81
Term Exam Paper Kit F.5 First Term Examination Mathematics (Paper 1) Name: Class: No.: Time Allowed: 90 minutes This paper consists of 3 sections. (Write your answers in the spaces provided.) Total Marks: 70 Answer ALL questions in this paper. Unless otherwise specified, all working steps MUST be clearly shown and numerical answers should be either exact or correct to 3 significant figures. Section A(1) (23 marks) 1. Solve x 4 13x 2 + 36 = 0. (3 marks) 2. If , find the values of real numbers a and b. (3 marks) NSS Mathematics in Action 1 © Pearson Education Asia Limited 2010 Marks

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Page 1: Chapter 1nsscore.smaths.ilongman.com/files/filemanagermodule/128460…  · Web viewFind the equation of the tangent to the circle C: (x ( 3)2 + (y + 1)2 = 10 at P(2, 2). A. x ( 3y

Term Exam Paper Kit

F.5 First Term Examination Mathematics (Paper 1)

Name: Class: No.:

Time Allowed: 90 minutesThis paper consists of 3 sections. (Write your answers in the spaces provided.)Total Marks: 70Answer ALL questions in this paper.Unless otherwise specified, all working steps MUST be clearly shown and numerical answers should be either exact or correct to 3 significant figures.

Section A(1) (23 marks)

1. Solve x4 13x2 + 36 = 0. (3 marks)

2. If , find the values of real numbers a and b.

(3 marks)

NSS Mathematics in Action 1 © Pearson Education Asia Limited 2010

Marks

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NSS Mathematics in Action

3. In the figure, O is the centre of the circle. AB is the tangent to the circle at Q. Find x and y.

(4 marks)

4. In the figure, BD and CA are two parallel chords of the circle. CD

intersects AB at F and . Find x.

(4 marks)

5. Consider all the circles passing through P(6, 7) and Q(–2, 3). Find the equation of the locus of their centres.

(4 marks)

NSS Mathematics in Action 2 © Pearson Education Asia Limited 2010

x

A

D

B

C

F

60

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Term Exam Paper Kit

6. The figure shows the circle with

centre C. It cuts the x-axis at A and B. Find (a) AB, (2 marks) (b) CAB. (3 marks)(Give your answers correct to 3 significant figures if necessary.)

(a)

(b)

Section A(2) (23 marks)

7. (a) It is known that , where a and b are constants. Find the values of a and b.

(3 marks)

NSS Mathematics in Action 3 © Pearson Education Asia Limited 2010

O

y

x

C

A B

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NSS Mathematics in Action

(b) Hence, solve x(x + 1)(x + 2)(x + 3) = 24. (Leave your answers in surd form if necessary.)

(4 marks)

8. The figure shows a semi-circle ABFE. AE and BF are produced to meet at D. (a) Show that △ABD ~ △FED.

(2 marks)

(b) If ED = EF, show that E is the mid-point of AD.(3 marks)

NSS Mathematics in Action 4 © Pearson Education Asia Limited 2010

A B

F

D

E

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Term Exam Paper Kit

9. In the figure, AB passes through the centre O and cuts the circle at E. CA and CB are tangents to the circle at D and B respectively. Find the lengths of(a) DA,

(2 marks)

(b) CD.(2 marks)

10. A straight line L passes through the point (2, 1) and has equal intercepts.(a) Find the equation of L.

(3 marks)

NSS Mathematics in Action 5 © Pearson Education Asia Limited 2010

AE

O

B

C

D1 cm

8 cm

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NSS Mathematics in Action

(b) Find the equations of the tangents to the circle x2 + y2 6x 2y – 8 = 0 which is perpendicular to L.

(4 marks)

Section B (24 marks)

11. (a) Expand (x + 1)4.(2 marks)

(b) Using the result of (a), answer the following questions.(i) If is a root of , find the possible values of the real

number b.(3 marks)

NSS Mathematics in Action 6 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

(ii) Solve .(Leave the radical sign ‘ ’ in your answers if necessary.)

(3 marks)

12. In Figure (1), a rectangle OPQR is inscribed in the circle. In Figure (2), a rectangular coordinate system is introduced in Figure (1) so that the coordinates of O, P and R are (0, 0), (8, 0) and (0, 6) respectively. C is the centre of the circle.

R

O P

Q

y

R

O P

Q

C

x

Figure (1) Figure (2)

(a) Find the coordinates of Q and C.(2 marks)

(b) Find the equations of the tangents to the circle at (i) P, (ii) R.

(2 marks) (2 marks)

(c) A student claims that a quadrilateral formed by the four tangents at O, P, Q and R is also a

NSS Mathematics in Action 7 © Pearson Education Asia Limited 2010

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NSS Mathematics in Action

rectangle. Is the student correct? Explain your answer.(2 marks)

13. In the figure, P(a, b) is a point on the straight line L1: x + y 2 = 0 and Q is a point on the y-axis such that PQ is horizontal. M1 is the mid-point of PQ.(a) Express a in terms of b.

(1 mark)

(b) It is given that when P moves along L1, the locus of M1 is a straight line L2. Find the equation of L2.

(3 marks)

NSS Mathematics in Action 8 © Pearson Education Asia Limited 2010

P(a, b)

y

O

M1

Q

x

L1

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Term Exam Paper Kit

(c) If the straight line L2 in (b) is a tangent to a circle centred at the origin, find the equation of the circle.

(4 marks)

End of paper

NSS Mathematics in Action 9 © Pearson Education Asia Limited 2010

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NSS Mathematics in Action

F.5 First Term ExaminationMathematics (Paper 1 - Extra Questions)

Section A(1) (23 marks)

1. Find the real root(s) of .

(3 marks)

2. Simplify and express each of the following in the form .

(a) (3 + 4i)(3 – 4i) (2 marks) (b) (2 marks)

3. In the figure, ACDE is inscribed in the circle. AE and CD are produced to meet at B.(a) Name a pair of similar triangles and give the reason.

(2 marks)(b) If BE = 3 cm, EA = 4.5 cm and BD = 4 cm, find CD.

(2 marks)

4. The figure shows two concentric circles with common centre O. RS is the tangent to the smaller circle at P. The radii of the larger circle and the smaller circle are 15 cm and 12 cm respectively. Find the length of RS.

(4 marks)

5. Given two points A(1, 3) and B(–3, 1), find the equation of the locus of a point P such that .

(3 marks)

6. It is given that the equation of a circle S is x2 + y2 + 4x 2y 5 = 0.(a) Show that P(1, 2) lies on S.

(1 mark)(b) Find the equation of the tangent to S at P.

(4 marks)

NSS Mathematics in Action 10 © Pearson Education Asia Limited 2010

A C

D

E

B

O

R P S

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Term Exam Paper Kit

Section A(2) (23 marks)

7. (a) Factorize x4 + 6x2 + 9. (1 mark)

(b) Hence, solve x4 + 5x2 + 9 = 0.(5 marks)

8. The figure shows a right-angled triangle ABC and a square ACDE. AOD and EOC are diagonals of the square.(a) Show that A, B, C and O are concyclic.

(2 marks)(b) Hence, show that OB bisects ABC.

(3 marks)

9. In the figure, AEB and ED are tangents to the circle at B and D respectively. ADC is a straight line.(a) Show that △ABD ~ △ACB.

(2 marks)(b) If AD = ED = 5 cm and DC = 4 cm, find the lengths of

(i) AB, (ii) AE.(2 marks) (2 marks)

(Give your answers correct to 3 significant figures.)

10. (a) In the figure, A(0, a) and B(b, 0) lie on the y-axis and the x-axis respectively. M(x, y) is the mid-point of AB.(i) Express a in terms of y, and b in terms of x.

(2 marks)(ii) If A and B move along the y-axis and the x-axis

respectively such that AB = 5 units, find the equation of the locus of M.

(2 marks)(b) In the figure, a ladder PQ of length 5 m, which stands vertically at

initial, is sliding away from a vertical wall. Describe and sketch the locus of the mid-point of the ladder.

(2 marks)

NSS Mathematics in Action 11 © Pearson Education Asia Limited 2010

B

CDA

E

P

Q

A B

C

D

E O

O

A(0, a)

y

B(b, 0)x

M(x, y)

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NSS Mathematics in Action

Section B (24 marks)

11. John and Mary try to solve the equation with given a and b. Both of them copy the equation wrongly. John copies it as while Mary copies it as

. Assume that the solutions for John’s and Mary’s equations are x = 4.76 and

x = 7.76 respectively.(a) Show that

(i) , (ii) . (2 marks) (2 marks)

(b) If a + b > 0, find the values of a and b.(4 marks)

12. In the figure, L1 and L2 are tangents to the circle S: (x 3)2 + (y 1)2 = 5 at D and E respectively. L1 and L2

intersect at the origin.(a) Find the equations of L1 and L2.

(4 marks)(b) Find ODE.

(4 marks)

13. The figure shows a parabola P: .

(a) Show that A(2a, a2) lies on P.(1 mark)

(b) It is given that M is the mid-point of A and B(0, 2). (i) Find the equation of the locus of M when A moves along P.

(3 marks)(ii) Sketch the locus of M on the figure. What kind of curve is it?

(3 marks)(iii) If N is the mid-point of A and C(0, 4), write down the equation of the locus of N

when A moves along P.(1 mark)

NSS Mathematics in Action 12 © Pearson Education Asia Limited 2010

O

y L1

L2

x

E

D

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Term Exam Paper Kit

F.5 First Term ExaminationMathematics (Paper 2)

Time Allowed: 1 hour***************************************************************************Instructions:(I) There are 36 questions in this paper and each question carries equal mark. Answer ALL

questions. (II) The diagrams in this paper may not be drawn to scale.***************************************************************************

1. Solve .

A. x = 0B. x = 0 or 1C. x = –1 or 1D. x = –1, 0 or 1

2. Solve .

A. x = 3 or 3

B. x = or

C. x = or

D. x = or

3. Solve (x2 2x)2 2x2 + 4x 3 = 0.A. x = 1, 1 or 3B. x = 1, 1 or 3C. x = 1, 3 or 3D. x = 1, 1, 3 or 3

4. Solve .

A. x = 9.5B. x = 4C. x = 4 or –5D. x = –4 or 5

NSS Mathematics in Action 13 © Pearson Education Asia Limited 2010

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NSS Mathematics in Action

NSS Mathematics in Action 14 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

5. Solve . A. x = a or bB. x = a2 or b2

C. x = a or bD. x = a, b, a2 or b2

6.A.B.C.D.

7. If , find the values of a and b.

A. a = 2 and b = 6B. a = 2 and b = 9C. a = –2 and b = 6D. a = 2 and b = –9

8. If is a real number, what is the value of a?

A. 3B. 3C. 6D. 6

9. In the figure, O is the centre of the circle. BOD is a diameter. If

, find .

A. 4 : 3B. 3 : 4C. 3 : 2D. 2 : 3

NSS Mathematics in Action 15 © Pearson Education Asia Limited 2010

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NSS Mathematics in Action

10. In the figure, AD and BC are produced to meet at E. Find CDE.A. 50B. 52C. 56D. 60

11. In the figure, O is the centre of the circle. AB = BC and AB BC. If the radius of the circle is 5 cm, find the length of AB, correct to the nearest integer.A. 6 cmB. 7 cmC. 8 cmD. It cannot be determined.

12. In the figure, AB and CE intersect at D, and AB CE. If BD = 3 cm, CD = 3.6 cm and AD = 1.4 cm, find the length of AE, correct to 3 significant figures.A. 1.82 cmB. 1.72 cmC. 1.62 cmD. 1.52 cm

13. In the figure, O is the centre of the circle. AOD and BOC are diameters. Find CED.A. 80 B. 50C. 40D. 20

NSS Mathematics in Action 16 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

14. In the figure, AB = BC = CD. Find ACB.A. 30B. 35C. 36D. 40

15. In the figure, and . Find CBD.

A. 50B. 60C. 70D. 80

16. In the figure, O is the centre of the two concentric circles. AOEB and AFDC are straight lines. If OF AC, which of the following must be true?I. △ADE ~ △ACBII. △AFO ~ △ACBIII. △AFO ~ △ADEA. I onlyB. II onlyC. II and III onlyD. I, II and III

17. In the figure, O is the centre of the circle. CB is the tangent to the circle at A and BO CO. Find the radius of the circle, correct to 3 significant figures.A. 6.50 cmB. 6.34 cmC. 5.36 cmD. 4.24 cm

NSS Mathematics in Action 17 © Pearson Education Asia Limited 2010

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NSS Mathematics in Action

18. In the figure, AB and AD are tangents to the circle at B and C respectively. If AB // CE, find x.A. 66B. 64C. 61D. 48

19. In the figure, CA and CB are tangents to the circle at A and B respectively. Find ADB.A. 130B. 100C. 65D. 60

20. In the figure, BA is the tangent to the circle at A. BC cuts the circle at D. If BA = 5 cm, BD = 3 cm and DA = 4 cm, find the radius of the circle, correct to 1 decimal place.A. 3.3 cmB. 3.5 cmC. 3.8 cmD. It cannot be determined.

21. In the figure, O is the centre of the circle. FG is the tangent to the circle at A. AOE and COD are diameters of the circle. Find BCD.A. 26B. 28C. 30D. 31

NSS Mathematics in Action 18 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

22. In the figure, two circles touch each other externally at B. CF, CG and DE are their common tangents, where H, K, M, N and B are the points of contact. FG is the tangent to the larger circle at A. Find the length of DE.A. 3.2 cmB. 3.3 cmC. 4.1 cmD. 5.4 cm

23. In the figure, O is the centre of the circle. CA and CE are tangents to the circle at A and B respectively. AODE is a straight line. Which of the following are true?I. BD // COII. △BDE ~ △COEIII. A, C, B and O are concyclic.A. I and II only B. II and III onlyC. I and III onlyD. I, II and III

24. What is the locus of the mid-points of all vertical chords of a circle?A. a circleB. a line segmentC. a pair of parallel linesD. a parabola

25. In the figure, a wheel is rolling along the horizontal ground. Which of the following may represent the locus of the centre of the wheel?

A.

B.

C.

NSS Mathematics in Action 19 © Pearson Education Asia Limited 2010

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NSS Mathematics in Action

D.

NSS Mathematics in Action 20 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

26. The figure shows a square with side 10 cm. A circle with radius 2 cm rolls inside along the sides of the square. Which of the following may represent the locus of the centre of the circle?

3 cm

10 cm

A.

3 cm

3 cm

B.

6 cm

C.

8 cm

6 cm

D.

6 cm

4 cm

27. Find the equation of the locus of a point P which maintains an equal distance from A(3, 4) and the origin.A.B.C.D.

NSS Mathematics in Action 21 © Pearson Education Asia Limited 2010

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NSS Mathematics in Action

28. Find the equation of the locus of a point P which maintains an equal distance from the origin and L: y = 4. A.B.C.D.

NSS Mathematics in Action 22 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

29. The figure shows a parabola P: y = 2 x2. A vertical line L cuts P and the x-axis at A and B respectively. If M is the mid-point of AB, find the equation of the locus of M when L moves along the x-axis.A.

B.

C.

D.

30. Find the centre and the radius of the circle 2x2 + 2y2 4x + 6y 1 = 0.A. centre = (2, 3), radius =B. centre = (2, 3), radius =

C. centre = (1, 1.5), radius =

D. centre = (1, 1.5), radius =

31. Find the equation of the circle passing through the origin, A(0, 3) and B(4, 0).A. x2 + y2 + 4x + 3y = 0B. x2 + y2 4x + 3y = 0C. x2 + y2 4x 3y = 0D. x2 + y2 + 4x 3y = 0

32. Given the circle C: x2 + y2 6x + 4y + 8 = 0, which of the following points lie(s) inside C?I. (1, 1)II. (1, 3)III. (2, 1)A. I onlyB. III onlyC. I and II onlyD. II and III only

NSS Mathematics in Action 23 © Pearson Education Asia Limited 2010

y

x

L

P: y = 2 – x2

A

B

M

O

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NSS Mathematics in Action

33. It is given that the circle S: and the straight line L: intersect at A and

B. Find the length of AB.A. 2B.C.D.

34. If the straight line L: 2x + y = k touches the circle C: x2 + y2 8x + 6y + 20 = 0, find the values of k.A. 0 or 5B. –5 or 5C. 0 or 10D. –5 or 10

35. Find the equation of the tangent to the circle C: (x 3)2 + (y + 1)2 = 10 at P(2, 2).A. x 3y 4 = 0B. x 3y + 4 = 0C. 3x y + 4 = 0D. 3x + y 4 = 0

36. If the tangent to a circle C: x2 + y2 = 13 at P(2, 3) cuts the x-axis and y-axis at A and B respectively, find the area of △OAB.A. 13 sq. units

B. sq. units

C. sq. units

D. sq. units

End of paper

NSS Mathematics in Action 24 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

F.5 First Term ExaminationMathematics (Paper 2 - Extra Questions)

1. If , then x = A. –3 or 1.B. –1 or 1. C. 0 or 1.D. 1 or 3.

2. If , find the values of a and b.

A. a = 1 and b = 1B. a = 1 and b = 1C. a = 2 and b = 1D. a = 2 and b = 2

3. In the figure, O is the centre of the circle. OF is perpendicular to both AB and CD. OF and AB intersect at E. If AB = 32 cm, CD = 24 cm and EF = 4 cm, find the radius of the circle.A. 16 cmB. 20 cmC. 24 cmD. 30 cm

4. In the figure, AB is a diameter of the circle. AB and CE are produced to meet at D. If AC = CE, find BCE.A. 20B. 18C. 16D. 14

5. In the figure, AC is the tangent to the circle at A. BDC is a straight line. If AC = 6 cm and BD = 5 cm, find the length of CD.A. 7 cm B. 5 cmC. 4 cmD. 3 cm

NSS Mathematics in Action 25 © Pearson Education Asia Limited 2010

5 cm

6 cm

B D C

A

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NSS Mathematics in Action

6. The figure shows a semi-circle. CE and CB are tangents to the semi-circle at D and B respectively. Find BCD.A. 54B. 56C. 58D. 62

7. Given two points A(5, 0) and B(0, 4), find the equation of the locus of a point P such that PB = 2PA.A. 3x2 + 3y2 40x + 8y + 84 = 0B. 3x2 + 3y2 + 40x 8y 84 = 0C. 3x2 + 3y2 + 10x 32y + 39 = 0D. 3x2 + 3y2 + 10x 32y 39 = 0

8. If the circle C: x2 + y2 6x 8y + F = 0 and the straight line L: 3x + 2y – 4 = 0 have no intersections, then A. F < –12.B. F > –12.C. F < 12.D. F > 12.

9. It is given that AB is a chord of the circle S: . If the equation of AB is and M(3.5, 2.5) is the mid-point of AB, find the value of a.

A. 4B. 3C. 2D. 1

End of paper

NSS Mathematics in Action 26 © Pearson Education Asia Limited 2010

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Term Exam Paper Kit

F.5 First Term Examination Mathematics (Paper 1)

Suggested Solutions and Marking Scheme

*******************************************************************General Instructions:(1) Marks will not be deducted for wrong spelling.(2) 1 mark will be deducted for poor expression or poor presentation.

(A maximum of 2 marks will be deducted for the whole paper.)(3) 1 mark will be deducted for wrong / no unit.

(A maximum of 1 mark will be deducted for the whole paper.)*******************************************************************

Suggested solutions Marks Remarks

Section A(1) (23 marks)

1.

2.

and

3. ( in alt. segment)(1M)

( at centre twice at ce

)

(1A)

NSS Mathematics in Action 27 © Pearson Education Asia Limited 2010

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(tangent radius)

(1M)In △PQR,

( sum of )△

(1A)

4. (s in the same segment)

(1M)(alt. s, CA // BD)

(1M)

∴ (arcs prop. to s at ce)(1M)

In △ABC,( sum of )△

(1A)

5. The required locus is the perpendicular bisector of PQ.

Coordinates of the mid-point of PQ (1M)

∵ Slope of PQ

∴ Slope of the locus (1M)

The equation of the required locus is

6. (a) When y = 0,(1M)

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∴ AB = (10 – 2) units = 8 units (1A)

(b) Coordinates of C (1M)

(1M)

∴ CAB = 36.9 (cor. to 3 sig. fig.) (1A)

Section A(2) (23 marks)

7. (a) When x = 0,

When x = –1,

(b)

(1A + 1A)

8. (a) Consider △ABD and △FED.BDA = EDF common angleDAB = DFE ext. , cyclic quad.∴ △ABD ~ △FED AAA (1A)

(b) ∵ △ABD ~ △FED proved in (a)

∴ corr. sides, ~ s△

Also, ED = EF given∴ BD = BA (1M)BEA = 90 in semi-circle (1M)∴ AE = ED prop. of isos. △ (1M)i.e. E is the mid-point of AD.

9. (a) OD = OE = 4 cm (radii)

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(1M)

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(tangent radius) (1M)

In △AOD,

(1A)(b) Let CD = x cm.

CB = CD = x cm (tangent properties)(tangent radius) (1M)

In △ABC,

∴ (1A)

10. (a) Let x-intercept = y-intercept = a.

Slope of L (1M)

∴ The equation of L is

(b) ∵ tangent L

∴ Slope of the tangent (1M)

Let y = x + c be the equation of the tangent.

By substituting (1) into (2), we have (1M)

(1M)

∴ The equations of the tangents to the circle are y = x – 8 and y = x + 4. (1A)

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Section B (24 marks)

11. (a)

(b) (i)

(from (a)) (1M)∵ is a root of .

(ii)

(1M)

(1A + 1A)

12. (a) Coordinates of Q (1A)∵ C is the mid-point of OQ.

∴ Coordinates of C (1A)

(b) (i) Slope of CP

∵ tangent at P CP (1M)

∴ Slope of the tangent at P

The equation of the tangent at P is

(1A)(ii) ∵ tangent at R // tangent at P (1M)

The equation of the tangent at R is

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(1A)

(c) Slope of OC

∵ tangent at O OC

∴ Slope of the tangent at O

(1M)∴ The tangent at O is not perpendicular to the tangent at P.∴ The quadrilateral formed is not a rectangle.∴ The student’s claim is not correct. (1A)

13. (a) ∵ P(a, b) is a point on L1.∴ By substituting (a, b) into the equation of L1, we have

(b) Let (x, y) be the coordinates of M1.Coordinates of P = (2 – b, b) (from (a))Coordinates of Q = (0, b)

∴ Coordinates of M

1

(1M)

By substituting (2) into (1), we have (1M)

∴ The equation of L2 is 2x + y – 2 = 0. (1A)(c) Let N be the point of contact.

∵ ON L2

∴ Slope of ON

∴ The equation of ON is . (1M)

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From (3), we have

By substituting (5) into (4), we have (1M)

By substituting into (5), we have

∴ Coordinates of N (1A)

Radius of the circle

∴ The equation of the circle is

(1A)

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F.5 First Term ExaminationMathematics (Paper 1 Extra Questions)Suggested Solutions and Marking Scheme

Suggested solutions Marks Remarks

Section A(1) (23 marks)

1.

Checking: When x = 0, When x = 3,

∴ The real root of the equation is 0. (1A)

2. (a)

(b)

(from (a))

3. (a) Consider △BAC and △BDE.ABC = DBE common angleBAC = BDE ext. , cyclic quad.△BAC ~ △BDE AAA (1A)

(b) ∵ △BAC ~ △BDE (proved in (a))

∴ (corr. sides, ~△s) (1M)

∴ (1A)4. Join OP and OR.

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(1M)

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O

R P S

OP RS (tangent radius) (1M)In △ORP,

(1A)∴ (line from centre chord bisects chord)

5. Let (x, y) be the coordinates of P.

∴ The equation of the required locus is . (1A)

6. (a) By substituting (1, 2) into the equation of S, we have (1M)

R.H.S. = 0∵ L.H.S. = R.H.S.∴ P(1, 2) lies on S.

(b) Let C be the centre of S.

Coordinates of C (1M)

∵ Slope of

∴ Slope of the tangent at P (1M)

The equation of the tangent at P is

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Section A(2) (23 marks)

7. (a)

(1A)

(b)

(1A + 1A)

8. (a) AOE = 90 prop. of square (1M)ABC = 90 given∵ AOE = ABC∴ A, B, C and O are concyclic. ext. = int. opp. (1A)

(b) OAC = 45 prop. of square (1M)

s in the same segment (1M)

∴ OBC = OBAi.e. OB bisects ABC.

9. (a) Consider △ABD and △ACB.BAD = CAB common angleABD = ACB in alt. segment∴ △ABD ~ △ACB AAA (1A)

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(1M)

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(b) (i) ∵ △ABD ~ △ACB (proved in (a))

∴ (corr. sides, ~△s) (1M)

(1A)(ii) (tangent properties)

(1M)∴

(1A)

10. (a) (i)

(1A)

(1A)

(ii) ∵ AB = 5 units∴ ...... (3) (1M)By substituting (1) and (2) into (3), we have

∴ The equation of the required locus is . (1A)(b) The locus of the mid-point of the ladder is a quarter of a circle with radius 2.5 m. (1A)

2.5 m

locus

(1A)

Section B (24 marks)

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11. (a) (i)

By substituting x = 4.76 into (1), we have (1M)

(ii)

By substituting x = 7.76 into (3), we have (1M)

(b) (2) + (4): (1M)

By substituting a + b = 1 into (2), we have (1M)

(1A)∴

(1A)

12. (a) The equations of L1 and L2 are in the form y = mx.(1M)

By substituting (1) into (2), we have

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∵ L1 and L2 are tangents to S.∴ For the equation (*),

(1M)

∴ The equation of L1 is .

The equation of L2 is .

(b) ∵

(1M)∴ L1 L2

i.e. DOE = 90 (1A)∵ OD = OE (tangent properties)∴ ODE = OED (base s, isos. )△ (1M)

( sum of )△

(1A)

13. (a) By substituting x = 2a into , we have

(1M)

∴ A(2a, a2) lies on P.(b) (i) Let (x, y) be the coordinates of M.

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(1M)

(1A)

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By substituting (1) into (2), we have (1M)

∴ The equation of the required locus is . (1A)

(ii)

(2A)The above curve is a parabola. (1A)

(iii) The equation of the required locus is . (1A)

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Answers

F.5 First Term Examination Mathematics (Paper 2)

1. D2. D3. A4. B5. C6. C7. B8. C9. A10. B11. B12. A13. C14. B15. D16. B17. D18. A

19. C20. A21. B22. A23. D24. B25. C26. B27. B28. C29. B30. D31. C32. B33. C34. C35. B36. D

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F.5 First Term Examination Mathematics (Paper 2 – Extra Questions)

1. B2. A3. B4. D5. C

6. B7. A8. D9. A

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F.5 First Term ExaminationMathematics (Paper 2)

Suggested Solutions

1. D

2. D

3. A

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4. B

5. C

6. C

7. B

and

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8. C

∵ is a real number.

9. AJoin OA.∵ ( at centre twice at ce)∴

∴ (arc prop. to s at centre)

10. BIn △ABE,

( sum of )△

(ext. , cyclic quad.)

11. BDraw OM and ON such that OM AB and ON BC.Also, AB = BC (given)∴ OM = ON (equal chords, equidistant from centre)∵ OMBN is a square.∴ OM = MBIn △OMB,

(Pyth. theorem)

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∵ OM AB (construction)∴ (line from centre chord bisects chord)

12. AConsider △ADE and △CDB.EAB = ECB (s in the same segment)i.e. EAD = BCDADE = CDB = 90 (vert. opp. s)∴ △ADE ~ △CDB (AAA)

∴ (corr. sides, ~ s)△

In △ADE,

(Pyth. theorem)

13. C∵ OB = OA (radii)∴ (base s, isos. )△

( sum of )△

(vert. opp. s)

( at centre twice at ce

)

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14. BLet ACB = x.∵ AB = BC = CD (given)

∴ (equal chords, equal arcs)

∴ ACB = BAC = CBD (arcs prop. to s at ce)In △ACB,

( sum of )△

∴ ACB = 35

15. DLet CBD = x.

∵ (given)

∴ (arcs prop. to s at ce)

∵ (given)

∴ (arcs prop. to s at ce

)

(opp.

s, cyclic quad.)

16. BConsider △AFO, △ADE and △ACB.OAF = EAD = BAC (common angle)∵ AFO = 90 (given),

ADE may not be 90,ACB = 90 ( in semi-circle)

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∴ Only II must be true.i.e. △AFO ~ △ACB (AAA)∴ The answer is B.

17. DOAC = OAB = 90 (tangent radius)In △OCA,

(Pyth. theorem)In △OAB,

(Pyth. theorem)In △OCB,

(Pyth. theorem)

∴ The radius of the circle is 4.24 cm.

18. A∵ AC = AB (tangent properties)∴ (base s, isos. )△

(alt. s, CE // AB)

(adj. s on st. line)

19. CJoin AB.∵ CA = CB (tangent properties)∴ (base s, isos. )△In △CBA,

( sum of )△

( in alt. segment)

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20. AIn △BAD,

∵∴ BDA = 90 (converse of Pyth. theorem)

(adj. s on st. line)

∴ AC is a diameter. (converse of in semi-circle)BAC = 90 (tangent radius)∵ △ABD ~ △CBA (AAA)

∴ (corr. sides, ~ s)△

∴ Radius

21. BJoin AD.∵ OE = OD (radii)∴ (base s, isos. )△

( in alt. segment)

(s in the same segment)

22. ALet DB = x cm and EB = y cm.DH = DM = DB = x cm (tangent properties)EK = EN = EB = y cm (tangent properties)

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∵ CH = CK and CM = CN (tangent properties)

and CE = CD = 6.6 cm

∵ GA = GH and FA = FK (tangent properties)

23. DJoin AB and mark F as the intersection of AB and OC.For I:∵ CA = CB and ACF = BCF tangent properties∴ CF AB prop. of isos. △i.e. BFC = 90DBA = 90 in semi-circle∴ DBA = BFC∴ BD // CO alt. s equalFor II:BED = CEO common angleDBE = OCE corr. s, BD // CO∴ △BDE ~ △COE AAAFor III:CBA = BDA in alt. segmentCOA = BDA corr. s, CO // BD∴ CBA = COA∴ A, C, B and O are concyclic. converse of s in the same segment

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24. BSince the mid-points of all vertical chords lie on the horizontal diameter (line joining centre to mid-pt. of chord chord), the required locus is a line segment.

25. C

26. B

27. BLet (x, y) be the coordinates of P.∴ The required equation is

28. CLet (x, y) be the coordinates of P.Distance between P and L = 4 – y ∴ The required equation is

29. BLet (x, y) be the coordinates of M.Then, coordinates of A = (x, 2y) ∴ The required equation is

(∵ A lies on a parabola P.)

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30. D

Centre

31. C∵ AOB = 90∴ AB is a diameter of the circle. (converse of in semi-circle)Centre = mid-point of AB

Radius = distance between O and the centre

∴ The equation of the circle is

32. B

Centre

RadiusFor (1, 1), distance from centreFor (1, 3), distance from centreFor (2, –1), distance from centre∴ Only (2, –1) lies inside C.∴ The answer is B.

33. C

From (1), we have

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By substituting (3) into (2), we have

When x = 0, y = 0.When x = 2, y = 2.

34. C

From (1), we have

By substituting (3) into (2), we have

∵ L is a tangent to C.∴ For the equation (*),

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35. BCentre = (3, 1)

Slope of line joining the centre and P

∴ Slope of the tangent

∴ The equation of the tangent to C at P is

36. D

Slope of OP

∴ Slope of the tangent

∴ The equation of the tangent to C at P is

Coordinates of A

Coordinates of B

∴ Area of △OAB

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F.5 First Term ExaminationMathematics (Paper 2 Extra Questions)

Suggested Solutions

1. B

2. A

From (2), we have b = –1 By substituting b = –1 into (1), we have

3. BJoin OA and OC.Let OE = x cm.∵ OE AB and OF CD (given)

∴ (line from centre chord bisects chord)

and (line from centre chord bisects chord)

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∵ OA = OC (radii)

(Pyth. theorem)

∴ Radius

4. DJoin AE.Let BCE = x.

(s in the same segment)

In △ADE,(ext. of )△

∵ CA = CE (given)∴ (base s, isos. )△

ACB = 90 ( in semi-circle)In △AEC,

( sum of △)

∴ BCE = 14

5. CConsider △ADC and △BAC.ACD = BCA (common angle)DAC = ABC ( in alt. segment)∴ △ADC ~ △BAC (AAA)

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Let CD = x cm.

(corr. sides, ~ s)△

∴ CD = 4 cm

6. B( in alt. segment)

ABC = 90 (tangent radius)∴

∵ CD = CB (tangent properties)∴ (base s, isos. )△

In △CDB,( sum of )△

7. ALet (x, y) be the coordinates of P.∴ The required equation is

8. D

From (1), we have

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By substituting (3) into (2), we have

∵ C and L have no intersections.∴ For the equation (*),

9. A

Slope of AB

Centre C = (a, a)∵ AB CM (line joining centre to mid-pt. of chord chord)

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