1 Attempts to find the correct angle D by doubling their angle ddM1 for 21 D .

⎛ 18 ⎞ eg. D = 2 tan−1 ⎜⎜ ⎟⎟ ⎝ 3 ⎠

109.5° or A1 awrt 109° or 1.91c

D = 109.47122...o

[6]

17

Question Number Aliter 7. (b) (i)

Marks

Scheme uuur c = OC = ± ( 3i + 3 j − 3k ) uuur AB = ± ( −i − j − 5 k )

Way 2 uuur OC =

As

(3) + (3) + ( −3) = 2

uuur uuur OC = AB =

2

2

uuur (1) + (1) + ( −5) = AB 2

2

2

27

then the diagonals are equal, and OACB is a rectangle.

A complete method of proving that the diagonals are equal.

M1

Correct result.

A1

diagonals are equal and OACB is a rectangle

A1 cso

[3] uuur uuur a = OA = 2i + 2 j + k ⇒ OA = 3 uuur uuur b = OB = i + j − 4 k ⇒ OB = 18 uuur uuur BC = ± ( 2i + 2 j + k ) ⇒ BC = 3 uuur uuur AC = ± ( i + j − 4 k ) ⇒ AC = 18 uuur uuur c = OC = ± ( 3i + 3 j − 3k ) ⇒ OC = 27 uuur uuur AB = ± ( −i − j − 5 k ) ⇒ AB = 27

Aliter 7. (b) (i)

(OA)2 + ( AC )2 = (OC )2 or (BC )2 + (OB )2 = (OC )2 or (OA)2 + (OB )2 = ( AB )2 or (BC )2 + ( AC )2 = ( AB )2

or equivalent

Way 3 ⇒ (3)2 + ( 18)2 =

(

27

)

A complete method of proving that Pythagoras holds using their values. Correct result

2

M1

A1

and therefore OA is perpendicular to OB or AC is perpendicular to BC and hence OACB is a rectangle.

perpendicular and OACB is a rectangle

A1 cso

[3] 14marks

18

Question Number

Scheme

Marks

8. (a) x

0

1

y or y

e1

e2

2 e

3 7

e

4

10

e

5

13

e4

2.71828… 7.38906… 14.09403… 23.62434… 36.80197… 54.59815… Either e 7 , e 10 and e 13 or awrt 14.1, 23.6 and 36.8 or e to the power awrt 2.65, 3.16, 3.61 (or mixture of decimals and e’s) At least two correct B1 All three correct B1 [2]

1 × 1 B1; 2 For structure of trapezium rule {.............} ; M1

(b)

{

(

1 I ≈ × 1 ; × e1 + 2 e2 + e 2

=

7

+e

10

+e

13

Outside brackets

)+e } 4

1 × 221.1352227... = 110.5676113... = 110.6 (4sf) 2

110.6

A1 cao [3]

Beware: In part (b) candidates can add up the individual trapezia:

(

(b) I ≈ 21 .1( e1 + e2 ) + 21 .1 e2 + e

7

) + .1( e 1 2

7

+e

19

10

) + .1( e 1 2

10

+e

13

) + .1( e 1 2

13

+ e4

)

Question Number

Scheme 1

t = (3x + 1) 2 ⇒

(c)

dt = dx

− 21

dt =3 dx

dt 3 3 = = 1 dx 2.(3x + 1) 2 2t

∫

∴I = e

∴I=

∫

2 3

(3 x +1)

A(3x + 1)

.3.(3x + 1)

1 2

… or t 2 = 3 x + 1 ⇒ 2t

so

Marks

∫

dx = et

⇒

3 2

dx 2t = dt 3

(3x + 1)

− 21

dt =A dx dt or 2t =3 dx

− 21

or t

M1 A1

Candidate obtains either dt or dx in terms of t … dx dt

∫

dx 2t . dt = e t . .dt dt 3

… and moves on to dM1 substitute this into I to convert an integral wrt x to an integral wrt t.

∫

t et dt

2 3

t et

A1

changes limits x → t so B1 that 0 → 1 and 5 → 4

change limits: when x = 0, t = 1 & when x = 5, t = 4 4

Hence I =

∫

2 3

tet dt ; where a = 1, b = 4, k =

2 3

1

[5] (d)

Let k be any constant for the first three marks of this part.

du ⎪⎧u = t ⇒ dt = 1 ⎪⎫ ⎨ dv t t⎬ ⎩⎪ dt = e ⇒ v = e ⎭⎪

(

∫

∫

k t et dt = k t e t − e t .1 dt

(

= k t et − et

4

∴

∫ 1

2 3

tet dt =

{(

)

Use of ‘integration by parts’ formula in the M1 correct direction. Correct expression with a A1 constant factor k.

)

Correct integration with/without A1 a constant factor k

+c

) (

2 4e4 − e4 − e1 − e1 3

Substitutes their changed limits into the integrand dM1 oe and subtracts oe.

)}

= 32 (3e4 ) = 2e 4 = 109.1963...

either 2e4 or awrt 109.2 A1 [5] 15 marks

• •

Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. 20