1 - Quod Erat Demonstrandum



F.4 ADDITIONAL MATHEMATICS FINAL REVISION QUIZ 1

Time allowed: 1[pic] hours

FORMULAS FOR REFERNCE

|[pic] |[pic] |

Section A (36 marks)

1. Expand (1 ( 2x)3. Suppose the coefficient of x2 in the expansion (1 ( 6x + 12x2 ( 8x3)n is 144, find

the value of n where n is a positive integer.

(6 marks)

2. Show, by mathematical induction, that 42n + 1 + 2(7n + 1) is divisible by 9 for all natural numbers n.

(6 marks)

3. Give the general solutions to the equation sinx + [pic]sin2x + sin3x = 0.

(5 marks)

4. | 2x ( 1 | = 2| x | ( 1 -------------------- (*)

(a) Solve (*) for 0 < x ( [pic].

(b) Solve (*).

(5 marks)

5. Let L1 : x + 2y ( 4 = 0

L2 : 3x ( 4y ( 10 = 0

(a) Write down an equation of the family of lines passing through the intersection of L1 and L2.

(b) Find the equation(s) of the line(s) in the family which makes an angle of ( with the line y = x,

where tan( = [pic].

(6 marks)

6. Let C : x2 + y2 + (t + 2)x + 4y + 4t = 0

L : 2x + y + 9 = 0

Given that L cuts C at two distinct points A and B.

(a) Without finding the coordinates of A and B, find, in terms of t, the x-coordinate of the

mid-point of A and B.

(b) Find the range of values of t.

(8 marks)

Section B (24 marks)

7. The figure shows a building with a horizontal base ABC such that AB = BC = CA = 10 m.

P, Q and R are points vertically above A, B and C respectively such that PA = 30 m, QB = 20 and

RC = 10 m.

(a) Show that PQR is an isosceles triangle. (3 marks)

(b) M and N are points on PR and PA respectively such that

QM and NM are both perpendicular to PR.

[Leave the answers of (i) and (ii) in surd form or exact value(s).]

(i) Find QM.

(ii) Find tan(APR and hence find MN.

Find tan(APQ and hence find QN.

(iii) Hence, show that the planes PQR and APRC are perpendicular to each other.

(9 marks)

8. Let L : 3x + 4y ( 10 = 0.

(a) (i) Find the distance between O and L.

(ii) Let M be a point on L such that OM ( L.

Show that the equation of the circle centred at

M with radius OM is 5x2 + 5y2 ( 12x ( 16y = 0.

(5 marks)

Let C be the circle in (a)(ii).

Suppose C cuts L at B and C and OA is a diameter of C.

(b) Find the coordinates of A. (2 marks)

Let C1 be the circle centred at O with radius OB.

Suppose C1 cuts the x-axis at D and E (as shown).

(c) (i) Find the x-coordinate of E.

(ii) Let F be a point such that AF ( DE and EF ( DA. Show that B, F and C are collinear.

(5 marks)

END OF PAPER

Marking Scheme of F.4 ADDITIONAL MATHEMATICS FINAL REVISION QUIZ 1

1. (1 ( 2x)3 ( 1 ( 3(2x) + 3(2x)2 ( (2x)3 ( 1 ( 6x + 12x2 ( 8x3 2A

(1 ( 6x + 12x2 ( 8x3)n ( (1 ( 2x)3n 1M

∵ Coefficient of x2 = 144

∴ 3nC2((2)2 = 144

[pic] = 144 1M

3n2 ( n ( 24 = 0

(n ( 3)(3n + 8) = 0

n = 3 or n = ([pic] (rejected) 2A

2. Let P(n) be the proposition that “42n + 1 + 2(7n + 1) is divisible by 9”.

When n = 1, 42n + 1 + 2(7n + 1) = 43 + 2(72) = 162 = 9(18 1

∴ P(1) is true.

Assume 42k + 1 + 2(7k + 1) = 9Q where Q is an integer 1

Consider 42k + 3 + 2(7k + 2) 1

= 16(42k + 1) + 14(7k + 1)

= 16(9Q ( 2(7k + 1)) + 14(7k + 1) 1

= 9(16Q) ( 18(7k + 1)

= 9[16Q ( 2(7k + 1)] 1

= 9Q’ where Q’ is an integer

P(k + 1) is also true.

By the principle of M.I., P(n) is true for all natural numbers n. 1

3. sinx + [pic]sin2x + sin3x = 0

sinx + sin3x +[pic]sin2x = 0

2sin2xcosx +[pic]sin2x = 0 1M

sin2x(2cosx + [pic]) = 0 1M

sin2x = 0 or cosx = ([pic] 1M

x = [pic] or x = 2n( [pic] [pic] (where n is any integer) 2A

4. (a) For 0 < x ( [pic], (*) will be

((2x ( 1) = 2x ( 1 1M

4x = 2 => x = [pic] 1A

(b) For x > [pic], (*) will be

2x ( 1 = 2x ( 1, x can be any value under the condition that x > [pic] 1

hence x > [pic].

For x ( 0, (*) will be

((2x ( 1) = (2x ( 1, there is no x satisfying this equation 1 Together with the result in (a), the solution to (*) is x ( [pic]. 1A

Alternatively, the common part of the curve y = |2x ( 1| and y = 2|x| ( 1 is corresponding to

x ( [pic], hence the solution to (*) is x ( [pic].

5.

(a) x + 2y ( 4 + k(3x ( 4y ( 10) = 0 (k is any real number) 1A

Alternatively, k(x + 2y ( 4) + (3x ( 4y ( 10) = 0 or y ( [pic] = m(x ( [pic])

(b) (1 + 3k)x + (2 ( 4k)y ( (4 + 10k) = 0

slope = [pic] 1A

tan( = [pic] 1M

[pic] = [pic]

7k ( 1 = 3((k + 3) or 7k ( 1 = (3((k + 3) 1M

k = 1 or k = (2

Hence the required lines are

[1 + 3(1)]x + [2 ( 4(1)]y ( [4 + 10(1)] = 0 and [1 + 3((2)]x + [2 ( 4((2)]y ( [4 + 10((2)] = 0

2x ( y ( 7 = 0 and 5x ( 10y ( 16 = 0 2A

6.

(a) Substitute y = ((2x + 9) into C.

x2 + (2x + 9)2 + (t + 2)x ( 4(2x + 9) + 4t = 0 1M

x2 + 4x2 + 36x + 81 + (t + 2)x ( 8x ( 36 + 4t = 0

5x2 + (t + 30)x + (4t + 45) = 0 -------------------------------- (*) 1

∵ sum of roots = ([pic] 1M

∴ x-coordinate of the mid-point of AB = ([pic] = ([pic] 1A

(b) Since L cuts C at two distinct points, ( of (*) > 0

(t + 30)2 ( 4(5)(4t + 45) > 0 2M

t2 + 60t + 900 ( 80t ( 900 > 0 => t2 ( 20t > 0

t > 20 or t < 0 2A

7.

(a) QR = [pic] = [pic] = 10[pic] m 1A

PQ = [pic] = [pic] = 10[pic] m 1A

∴ QR = PQ and hence (PQR is isosceles 1

(b)(i) PR = [pic]= [pic] = 10[pic] m

∴ PM = [pic] = 5[pic] m

QM = [pic] = [pic] = [pic] = 5[pic] m 1A

(b)(ii) tan(APR = [pic] = [pic] 1A

tan(APR = [pic] => MN = [pic](5[pic] = [pic] m 1A

tan(APQ = [pic] = 1 1A

tan(APR = [pic] => cos(APR = [pic] 1M

cos(APR = [pic] => PN = 5[pic]([pic] = [pic] m

QN 2 = PN 2 + PQ 2 ( 2(PN)(PQ)cos(APQ

QN = [pic] = [pic] m 1M+ 1A

(b)(iii) QM 2 + MN 2 = (5[pic])2 + ([pic])2 = 75 + [pic] = [pic]

QN 2 = ([pic])2 = [pic] 1A

∴ QM 2 + MN 2 = QN 2 and hence PQR ( APRC. 1

8.

(a)(i) Distance = [pic] = 2 1A

(a)(ii) 5x2 + 5y2 ( 12x ( 16y = 0 => x2 + y2 ( [pic]x ( [pic]y = 0

centre M = [pic] and 3([pic]) + 4([pic]) ( 10 = 0 ∴ M lies on L. 1A centre+1M

radius = [pic][pic] = 2 = OM 1M+1M

Hence the equation represents the circle required.

Alternatively,

∵ M lies on L ∴ let M(a , [pic]) 1M

∵OM ( L ∴ slope of OM = [pic] 1M

∴ a = [pic] and hence M([pic] , [pic]) 1A

∴ Equation of the circle is [pic] or 5x2 + 5y2 ( 12x ( 16y = 0 1M

(b) ∵ M is the mid-point of OA, let A(h , k)

∴ [pic] and [pic] 1M

∴ A[pic] 1A

(c) (i) OB2 = OM2 + MB2

OB = [pic] = 2[pic] 1M

∵ OD = OB ∴ x-coordinate of E = 2[pic] 1A

(ii) ∵AF ( DE ∴ let F[pic]

∵EF ( DA ∴ slope of EF ( slope of DA = [pic]([pic] = (1 1M

[pic]t = [pic] => t = [pic] ∴ F[pic] 1A

∵ 3([pic]) + 4([pic]) ( 10 = 0 1M

∴ F lies on BC and hence B, F and C are collinear.

(Method 2) Consider a point G such that AG ( DE. If G[pic] lies on L,

3([pic]) + 4t ( 10 = 0 => t = [pic] ; slope of EG ( slope of DA = [pic]([pic] = (1

( EG ( DA and hence F = G showing that B, F and C are collinear. 3M

(Method 3) On solving [pic], we have

B[pic] and C[pic]; slope of BF = [pic] = ([pic], slope of FC = [pic] = ([pic]

∴slope of BF = slope of FC and hence B, F and C are collinear.

1M for correct coordinates of B and C and appropriate way of showing B, F, C being collinear

END OF SOLUTION

F.4 ADDITIONAL MATHEMATICS FINAL REVISION QUIZ 2

Time allowed: 1[pic] hours

FORMULAS FOR REFERNCE

|[pic] |[pic] |

Section A (42 marks)

1. Solve |x| = 6 ( x2 .

(4 marks)

2. Let F : x2 + y2 + (k + 2)x ( (2k + 6)y + (4k ( 26) = 0 be a family of circles.

(a) Show that the radius of circles in F is [pic].

(b) Find the radius of the smallest circle in F.

(5 marks)

3. By using mathematical induction, show that

[pic][pic]

for any positive integer n.

(7 marks)

4. If the acute angle between x ( 2y = 0 and ax + 3y = 0 is ( where sin( = [pic] ,

evaluate a.

(6 marks)

5. Let y = [pic], determine [pic] from first principles.

(5 marks)

6. (a) Show that 4sin(sin(60( + ()sin(60( ( () ( 3sin( ( 4sin3( .

(b) Given that sin3( ( 3sin( ( 4sin3( . Give the general solution to the following equation

sin(sin(60( + ()sin(60( ( () = [pic].

(7 marks)

7. Let C1 : [pic] and C2 : [pic].

(a) Suppose the graphs of C1 and C2 intersect at A and B, determine the equation of AB.

(b) Let L : [pic]. Suppose L is parallel to the line AB.

Suppose L cuts the circle [pic] at C and D.

(i) Find the mid-point of C and D in terms of c.

(ii) Show that, no matter what value of c is, the mid-point of C and D must lie on the same

straight line. (8 marks)

Section B (24 marks)

8. In the figure, the circle C1 with centre at

A(2,3) touches the line L : 3x + 4y ( 8 = 0.

L’ is a straight line passing the origin and

also touches C1.

(a) Find the equation of C1.

(2 marks)

(b) Show that C1 touches the y-axis.

(2 marks)

(c) Find the equation of L’.

(4 marks)

(d) If C2 is another circle with the same

radius as C1 and touches L and L’

as shown in the figure, find the equation

of C2 and give your answer in the form

of (x ( h)2 + (y ( k)2 = r2 .

(4 marks)

9. The following figure shows two identical triangular planes ADC and ADB placed on the horizontal

table such that the edges AB and AC lie on the table and DA is vertical. Given that AD = 2, AB = 4

and the angle between the planes ABC and DBC is 30(

(a) Show that (ABC is equilateral.

(4 marks)

(b) M is a point on AC such that BM is perpendicular

to AC. N is the mid-point of AD.

(i) Find (, the angle between the planes DMB and

the ground ABC.

(ii) Let ( be the angle between NMB and the ground

ABC. A student claimed that the angle between

the planes NMB and DMB is ( ( (.

Explain whether the student’s claim is correct.

(iii) R is a point on CD such that the planes NMB and RMB are perpendicular. Find (CRM.

(8 marks)

END OF PAPER

Marking Scheme of F.4 ADDITIONAL MATHEMATICS FINAL REVISION QUIZ 2

1. |x| = 6 ( x2

x2 + |x| ( 6 = 0 => |x|2 + |x| ( 6 = 0 1M

(|x| ( 2)(|x| + 3) = 0 1M

|x| = 2 or |x| = (3 (rejected) 1M

x = (2 1A/4

Alternatively, the equation is equivalent to x = 6 ( x2 or (x = 6 ( x2 . 1M

x = 6 ( x2 => x = 2 or (3 (rejected ∵ RHS < 0) 1M

(x = 6 ( x2 => x = (2 or 3 (rejected ∵ RHS < 0) 1M

Therefore, x = (2 1A/4

2.

(a) radius = [pic] 1M

= [pic]

= [pic] 1/2

(b) 5k2 + 12k + 144 = 5(k2 + [pic]k) + 144

= 5[k2 + [pic]x + ([pic])2 ( ([pic])2] + 144 1M

= 5(k + [pic])2 ( [pic] + 144 = 5(k + [pic])2 + [pic] 1A

( Radius attains its minimum when k = ([pic]

Thus, radius of smallest circle = [pic] = [pic] 1A/3

Alternatively, the center of the smallest circle lies on 2x ( 6y + 4 = 0, hence

2(([pic]) ( 6([pic]) + 4 = 0 2M

k = ([pic]

Thus, radius of the smallest circle = [pic] = [pic] 1A/3

3. Let P(n) be the proposition.

When n = 1, LHS = [pic] = (1, RHS = 1 ( [pic] = (1 1

∴ P(1) is true.

Suppose P(k) is true. 1

Consider P(k + 1).

LHS = [pic] 1M

= 1 ( [pic] + [pic] [M.I. assumption] 1M

= 1 + [pic]

= 1 + [pic] 1M

= 1 ( [pic] 1

∴ P(k + 1) is also true.

By the principle of mathematical induction, P(n) is true for any positive integer n. 1/7

4. sin( = [pic] => tan( = [pic] 1A

tan( = [pic] 2M

[pic] = [pic]

[pic] or [pic] 1M

6 + 4a = 6 ( a or 6 + 4a = a ( 6

a = 0 or (4 2A/6

5. [pic] 1M

= [pic]

= [pic] 1M

= [pic]

= [pic] 1M

= [pic] 1M

= ([pic] 1A/5

6.

(a) 4sin(sin(60( + ()sin(60( ( ()

( 4sin( (([pic])[cos(60( + ( + 60( ( () ( cos(60( + ( ( (60( ( ())] 1M

( (2sin( (cos120( ( cos2() 1M

( (2sin( [([pic] ( (1 ( 2sin2()] ( 3sin( ( 4sin3( 1M/3

(b) sin(sin(60( + ()sin(60( ( () = [pic]

4sin(sin(60( + ()sin(60( ( () = [pic] 1M

3sin( ( 4sin3( = [pic]

sin3( = [pic] 1M

3( = 180(n + ((1)n(30( 1M

( = 60(n + ((1)n(10( (where n is any integer) 1A/4

7.

(a) AB is a common chord of two circles, its equation can be found by C1 ( C2 = 0, i.e.

([pic]) ( ([pic]) = 0 1M

[pic] 1A

(b) (i) By (a), m = 2

Substitute [pic] into [pic], yield 1M

[pic]

[pic] - - - - - - (*) 1

Let (h , k) be the mid-point of C and D, then

[pic] - - - - - - (1) 1M

∵(h , k) lies on [pic]

( [pic] - - - - - - (2)

Hence the mid-point of C and D is [pic] 1A

(b) (ii) (1) ( [pic]

(2) ( [pic]

Hence [pic] ( [pic] 1M

Thus, [pic] must lie on [pic] 1A

Alternative method, centres must lie on the perpendicular bisector of the chord CD, done.

8.

(a) Radius of C1 = [pic] = 2 1A

Equation of C1 is (x ( 2)2 + (y ( 3)2 = 22

x2 + y2 ( 4x ( 6y + 9 = 0 1A/2

(b) Perpendicular distance from A(2,3) to the y-axis = 2 = Radius of C1 1M

( C1 touches the y-axis 1/2

(c) Since L’ passes through the origin, we can let the equation of L’ be

y = mx => mx ( y = 0 1M

[pic] = 2 1M

(2m ( 3)2 = 4(m2 + 1)

4m2 ( 12m + 9 = 4m2 + 4 1M

( m = [pic] Hence the equation of L’ is y = [pic]x. 1A/4

(d) To determine the point of intersection of L and L’, we solve [pic]

3x + 4([pic]x) ( 8 = 0 1M

[pic]x = 8

x = [pic] and y = [pic]([pic] = [pic]

( The point of intersection of L and L’ is ([pic],[pic]) 1A

Let (a,b) be the centre of C2, then [pic] 1M

Hence a = [pic] , b = ([pic]

( Equation of C2 is [pic] 1A/4

9.

(a) Y is a point on BC such that YD ( BC and YA ( BC.

Then, angle between ABC and DBC = (AYD = 30(. 1M

In (DAY , tan30( = [pic] = [pic] 1M

AY = 2[pic]

In (BYA, BY = [pic] = 2 1M

( BC = 4

((ABC is equilateral. 1/4

(b)(i) Note that DM ( MB and AM ( MB, hence ( = (DMA. 1M

In (DMA, tan(DMA = [pic] = [pic] = 1 1M

( (DMA = 45(

(( = 45( 1A/3

(b)(ii) Note that NM ( MB and AM ( MB, hence ( = (NMA.

Angle between planes NMB and DMB = (DMN 1M

= (DMA ( (NMA

= ( ( ( 1M

( The student’s claim is correct. 1/3

(b)(iii) If the planes NMB and RMB are perpendicular,

then (NMR = 90( 1M

As AM = MC and AN = ND, by mid-point theorem,

we have MN // CD.

( (CRM = 90( 1/2

END OF SOLUTION

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