form 4 set 1(2)

3

USAHA +DOA+TAWAKAL

3781/1 2014 Maths Catch Network [Lihat halaman sebelah]SULIT

SKEMA JAWAPANEDISI MID TERM PILIHAN 1

PAPER 1 /KERTAS 1Jawapan1 (a) 4, 10

(b) {4, 10}

2 (a) Langkah 1: fLangkah 1:

(b) f(xLangkah 1: 8xLangkah 2: 8xLangkah 3: x

3Given f(x) =

7x and gf(x) =

xx .

Diberi f(x) =7x dan gf(x) = x .

Langkah 1: gf(x) = g(f(x))

Langkah 2: = g(7x)

Langkah 3: Let y =7x

Langkah 4: So, x =7y

Langkah 5: g(y) =( )7

y

7y

Langkah 6: = ( )7y ( )y

7

Langkah 7:87 y

Langkah 8: Therefore, g : x87 x

4 Given f(x) = x + 6 and gf(x) = 5x + 4.Diberi f(x) = x + 6 dan gf(x) = 5x + 4.Langkah 1: gf(x) = g(f(x))Langkah 2: = g(x + 6)Langkah 3: g(x + 6) = 5x + 4Let y = x + 6,So,Langkah 4: x = yLangkah 5: g(y) = 5( )y + 4Langkah 6: = 5yLangkah 7: Therefore, g : x x

5 Given f(x) = x g(x x.

Langkah 1: gf(x) = g(f(x))Langkah 2: = g(xLangkah 3: xLangkah 4: x + 13Langkah 5: fg(x) = f(g(x))Langkah 6: = f x)Langkah 7: xLangkah 8: x

6 (a) Let f (4) = kSoLangkah 1: f(k) = 4Langkah 2: k + 1 = 4Langkah 3: k = 3Langkah 4: Therefore, f (4) = 3

(b) Let f (x) = ySoLangkah 1: f(y) = xLangkah 2: y + 1 = xLangkah 3: y = xLangkah 4: Therefore, f (x) = x

7 9(1)2 p = 0Langkah 1: p = 0Langkah 2: p = 1

8 (x a)(x b) = 0Langkah 1: (x + 3)(xLangkah 2: x2 xLangkah 3: 5x2 xLangkah 4: Therefore, m n

9 y xy x2 x + pLangkah 1: x2 x + p xLangkah 2: x2 + pThe equation does not have real rootsb2 ac < 0Langkah 3: (0)2 pLangkah 4: 4pLangkah 5: 4p < 24Langkah 6: p < 6

10 x2 + 9xLangkah 1: x2 x + 3)

Langkah 2: x2 x32 )2 3

2 )2 + 3)

Langkah 3: x32 )2 9

4 + 3)

Langkah 4: x32 )2 +

34 )

Langkah 5: x32 )2 9

4

Langkah 6: Therefore, m32 and n

94

11 y x2 + 2qx qLangkah 1: x2 qx + 9q)

Langkah 2: x2 qx + (q

2 )2 q2 )2 + 9q)

Langkah 3: x q)2 q2 + 9q)Langkah 4: x q)2 + q2 qMaximum value = q2 qLangkah 5: q2 qLangkah 6: q2 q + 8 = 0Langkah 7: (q qLangkah 8: q = 1 or q = 8

12 (a) s(b) t = 5(c) x = 7

SKEMA JAWAPAN

4

USAHA +DOA+TAWAKAL

3781/1 2014 Maths Catch Network [Lihat halaman sebelah]SULIT

13 (3x + 5)(x x + 16Langkah 1: 3x2 + 11x x + 16Langkah 2: 3x2 + 3xLangkah 3: (x2 + xLangkah 4: 3(x + 2)(x

Langkah 5: The range of values of x x

14 2x + 8 x + 7 = 2Langkah 1: 2x28 x27 = 21

Langkah 2: 256(2x x) = 21

Langkah 3: 128(2x) = 21

Langkah 4: 2x + 7 = 21

Langkah 5: x + 7 = 1Langkah 6: x

15 25 x = 6253x

Langkah 1: 5 x = 54(3x

Langkah 2: x xLangkah 3: x

Langkah 4: x1

13

16 log2 (7x 2 (5x + 3) = 4

Langkah 1: log27x + 45x + 3 = 4

Langkah 2:7x + 45x + 3 = 24

Langkah 3: 7x + 4 = 16(5x + 3)Langkah 4: 7x + 4 = 80x + 48

Langkah 5: x4473

17 log49 x = log7 2

Langkah 1:log7 x

log7 49 = log7 2

Langkah 2:log7 x

2 log7 7 = log7 2

Langkah 3: log7 x = 2 log7 2Langkah 4: log7 x = log7 22

Langkah 5: x = 4

18 AB = 5Langkah 1: (7c c)2 + ( )2 = 5Langkah 2: (6c + 5)2 2 = 5Squaring both sides,Langkah 3: (6c + 5)2 2 = 52

Langkah 4: (6x + 5)2

Langkah 5: (6x + 5)2 = 16Langkah 6: 6x + 5 = ±4

Langkah 7: c16 or c

32

19(1, 7) = (

j2 ,

k2 )

Langkah 1:j

2 = 1

Langkah 2: j

Langkah 3: j = 12

Langkah 4:k

2 = 7

Langkah 5: kLangkah 6: k = 15

20 Area of A

Langkah 1: =12 | 7 4 10 7 |=

12

Langkah 2: =12 |0|

Langkah 3: = 0 unit2

Langkah 4: Hence, A, B and C lie on a straight line.

21 Gradient of JK

Langkah 1: = 1 + 9

Langkah 2:35

Langkah 3:35 )m2

Langkah 4: m2 =53

Midpoint of JK

Langkah 1: = ( 2 ,9 + 3

2 )

Langkah 2:

Equation:

Langkah 1: y53 (x + 4)

Langkah 2: 3y x + 20Langkah 3: 5x y + 38 = 0

22 (a)Mean, x̄ =

xN

Langkah 1: =3 + 2 + 7 + 5 + 8 + 1 + 2

7

Langkah 2: =287

Langkah 3: = 4Median = 3Mode = 2

(b) New mean = 4(4) + 3 = 19New median = 4(3) + 3 = 15New mode = 4(2) + 3 = 11

23 Langkah 1: Rearrange the data,38 41 45 48 49 50 62Langkah 2: First quartile = 41Langkah 3: Third quartile = 50Langkah 4: Interquartile rangeLangkah 5: = 9

24 (a)Langkah 1: Mean =

246

Langkah 2: = 4(b)

Langkah 1: Standard deviation =9006 - 42

Langkah 2: = 134Langkah 3: = 11.576

25 Let C = (x, y)Given MC = 7