maths rev ex for s4

8/18/2019 Maths Rev Ex for S4

http://slidepdf.com/reader/full/maths-rev-ex-for-s4 1/30

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 =−− x x

(b) 0342 2 =−− x x (Leave your answers in surd form.)

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

3. Given that f ( x) =1

52

+

−

x

x, find

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

4. Simplify

ba

ba

4

12

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 + 2 x2 + 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. Simplify2 2

1 2

1 2 x x x+

− − −.

9. Solve the simultaneous equations2

5

3 2

x y

y x x

+ =⎧⎨

= − +⎩.

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

(Leave your answers in surd form if necessary.)

11. Solve θ θ θ θ sin3cos4sin2cos −=− 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 =−− x x .

(a) Find the value ofα

β

β

α + .

(b) Form a quadratic equation in x with roots β

α and

α

β .

14. (a) Solve the equation 15

12

8

+ x

= 2

7

4

− x

.

(b) Using the value of x obtained in (a), find the value of 15

12

8

+ x

.




21. In the figure, L1 is a straight line passing through A(−2, −2) of slope3

4. The equation of straight line L3 is x −

7 y − 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 = 2 x + c intersects the quadratic curve

542 2 ++= x x y at two points P and Q.

(a) Find the range of possible values ofc.(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 ofP and Q.

Multiple Choice

1. Which of the following quadratic equations has roots 3 and2

1− ?

A. ( x − 3)(2 x + 1) = 0

B. ( x + 3)(2 x − 1) = 0C. ( x + 3)(2 x + 1) = 0

D. ( x − 3)(2 x − 1) = 0

2. If the equation 02102=++ xkx has two distinct real roots, then

A. k >2

25.

B. k <2

25.

C. k >2

25− .

D. k <2

25− .

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




13. If one root of the equation 012 2 =++ bx x doubles the other, find the value(s) ofb.

A. ±3B. 3

C. −3

D.2

3±

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) =13

13

−

+− x

x

, then f (−1) =

A. −1.

B. 3

2

.C. 1.

D.3

1.

16. If f ( x) =1

1

+

−

x

x, then )

1( x

f =

A. f ( x).

B. − f ( x).

C.

)(

1

x f

.

D.)(

1

x f − .

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. Aftert second, its height (h m) above the ground is given by h = (24t

− 2t 2). When will the stone attain its maximum height?

A. 3 sB. 6 sC. 15 sD. 30 s

19. Find the coordinates of the vertex of the graph y = x2 + 8 x + 15.

A. (−4, 15)

B. (−8, −49)C. (4, −1)

D. (−4, −1)

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

62 +− x x

.

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




21. 4

1

)64( =

A. 16B. 4

C. 22 D. 2

22. 11)3( −+ x x =

A. x23

B.2

3 x

C. 332

− x

D.3

32 x

23.nn

nn

55

551

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. nmnmaaa +=

B. nm

n

m

aa

a −=

C. na− =n

a1

D. m

n

n maa =

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

y2

1= ?

A.

B.

C.

D.

26.nn

n

93

)3( 26

=

A. 1

B. 23n

3

C.n

3D. n9




27. 10log50log20log −+ =

A. 60log

B. 2C. 40log20log +

D.10log

70log

28. 4019

20102009

222log

⋅ =

A. −1B. 0C. 1D. Cannot be determined

29. If y x =8log , then

A. x y 8=

B. 8 x y =

C. y

x=8D. x

y=8

30.4log

32log

3

3 =

A. 8

B. 8log3

C. 28log3

D.2

5

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

A. B. C. D.

32. Find the equation of the straight line passing through the point (0,−2) and with slope2

3.

A. 22

3−=

x y

B. 3 x + 2 y − 4 = 0

C. 3 x − 2 y − 2 = 0

D. 2 x − 3 y − 6 = 0

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

34. The figure shows the straight lineax + by + c = 0 passing through the origin.Which of the following must be FALSE?A. c = 0B. ab < 0C. a < 0D. b < 0




35. If the two straight lines 04:1 =+ ymx L and 09:2 =− ymx L are perpendicular to each other, find the

value(s) of m.A. 0 onlyB. 6 only

C. −6 only

D. ±6

36. Which of the following straight lines does not intersect the line x + 2 y = 3?

A. 2 x + 4 y + 7 = 0B. 6 x + 3 y − 5 = 0

C. x − 2 y = 3

D. 2 x − y = 0

37. When 3 x3 + 2 x2 – 1 is divided by 3 x + 1, the remainder is

A.2

3− .

B.4

3− .

C.8

9− .

D.16

9− .

38. Let P( x) = 2 x3 – 3 x2 + kx – 1. When P( x) is divided by 2 x + 1, the remainder is 1. Find the remainder when P( x)is divided by x + 2.A. –39B. –17C. –13D. –9

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

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

40. Factorize 6 x3 + x

2 – 5 x – 2.

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

D. ( x – 1)(2 x + 1)(3 x + 2)

41. When P( x) is divided by x + 1, the remainder is –2. When P( x) is divided by 2 x + 2, the remainder isA. –1.B. –2.C. –3.D. –4.

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

true?I. P(–2) = 7

II. P(2) = –5III. P(3) = –8A. I and II onlyB. I and III onlyC. II and III onlyD. I, II and III

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

2 f

⎛ ⎞⎜ ⎟⎝ ⎠

= f (1) = –2, then f ( x) =

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

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




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

3 and ( x + 1)(2 x + 1)

2( x – 2)

2.

H.C.F. L.C.M.A. ( x + 1)( x – 2) ( x + 1)

2(2 x + 1)

2( x – 2)

3

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

2( x + 1)

2(2 x – 1)(2 x + 1)

2( x – 2)

3

45. Find the H.C.F. and L.C.M. of 4 x2 y

2, 6 x3 y and 8 xyz.

H.C.F. L.C.M.A. 4 xy 24 x3 y2 B. 4 xy 48 x3

y2 z

C. 2 xy 24 x3 y

2 z

D. 2 xy 48 x3 y

2 z

46. The H.C.F. and L.C.M of P and Q are 2ab and 12a3b

2c respectively. If P is 4ab

2, then Q isA. 3a

2bc.

B. 6a2bc.

C. 6a3bc.

D. 6a3b

2c.

47. =

−

−

y x

y

x

x

y

11

A. y – x

B. x + y

C.1

x y+

D.1

y x−

48. If2 4

3 4

y x x

y x

⎧ = + −⎨

= +⎩, then y =

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

49. Solve the simultaneous equations22 4 1

4

y x x

x y

⎧ = − −⎨

+ =⎩

.

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

D. (–1, 5) or5 3

,2 2

⎛ ⎞⎜ ⎟⎝ ⎠

50. If the simultaneous equations22 3 2 y x x

y k x

⎧ = + +⎨

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

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

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

A. 1B. –1C. 1 or –1D. 1 or –11




59. Ifm

1 cos =θ and 0° < θ < 90°, then =−°− )360tan()sin( θ θ

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 x x 2cos2sin52 =+ have?

A. 1

B. 2C. 3D. 4

61. Solve 3 tan 1 0 x + = 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 0 x 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 0 x 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 =−− x x

( x −14)( x + 2) = 0 x = 14 or x = –2

(b) 0342 2 =−− x x

x =)2(2

)3)(2(4)4()4( 2 −−−±−−

=4

404 ±

=2

102 ±

2. 0))(1(482 <− k

4k > 64k > 16

3. (a) f (2) = 12

5)2(2

+

− = 3

1

−

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

5)1(2

++

−+

t

t

=2

32

+

−

t

t

(c) When x = −1, f ( x) is undefined.

4.

ba

ba

4

1

2

212

)(

)( −

=

ba

ba

2

1

24 −

=21

2

14

+

−

b

a

=3

2

7

b

a

5. (a) 25log4log +

=

2

5log4log +

= )54log( 2×

= 100log

= 2(b) 2log18log90log +−

= ⎟ ⎠

⎞⎜⎝

⎛ ×

18

290log

= 10log

= 1

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

31

−−

−=

2

1−

(b) The equation of the straight line:

)2(

2

1

11 −

⎟ ⎠

⎞⎜⎝

⎛ −

−=− x y

y – 1 = 2( x – 2)

2 x – y − 3 = 0

7. (a) ∵ x – 3 is a factor of p( x).∴ By the converse of the factor theorem,

0)3( = p




513

0631827

06)3()3(2)3( 23

−=

=+++

=+++

k

k

k

17−=k

(b) By the remainder theorem,)2(remainder −= p

63488

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

+++−=

+−−−+−=

40=

8. 2

2

1

122 −−

+− x x x

)2)(1(

2

)1)(1(

1

−++

−+=

x x x x

)2)(1)(1(

)1(22

−−+

−+−=

x x x

x x

)2)(1)(1(222−−+

−+−= x x x

x x

)2)(1)(1(

43

−−+

−=

x x x

x

9. ⎩⎨⎧

+−=

=+

23

5

2 x x y

y x

)2(

)1(

KK

KK

From (1), we have y = 5 – x ……(3)

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

235 2 +−=− x x x

0322 =−− x x 0)1)(3( =+− x x

1or3 −== x x By substituting x = 3 into (3), we have

2

35

=

−= 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 ofθ and OP = r .

∵ tan θ = 2

∴ Let 1−= x and 2−= y .

5

)2()1( 22

=

−+−=r

∴ 5

2sin −=θ

θ θ cos)sin(90 =+°

5

1−=




11.

3tan

3 cos

sin

cos3sin

coscos4sin2sin3

sin3cos4sin2cos

=

=

=

−=−

−=−

θ

θ

θ

θ θ

θ θ θ θ

θ θ θ θ

∴

fig.)sig.3to(cor.252orfig.)sig.3to(cor.71.6

71.5651180or71.5651

°°=

°+°°≈θ

12. Semi-perimeter of △ ABD

cm24

cm2

211710

=

++=

∴ Area of △ ABD

2

2

cm371424

cm)2124)(1724)(1024(24

×××=

−−−=

2

cm 84= ∴ Area of parallelogram ABCD

ABD△of area2 ×=

2

2

cm168

cm842

=

×=

Section A(2)

13. (a) α + β =2

6 = 3, αβ =

2

3−

α

β

β

α

+ = αβ

β α 22 +

=αβ

αβ β α 2)( 2 −+

=

2

3

39

−

+

= −8

(b)α

β

β

α ⋅ = 1

The required quadratic equation is:

0182 =++ x x

14. (a) 15

12

8

+ x

= 2

7

4

− x

15

)12(3

2

+ x

= 72 − x

5

12 + x = x −7

2 x + 1 = 5 x − 35

x = 12

(b) When x = 12, 15

12

8

+ x

= 15

25

8 = 32

15. (a)

x −1 0 1 2 3 4 5

y 3 −2 −5 −6 −5 −2 3(b) graph of y = x

2 − 4 x −2

Alternatively,

when x = 12, 15

12

8

+ x

= 2

7

4

− x

= 2

5

4 = 32




(c) (i) From the graph, the roots of x2 − 4 x − 2 = 0 are −0.5 and 4.5.

(ii) From the graph, the minimum value of y = x2 − 4 x − 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 − 2 x) cm222 )25.34()5.10( x x x −=++

22 413825.119025.110212 x x x x +−=++

010801592 2 =+− x x 0)72)(152( =−− x x

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 = 2 x – 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) 2 x + 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) 8508.2or3508.0

4

415

)2(2

)2)(2(455sin

02sin5sin2

2sin5sin2

2

2

2

−≈

±−=

−−±−=

=−+

=+

θ

θ θ

θ θ

When 3508.0sin ≈θ ,

fig.)sig.3to(cor. 159orfig.)sig.3to(cor. 20.5 i.e.

5363.20180or20.5363

°°=

°−°°≈

θ

θ

(b) ∵ °≤≤° 72020 θ

From (a), we have

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

fig.)sig.3to(cor.7.79orfig.)sig.3to(cor.10.3

159.4637360or20.5363360or

159.4637or20.53632

°°

°°=

°+°°+°

°°≈

θ

θ

Section B

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

Then CD = (200 − 2 x) m.

∴ A = x(200 − 2 x)

= 200 x − 2 x2

= −2( x2 − 100 x)

= −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 23 ⋅

= 6log5log 32

3 +

= 6log5log2 33 +

= ba +2

(b) 10log3 = ⎟ ⎠

⎞⎜⎝

⎛

15

150log3

= 15log150log 33 −

= )5log3(log2 33 +−+ ba = aba −−+ 12

= a + b − 1

(c)

a

b+=

+=

×=

=

1

5log

6log3log

5log

)63(log

5log

18log

18log

3

33

3

3

3

3

5

Alternatively,

10log3 =3

30log3

= 3log6log5log 333 −+

= a + b − 1




21. (a) Equation of L1: )2(3

42 +=+ x y

4 x − 3 y + 2 = 0

(b) Slope of L3 =7

1

Slope of L4 = −7

Equation of L4: )7

12(73 −−=+ x y

7 x + y − 9 = 0

(c) Solving⎩⎨⎧

=−+

=+−

097

0234

y x

y x, 4 x − 3(9 − 7 x) + 2 = 0

25 x − 25 = 0, i.e. x = 1, y = 2

∴ B = (1, 2)

Slope of L5 = slope of L2 =

3

4

1− =

4

3−

∴ Equation of L5: )1(

4

32 −−=− x y

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

3 x + 4 y − 12 = 0or

22. (a)⎩⎨⎧

+=

++=

c x y

x x y

2

542 2

)2......(

)1......(


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

5422

2

2

=−++

++=+

c x x

x xc x

∵ The straight line y = 2 x + c intersects the quadratic curve 542 2 ++= x x y at two points.

∴ Δ > 0

0)5)(2(422 >−− c

2

9

368

08404

>

>

>+−

c

c

c

∴ The range of possible values of c is2

9>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

0122

0)65(22

2

2

2

±−=

−−±−=

=−+

=−++

x

x x

x x

3660.1or0.3660 −≈

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

fig.)sig.3to(cor.73.6

6)3660.0(2

=

+≈ y

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

fig.)sig.3to(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.

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




Multiple Choice

1. A

2

1 or3

012or03

0)12)(3(

−==

=+=−

=+−

x x

x x

x x

∴ The answer is A.

2. B

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

∴ Δ> 0

i.e.

2

25

1008

0)2)((4102

<

<

>−

k

k

k

3. D

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

3or2

0)3)(2(

06

6

)1()2(3

)1()31)(21(

2

2

−=

=−+

=−−

−=−

−=−

−=−+

k

k k

k k

k k

k k

k k

4. D

∵ The graph of y = ax

2

+ bx + c opens upwards.∴ a > 0

∵ The graph has two x-intercepts.

∴ Δ> 0

∴ The answer is D.

5. A

3or1

0)3)(1(

0)3)(12(

0)3()3)(2(

3)3)(2(

−==−+

=−−+

=−−−+

−=−+

x

x x

x x

x x x

x x x

6. B

∵ It is obvious that 2 is a root of the equation2

12

1−=−

x x .

∴ The answer may be A or B.

By substituting2

1−= x into

2

12

1−=−

x x ,

L.H.S. = 2

1

222

1

2

1

1

2

1

−=+−=⎟

⎠

⎞⎜⎝

⎛ −

−− = R.H.S.

∴ 2

1− 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

cm2

)7(749

cm

22

)7(7

2

)7(77

⎥⎦

⎤⎢⎣

⎡−−−=

⎥

⎦

⎤⎢

⎣

⎡−

−−

−−=

x x

x x x




∵ Area of △ BPQ 2cm20=

∴

(rejected)10or4

0)10)(4(

04014

02072

202

)7(749

2

2

2

=

=−−

=+−

=+−

=−−−

x

x x

x x

x x

x x

8. D

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

∴ Δ = 0

i.e.

4or(rejected) 0

0)4(

04

0))(1(4

2

2

−=

=+

=+

=−−

b

bb

bb

bb

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

9. A

By the quadratic formula,

3

31

3

33

6

126

)3(2

)2)(3(466 2

±−=

±−=

±−=

−±−= x

10. D

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

12

)4(3

)(

3

1

3 and

41

4

22

=

−−=

+=+

−=−

=

−=−=+

β α αβ αβ β α

αβ

β α

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(

=

+−=−+−=

+−−−−−=

k ak

k ak

k aak a

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

∴ The answer is C.

12. C

Let the other root be α .

3

1

352

3

)5(2

−=

=+

−−=+

α

α

α




∴ The other root is3

1− .

13. A

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

⎪⎪⎩

⎪⎪⎨

⎧

=

−=+

)2( 2

1

)(2

(1) 2

2

KK

KK

α α

α α b

From (2),

2

1

4

12

±=

=

α

α

From (1), we have

)3( 6

23

KKα

α

−=

−=

b

b

By substituting 2

1±=α into (3), we have

3

3or3

2

16or

2

16

±=

−=

⎟ ⎠

⎞⎜⎝

⎛ −−⎟

⎠

⎞⎜⎝

⎛ −=b

14. C

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

a

c

a

b

=

−=+

αβ

β α

For the required quadratic equation,

sum of roots

a

b=

+−=

−+−=

)(

)(

β α

β α

product of roots

a

c=

=

−−=

αβ

β α ))((

∴ The required quadratic equation is

0

0

2

2

=+−

=+⎟ ⎠

⎞⎜⎝

⎛ −

cbxax

a

c x

a

b x

15. B

3

2

2

3

4

13

13

1

13

13)1(

1

)1(

1

=

=

−

+=

−

+=− −−

−

f




16. B

)(

1

1

1

1

1

1

11

11

1

x f

x

x

x

x

x

x x

x

x

x

x 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

+−−=

−−−=

−+−−=

−−=

−=

t

t

t t

t t

t t h

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

19. D

1)4(

1516)4(

154)48(

158

2

2

222

2

−+=

+−+=

+−++=

++=

x

x

x x

x x y

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

20. C

Consider 784 2 +− x x .

3)1(4

74)12(4

7)2(4784

2

22

22

+−=

+−+−=

+−=+−

x

x x

x x x x

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

∴ The maximum value of784

62 +− x x

is 23

6= .

21. C

22

2

2

)2(64

2

3

4

6

4

1

64

1

=

=

=

=




22. D

3

3

3

3)3(

2

)1(

)1)(1(11

2

x

x

x x x x

=

=

=

−

−+−+

23. D

3

13

6

26

)15(5

)125(5

555

555

55

55 2

1

2

=

=

+

+=

+⋅+⋅=

++

+

+

n

n

nn

nn

nn

nn

24. D

n

m

nmn m

a

aa

=

=1

)(


25. B

When x = 0, y = =02

11.

∴ Both C and D do not represent the graph of x y 2

1

= .

When x increases, y decreases.

∴ A does not represent the graph of x

y2

1= .


26. A

1

3

3

3

3

33

3

)3(3

3

93

)3(

3

3

2

3

2

3

2

26

26

=

=

=

=

=

+

×

n

n

nn

n

nn

n

nn

n

nn

n

27. B

2

100log

105020log10log50log20log

=

=

⎟ ⎠ ⎞⎜

⎝ ⎛ ×=−+

28. B

0

1log

2

2log

2

2log

2

22log

4019

4019

4019

20102009

4019

20102009

=

=⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ =

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅ +




29. C

By the definition of logarithms to an arbitrary base,

if y x =8log ,

then y x=8 .

30. D

2

5

2log2

2log5

2log

2log

4log

32log

3

3

2

3

5

3

3

3

=

=

=

31. D

For the straight line with equation 0743 =+− y x ,

04

7

)4(

7intercept-

04

3

)4(

3slope

>=−

−=

>=−

−=

y


32. A

Slope of the straight line =2

3

2linestraighttheof intercept- −= y

∴ The equation of the straight line is

22

3

−= x y


33. C

Let M be the mid-point of AC .

Coordinates of M

5),1(

2

73 ,

2

35

−=

⎟ ⎠

⎞⎜⎝

⎛ ++−=

∵ 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

=+

−=

y x

x y

34. B

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

slope 0<−= ba

0

0)(

0

2

>

>

>

ab

bb

a

b

a

∴ B must be false.

35. D

Slope of4

1 m L −=

Slope of9)9(

2

mm L =

−−=




∵ L1 ⊥ L2

∴ Slope of L1 × slope of L2 = −1

6

36

136

194

2

2

±=

=

=

−=⎟ ⎠

⎞⎜⎝

⎛ ⎟

⎠

⎞⎜⎝

⎛ −

m

m

m

mm

36. A

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

1−

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

2

2

1−

∴ Slopes of the two straight lines are the same.

y-intercept of the straight line x + 2 y = 3 is2

3.

y-intercept of the straight line 2 x + 4 y + 7 = 0 is4

7− .

∴ y-intercepts of the two straight lines are different.

∴ The straight line 2 x + 4 y + 7 = 0 does not intersect the line x + 2 y = 3.

37. C

Let 123)( 23 −+= x x x f .

By the remainder theorem, we have

9

8

19

2

9

1

13

12

3

13

3

1remainder

23

−=

−+−=

−⎟ ⎠

⎞⎜⎝

⎛ −+⎟

⎠

⎞⎜⎝

⎛ −=

⎟ ⎠

⎞⎜⎝

⎛ −= f

38. B

By the remainder theorem, we have

6

32

1124

3

4

1

112

1

2

13

2

12

12

1

23

−=

=−

=−−−−

=−⎟ ⎠

⎞⎜⎝

⎛ −+⎟

⎠

⎞⎜⎝

⎛ −−⎟

⎠

⎞⎜⎝

⎛ −

=⎟ ⎠

⎞⎜⎝

⎛ −

k

k

k

k

P

∴ 1632)( 23−−−= x x x xP

By the remainder theorem, we havethe required remainder

17

1121216

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

)2(

23

−=

−+−−=

−−−−−−=

−= P

39. A

5

12525

040)5(58)5()5(3

0)5(

23

=

=

=+−−

=

k

k

k

f




∵ 0)4()5( =−= f f

∴ By the factor theorem, x – 5 and x + 4 are the factors of f ( x).

20)4)(5( 2 −−=+− x x x x

By long division,

)23)(4)(5(

)23)(20(405853 223

−+−=

−−−=+−−

x x x

x x x x x x

40. DLet 256)( 23 −−+= x x x x f .

∵

0

2516

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

++−=

++−=−−+

x x x

x x x x x x

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(

2

2

−=

−=

⎟ ⎠

⎞⎜⎝

⎛ −=

P

P

42. B

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

13)()2)(3(

13)()6()( 2

+−+−=

+−−−=

•

•

x xQ x x

x xQ x x xP

7

1)2(3)2()22)(32()2(

=

+−−−+−−−=− • QP

∴ I is true.

5)2(4

1)2(3)2()22)(32()2(

−−=

+−+−= •

Q

QP

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

8

1)3(3)3()23)(33()3(

−=

+−+−= • QP

∴ III is true.


43. B

∵ 2)1(2

1)1( −==⎟

⎠

⎞⎜⎝

⎛ =− f f f

∴ 02)1(22

12)1( =+=+⎟

⎠ ⎞⎜

⎝ ⎛ =+− f f f

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

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

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

−−−+=

−−+=+

•

•

xQ x x x x f

xQ x x x x f

∵ f ( x) is a cubic polynomial with leading coefficient 2.

∴ Q( x) = 1

∴ 2)12)(1)(1()( −−−+= x x x x f




44. D

2222

3232

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

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

−×+×+=−++

−×−×+=−−+

x x x x x x

x x x x x x

∴ H.C.F. 2)2)(1( −+= x x

L.C.M. 322 )2()12)(12()1( −+−+= x x x x

45. C

z y x xyz

y x y x

y x y x

×××=

×××=

××=

28

3 26

24

3

33

22222

∴ H.C.F. xy2=

L.C.M.

z y x

z y x

23

233

24

32

=

××××=

46. C

cbacba

baab

baabP

××××==

××==

××==

23223

222

3212L.C.M.

2 2H.C.F.

2 4

∴ bcacbaQ323 632 =××××=

47. B

y x

x y

x y x y

x y

x y

xy

x y

xy

x y

y x

y

x

x

y

+=

−

+−=

−

−=

−

−

=−

−

))((

11

22

22

48. D

⎩⎨

⎧

+=

−+=

......(2) 43

......(1) 42

x y

x x y


4or2

0)4)(2(

082

4432

2

−=

=−+

=−−

−+=+

x

x x

x x

x x x


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

y x

x x y





2

5or1

0)52)(1(

0532

4142

2

2

−=

=−+

=−−

=−−+

x

x x

x x

x x x


5

41

=

=+−

y

y

By substituting2

5= x into (2), we have

2

3

42

5

=

=+

y

y

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

⎞⎜⎝

⎛

2

3 ,

2

5.

50. B

⎩⎨⎧

−=

++=

......(2)

......(1) 232 2

xk y

x x y


0)2(42

232

2

2

=−++

++=−

k x x

x x xk

∵ The simultaneous equations have two distinct real solutions.

∴

0

080)2)(2(44

02

>

>>−−

>Δ

k

k k

51. D

⎩⎨⎧

+−=

−=

......(2) 153

......(1) 22

x x y

kx y


03)5(3

15322

2

=++−

+−=−

xk x

x xkx

∵ The line 2−= kx y touches the quadratic curve 153 2 +−= x x y at one point.

∴

11or1

0)11)(1(

01110

0)3)(3(4)]5([

0

2

2

−=

=+−

=−+

=−+−

=Δ

k

k k

k k

k

52. A

Let the equation of the straight line be y

=c, where

c is a constant.

⎩⎨⎧

−+−=

=

......(2) 13

......(1) 2

x x y

c y


0)1(3

132

2

=++−

−+−=

c x x

x xc

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

∴

4

5054

0)1)(1(4)3(

0

2

>

<+−

<+−−

<Δ

c

c

c





53. C

⎩⎨⎧

+=

−−=

......(2) 1

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

x y

x x y

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

024

0482

23242

23)12(2

1)1(3)1(2

2

2

2

2

2

=+−

=+−

+−+−=

+−+−=

−−−−=

y y

y y

y y y y

y y y y

y y y

∵ α and β are the y-coordinates of the intersections of the two graphs.

∴ α and β are the roots of 0242 =+− x x .

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) 28

56)(2

=+

=+

y x

y x

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

∴

.......(2) 400

20

22

222

=+

=+

y x

y x

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

16or12

0)16)(12(

019228

0384562

40056784

400)28(

2

2

22

22

=

=−−

=+−

=+−

=+−+

=−+

x

x x

x x

x x

x x x

x x


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° < 2 x < 540°, sin 2 x > 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




270

1sin

=

−=°

α

α

57. B

∵ 12cos1 ≤≤− x

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

= 4The minimum value of (3 – cos 2 x) = 3 – 1

= 2

∴ The maximum value of

1

2

2

2cos3

2

=

=− x

The minimum value of

2

1

4

2

2cos3

2

=

=− x

58. D

θ θ

θ

θ θ

θ θ

θ θ θ

θ θ θ

θ θ θ

cossin

sin

cossin

tan

1sin

)sin)(90tan(sin

)sin)](90(180tan[sin

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

2

2

=

⎟ ⎠

⎞⎜⎝

⎛ =

⎟ ⎠

⎞⎜⎝

⎛ =

−−°−=

−−°+°−=

+°−°+°

59. B

m

m

mm

1

1

coscos

1

cos

cos1

cos

sin

cos

sinsin

)tan(sin)360tan()sin(

2

2

2

−=

−=

−=

−=

=

⎟ ⎠

⎞⎜⎝

⎛ =

−−=−°−

θ θ

θ

θ

θ

θ

θ

θ θ

θ θ θ θ

60. C

(rejected)2

5sinor0sin

0)5sin2(sin

0sin5sin2)sin1(2sin52

cos2sin52

2

2

2

−==

=+

=+−=+

=+

x x

x x

x x x x

x x

When 0sin = x ,

x = 0°, 180° or 360°

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



61. C

°°=

°−°°−°=

°−=

−=

=+

330or150

30360or30180

30tantan

3

1tan

01tan3

x

x

x

x

62. C

2

1cosor

3

2cos

0)1cos2)(2cos3(

02coscos6 2

=−=

=−+

=−+

x x

x x

x x

When3

2cos −= x ,

integer)nearesttheto(cor. 228orinteger)nearesttheto(cor.1321897.48801or1897.48180

1897.48coscos

°°=°+°°−°≈

°−≈

x

x

When2

1cos = x ,

°°=

°−°°=

°=

300or60

60360or60

60coscos

x

x

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

63. B

1tanor5

1tan

0)1)(tan1tan5(

01tan4tan5

0tan5tan41

0sin5cossin4cos

2

2

22

−==

=+−

=−+

=−−

=−−

x x

x x

x x

x x

x x x x

When5

1tan = x ,

integer)nearesttheto(cor. 11°= x (∵ °≤≤° 1800 x )

When 1tan −= x ,

°=

°−°=

°−=

135

45180

45tantan

x

x

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

maths rev ex for s4

Documents