WRP Solutions Extension 1 2012

WESTERN REGION

2012

Preliminary Final

EXAMINATION

Mathematics - Extension 1

SOLUTIONS

Preliminary Examination – Mathematics Extension 1 2012

Multiple Choice Answer Sheet

Name ______________________________

Completely fill the response oval representing the most correct answer.

1. A B C D

2. A B C D

3. A B C D

4. A B C D

5. A B C D

6. A B C D

7. A B C D

8. A B C D

9. A B C D

10. A B C D

Question 11 Preliminary HSC Trial Examination-

Mathematics Extension 1

2012

Part Solution Marks Comment

(a) 16

4 – x=

1

8x

2

44 – x

= 2-3x

216 – 4x

= 2-3x

16 – 4x = -3x

x = 16

2

1

1

(b)

3y = 2x + 4 y = 3x + 7

y = 23

x + 43

m2 = 3

m1 = 23

tan = m1 – m2

1 + m1 m2

tan =

2

3– 3

1 + 23 3

tan = – 7

9

= 37°52 = 38°(nearest degree)

Obtuse angle is 142°

2

1 for correctly

substituting

gradients into

formulae.

1 for obtuse

angle

(c)

x2– 4x

> 0

x2x

2– 4

x > 0 x

2

x(x – 2)(x + 2) > 0

– 2 < x < 0, x > 2

2

1

1 for correct

solution



2012


(d)

y = uv + vu

y = x 1

2 x – 1 + x – 1 1 at x = 5, y = 10

m =5

2 4+ 4

m = 134

y – 10 = 134

(x – 5)

4y – 40 = 13x – 65

13x – 4y – 25 = 0

2

1 for gradient

1 for equation

(e) P(2) = 23

+ 22– a = 4 P(x) = x

3+ x

2– 8

12 – a = 4 P(0) = -8

a = 8

2 1 for a

1 for P(0)

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2012


(a) (i)

(ii)

(iii)

16C8 = 12870

6C410C4 = 3150

5C39C4 = 1260

1

1

1

(b) Couples sitting together = 4! 24

= 384

Total arrangements of Ben and Gaby not as couple

= 5! 23

= 960

Ben and Gaby apart = 960-384 =576

2 1 for

arrangements

sitting

together

1 for answer

(c)

+ + = – a1

= – a

1

+

1

+

1

= + +

= b

– ab= – 1

a

( + + ) 1

+

1

+

1

= – a – 1

a= 1

2

1

1

Or any other

correct

alternate

method

(d)(i)

(ii)

x4

+ 4 x

2+ a

2

– (bx)2

x4

+ 2a – b

2x

2+ a

2

2a – b2

= 0 equating coeff of x2

a2

= 4 constant term

from Since b2

= 2a, a = 2 since a > 0

b = ±2

x4

+ 4 x

2+ 2

2

– (2x)2

=

x

2+ 2

+ 2x

x

2+ 2

– 2x

= x

2+ 2x + 2

x

2– 2x + 2

2

1

1 for a

1 for b

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2012


(a)

prove true for n = 1

72– 3

3= 49 – 27 = 22 div by 11

assume true for n = k

72k

– 33k

= 11M where M is an integer

72k

= 11M + 33k

prove true for n = k+1

72(k + 1)

– 33(k + 1)

= 72k + 2

– 33k + 3

= 49 72k

– 27 33k

= 4911M + 3

3k– 27 3

3k

= 11 49M + 49 33k

– 27 33k

= 11 49M + 22 33k

= 1149M + 2 3

3k

= 11P where P is an integer true for n=k+1

It is true when n=1 and so it is true when n=2 and so on.

Hence it is true for all n, a positive integer.

3

1

1

1



2012


(b)

(i)

The exterior angle of a triangle ( ALC) is equal to the sum of the

opposite interior angles.

3

1

(ii) ABC = BAC = (base angles of isosACB)

AYC = ABC = (angles standing on the same arc are equal )

AYX = ACX = (angles standing on the same arc are equal )

1

(iii) XYM = + (all shown above)

= CLB

= + YMLX is a cyclic quadrilateral.

(exterior angle of cyclic quad is equal to opp interior angle)

1



2012


(c)

let YXN = ZXN = (given)

let MXZ =

MXZ = XYN ( in alt segment equal to

= between chord and tangent )

XNZ = + (ext of XYN equal to sum of interior opp s)

MXN = MXZ + ZXN (adjacent s)

= +

MXN = XNZ (bases isosceles MXN)

MX = MN (sides of isosceles )

4

1

1

1

1

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2012


(a)

tan 60° = CBh

tan 30° = DBh

CB = htan 60° DB = htan 30°

CD2

= CB2

+ BD2– 2CB. BD cos 110°

CD2

= htan 60°

2

+ htan 30°

2

– 2htan 60°

htan 30°

cos 110°

CD2

= h2tan

2 60° + tan

2 30° – 2tan 60°tan 30°cos 110°

h2

= 8502

4.01737362

h2

= 179843.8653

h = 424.08metres

3

1 for CB

and DB

1 for Cosine

Rule

1 answer



2012


(b)

LHS= tan

45° + 2

=

tan 45° + tan 2

1 – tan 45°tan 2

= 1 + t1 – t

since t = tan 2

RHS= 1 + sin

cos =

1 +2t

1 + t2

1 – t2

1 + t2

=

1 + t2

+ 2t

1 + t2

1 – t2

1 + t2

= (1 + t)2

(1 – t)(1 + t)

= 1 + t1 – t

tan

45° + 2

=1 + sin

cos

3

1 for

definition

1 for

definitions

1 for

simplificati

on



2012


(c)(i)

1 – cos 2

1 + cos 2 =

1 – 1 – 2sin

2

1 + 2cos

2 – 1

=2sin

2

2cos2

= tan2

2

1

1

(ii)

tan2 22 1

2° =

1 – cos 45°

1 + cos 45°

=

1 –1

2

1 +1

2

=

2 – 1

2

2 + 1

2

=2 – 1

2 + 1

2 – 1

2 – 1

tan2 22 1

2° =

2 – 12

2 – 1

=

2 – 12

tan 22 12

° = 2 – 1

2 1

1

Or alternate

method eg

forming

quadratic

equation

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WRP Solutions Extension 1 2012

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