C4 MS Jan 2007

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 Âą 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