o-level additional mathematics mock ... - sap …sapgrp.com/shop/images/pdf/3210066_preface.pdf ·...

DURATION: 2 hours

INSTRUCTIONSTOCANDIDATES

Answer all questions.

The number of marks is given in brackets [ ] at the end of each question or part question.The total number of marks in this paper is 80.

The use of approved scientific calculator is expected, where appropriate.If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.For �, use either your calculator value or 3.142, unless the question requires the answer in terms of �.

(A list of mathematical formulae is given on the last page of the solutions booklet.)

O-LEVELadditiOnaL MatHEMatiCS

MOCK EXaMinatiOn 1PaPEr 1

Name Class Date

80

01 MExam 1_OL Add Math MExam.ind1 1 8/1/2016 10:26:15 AM

O-Level Additional Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL rIGhTs rEsErvEd.

�Mock Examination 1 (Paper 1)

1 (a) Are the points A(–130, – 127), B(2, 5), C(7, 10) and D(100, 97) collinear? [3] (b) Prove that the points A(0, 3), B(4, 0), C(1, –4) and D(–3, –1) are the vertices of

a square. [4]

� Solve the following equations. (i) 2x + 5 √

__ x – 3 = 0 [3]

(ii) 2(lg x)2 + lg x5 – 3 = 0 [3] (iii) 2e2x + 5ex – 3 = 0 [3]


DURATION: 2 hours 30 minutes







MOCK EXaMinatiOn 1PaPEr �

Name Class Date

100



10Mock Examination 1 (Paper 2)

1 Find the equation of the circle if A(–3, –4), B(–3, 6) and C (8, 6) are on the circumference of the circle. [5]

� Sketch the graph of y = –2 sin 2x – 1 for –� < x < �. [5]


DURATION: 2 hours






Name Class Date



80




1 The equation of a curve is y = 1 + x _____ 1 – x . Find the values of d2y

___ dx2 when the rate at which y

is changing is twice the rate at which x is changing. [5]

2 Evaluate ∫ (ex + 1)2

_______ ex – 1 dx. [4]


Name Class Date









100




1 (a) Without using a calculator, find the value of

( 4 0 ) – ( 4 1 ) 0.5 + ( 4 2 ) 0.25 – ( 4 3 ) 0.125 + ( 4 4 ) 0.0625. [4]

(b) Simplify ( 1 + √_____

x + 1 ) 4 + ( 1 – √_____

x + 1 ) 4 . [4]

2 (a) Without using a graph, solve the equation ||5x – 5| – 5| = 10. [4] (b) solve the inequality 6 – x – x2 < 0. [3]


DURATION: 2 hours






Name Class Date



80




1 (i) solve the simultaneous inequalities. [4] (x – 1)2 36 | x – 6 | < 5 (ii) Sketch the graphs of y = (x – 1)2 and y = |x – 6| on the same diagram and label

your answers in (i). [4]

� (a) Prove that sin2 4A = 16 sin2 A(1 – sin2A)(1 – 2 sin2A)2. [4]

(b) Prove that √_________

1 + sin 2A = sin A + cos A. [4]


Name Class Date









100




1 Solve the simultaneous equations. [5]

10 ___ x + y __ 6 = 7

4 __ x + y __ 4 = 5

� Simplify 1 _________ ( √__

x – √__

y )3 – 1 ___________ ( √__

x )3 – ( √__

y )3 . [5]


Name Class Date

DURATION: 2 hours








80




1 (i) Find the value of m for which the coefficient of x3 in the expansion of (1 + x)5 + (3 + mx)4 + (2 – x)3 is 105. [4] (ii) What is the coefficient of x4 in the expansion? [2]

� Prove that cos A – sin A ___________ sin A + cos A = tan 2A – sec 2A. [4]


Name Class Date









100




1 Expand (x3 + x2 + x)3. [6]

� Prove that 1 + tan2A ________ 1 – tan2A = sec 2A. [4]


Name Class Date

DURATION: 2 hours








80




1 Solve the simultaneous equations. [5]

2 __ x + y = 9

x + 5 __ y = 3 __ 2

� (a) Express sin 8A in terms of sin A. [4] (b) Express cos 4A in terms of cos A. [4]


Name Class Date









100




1 Find the coefficient of x2 in the expansion of (1 – x)5(2 + x)4. [5]

� (i) differentiate ln(sin x) with respect to x. [3]

(ii) hence, evaluate ∫ 0.1 0.2 (cot 3x + sec2x) dx. [4]


Solutions to O-Level Additional Mathematics Mock Examinations© Singapore Asia Publishers Pte LtdALL rightS rESErvEd.

S�Mock Examination 1 (Paper 1)

mock examination �

Paper �1 (a) Equation of the straight line BC:

y – 5 _____ x – 2 = 10 – 5 ______ 7 – 2

y – 5

_____ x – 2 = 1

y – 5 = x – 2 y = x + 3 When x = –130, y = –130 + 3 = –127 Points A, B and C are collinear. When x = 100, y = 100 + 3 = 103 ≠97 Points A,B,CandDarenotcollinear.

(b) Length of AB = √______________

(0 – 4)2 + (3 – 0)2 = 5 Length of BC = √

_______________ (4 – 1)2 + (0 + 4)2

= 5 Length of CD = √

________________ (1 + 3)2 + (–4 + 1)2

= 5 Length of DA = √

________________ (–3 – 0)2 + (–1 – 3)2

= 5 Since all the sides are equal in length, ABCD is either

a rhombus or a square. gradient of AB × gradient of BC

= 0 – 3 _____ 4 – 0 × ( –4 – 0 ______ 1 – 4 ) = – 3 ___ 4 × –4 ___ –3

= –1 AB is perpendicular to BC ⇒ ∠ABC=90º ABCD is a square. Points A, B, C and D are the vertices of a square.

(proven)

2 (i) 2x + 5 √_ x – 3 = 0

2x – 3 = –5 √_ x

Squaring both sides, 4x2 – 12x+9=25x 4x2 – 37x+9=0 (x–9)(4x – 1) = 0 x–9=0 or 4x – 1 = 0

x=9 x = 1 __ 4

x=1__4,9

(ii) 2(lg x)2 + lg x5 – 3 = 0 2(lg x)2 + 5 lg x – 3 = 0 (2 lg x – 1)(lg x + 3) = 0 2 lg x – 1 = 0 or lg x + 3 = 0 2 lg x = 1 lg x = –3

lg x = 1 __ 2 x = 10–3

x = 100.5 = 0.001 ≈3.16 x=0.001,3.16

(iii) 2e2x + 5ex – 3 = 0 2(ex)2 + 5ex – 3 = 0 (2ex – 1)(ex + 3) = 0 2ex – 1 = 0 or ex + 3 = 0 2ex = 1 ex = –3 (N.A.) ex = 1 __ 2

x = ln 1 __ 2

≈–0.693 x=–0.693

3 42 × 5x – 2 = 10x – 2

42 · 5x

_____ 52 = 10 x___ 102

42 × 102

_______ 52 = 10 x___ 5x

64=2x

26 = 2x

x =6

4 (i) Let f(x) = x4 + x3 – 13x2 – x + 12 When x = 1, f(1) = 1 + 1 – 13 – 1 + 12 = 0 When x = –1, f(–1) = 1 – 1 – 13 + 1 + 12 = 0 (x + 1)(x – 1) or x2 – 1 is a factor of f(x). By inspection, x4 + x3 – 13x2 – x + 12 = (x2 – 1)(x2 + x – 12)

–x2

–12x2 _____ –13x2

(x2 – 1)(x2 + x – 12) = 0 (x – 1)(x + 1)(x + 4)(x – 3) = 0 x=–4,–1,1,3 (ii) tan4 x + tan3 x – 13tan2 x – tan x+ 12 = 0 From (i), (tan x – 1)(tan x + 1)(tan x + 4)(tan x – 3) = 0 When tan x = –4, x=180º–75.96º,360º–75.96º ≈104.0º,284.0º When tan x = –1, x=180º–45º,360º–45º =135º,315º When tan x = 1, x=45º,180º+45º =45º,225º When tan x = 3, x=71.57º,180º+71.57º ≈71.6º,251.6º x=45º,71.6º,104.0º,135º,225º,251.6º,284.0º,315º

5 y = x3 + x2 + 3

dy

___ dx = 3x2 + 2x When x = 2,

dy___ dx = 3(22) + 2(2)

=16 At Q, 3x2 + 2x = 1 __ 2×16 3x2 + 2x = 8 3x2 + 2x – 8 = 0 (3x – 4)(x + 2) = 0 When 3x – 4 = 0,

x = 4 __ 3 = 1 1 __ 3

y = ( 4 __ 3 ) 3 + ( 4 __ 3 ) 2 + 3

= 7 4 ___ 27 When x + 2 = 0, x = –2 y = (–2)3 + (–2)2 + 3 = –1 the coordinates of Q are(11__3,74___27)or(–2,–1).

SoLUtionS

06 Ans_OL Add Math MExam.indd 1 8/1/2016 10:24:33 AM

o-level additional mathematics mock ... - sap …sapgrp.com/shop/images/pdf/3210066_preface.pdf ·...

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