Mark Scheme (Results) Summer 2009
GCE
GCE Mathematics (6665/01)
June 2009 6665 Core Mathematics C3 Mark Scheme Question Number Q1
(a)
Scheme Iterative formula: xn +1 =
Marks
2 + 2 , x0 = 2.5 ( xn ) 2
An attempt to substitute x0 = 2.5 into the iterative formula. M1 Can be implied by x1 = 2.32 or 2.320 Both x1 = 2.32(0) A1 and x2 = awrt 2.372 Both x3 = awrt 2.356 A1 cso and x4 = awrt 2.360 or 2.36
2 x1 = +2 (2.5) 2
x1 = 2.32 x2 = 2.371581451... x3 = 2.355593575... x4 = 2.360436923...
(3) (b)
Let f ( x) = − x 3 + 2 x 2 + 2 = 0 f (2.3585) = 0.00583577... f (2.3595) = − 0.00142286... Sign change (and f ( x) is continuous) therefore a root
α is such that α ∈ ( 2.3585, 2.3595 ) ⇒ α = 2.359 (3 dp)
Choose suitable interval for x, M1 e.g. [2.3585, 2.3595] or tighter any one value awrt 1 sf dM1 or truncated 1 sf both values correct, sign change A1 and conclusion At a minimum, both values must be correct to 1sf or truncated 1sf, candidate states “change of sign, hence root”.
(3)
[6]
6665/01 GCE Mathematics June 2009
2
Question Number Q2
(a)
Scheme
cos 2 θ + sin 2 θ = 1
Marks
( ÷ cos θ ) 2
cos 2 θ sin 2 θ 1 + = 2 2 cos θ cos θ cos 2 θ
Dividing cos 2 θ + sin 2 θ = 1 by M1 cos 2 θ to give underlined equation.
1 + tan 2 θ = sec 2 θ
tan 2 θ = sec 2 θ − 1 (as required)
Complete proof. No errors seen. A1 cso
AG
(2) (b)
2 tan 2 θ + 4secθ + sec 2 θ = 2,
( eqn *)
0 ≤ θ << 360o Substituting tan 2 θ = sec 2 θ − 1 into eqn * to get a M1 quadratic in secθ only
2(sec 2 θ − 1) + 4secθ + sec 2 θ = 2
2sec 2 θ − 2 + 4secθ + sec 2 θ = 2
Forming a three term “one sided” M1 quadratic expression in secθ .
3sec 2 θ + 4secθ − 4 = 0
( secθ
+ 2 )( 3secθ − 2 ) = 0
secθ = −2 or secθ =
Attempt to factorise M1 or solve a quadratic.
2 3
1 1 2 = −2 or = cos θ cos θ 3
cos θ = − 12 ; or cos θ =
α = 120o
cos θ = − 12
3 2
A1;
or α = no solutions
θ1 = 120o
120o
A1
240o or θ 2 = 360o − θ1 when B1 solving using cos θ = ...
θ 2 = 240o
θ = {120o , 240o }
Note the final A1 mark has been changed to a B1 mark.
(6) [8]
6665/01 GCE Mathematics June 2009
3
Question Number
Scheme
Marks
t
P = 80 e 5
Q3 (a)
80 B1
0
t = 0 ⇒ P = 80e 5 = 80(1) = 80
(1)
(b)
Substitutes P = 1000 and
t 1000 P = 1000 ⇒ 1000 = 80e ⇒ = e5 80 t 5
t rearranges equation to make e 5 the M1 subject.
⎛ 1000 ⎞ ∴ t = 5ln ⎜ ⎟ ⎝ 80 ⎠ t = 12.6286...
awrt 12.6 or 13 years A1 Note t = 12 or t = awrt 12.6 ⇒ t = 12 will score A0
(2)
1
(c)
t ke 5 and k ≠ 80. M1
t dP = 16e 5 dt
1t
16e 5
A1 (2)
(d)
t
50 = 16e 5 ⎛ 50 ⎞ ∴ t = 5ln ⎜ ⎟ ⎝ 16 ⎠
P = 80e P =
1⎛ ⎛ 50 ⎞ ⎞ ⎜ 5ln ⎜ ⎟ ⎟ 5⎝ ⎝ 16 ⎠ ⎠
Using 50 = ddPt and an attempt to solve M1 to find the value of t or 5t .
{= 5.69717...} 1
or P = 80e 5
Substitutes their value of t back dM1 into the equation for P.
( 5.69717...)
80(50) = 250 16
250 or awrt 250 A1 (3) [8]
6665/01 GCE Mathematics June 2009
4
Question Number Q4 (i)(a)
Scheme
Marks
y = x 2 cos3 x
⎧ u = x2 ⎪ Apply product rule: ⎨ du ⎪ = 2x ⎩ dx
v = cos3x ⎫ ⎪ ⎬ dv = − 3sin 3x ⎪ dx ⎭ Applies vu ′ + u v′ correctly for their u , u ′ , v , v′ AND gives an M1 expression of the form α x cos3x ± β x 2 sin 3 x Any one term correct A1 Both terms correct and no further simplification to terms in A1 cos α x 2 or sin β x 3 .
dy = 2 x cos3x − 3 x 2 sin 3x dx
(3) (b)
y =
ln( x 2 + 1) x2 + 1
u = ln( x + 1) 2
something x2 + 1 2x ln( x 2 + 1) → 2 x +1
ln( x 2 + 1) →
du 2x ⇒ = 2 dx x + 1
⎧ u = ln( x 2 + 1) ⎪ Apply quotient rule: ⎨ du 2x ⎪ = 2 ⎩ dx x + 1
M1 A1
v = x 2 + 1⎫ ⎪ ⎬ dv = 2x ⎪ dx ⎭ v u ′ − u v′ M1 v2 Correct differentiation with correct A1 bracketing but allow recovery.
⎛ 2x ⎞ 2 2 ⎜ 2 ⎟ ( x + 1) − 2 x ln( x + 1) dy 1 + x ⎠ = ⎝ 2 dx ( x2 + 1)
Applying
(4)
⎧ ⎫ 2 x − 2 x ln( x 2 + 1) ⎪ ⎪ dy = ⎨ ⎬ 2 ⎪⎩ dx ⎪⎭ ( x2 + 1)
6665/01 GCE Mathematics June 2009
{Ignore subsequent working.}
5
Question Number
Scheme
Marks
y = 4 x + 1, x > − 14
(ii)
At P, y = 4(2) + 1 =
9 = 3
At P , y = 9 or 3
dy 1 −1 = ( 4 x + 1) 2 (4) dx 2
B1
± k (4 x + 1)
− 12
M1*
2(4 x + 1)
− 12
A1 aef
dy 2 = 1 dx (4 x + 1) 2 dy 2 = 1 dx ( 4(2)+1) 2
At P,
Hence m(T) =
Substituting x = 2 into an equation M1 involving ddyx ;
2 3
y − y1 = m ( x − 2) or y − y1 = m ( x − their stated x ) with ‘their TANGENT gradient’ and their y1; dM1*; or uses y = mx + c with ‘their TANGENT gradient’, their x and their y1.
Either T: y − 3 = ( x − 2) ; 2 3
or y = 23 x + c and 3=
2 3
( 2) + c
Either T:
⇒ c=3−
4 3
=
5 3
;
3 y − 9 = 2( x − 2) ;
T:
3 y − 9 = 2x − 4
T:
2x − 3 y + 5 = 0
2x − 3 y + 5 = 0
A1
Tangent must be stated in the form ax + by + c = 0 , where a, b and c are integers. (6)
or T:
y = 23 x + 53
T:
3y = 2x + 5
T:
2x − 3 y + 5 = 0 [13]
6665/01 GCE Mathematics June 2009
6
Question Number Q5
Scheme
(a)
Marks
y
Curve retains shape when x > 12 ln k B1
Curve reflects through the x-axis when x < 12 ln k B1
( 0, k − 1) O
( 12 ln k , 0 )
x
( 0, k − 1)
and
( 12 ln k , 0 )
marked
B1
in the correct positions. (3) (b)
y
Correct shape of curve. The curve should be contained in B1 quadrants 1, 2 and 3 (Ignore asymptote)
( 0, 12 ln k )
(1 − k , 0 ) O
x
(1 − k , 0 )
and ( 0, 12 ln k )
B1
(2)
Either f ( x) > − k or y > − k or (c) Range of f: f ( x) > − k or y > − k or (− k , ∞)
(− k , ∞) or f > − k or B1 Range > − k . (1)
(d)
Attempt to make x M1 (or swapped y) the subject
y = e2 x − k ⇒ y + k = e2 x ⇒ ln ( y + k ) = 2 x
Makes e 2 x the subject and M1 takes ln of both sides
⇒ 12 ln ( y + k ) = x Hence f −1 ( x) = 12 ln( x + k )
1 2
ln( x + k ) or ln ( x + k )
A1 cao (3)
Either x > −k or (− k , ∞) or (e)
Domain > − k or x “ft one sided B1 inequality” their part (c) RANGE answer
f −1 ( x) : Domain: x > −k or (− k , ∞)
(1) [10]
6665/01 GCE Mathematics June 2009
7
Question Number Q6
(a)
Scheme
A = B ⇒ cos ( A + A ) = cos 2 A = cos A cos A − sin A sin A
Marks
Applies A = B to cos ( A + B ) to give the underlined equation or M1 cos 2 A = cos 2 A − sin 2 A
cos 2 A = cos 2 A − sin 2 A and cos 2 A + sin 2 A = 1 gives
cos 2 A = 1 − sin 2 A − sin 2 A = 1 − 2sin 2 A (as required)
(b)
Complete proof, with a link between LHS and RHS. No errors A1 AG seen.
C1 = C2 ⇒ 3sin 2 x = 4sin 2 x − 2cos 2 x
(2)
Eliminating y correctly. M1 Using result in part (a) to substitute for sin 2 x as ± 1 ± cos 2 x or k sin 2 x as M1 2 ⎛ ± 1 ± cos 2 x ⎞ k⎜ ⎟ to produce an 2 ⎝ ⎠ equation in only double angles.
⎛ 1 − cos 2 x ⎞ 3sin 2 x = 4 ⎜ ⎟ − 2cos 2 x 2 ⎝ ⎠
3sin 2 x = 2 (1 − cos 2 x ) − 2cos 2 x 3sin 2 x = 2 − 2cos 2 x − 2cos 2 x 3sin 2 x + 4cos 2 x = 2
Rearranges to give correct result A1 AG (3)
(c)
3sin 2 x + 4cos 2 x = R cos ( 2 x − α ) 3sin 2 x + 4cos 2 x = R cos 2 x cos α + R sin 2 x sin α Equate sin 2 x : 3 = R sin α Equate cos 2 x : 4 = R cos α
R=
R=5
32 + 42 ; = 25 = 5 tan α = ±
tan α =
3 4
⇒ α = 36.86989765...
sin α = ±
o
3 their R
3 4
or tan α = ± or cos α = ±
4 3
or
4 their R
B1
M1
awrt 36.87 A1 Hence, 3sin 2 x + 4cos 2 x = 5cos ( 2 x − 36.87 ) (3)
6665/01 GCE Mathematics June 2009
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Question Number (d)
Scheme
Marks
3sin 2 x + 4cos 2 x = 2 5cos ( 2 x − 36.87 ) = 2 cos ( 2 x − 36.87 ) =
cos ( 2 x ± their α ) =
2 5
( 2 x − 36.87 )
= 66.42182...o
( 2 x − 36.87 )
= 360 − 66.42182...o
2 their R
M1
awrt 66
A1
One of either awrt 51.6 or awrt A1 51.7 or awrt 165.2 or awrt 165.3 Both awrt 51.6 AND awrt 165.2 A1
Hence, x = 51.64591...o , 165.22409...o
(4)
If there are any EXTRA solutions inside the range 0 ≤ x < 180° then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 ≤ x < 180°. [12]
6665/01 GCE Mathematics June 2009
9
Question Number Q7
Scheme
f ( x) = 1 −
Marks
2 x−8 + ( x + 4) ( x − 2)( x + 4)
x ∈ ℝ, x ≠ −4, x ≠ 2. (a)
An attempt to combine to one M1 fraction
( x − 2)( x + 4) − 2( x − 2) + x − 8 f ( x) = ( x − 2)( x + 4)
=
x2 + 2 x − 8 − 2 x + 4 + x − 8 ( x − 2)( x + 4)
=
x 2 + x − 12 [( x + 4)( x − 2)]
=
( x + 4)( x − 3) [( x + 4)( x − 2)]
=
( x − 3) ( x − 2)
Correct result of combining all A1 three fractions
Simplifies to give the correct numerator. Ignore omission of A1 denominator An attempt to factorise the dM1 numerator. Correct result A1 cso AG (5)
(b)
g( x) =
ex − 3 ex − 2
x ∈ ℝ, x ≠ ln 2.
⎧ u = ex − 3 ⎪ Apply quotient rule: ⎨ du x ⎪ =e ⎩ dx
g′( x) =
v = ex − 2 ⎫ ⎪ ⎬ dv x =e ⎪ dx ⎭ v u ′ − u v′ M1 v2 Correct differentiation A1
e x (e x − 2) − e x (e x − 3) (e x − 2) 2
=
e 2 x − 2e x − e 2 x + 3e x (e x − 2) 2
=
ex (e x − 2) 2
Applying
Correct result
A1 AG cso (3)
6665/01 GCE Mathematics June 2009
10
Question Number (c)
Scheme
g′( x) = 1 ⇒
Marks
ex =1 (e x − 2) 2 Puts their differentiated numerator M1 equal to their denominator.
e x = (e x − 2) 2 e x = e 2 x − 2e x − 2e x + 4
e 2 x − 5e x + 4 = 0
e2 x − 5e x + 4
A1
Attempt to factorise M1 or solve quadratic in e x
(e x − 4)(e x − 1) = 0 e x = 4 or e x = 1
x = ln 4 or x = 0
both x = 0, ln 4
A1 (4) [12]
6665/01 GCE Mathematics June 2009
11
Question Number Q8
(a)
Scheme
sin 2 x = 2sin x cos x
Marks
2sin x cos x
B1 aef (1)
cosec x − 8cos x = 0 ,
(b)
0< x<π
1 − 8cos x = 0 sin x
Using cosec x =
1 sin x
M1
1 = 8cos x sin x
1 = 8sin x cos x 1 = 4 ( 2sin x cos x ) 1 = 4sin 2x
sin 2x =
sin 2 x = k , where −1 < k < 1 and M1 k≠0 sin 2x = 14 A1
1 4
Radians
2 x = {0.25268..., 2.88891...}
Degrees
2 x = {14.4775..., 165.5225...}
Radians
x = {0.12634..., 1.44445...}
Degrees
x = {7.23875..., 82.76124...}
6665/01 GCE Mathematics June 2009
Either arwt 7.24 or 82.76 or 0.13 or 1.44 or 1.45 or awrt 0.04 π or A1 awrt 0.46 π . Both 0.13 and 1.44 A1 cao (5) Solutions for the final two A marks must be given in x only. If there are any EXTRA solutions inside the range 0 < x < π then withhold the final accuracy mark. Also ignore EXTRA solutions outside the range 0 < x < π . [6]
12
