2007 Ce A Math Solution

CE Additional Mathematics 2007 Suggested Solution Section A 1. [3 marks]

∫

x4 + 1 dx = x2 =

2.

∫

x 2 + x −2 dx =

x3 x −1 + +C 3 −1

x3 1 − + C , where C is a constant 3 x

[3 marks] 0 −2 1 −3 1 1 Area of the quad. = 2 0 = (2k − 2k + 2 + 6) = 4 sq. units 2 2 k k 0 −2

3.

[4 marks] cos x − 2 cos 2 x + cos 3 x = 0 cos x + cos 3 x = 2 cos 2 x

2 cos 2 x cos(− x) = 2 cos 2 x 2 cos 2 x cos x = 2 cos 2 x

(

)

cos 2 x 2 cos x − 2 = 0 2 2 D D 2 x = 360 n ± 90 or x = 360D n ± 45D x = 180D n ± 45D or x = 360D n ± 45D ∴ x = 180D n ± 45D cos 2 x = 0 or cos x =

4.

where n is any integer

[4 marks] ⎡⎣( x + Δx) 2 + 1⎤⎦ − ⎡⎣ x 2 + 1⎤⎦ d 2 ( x + 1) = lim Δx → 0 dx Δx x 2 + 2 xΔx + Δx 2 + 1 − x 2 − 1 = lim Δx → 0 Δx

= lim (2 x + Δx) Δx → 0

= 2x

1

5.

[5 marks] Let P (n) :

1 1 1 1 1 − − 2 − ⋅⋅⋅ − n = n a −1 a a a a (a − 1)

When n = 1 L.S. of P (1) =

1 1 a − a +1 1 − = = a − 1 a a(a − 1) a (a − 1)

R.S. of P (1) =

1 a(a − 1)

∴ P(1) is true. Assume P (k ) is true, where k is any positive integer

i.e.

1 1 1 1 1 − − 2 − ⋅⋅⋅ − k = k a −1 a a a a (a − 1)

1 1 1 1 1 1 1 − − 2 − ⋅⋅⋅ − k − k +1 = k − k +1 a −1 a a a a a (a − 1) a =

1 1 − k +1 a (a − 1) a

=

a − a +1 1 = k +1 k +1 a (a − 1) a (a − 1)

k

∴ P(k + 1) is true. By the principle of M.I., P (n) is true for all positive integer n . 6.

[5 marks] Equate cos x = sin 2 x cos x = 2sin x cos x cos x(1 − 2sin x) = 0

∴x =

π 6

or

π 2

(Rejected)

Area of the shaded region π

π

cos 2 x ⎤ 6 ⎡ = 6 cos x − sin 2 xdx = ⎢sin x + 0 2 ⎥⎦ 0 ⎣

∫

π ⎞ ⎛ 1⎞ 1 1 1 1 ⎛ π 1 = ⎜ sin + cos ⎟ − ⎜ 0 + ⎟ = + − = 6 2 3⎠ ⎝ 2⎠ 2 4 2 4 ⎝ 7.

[5 marks] (a)

⎛ 1 ⋅ (−3) + 2 ⋅ (2) 1 ⋅ 4 + 2 ⋅1 ⎞ ⎛ 2 ⎞ C =⎜ , ⎟ = ⎜ ,2⎟ 2 +1 2 +1 ⎠ ⎝ 3 ⎠ ⎝ 2

40 2 5 ⎛2⎞ , AC = (2 − ) 2 + (2 − 1) 2 = (b) OA = 2 + 1 = 5 , OC = ⎜ ⎟ + 22 = 9 3 3 ⎝3⎠ 2

2

2

40 25 − 9 9 = 1 cos ∠AOC = 40 2 2 5 9 5+

OC =

40 2 10 , OB = 22 + 42 = 20 , BC = (−2 − ) 2 + (4 − 2) 2 = 9 3 3

40 100 + 20 − 9 = 1 cos ∠BOC = 9 40 2 2 20 9

∴∠AOC = ∠BOC 8.

[5 marks] JJJG JJJG JJJG (a) BC = OC − OB JJJG = kOA − (2i + 6 j) = k (6i + 3 j) − (2i + 6 j)

(b)

= (6k − 2)i + (3k − 6) j JJJG JJJG BC ⋅ OA = 0

[(6k − 2)i + (3k − 6) j] ⋅ [6i + 3j] = 0 36k − 12 + 9k − 18 = 0

k=

9.

2 3

[5 marks] 2

(a)

⎛ x⎞ y + ⎜ ⎟ = 25 ⎝2⎠ 2

x 2 + 4 y 2 = 100 (b) Differentiate above equation w.r.t. t , we have 2 x dy dx 16 y 8 y . = −2 , ∴ = = dt dt 2x x When y = 3 , x = 8

We have

∴

dx 8 ⋅ 3 = = 3 m/s dt 8

3

dx dy + 8y = 0. dt dt

10. [5 marks] (a) By similar triangle, (b)

f '(1) 2 3 = , ∴ f '(1) = 2 3 4

x -coordinate of turning point = 4

1 ∴ f "(4) = − ve 2 ∴ f ( x) attains its maximum when x = 4 .

As f "( x) = −

11. [5 marks] (a) When 0 ≤ x ≤ 1 , we have x − 1 ≤ 0 and x ≥ 0 ∴− ( x − 1) = x − 1 ∴− x + 1 = x − 1 ∴x =1 (b) When x < 0 , we have x − 1 < 0 ∴− ( x − 1) = − x − 1 ∴− x + 1 = − x − 1 Rejected When x > 1 , x − 1 , x > 0 ∴ x −1 = x −1 ∴ x > 1 is the solution Hence, the overall solution is x ≥ 1 . 12. [6 marks] (1 − 2 x + x 2 ) n = ⎡⎣1 − (2 x − x 2 ) ⎤⎦

n

= 1 − C1n (2 x − x 2 ) + C2n (2 x − x 2 ) 2 − C3n (2 x − x 2 )3 + ⋅⋅⋅ = 1 − n(2 x − x 2 ) +

n(n − 1) n(n − 1)(n − 2) (4 x 2 − 4 x3 + ⋅⋅⋅) − (8 x3 + ⋅⋅⋅) + ⋅⋅⋅ 2 6

Coefficient of x 2 = n + 2n(n − 1) = 66

2n 2 − n − 66 = 0 (2n + 11)(n − 6) = 0 ∴ n = 6 or n = −

11 (Rejected) 2

∴ coefficient of x3 =

6⋅5 6⋅5⋅ 4 (−4) − ⋅ 8 = −60 − 160 = −220 2 6

13. [7 marks] (a)

⎧ y = x2 ⎨ ⎩ y = mx − 2m

x 2 − mx + 2m = 0 ...........(*) 4

Δ = m 2 − 4 ⋅ 2m = m 2 − 8m Δ>0 m 2 − 8m > 0 m(m − 8) > 0 ∴ m > 8 or m < 0 (b) (i) Let the mid-point of AB be M (a, b) From (*) ⎛ m⎞ m a = −⎜− ⎟ = ⎝ 2⎠ 2 b = m⋅

m m 2 − 4m − 2m = 2 2

⎛ m m 2 − 4m ⎞ (ii) As x + y = 5 bisect AB , ∴ ⎜ , ⎟ lies on x + y = 5 2 ⎝2 ⎠

m m 2 − 4m + =5 2 2

m 2 − 3m − 10 = 0 m = −2 or m = 5 by (a) m = 5 is rejected, ∴ m = −2 . Section B 14. (a) [3 marks] Let BC = h , BX = b and AB = a ∴ XC = b 2 + h 2 and AC = a 2 + h 2 also AX = a 2 − b AX 2 + XC 2 = (a 2 − b 2 ) + (b 2 + h 2 ) = a 2 + h 2 ∴ AX 2 + XC 2 = AC 2 ∴ AX ⊥ XC (b) [9 marks] (i) Let FB = A , AX = h 1 + (3 2) 2 − A 2 cos135 = 2 ⋅1 ⋅ 3 2 1 1 + 18 − A 2 − = 2 6 2 2 A = 25 , ∴ A = 5 D

h⋅A 1 = 1⋅ 3 2 sin135D 2 2 1 5h = 3 2 ⋅ 2 5

h=

3 5 2

9 441 ⎛3⎞ XB = (3 2) − ⎜ ⎟ = 18 − = 25 25 ⎝5⎠ 2

2

XB =

21 5

(ii) ∠CXB = tan −1 ∠EXF = tan −1

7 1 = tan −1 21 3 7 4

1 7 tan θ = tan(180D − tan −1 − tan −1 ) = − 3 4

1 7 + 3 4 = −5 ⎛ 1 ⎞⎛ 7 ⎞ 1− ⎜ ⎟ ⎜ ⎟ ⎝ 3 ⎠⎝ 4 ⎠

15. (a) [3 marks] Let the distance from P to L be d | 2a − b | | 2a − b | d= = 5 22 + 12 2

⎛8 5 ⎞ 2 d + ⎜⎜ ⎟⎟ = r ⎝ 2 ⎠ 2

(By Pythagoras Theorem)

(2a − b) 2 + 80 = r 2 5 4a 2 − 4ab + b 2 + 400 5 (b) [9 marks] (i) As L2 : 2 x + y = 0 is tangent to the circle r2 =

r=

| 2a + b | 22 + 12

=

| 2a + b | 5

(2a + b) 2 4a 2 − 4ab + b 2 + 400 = 5 5 2 2 2 ∴ 4a + 4ab + b = 4a − 4ab + b 2 + 400 ab = 50 ∴ Locus of P : xy = 50 By (a)

(ii) Let A = Area of C = π r 2 =π

4a 2 − 4ab + b 2 + 400 5

π⎡

2 ⎤ ⎛ 50 ⎞ = ⎢ 4a − 4 ⋅ 50 + ⎜ ⎟ + 400 ⎥ 5 ⎢⎣ ⎝ a ⎠ ⎦⎥ 2

6

2π 5

1250 ⎤ ⎡ 2 ⎢⎣ 2a + 100 + a 2 ⎥⎦ dA π ⎛ 2 ⋅ 2500 ⎞ = ⎜ 8a − ⎟=0 da 5 ⎝ a3 ⎠ =

5000 = 625 8 ∴ a = ±5 a4 =

b=±

50 = ±10 5

1⎛ 2500 ⎞ r 2 = ⎜ 4 ⋅ 25 + 200 + ⎟ = 80 5⎝ 25 ⎠ Hence the equations of C : ( x ± 5) 2 + ( y ± 10) 2 = 80 16. (a) [5 marks] (i) Radius of the inner circle = 1 ⋅ cos θ = cos θ A=

1 2 ⎡ cos θ tan θ 1 ⎤ − (cos 2 θ ) ⋅ θ ⎥ (1 − cos 2 θ )(2π − 2θ ) + 2 ⎢ 2 2 2 ⎣ ⎦ = (sin 2 θ )(π − θ ) + cos θ sin θ − θ cos 2 θ = cos θ ⋅

sin θ + π sin 2 θ − θ (sin 2 θ + cos 2 θ ) cos θ

= π sin 2 θ − θ + cos θ sin θ 1 = π sin 2 θ − θ + sin 2θ sq. units 2 (ii)

dA 1 = 2π sin θ cos θ − 1 + ⋅ 2 ⋅ cos 2θ dθ 2 = 2π sin θ cos θ − 1 + cos 2θ

= π sin 2θ + 1 − 2sin 2 θ − 1 sin θ = π sin 2θ − 2sin θ cos θ = sin 2θ (π − tan θ ) cos θ (b) [3 marks] When

dA =0 dθ

sin 2θ = 0 (rejected, as 0 < θ <

π 2

) or tan θ = π

(c) [4 marks] PR = 2sin θ Let P be the perimeter of the shaded region P = 1⋅ (2π − 2θ ) + 2sin θ + (2 cos θ )π = 2π − 2θ + 2sin θ + 2π cos θ

∴

dP = −2 + 2 cos θ − 2π sin θ dθ 7

But

dP dθ

= −7.38 ≠ 0 θ = tan −1 π

Hence the student's guess is incorrect. 17. (a) [1 mark] JJJJG 1 G G OM = (a + b) 2 (b) [4 marks] JJJG G JJJG 2 G (i) OP = a and OQ = kb 3 G G JJJG JJJG 3 ⎛⎜ 2 a + 4kb ⎞⎟ JJJG 3OP + 4OQ 3 ⎠ = 2 aG + 4k bG = ⎝ OG = 7 7 7 7

1 ⎛ 4k ⎞ ⎜ ⎟ JJJJG JJJG 1 7 (ii) As OM // OG , ∴ 2 = ⎝ ⎠ , ∴ k = 1 2 ⎛2⎞ 2 ⎜⎝ 7 ⎟⎠ JJJG G JJJG 2 G From the givens, we have OP = a and OQ = kb 3 JJJG JJJG JJJG G 2G ∴ PQ = OQ − OP = kb − a 3 1G 2G = b− a 2 3 (c) [7 marks] G G G G 1 1 (i) a ⋅ b =| a || b | cos 60D = 1⋅1 ⋅ = 2 2 JJJG JJJG JJJG 1 G 2 G 1 G 2 G 1 2 1 4 13 | PQ |2 = PQ ⋅ PQ = ( b − a ) ⋅ ( b − a) = − ( ) + = 2 3 2 3 4 3 2 9 36 JJJG 13 | PQ | = 6 JJJJG JJJJG JJJJG 1 G G 1 G G 1 3 (ii) | OM |2 = OM ⋅ OM = (a + b) ⋅ (a + b) = (1 + 1 + 1) = 2 2 4 4 JJJJG 3 | OM | = 2 JJJG JJJJG JJJG JJJJG PQ ⋅ OM = | PQ || OM | cos ∠QGM 13 3 ⎛1 G 2 G⎞ 1 G G cos ∠QGM ⎜ b − a⎟⋅ a + b = 3 ⎠ 2 6 2 ⎝2

(

)

1 1 1 1 13 3 cos ∠QGM + − − = 8 4 3 6 6 2 ∴∠QGM = 104D 8

18. (a) [2 marks] dy 1 −1 1 = 2 ⋅ x 2 −1 = −1 dx 2 x dy =0 dx x = r

1 −1 = 0 r ∴r = 1 ∴

(b) [10 marks] (i)

⎡ S = 2⎢ ⎣

∫

1

2 x − xdx + 1 ⋅1 +

0

By Symmetry ∴ S =4

∫

1

∫

1

∫

⎤ 2 3 − x − (3 − x)dx ⎥ 2 ⎦ 3

2 x − xdx =

0

∫

3

2 3 − x − (3 − x)dx

2

2 x − xdx + 2

0 1

⎡ 3 ⎤ ⎢ 2 x 2 x2 ⎥ 16 ⎛4 1⎞ ⎛8−3⎞ − ⎥ + 2 = 4⎜ − ⎟ + 2 = 4⎜ +2= = 4⎢ sq. units ⎟ 3 2 3 2 6 3 ⎝ ⎠ ⎝ ⎠ ⎢ ⎥ ⎣ 2 ⎦0

1 ⎡ (ii) Volume = 2 π ⎢ 2 ⎣ =π

∫

1

∫

⎤ π 1 ⋅1 (2 x − x) 2 dx ⎥ + 0 2 ⎦ 1

3

4 x + x 2 − 4 x 2 dx +

0

π 2

1

5 ⎤ ⎡ ⎢ 2 x3 x2 ⎥ π = π ⎢2 x + − 4 ⋅ ⎥ + 5 3 2 ⎢ ⎥ 2 ⎦0 ⎣

1 8 π = π (2 + − ) + 3 5 2 ⎛ 30 + 5 − 24 ⎞ π 22π + 15π 37π =π ⎜ = cubic units ⎟+ = 15 30 30 ⎝ ⎠ 2 (iii) Let OQ = 1 + k

1 ⎛ 37π Volume of the first portion = ⎜ 3 ⎝ 30 1 ⎛ 11π ⎞ 1 37π 2 ⎜ ⎟ + π ⋅1 ⋅ k = 2 ⎝ 15 ⎠ 2 90

⎞ 37π ⎟= ⎠ 90

11 k 37 + = 30 2 90 ∴k =

4 , 45

OQ : OP = 1 +

4 : 3 = 49 :135 45 9