Mark Scheme (Results) Summer 2007 - Edexcel

6
4
2
Question Scheme Marks
number
7. (a) 5
(b)
2 4 6
Shape (closed curve, approx. symmetrical about
the initial line, in all ‘quadrants’ and
x
−4
5 − √3 5 + √3 ‘centred’ to the right of the pole/origin). B1
−2
5 Scale (at least one correct ‘intercept’ r value…
−6
shown on sketch or perhaps seen in a table). B1 (2)
(Also allow awrt 3.27 or awrt 6.73).
y = r sinθ
= 5sinθ
+ √ 3sinθ
cosθ
M1
d y
2
= 5cosθ
− √ 3sin θ + √ 3cos
θ
dθ
( = 5cosθ
+ √ 3cos 2θ
)
A1
2
2
5cosθ − √ 31−
cos θ + √ 3cos θ =
M1
2
−2
−4
y
( ) 0
2√
3cos θ + 5cosθ
− √ 3 = 0
2 √ 3cosθ
−1
cosθ
+ √ 3 = 0 cosθ
=
( )( ) ...
(0.288…)
⎛
1 ⎞
θ = 1.28 and 5.01 (awrt) (Allow ±1.28 awrt) ⎜Also allow ± arccos ⎟ A1
⎝
2√
3 ⎠
⎛ 1 ⎞ 11
r = 5 + √ 3⎜
⎟ = (Allow awrt 5.50) A1 (6)
⎝ 2√
3 ⎠ 2
2
2
(c) r = 25 + 10√
3cosθ
+ 3cos θ
B1
2 53θ
⎛ sin 2θ
⎞
25 + 10√
3cosθ + 3cos θ dθ
= + 10√
3sinθ
+ 3
∫
⎜ ⎟
M1 A1ft A1ft
2
⎝ 4 ⎠
(ft for integration of ( ) bcosθ
a + and c cos2θ
respectively)
2π
1 ⎡
3sin 2θ
3θ
⎤
25 10 3sin
2 ⎢ θ + √ θ + +
4 2 ⎥
⎣
⎦ 0
= ......
M1
53
= 1 ( 50π + 3π
) =
π or equiv. in terms of π.
2
2
A1 (6)
14
(b) 2 nd M: Forming a quadratic in cos θ .
3 rd M: Solving a 3 term quadratic to find a value of cos θ (even if called θ ).
Special case: Working with r cosθ
instead of r sinθ
:
1 st 2
M1 for r cosθ
= 5cosθ
+ √ 3cos θ
1 st A1 for derivative − 5sinθ
− 2√
3sinθ
cosθ
, then no further marks.
(c) 1 st M: Attempt to integrate at least one term.
2 nd M: Requires use of the 2
1 , correct limits (which could be 0 to 2π, or
−π to π, or ‘double’ 0 to π), and subtraction (which could be implied).
M1