Question 8:
8 | – | 6 | – | 4 | – | 2 | 2 | 4
PMT
v 1
(b) ∫ dv= ∫ dx M1
11+2v2 x
ln(1+2v2),=lnx(+C)
4
dM1A1, B1
Ax4 = 1 + 2  v2 y  2 Ax6 −x2 Ax4 −1 dM1
2  so y = or y = x
Ax4 = 1 +  x 2 2
1 1
x  e4lnx+4c − 
2 2
or y = M1A1 7
1 1
First M1 accept ∫ dv=∫ dx
2v+1 x
v
Second M1 requires an integration of correct form ¼ may be missing
A1 for LHS correct with ¼ and B1 is independent and is for lnx
Third M1 is dependent and needs correct application of log laws
Fourth M1 is independent and merely requires return to y/x for v
N.B. There is an IF method possible after suitable
rearrangement – see note.
A−1
(c) x = 1 at y = 3: 3 = A = ... M1
2
19x6 −x2 19x4 −1
y = or y = x A1 2
2 2
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8. (a) r cos θ = 4(cos θ – cos2 θ) or rcos θ = 4 cos θ – 2 cos 2θ – 2 B1
d(rcosθ) d(rcosθ)
=4(−sinθ+2cosθsinθ) or =4(−sinθ+sin2θ) M1A1
dθ dθ
1
4(–sin θ + 2 cos θ sin θ) = 0 ⇒ cos θ = which is
2
π
satisfied by θ = and r = 2(*) dM1A1 5
3
Alternative for first 3 marks:
dr
=4sinθ B1
dθ
dx dr
=−rsinθ+cosθ =−4sinθ+8sinθcosθ M1A1
dθ dθ
Substituting r = 2 and 0 = f into original equation scores 0 marks.
PMT
1
(b) ∫r2dθ=(8)∫(1−2cosθ+cos2θ)dθ M1
2  sin2θ θ
=(8) θ−2sinθ+ +
 
 4 2 M1A1
3θ sin2θ π 2  3π  π 3
8 −2sinθ+ =8   −2− − 3+ =2π−16+7 3
    
 2 4 π   4   2 8  M1
3
1 1 3
×1× 3 =
Triangle: 2 (rcos θ)(r sin θ) = 2 2 M1A1
( ) 3 15 3
2π−16+7 3 + =(2π−16)+
Total area: 2 2 (A1)A1 8
M1 needs attempt to expand (1 – cos θ)2 giving three
terms (allow slips)
Second M1 needs integration of cos2 θ using cos 2θ ± 1
Third M1 needs correct limits– may evaluate two areas and subtract
M1 needs attempt at area of triangle and A1 for cao
Next A1 is for value of area within curve, then final A1 is cao,
must be exact but allow 4 terms and isw for incorrect collection of terms
Special case for use of r sin θ gives B0M1A0M0A0
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PMT
d3y d2y dy d2y dy
9. (a) (x2 +1) +2x =4y +(1−2x) −2 M1A1
dx3 dx2 dx dx2 dx
d3y d2y dy
(x2 +1) =(1−4x) +(4y−2) (*) A1 3
dx3 dx2 dx
M: Use of product rule (at least once) and implicit
differentiation (at least once).
d2y
(b)   =3 B1
 
dx2

0
d3y d3y d2y
  =5 Follow through: = +2 B1ft
 
dx3

dx3 dx2
0
3 5
y = 1 + x + x2 + x3... M1A1 4
2 6
M: Use of series expansion with values for the derivatives
(can be allowed without the first term 1, and can also
be allowed if final term uses 3 rather than 3!)
PMT
(c) x = –0.5, y ≈ 1 – 0.5 + 0.375 – 0.104166…
= 0.77 (2 d.p.) [awrt 0.77] B1 1
[8]
10. (a) |(x – 3) + iy | = 2|x + iy| ⇒ (x – 3)2 + y2 = 4x2 + 4y2 M1A1
∴x2 + y2 + 2x – 3 = 0
(x + 1)2 + y2 = 4 M1
Centre (–1, 0), .radius 2 A1, A1 5
1st M: Use z = x + iy, and attempt square of modulus of each side.
Not squaring the 2 on the RHS would be M1 A0.
2nd M: Attempting to express in the form (x – a)2 + (y – b)2 =k,
or attempting centre and radius from the form
x2 + y2 + 2gx + 2fy + c = 0
(b)
y
x
–3 –1 O 1
–
√3
Circle, centre on x-axis B1
C (–1, 0), r = 2 dB1ft
Follow through centre and radius, but dependent on first B1.
There must be indication of their ‘–3’, ‘–1’ or ‘1’ on the x-axis
and no contradictory evidence for their radius.
Straight line B1
Straight line through (–1, 0), or perp. bisector of (–3, 0) and
(0, 3). B1
Straight line through point of int. of circle & –ve y-axis, or
through (0, − 3) B1 5
(c) Shading (only) inside circle B1
Inside correct circle and all of the correct side of the correct line…
this mark is dependent on all the previous B marks in parts (b)
and (c). dB1 2
[12]
PMT
11. (a) (cos θ + i sin θ)1 = cos θ + i sin θ ∴true for n = 1 B1
Assume true for n = k, (cos θ + i sin θ)k = cos kθ + i sin kθ
(cos θ + i sin θ)k+1 = (cos kθ + i sin kθ)(cos θ + i sin θ) M1
= coskθ cos θ – sin kθ sin θ + i(sin kθ cosθ + cos kθ sin θ) M1
(Can be achieved either from the line above or the line below)
= cos(k + 1)θ + isin(k + 1)θ A1
Requires full justification of (cos θ + i sin θ)k+1
= cos(k + 1)θ + i sin(k + 1)θ
(∴ true for n = k + 1 if true for n = k) ∴ true for n ∈ + by induction A1cso 5
Alternative:
For the 2nd M mark: (ei kθ )(eiθ ) = eiθ(k+1)
(b) cos 5θ = Re[(cosθ + i sin θ)5]
= cos5 θ + 10 cos3 θ i2 sin2 θ + 5 cos θ i4 sin4 θ M1A1
= cos5 θ – 10 cos3 θ sin2 θ + 5 cos θ sin4 θ M1
= cos5 θ – 10 cos3 θ(1 – cos2 θ) + 5 cos θ(1 – cos2 θ)2 M1
cos 5θ = 16 cos5 θ – 20 cos3 θ + 5 cos θ (*) A1cso 5
Alternative:
5 2 3 4 5
 1 1 1 1 1 1
z+  = z5 +5z4 +10z3  +10z2  +5z  +  M1
 z z z z z z
= 2 cos 5θ + 10 cos 3θ + 20 cosθ A1
(2 cos θ)5 = ........and attempt to put cos 3θ in powers of cos θ M1
Correct method (or formula) for putting cos 3θ in powers of cos θ M1
cos 5θ = 16 cos5 θ – 20 cos3 θ + 5 cos θ A1cso
cos5θ
(c) =0⇒cos5θ=0 M1
cosθ
π π
5θ= θ= A1
2 10
π
x = 2cos θ, x = 2 cos is a root (*) A1 3
10
PMT
Alternatives:
(i) Substitute given root into x4 – 5x2 + 5:
 π 4  π 2  π 4  π 2
2cos  −52cos  +5=24cos  −5×22cos  +5 M1
 10  10  10  10
5π
‘Multiply by cos θ’ and use result from part (b): ... = cos A1
10
= 0 and conclusion A1
π
(ii) Use 5θ= in result from part (b) M1
2
 π 5  π 3  π
16cos  −20cos  +5cos  and divide by cos θ A1
 10  10  10
= 0 and conclusion A1
[13]