rev ex for s4

TSW/Revision Exercise for S4 final exam (10-11)/1 of 30

Revision Exercise for S.4 Mathematics Final Examination Section A(1) 1. Solve the following quadratic equations.

(a) 028122 =−− xx (b) 0342 2 =−− xx (Leave your answers in surd form.)

2. Find the range of possible values of k so that the quadratic equation 082 =++ kxx has no real roots.

3. Given that f(x) = 152

+−

xx , find

(a) f(2), (b) f(t + 1), (c) the value of x for which f(x) is undefined.

4. Simplify

ba

ba

41

2

212

)(

)( −and express your answer with positive indices.

5. Evaluate the following without using a calculator. (a) 25log4log + (b) 2log18log90log +− 6. Given A(2, 1) and B(−2, 3), find (a) the slope of AB, (b) the equation of the straight line passing through A and perpendicular to AB. 7. It is given that p(x) = x3 + 2x2 + kx + 6 and x – 3 is a factor of p(x).

(a) Find the value of k. (b) Hence, find the remainder when p(x) is divided by x + 2.

8. Simplify 2 2

1 21 2x x x+

− − −.

9. Solve the simultaneous equations 2

53 2

x yy x x+ =⎧

⎨= − +⎩

.

10. If tan 2θ = and 180° < θ < 270°, find θsin and sin(90 )θ°+ .

(Leave your answers in surd form if necessary.) 11. Solve θθθθ sin3 cos 4sin 2 cos −=− for 0° ≤ θ ≤ 360°. 12. Find the area of parallelogram ABCD.

Section A(2) 13. Let α and β be the roots of the quadratic equation 0362 2 =−− xx .

(a) Find the value of αβ

βα+ .

(b) Form a quadratic equation in x with roots βα and

αβ .

14. (a) Solve the equation 1512

8+x

= 27

4−x

.

(b) Using the value of x obtained in (a), find the value of 1512

8+x

.


15. Let y = x2 − 4x − 2. (a) Complete the following table.

x −1 0 1 2 3 4 5 y

(b) Plot the graph of y = x2 − 4x − 2 on the following graph paper.

(c) Using the graph,

(i) solve the equation x2 − 4x − 2 = 0, (ii) find the minimum value of y = x2 − 4x − 2. 16. The perimeter of a right-angled triangle is 45 cm and the lengths of the two arms of the right angle differ by

10.5 cm. Find the length of the hypotenuse of the triangle. 17. The figure shows the graph of y = x2 + k for 5.35.3 ≤≤− x .

(a) Find the value of k. (b) Solve each of the following simultaneous equations by adding a suitable straight line on the graph.

(i) ⎩⎨⎧

−=+=

62

2

xykxy

(ii) ⎩⎨⎧

−=++=

12

2

yxkxy

18. (a) Solve 22sin 5sin 2θ θ+ = for 0° ≤ θ ≤ 360°.

(b) Hence, solve 22sin 2 5sin 2 2θ θ+ = for 0° ≤ θ ≤ 360°. Section B 19. A farmer wants to enclose a rectangular plot ABCD next to a brick

wall AB with fencing along BC, CD and DA to keep his animals. If he has got 200 m of fencing, find the dimensions of the plot so that its area is a maximum.

20. If a=5log3 and b=6log3 , express the following in terms of a and b. (a) 150log3 (b) 10log3 (c) 18log5


21. In the figure, L1 is a straight line passing through A(−2, −2) of slope 34 . The equation of straight line L3 is x −

7y − 12 = 0. The straight line L4 passes through E )3 ,7

12( − and is perpendicular to L3. It intersects L1 at B. L2

is a straight line perpendicular L1 with positive intercepts and does not pass through B.

(a) Find the equation of L1. (b) Find the equation of L4. (c) Find the equation of straight line L5 passing through B and parallel to L2. (d) Find a possible equation of L2. 22. In the figure, the straight line y = 2x + c intersects the quadratic curve

542 2 ++= xxy at two points P and Q. (a) Find the range of possible values of c. (b) It is given that c is an even number.

(i) Find the least value of c. (ii) Using the result of (b)(i), find the coordinates of P and Q.

Multiple Choice

1. Which of the following quadratic equations has roots 3 and 21

− ?

A. (x − 3)(2x + 1) = 0 B. (x + 3)(2x − 1) = 0 C. (x + 3)(2x + 1) = 0 D. (x − 3)(2x − 1) = 0 2. If the equation 02102 =++ xkx has two distinct real roots, then

A. k > 225 .

B. k < 225 .

C. k > 225

− .

D. k < 225

− .

3. If 1 is a root of the equation (x + 2)(x − 3) = k(1 − k), where k is a constant, then k =

A. −2 or −3. B. 2 or 3. C. 2 or −3. D. −2 or 3.

4. The figure shows the graph of y = ax2 + bx + c.

Which of the following is true? A. a < 0 and Δ < 0 B. a < 0 and Δ > 0 C. a > 0 and Δ < 0 D. a > 0 and Δ > 0


5. The root(s) of the equation (x + 2)(x − 3) = x − 3 is/are A. −1 or 3. B. 3 only. C. −2 only. D. −2 or 3.

6. If 2121

−=−x

x , then x =

A. 2.

B. 2 or 21

− .

C. −2 or 21 .

D. −2 or 21

− .

7. In the figure, ABCD is a square of side 7 cm. P and Q are points on AD and CD

respectively such that PD = QD = x cm. If the area of ΔBPQ is 20 cm2, find the value(s) of x.

A. 4 or 10 B. 3 or 5 C. 4 D. 3 or 4

8. In the figure, the graph of y = −x2 + bx + b touches the x-axis. Find the value(s) of b. A. 0 or 4 B. 0 or −4 C. 4 D. −4 9. The roots of the equation 0263 2 =++ xx are

A. 331±− .

B. 3

322 ±− .

C. 331± .

D. 6

606 ±− .

10. Let α and β be the roots of the equation 0342 =−+ xx .

Find the value of 22 αββα + . A. −22 B. −12 C. 22 D. 12

11. If (x − a)(x + 1) = (k − a)(k + 1), then x = A. k. B. a. C. k or a − 1 − k. D. a or a − 1 − k. 12. If 2 is a root of the equation 053 2 =+− cxx , find the other root. A. c

B. 35

C. 31

−

D. Cannot be determined


13. If one root of the equation 012 2 =++ bxx doubles the other, find the value(s) of b. A. ±3 B. 3 C. −3

D. 23

±

14. If α and β are the roots of the quadratic equation ax2 + bx + c = 0, which of the following is a quadratic

equation in x with roots α− and β− ? A. ax2 + bx + c = 0 B. ax2 + bx − c = 0 C. ax2 − bx + c = 0 D. ax2 − bx − c = 0

15. If f (x) = 1313−+

−x

x

, then f (−1) =

A. −1.

B. 32 .

C. 1.

D. 31 .

16. If f (x) = 11

+−

xx , then )1(

xf =

A. f (x). B. −f (x).

C. )(

1xf

.

D. )(

1xf

− .

17. Which of the following CANNOT be the graph of a function y = f (x)?

A. B. C. D.

18. A stone is projected vertically upwards. After t second, its height (h m) above the ground is given by h = (24t

− 2t2). When will the stone attain its maximum height? A. 3 s B. 6 s C. 15 s D. 30 s 19. Find the coordinates of the vertex of the graph y = x2 + 8x + 15. A. (−4, 15) B. (−8, −49) C. (4, −1) D. (−4, −1)

20. Find the optimum value of the function y = 784

62 +− xx

.

A. Maximum value = 3 B. Minimum value = 3 C. Maximum value = 2 D. Minimum value = 2


21. 41

)64( = A. 16 B. 4 C. 22 D. 2 22. 11)3( −+ xx = A. x23 B.

2

3x C. 33

2

−x

D. 3

32x

23. nn

nn

5555

1

2

++

+

+

=

A. 1

B. 12)1(2

++

nn

C. 5

D. 3

13

24. Let a be a non-zero number. Which of the following statements is FALSE? A. nmnm aaa +=

B. nmn

m

aaa −=

C. na− = na1

D. mn

n m aa =

25. Which of the following may represent the graph of xy21

= ?

A.

B.

C.

D.

26. nn

n

93)3( 26

=

A. 1

B. 23n

3 C. n3 D. n9


27. 10log50log20log −+ = A. 60log B. 2 C. 40log20log +

D. 10log70log

28. 4019

20102009

222log ⋅ =

A. −1 B. 0 C. 1 D. Cannot be determined 29. If yx =8log , then A. xy 8= B. 8xy = C. yx=8 D. xy=8

30. 4log

32log

3

3 =

A. 8 B. 8log3 C. 28log3

D. 25

31. Which of the following may represent the graph of 3x − 4y + 7 = 0 ?

A. B. C. D.

32. Find the equation of the straight line passing through the point (0, −2) and with slope 23 .

A. 22

3−=

xy

B. 3x + 2y − 4 = 0 C. 3x − 2y − 2 = 0 D. 2x − 3y − 6 = 0 33. In the figure, OABC is a parallelogram. O, A(3, 7), B and C(−5, 3) are the

vertices of the parallelogram. Find the equation of OB. A. 2x + y = 0 B. x + 2y = 0 C. 5x + y = 0 D. x + 5y = 0

34. The figure shows the straight line ax + by + c = 0 passing through the origin.

Which of the following must be FALSE? A. c = 0 B. ab < 0 C. a < 0 D. b < 0


35. If the two straight lines 04:1 =+ ymxL and 09:2 =− ymxL are perpendicular to each other, find the value(s) of m. A. 0 only B. 6 only C. −6 only D. ±6

36. Which of the following straight lines does not intersect the line x + 2y = 3? A. 2x + 4y + 7 = 0 B. 6x + 3y − 5 = 0 C. x − 2y = 3 D. 2x − y = 0 37. When 3x3 + 2x2 – 1 is divided by 3x + 1, the remainder is

A. 23

− .

B. 43

− .

C. 89

− .

D. 169

− .

38. Let P(x) = 2x3 – 3x2 + kx – 1. When P(x) is divided by 2x + 1, the remainder is 1. Find the remainder when P(x)

is divided by x + 2. A. –39 B. –17 C. –13 D. –9

39. Let f(x) = 3x3 − kx2 – 58x + 40. If f(5) = f(–4) = 0, f(x) can be factorized as

A. (x – 5)(x + 4)(3x – 2). B. (x + 5)(x – 4)(3x + 2). C. (x – 5)(x + 4)(3x + 2). D. (x + 5)(x – 4)(3x – 2).

40. Factorize 6x3 + x2 – 5x – 2. A. (x + 1)(2x – 1)(3x + 2) B. (x + 1)(2x + 1)(3x – 2) C. (x – 1)(2x – 1)(3x – 2) D. (x – 1)(2x + 1)(3x + 2)

41. When P(x) is divided by x + 1, the remainder is –2. When P(x) is divided by 2x + 2, the remainder is

A. –1. B. –2. C. –3. D. –4.

42. When a polynomial P(x) is divided by x2 – x – 6, the remainder is –3x + 1. Which of the following must be

true? I. P(–2) = 7 II. P(2) = –5 III. P(3) = –8 A. I and II only B. I and III only C. II and III only D. I, II and III

43. If f(x) is a cubic polynomial with leading coefficient 2 and f(–1) = 12

f ⎛ ⎞⎜ ⎟⎝ ⎠

= f(1) = –2, then f(x) =

A. (x + 1)(x – 1)(2x + 1) – 2. B. (x + 1)(x – 1)(2x – 1) – 2. C. (x + 1)(x – 1)(2x – 1) + 2. D. (x + 1)(x – 1)(2x + 1) + 2.


44. Find the H.C.F and L.C.M. of (x + 1)2(2x – 1)(x – 2)3 and (x + 1)(2x + 1)2(x – 2)2. H.C.F. L.C.M.

A. (x + 1)(x – 2) (x + 1)2(2x + 1)2(x – 2)3

B. (x + 1)(x – 2) (x + 1)2(2x – 1)(2x + 1)(x – 2)3 C. (x + 1)2(x – 2)3 (x + 1)2(2x – 1)(2x + 1)2(x – 2)3 D. (x + 1)(x – 2)2 (x + 1)2(2x – 1)(2x + 1)2(x – 2)3

45. Find the H.C.F. and L.C.M. of 4x2y2, 6x3y and 8xyz.

H.C.F. L.C.M. A. 4xy 24x3y2 B. 4xy 48x3y2z C. 2xy 24x3y2z D. 2xy 48x3y2z

46. The H.C.F. and L.C.M of P and Q are 2ab and 12a3b2c respectively. If P is 4ab2, then Q is

A. 3a2bc. B. 6a2bc. C. 6a3bc. D. 6a3b2c.

47. =−

−

yx

yx

xy

11

A. y – x B. x + y

C. 1x y+

D. 1y x−

48. If 2 4

3 4y x xy x

⎧ = + −⎨

= +⎩, then y =

A. –2 or 4. B. –4 or 2. C. –8 or 10. D. –2 or 16.

49. Solve the simultaneous equations 22 4 14

y x xx y

⎧ = − −⎨

+ =⎩.

A. (1, 3) B. (–1, 5) C. (1, 3) or (–1, 5)

D. (–1, 5) or 5 3,2 2

⎛ ⎞⎜ ⎟⎝ ⎠

50. If the simultaneous equations 22 3 2y x x

y k x⎧ = + +⎨

= −⎩ have two distinct real solutions, find the range of possible

values of k. A. k > 8 B. k > 0 C. k < 0 D. k < –8

51. If the line y = kx – 2 touches the quadratic curve y = 3x2 – 5x + 1 at one point, find the value(s) of k.

A. 1 B. –1 C. 1 or –1 D. 1 or –11


52. Which of the following lines has no intersections with the quadratic curve 2 3 1y x x= − + − ? A. y = 4 B. y = 1 C. y = –1 D. y = –4

53. The figure shows the graphs of 22 3 1y x x= − − and y = x + 1.

Which of the following equations whose roots are α and β? A. x2 – 2x – 2 = 0 B. x2 – 2x – 1 = 0 C. x2 – 4x + 2 = 0 D. x2 – x – 4 = 0

54. If the perimeter and the length of the diagonal of a rectangle are 56 cm and 20 cm respectively, find the

dimensions of the rectangle. A. 10 cm × 18 cm B. 11 cm × 17 cm C. 12 cm × 16 cm D. 13 cm × 15 cm

55. If 180° < x < 270°, which of the following must be true?

I. sin x < cos x II. sin x⋅cos x > 0 III. sin 2x > 0 A. I and II only B. I and III only C. II and III only D. I, II and III

56. The figure shows the graph of 3 2siny x= − . What are the values of α

and k? A. α = 180, k = 3 B. α = 180, k = 5 C. α = 270, k = 3 D. α = 270, k = 5

57. What are the maximum and minimum values of x2cos3

2−

?

Maximum value Minimum value

A. 1 23

B. 1 12

C. 2 12

D. 2 25

58. cos(90 ) tan(270 )sin(180 )θ θ θ°+ °− °+ =

A. 2sin θ B. 2sin θ− C. 2cos θ D. sin cosθ θ


59. If m1 cos =θ and 0° < θ < 90°, then =−°− )360tan()sin( θθ

A. m

m 1− .

B. m

m 12 − .

C. m

m 12 − .

D. m

m21− .

60. For 0° ≤ x ≤ 360°, how many solutions does the equation xx 2cos2sin52 =+ have?

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

61. Solve 3 tan 1 0x + = for 0° ≤ x ≤ 360°.

A. 30° or 210° B. 150° or 210° C. 150° or 330° D. 30° or 330°

62. Solve 26cos cos 2 0x x+ − = for 0° ≤ x ≤ 360°.

A. 60° or 132° B. 48° or 120° C. 60°, 132°, 228° or 300° D. 60°, 132°, 240° or 312°

63. Solve 2 2cos 4sin cos 5sin 0x x x x− − = for 0° ≤ x ≤ 180°.

A. 11° or 45° B. 11° or 135° C. 45° or 169° D. 45° or 135°


Section A(1) 1. (a) 028122 =−− xx (x −14)(x + 2) = 0 x = 14 or x = –2 (b) 0342 2 =−− xx

x = )2(2

)3)(2(4)4()4( 2 −−−±−−

= 4

404 ±

= 2

102 ±

2. 0))(1(482 <− k 4k > 64 k > 16

3. (a) f(2) = 12

5)2(2+− =

31

−

(b) f(t + 1) =1)1(5)1(2

++−+

tt

= 232

+−

tt

(c) When x = −1, f(x) is undefined.

4.

ba

ba

41

2

212

)(

)( − =

ba

ba

21

24 −

= 21

214

+

−

ba

= 3

27

ba

5. (a) 25log4log +

= 25log4log +

= )54log( 2× = 100log = 2

(b) 2log18log90log +−

= ⎟⎠⎞

⎜⎝⎛ ×

18290log

= 10log = 1

6. (a) Slope of AB =)2(2

31−−− =

21

−

(b) The equation of the straight line:

)2(

21

11 −⎟⎠⎞

⎜⎝⎛−

−=− xy

y – 1 = 2(x – 2) 2x – y − 3 = 0 7. (a) ∵ x – 3 is a factor of p(x).

∴ By the converse of the factor theorem, 0)3( =p


513063182706)3()3(2)3( 23

−==+++=+++

kk

k

17−=k (b) By the remainder theorem,

)2(remainder −= p

63488

6)2(17)2(2)2( 23

+++−=+−−−+−=

40=

8. 2

21

122 −−

+− xxx

)2)(1(2

)1)(1(1

−++

−+=

xxxx

)2)(1)(1()1(22−−+

−+−=

xxxxx

)2)(1)(1(222−−+

−+−=

xxxxx

)2)(1)(1(43

−−+−

=xxx

x

9. ⎩⎨⎧

+−=

=+

235

2 xxyyx

)2()1(

KK

KK

From (1), we have y = 5 – x ……(3) By substituting (3) into (2), we have

235 2 +−=− xxx 0322 =−− xx

0)1)(3( =+− xx 1or 3 −== xx By substituting x = 3 into (3), we have

235

=−=y

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

6

)1(5=

−−=y

∴ The solutions of the simultaneous equations are (3, 2) and (−1, 6). 10. ∵ 180° < θ < 270°

∴ θ lies in quadrant III. Let P(x, y) be a point on the terminal side of θ and OP = r. ∵ tan θ = 2 ∴ Let 1−=x and 2−=y .

5

)2()1( 22

=

−+−=r

∴ 5

2sin −=θ

θθ cos)sin(90 =+°

5

1−=


11.

3 tan

3 cos sin

cos3 sin cos cos4 sin2 sin3 sin3 cos 4sin 2 cos

=

=

=−=−−=−

θθθ

θθθθθθθθθθ

∴ fig.) sig. 3 to(cor. 252or fig.) sig. 3 to(cor. 71.6

71.5651180or 71.5651°°=

°+°°≈θ

12. Semi-perimeter of △ABD

cm 24

cm 2

211710

=

++=

∴ Area of △ABD

2

2

cm 371424

cm )2124)(1724)(1024(24

×××=

−−−=

2cm 84= ∴ Area of parallelogram ABCD

ABD△ of area2×=

2

2

cm 168cm 842

=

×=

Section A(2)

13. (a) α + β = 26 = 3, αβ =

23

−

αβ

βα+ =

αββα 22 +

=αβ

αββα 2)( 2 −+

=

2339

−

+

= −8

(b) αβ

βα⋅ = 1

The required quadratic equation is: 0182 =++ xx

14. (a) 1512

8+x

= 27

4−x

15)12(3

2+x

= 72 −x

5

12 +x = x −7

2x + 1 = 5x − 35 x = 12

(b) When x = 12, 1512

8+x

= 1525

8 = 32 15. (a)

x −1 0 1 2 3 4 5 y 3 −2 −5 −6 −5 −2 3

(b) graph of y = x2 − 4x −2

Alternatively,

when x = 12, 1512

8+x

= 27

4−x

= 25

4 = 32


(c) (i) From the graph, the roots of x2 − 4x − 2 = 0 are −0.5 and 4.5. (ii) From the graph, the minimum value of y = x2 − 4x − 2 is −6. 16. Let the length of the shortest side be x cm.

Then, the other side = (x + 10.5) cm and the hypotenuse = [45 − x − (x + 10.5)] cm = (34.5 − 2x) cm

222 )25.34()5.10( xxx −=++

22 413825.119025.110212 xxxx +−=++ 010801592 2 =+− xx 0)72)(152( =−− xx x = 7.5 or x = 72 (rejected) ∴ The length of the hypotenuse is [34.5 − 2(7.5)] cm = 19.5 cm. 17. (a) ∵ The y-intercept of the graph of y = x2 + k is –5.

∴ k = –5 (b)

(i) y = 2x – 6 x 0 1 2 y –6 –4 –2

∵ The two graphs intersect at only one point (1, –4). ∴ The solution of the simultaneous equations is (1, –4).

(ii) 2x + y = –1 x –2 0 2 y 3 –1 –5

∵ The two graphs intersect at (–3.2, 5.4) and (1.2, –3.4). ∴ The solutions of the simultaneous equations are (–3.2, 5.4) and (1.2, –3.4).


18. (a)

(rejected) 8508.2or 3508.04

415

)2(2)2)(2(455

sin

02sin5sin22sin5sin2

2

2

2

−≈

±−=

−−±−=

=−+

=+

θ

θθ

θθ

When 3508.0sin ≈θ ,

fig.) sig. 3 to(cor. 159or fig.) sig. 3 to(cor. 20.5 i.e.5363.20180or 20.5363

°°=°−°°≈

θθ

(b) ∵ °≤≤° 72020 θ From (a), we have

fig.) sig. 3 to(cor. 260or fig.) sig. 3 to(cor. 190or

fig.) sig. 3 to(cor. 7.79or fig.) sig. 3 to(cor. 10.3159.4637360or 20.5363360or

159.4637or 20.53632

°°

°°=°+°°+°

°°≈

θ

θ

Section B 19. Let BC = x m and area of ABCD = A m2.

Then CD = (200 − 2x) m. ∴ A = x(200 − 2x)

= 200x − 2x2 = −2(x2 − 100x) = −2(x − 50)2 + 5000 ∴ A is a maximum when x = 50. ∴ For the area to be maximum, the dimensions of the plot are 50 m × 100 m. 20. (a) 150log3 = )65(log 2

3 ⋅

= 6log5log 32

3 + = 6log5log2 33 + = ba +2

(b) 10log3 = ⎟⎠⎞

⎜⎝⎛

15150log3

= 15log150log 33 − = )5log3(log2 33 +−+ ba = aba −−+ 12 = a + b − 1

(c)

ab+

=

+=

×=

=

15log

6log3log5log

)63(log5log

18log18log

3

33

3

3

3

35

Alternatively,

10log3 = 3

30log3

= 3log6log5log 333 −+ = a + b − 1


21. (a) Equation of L1: )2(342 +=+ xy

4x − 3y + 2 = 0

(b) Slope of L3 = 71

Slope of L4 = −7

Equation of L4: )7

12(73 −−=+ xy

7x + y − 9 = 0

(c) Solving ⎩⎨⎧

=−+=+−

0970234

yxyx

, 4x − 3(9 − 7x) + 2 = 0

25x − 25 = 0, i.e. x = 1, y = 2 ∴ B = (1, 2)

Slope of L5 = slope of L2 =

341− =

43

−

∴ Equation of L5: )1(432 −−=− xy

i.e. 3x + 4y − 11 = 0 (d) A possible equation of L2: 3x + 4y − 12 = 0 or

22. (a) ⎩⎨⎧

+=++=

cxyxxy

2542 2

)2......()1......(

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

......(3) 0)5(22

54222

2

=−++

++=+

cxxxxcx

∵ The straight line y = 2x + c intersects the quadratic curve 542 2 ++= xxy at two points. ∴ Δ > 0

0)5)(2(422 >−− c

2936808404

>

>>+−

c

cc

∴ The range of possible values of c is 29

>c .

(b) (i) The least value of c is 6. (ii) By substituting c = 6 into (3), we have

4122

)2(2)1)(2(422

01220)65(22

2

2

2

±−=

−−±−=

=−+

=−++

x

xxxx

3660.1or 0.3660 −≈ By substituting 3660.0≈x and c = 6 into (2), we have

fig.) sig. 3 to(cor. 73.6

6)3660.0(2=

+≈y

By substituting 3660.1−≈x and c = 6 into (2), we have

fig.) sig. 3 to(cor. 27.3

6)3660.1(2=

+−≈y

∴ The coordinates of P and Q are (–1.37, 3.27) and (0.366, 6.73) respectively.

3x + 4y + c = 0, where c < 0 and c ≠ −11


Multiple Choice 1. A

21 or 3

012or 030)12)(3(

−==

=+=−=+−

xx

xxxx

∴ The answer is A.

2. B ∵ The equation kx2 + 10x + 2 = 0 has two distinct real roots.

∴ Δ> 0

i.e.

22510080)2)((4102

<

<>−

k

kk

3. D

∵ 1 is a root of the equation (x + 2)(x – 3) = k(1 – k). ∴

3or 20)3)(2(06

6)1()2(3)1()31)(21(

2

2

−==−+=−−

−=−

−=−−=−+

kkkkk

kkkkkk

4. D ∵ The graph of y = ax2 + bx + c opens upwards.

∴ a > 0 ∵ The graph has two x-intercepts. ∴ Δ> 0 ∴ The answer is D.

5. A

3or 10)3)(1(0)3)(12(0)3()3)(2(

3)3)(2(

−==−+=−−+=−−−+

−=−+

xxxxxxxx

xxx

6. B

∵ It is obvious that 2 is a root of the equation 2121

−=−x

x .

∴ The answer may be A or B.

By substituting 21

−=x into 2121

−=−x

x ,

L.H.S. = 2122

21

21

121

−=+−=⎟⎠⎞

⎜⎝⎛−

−− = R.H.S.

∴ 21

− is another root of the equation.

∴ The answer is B.

7. C Area of △BPQ = Area of ABCD – area of △ABP – area of △BCQ – area of △PQD

22

22

2

cm 2

)7(749

cm 22

)7(72

)7(77

⎥⎦

⎤⎢⎣

⎡−−−=

⎥⎦

⎤⎢⎣

⎡−

−−

−−=

xx

xxx


∵ Area of △BPQ 2cm 20=

∴

(rejected) 10or 40)10)(4(04014

02072

202

)7(749

2

2

2

==−−=+−

=+−

=−−−

xxxxx

xx

xx

8. D

∵ The graph of y = −x2 + bx + b touches the x-axis. ∴ Δ = 0

i.e.

4or (rejected) 00)4(040))(1(4

2

2

−==+=+

=−−

bbb

bbbb

[If b = 0, then y = –x2, which is obviously not the equation represented by the graph.] 9. A By the quadratic formula,

331

333

6126

)3(2)2)(3(466 2

±−=

±−=

±−=

−±−=x

10. D

Since α and β are the roots of the equation x2 + 4x – 3=0, we have

12)4(3

)(

313 and

414

22

=−−=+=+

−=−

=

−=−=+

βααβαββα

αβ

βα

11. C

∵ It is obvious that x = k is a solution of the equation (x – a)(x + 1) = (k – a)(k + 1). ∴ The answer may be A or C. By substituting x = a – 1 – k into (x – a)(x + 1) = (k – a)(k + 1), L.H.S.

R.H.S.)1)((

))(1()11)(1(

=+−=−+−=

+−−−−−=

kakkak

kaaka

∴ x = a – 1 – k is another solution of the equation. ∴ The answer is C.

12. C

Let the other root be α.

31

352

3)5(2

−=

=+

−−=+

α

α

α


∴ The other root is 31

− .

13. A

Let α and 2α be the roots of the equation 2x2 + bx + 1 = 0.

⎪⎪⎩

⎪⎪⎨

⎧

=

−=+

)2( 21)(2

(1) 2

2

KK

KK

αα

αα b

From (2),

21

412

±=

=

α

α

From (1), we have

)3( 62

3

KKα

α

−=

−=

b

b

By substituting 21

±=α into (3), we have

33or 3

216or

216

±=−=

⎟⎠⎞

⎜⎝⎛−−⎟

⎠⎞

⎜⎝⎛−=b

14. C

Since α and β are the roots of 02 =++ cbxax , we have

ac

ab

=

−=+

αβ

βα

For the required quadratic equation, sum of roots

ab

=

+−=−+−=

)()(

βαβα

product of roots

ac

=

=−−=

αββα ))((

∴ The required quadratic equation is

0

0

2

2

=+−

=+⎟⎠⎞

⎜⎝⎛−

cbxaxacx

abx

15. B

32234

13

131

1313)1(

1

)1(

1

=

=

−

+=

−+

=− −−

−

f


16. B

)(11

11

1

1

11

111

xfxxxx

xx

xx

x

xx

f

−=

⎟⎠⎞

⎜⎝⎛

+−

−=

+−

=

+

−

=

+

−=⎟

⎠⎞

⎜⎝⎛

17. C

Consider the graph in C. Let us say, when x = 0, there are two corresponding values of y. ∴ y is not a function of x. ∴ It cannot be the graph of a function y = f (x).

18. B

72)6(2]36)6[(2

]6)612[(2)12(2

224

2

2

222

2

2

+−−=

−−−=

−+−−=

−−=

−=

tt

tttt

tth

∴ After 6s, the stone will attain its maximum height. 19. D

1)4(1516)4(

154)48(158

2

2

222

2

−+=

+−+=

+−++=

++=

xx

xxxxy

∴ The coordinates of the vertex are (−4, −1). 20. C

Consider 784 2 +− xx .

3)1(474)12(4

7)2(4784

2

22

22

+−=

+−+−=

+−=+−

xxxxxxx

∴ The minimum value of 784 2 +− xx is 3.

∴ The maximum value of 784

62 +− xx

is 236= .

21. C

22

2

2

)2(64

23

46

41

641

=

=

=

=


22. D

33

3

3)3(

2

)1(

)1)(1(11

2

x

x

xxxx

=

=

=−

−+−+

23. D

313626

)15(5)125(5

555555

5555 2

1

2

=

=

++

=

+⋅+⋅

=++

+

+

n

n

nn

nn

nn

nn

24. D

nm

nmn m

a

aa

=

=1

)(

∴ The answer is D. 25. B

When x = 0, y = =021 1.

∴ Both C and D do not represent the graph of xy21

= .

When x increases, y decreases.

∴ A does not represent the graph of xy21

= .

∴ The answer is B. 26. A

13333

333

)3(33

93)3(

3

3

2

3

2

3

2

26

26

=

=

=

=

=

+

×

n

n

nn

n

nn

n

nn

n

nn

n

27. B

2100 log

105020log10 log 50 log20 log

==

⎟⎠⎞

⎜⎝⎛ ×

=−+

28. B

01 log22log

22log

222log

4019

4019

4019

20102009

4019

20102009

==

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅ +


29. C By the definition of logarithms to an arbitrary base, if yx =8log , then yx=8 .

30. D

25

2log 22log 5

2log2log

4log32log

3

3

23

53

3

3

=

=

=

31. D For the straight line with equation 0743 =+− yx ,

047

)4(7intercept-

043

)4(3slope

>=−

−=

>=−

−=

y

∴ The answer is D. 32. A

Slope of the straight line = 23

2linestraight theofintercept - −=y ∴ The equation of the straight line is

223

−= xy

∴ The answer is A. 33. C

Let M be the mid-point of AC.

Coordinates of M 5) ,1(

273 ,

235

−=

⎟⎠⎞

⎜⎝⎛ ++−

=

∵ OABC is a parallelogram. ∴ M is also the mid-point of OB. (diags. of //gram) i.e. M lies on OB.

Slope of OB = slope of OM 501

05−=

−−−

=

∴ The equation of OB is

05

5=+−=

yxxy

34. B

For the straight line ax + by + c = 0,

slope 0<−=ba

0

0)(

0

2

>

>

>

ab

bba

ba

∴ B must be false. 35. D Slope of

41mL −=

Slope of 9)9(2mmL =

−−=


∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = −1

636

136

194

2

2

±==

=

−=⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−

mm

m

mm

36. A

For the straight line x + 2y = 3, slope = 21

−

For the straight line 2x + 4y + 7 = 0, slope = =−42

21

−

∴ Slopes of the two straight lines are the same.

y-intercept of the straight line x + 2y = 3 is 23 .

y-intercept of the straight line 2x + 4y + 7 = 0 is 47

− .

∴ y-intercepts of the two straight lines are different. ∴ The straight line 2x + 4y + 7 = 0 does not intersect the line x + 2y = 3.

37. C

Let 123)( 23 −+= xxxf . By the remainder theorem, we have

98

192

91

1312

313

31remainder

23

−=

−+−=

−⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛−=

⎟⎠⎞

⎜⎝⎛−= f

38. B

By the remainder theorem, we have

6

32

1124

341

1121

213

212

121

23

−=

=−

=−−−−

=−⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛−−⎟

⎠⎞

⎜⎝⎛−

=⎟⎠⎞

⎜⎝⎛−

k

k

k

k

P

∴ 1632)( 23 −−−= xxxxP By the remainder theorem, we have

the required remainder

171121216

1)2(6)2(3)2(2)2(

23

−=−+−−=

−−−−−−=

−= P

39. A

512525040)5(58)5()5(30)5(

23

===+−−

=

kk

kf


∵ 0)4()5( =−= ff ∴ By the factor theorem, x – 5 and x + 4 are the factors of f(x).

20)4)(5( 2 −−=+− xxxx By long division,

)23)(4)(5()23)(20(405853 223

−+−=−−−=+−−

xxxxxxxxx

40. D

Let 256)( 23 −−+= xxxxf . ∵

02516

2)1(5)1()1(6)1( 23

=−−+=

−−+=f

∴ By the factor theorem, x – 1 is a factor of f(x). By long division,

)23)(12)(1()276)(1(256 223

++−=++−=−−+xxx

xxxxxx

41. B

∵ When P(x) is divided by x + 1, the remainder is –2. ∴ By the remainder theorem, 2)1( −=−P

The required remainder

2)1(22

−=−=

⎟⎠⎞

⎜⎝⎛−=

P

P

42. B

Let Q(x) be the quotient. By division algorithm, we have

13)()2)(3(13)()6()( 2

+−+−=+−−−=

•

•

xxQxxxxQxxxP

71)2(3)2()22)(32()2(

=+−−−+−−−=− •QP

∴ I is true.

5)2(41)2(3)2()22)(32()2(

−−=+−+−= •

QQP

∵ Q(2) can be non-zero. ∴ II may be false.

81)3(3)3()23)(33()3(

−=+−+−= •QP

∴ III is true. ∴ The answer is B.

43. B

∵ 2)1(21)1( −==⎟⎠⎞

⎜⎝⎛=− fff

∴ 02)1(2212)1( =+=+⎟⎠⎞

⎜⎝⎛=+− fff

By the factor theorem, x + 1, 2x – 1 and x – 1 are the factors of f(x) + 2. By division algorithm, we have

2)()12)(1)(1()(

)()1)(12)(1(2)(−−−+=

−−+=+•

•

xQxxxxfxQxxxxf

∵ f(x) is a cubic polynomial with leading coefficient 2. ∴ Q(x) = 1 ∴ 2)12)(1)(1()( −−−+= xxxxf


44. D

2222

3232

)2()12( )1()2()12)(1()2( )12()1()2)(12()1(

−×+×+=−++

−×−×+=−−+

xxxxxxxxxxxx

∴ H.C.F. 2)2)(1( −+= xx

L.C.M. 322 )2()12)(12()1( −+−+= xxxx

45. C

zyxxyzyxyxyxyx

×××=

×××=

××=

283 26

24

3

33

22222

∴ H.C.F. xy2=

L.C.M.

zyxzyx

23

233

2432

=

××××=

46. C

cbacbabaabbaabP

××××==

××==××==

23223

222

3212L.C.M. 2 2H.C.F. 2 4

∴ bcacbaQ 323 632 =××××=

47. B

yxxy

xyxyxyxy

xyxy

xyxy

yx

yx

xy

+=−

+−=

−−

=

−

−

=−

−

))((

11

22

22

48. D

⎩⎨⎧

+=−+=

......(2) 43

......(1) 42

xyxxy


4or 20)4)(2(082

4432

2

−==−+=−−

−+=+

xxxxx

xxx


2

4)2(3−=

+−=y

By substituting x = 4 into (2), we have

16

4)4(3=

+=y

∴ y = –2 or 16 49. D

⎩⎨⎧

=+−−=

......(2) 4......(1) 142 2

yxxxy



25or 1

0)52)(1(05324142

2

2

−=

=−+=−−

=−−+

x

xxxxxxx


541

==+−

yy

By substituting 25

=x into (2), we have

23

425

=

=+

y

y

∴ The solutions of the simultaneous equations are (–1, 5) and ⎟⎠⎞

⎜⎝⎛

23 ,

25 .

50. B

⎩⎨⎧

−=++=

......(2) ......(1) 232 2

xkyxxy


0)2(42232

2

2

=−++

++=−

kxxxxxk

∵ The simultaneous equations have two distinct real solutions. ∴

0080)2)(2(440

2

>>>−−

>Δ

kk

k

51. D

⎩⎨⎧

+−=−=

......(2) 153......(1) 2

2 xxykxy


03)5(31532

2

2

=++−

+−=−

xkxxxkx

∵ The line 2−= kxy touches the quadratic curve 153 2 +−= xxy at one point. ∴

11or 10)11)(1(011100)3)(3(4)]5([0

2

2

−==+−=−+

=−+−

=Δ

kkkkk

k

52. A

Let the equation of the straight line be y = c, where c is a constant.

⎩⎨⎧

−+−==

......(2) 13......(1)

2 xxycy


0)1(313

2

2

=++−

−+−=

cxxxxc

Suppose y = c and 132 −+−= xxy have no intersections. ∴

450540)1)(1(4)3(0

2

>

<+−<+−−

<Δ

c

cc


∴ The answer is A. 53. C

⎩⎨⎧

+=−−=

......(2) 1......(1) 132 `2

xyxxy

From (2), x = y – 1 ……(3) By substituting (3) into (1), we have

0240482

2324223)12(21)1(3)1(2

2

2

2

2

2

=+−

=+−

+−+−=

+−+−=

−−−−=

yyyy

yyyyyyyy

yyy

∵ α and β are the y-coordinates of the intersections of the two graphs. ∴ α and β are the roots of 0242 =+− xx .

54. C

Let x cm and y cm be the length and the width of the rectangle respectively. ∵ The perimeter of the rectangle is 56 cm. ∴

.......(1) 2856)(2

=+=+

yxyx

∵ The length of the diagonal of the rectangle is 20 cm. ∴

.......(2) 40020

22

222

=+

=+

yxyx

From (1), y = 28 – x ……(3) By substituting (3) into (2), we have

16or 120)16)(12(019228038456240056784400)28(

2

2

22

22

==−−=+−

=+−

=+−+

=−+

xxxxxxx

xxxxx


16

1228=

−=y


12

1628=

−=y

∴ The dimensions of the rectangle are 12 cm × 16 cm. 55. C

For I: When x = 200°, sin 200° ≈ –0.342 and cos 200° ≈ –0.940 i.e. sin 200° > cos 200° ∴ I may not be true.

For II: When 180° < x < 270°, sin x < 0 and cos x < 0. ∴ sin x cos x > 0 ∴ II is true.

For III: When 360° < 2x < 540°, sin 2x > 0. ∴ III is true.

∴ The answer is C. 56. D

From the graph, k is the maximum value of y when x = α. ∵ 1sin1 ≤≤− x ∴ k = maximum value of y = 3 – 2(–1) = 5


2701sin

=−=°

αα

57. B

∵ 12cos1 ≤≤− x ∴ The maximum value of (3 – cos 2x) = 3 – (–1) = 4

The minimum value of (3 – cos 2x) = 3 – 1 = 2

∴ The maximum value of

122

2cos32

=

=− x

The minimum value of

2142

2cos32

=

=− x

58. D

θθθθθ

θθ

θθθθθθθθθ

cossinsincossin

tan1sin

)sin)(90tan(sin)sin)](90(180tan[sin

)180sin()270tan()90cos(

2

2

=

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

−−°−=−−°+°−=

+°−°+°

59. B

mm

mm

1

1

coscos

1coscos1

cossin

cossinsin

)tan(sin)360tan()sin(

2

2

2

−=

−=

−=

−=

=

⎟⎠⎞

⎜⎝⎛=

−−=−°−

θθ

θθ

θθ

θθθ

θθθθ

60. C

(rejected) 25sinor 0sin

0)5sin2(sin0sin5sin2

)sin1(2sin52cos2sin52

2

2

2

−==

=+=+

−=+

=+

xx

xxxx

xxxx

When 0sin =x , x = 0°, 180° or 360°

∴ The equation xx 2cos2sin52 =+ has 3 solutions for °≤≤° 3600 x .


61. C

°°=°−°°−°=

°−=

−=

=+

330or 15030360or 30180

30tantan3

1tan

01tan3

xx

x

x

62. C

21cosor

32cos

0)1cos2)(2cos3(02coscos6 2

=−=

=−+=−+

xx

xxxx

When 32cos −=x ,

integer)nearest the to(cor. 228or integer)nearest the to(cor. 1321897.48801or 1897.48180

1897.48coscos

°°=°+°°−°≈

°−≈xx

When 21cos =x ,

°°=°−°°=

°=

300or 6060360or 60

60coscosxx

∴ x = 60°, 132° (cor. to the nearest integer), 228° (cor. to the nearest integer) or 300° 63. B

1tanor 51tan

0)1)(tan1tan5(01tan4tan50tan5tan410sin5cossin4cos

2

2

22

−==

=+−=−+

=−−

=−−

xx

xxxx

xxxxxx

When 51tan =x ,

integer)nearest the to(cor. 11°=x (∵ °≤≤° 1800 x ) When 1tan −=x ,

°=°−°=

°−=

13545180

45tantanxx

∴ x = 11° (cor. to the nearest integer) or 135°

rev ex for s4

Documents