MATHEMATICS T

Bertujuan membantu calon STPM yang gagal atau berprestasi lulus sebahagian ke arah meningkatkan pencapaian agar mencapai tahap LULUS PENUH Digubal oleh tenaga pengajar STPM berpengalaman di Johor Darul Takzim

1

PANEL PENGGUBAL

1. BOOPATHY RAMASAMY SMK TUN HABAB, KOTA TINGGI 2. LIM POH THAI SMK DATO PENGGAWA BARAT, PONTIAN 3. KOH AH LEH SMK DATO BENTARA LUAR, BATU PAHAT 4. GOH KIM HENG SMK SERI MERSING, MERSING 5. LEE KIM SING SEKOLAH TINGGI SEGAMAT, SEGAMAT

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POLYNOMIALSCHECKLIST:AT THE END OF THIS TOPIC STUDENTS SHOULD BE ABLE TO:1. 2. 3. 4. 5. 6. 7. 8. 9. Understand the meaning of the degree and coefficients of polynomials. Carry out elementary operations on polynomials. Use the condition for the equality of two polynomials. Find the factors and zeroes of polynomials Prove and use the remainder and factor theorems. Use the process of completing the square for a quadratic polynomial Derive the quadratic formula Solve linear, quadratic, and cubic equations that can be transformed into quadratic or cubic equations. Use the discriminant of a quadratic equation to determine the properties of its roots. 10. Prove and use the relationships between the roots and coefficients of a quadratic equation. 11. Solve inequalities involving polynomials of degrees not exceeding three, rational functions and the modulus sign. 12. Solve a pair of simultaneous equat ions involving polynomials of degree not exceeding three. 13. Express rational functions in partial fractions.

3

SMART TIPSNo. 1 Formula / Termsa0 x n a1x n 1 a2 x n 2 . ar x n r . an is a polynomial of degree n , such that n is a real number and coefficients a i , i ! 0,1,2, . , n with a0 { 0.

Points to Note Types of polynomials : linear, quadratic, cubic, quartic etc.

2

Equality of Two Polynomials

Two polynomials are equal if their degrees are the same, having same number of terms and all the like terms are having equal coefficients. To add or subtract two polynomials, the coefficients of the like terms are added or subtracted. To multiply two polynomials, all the terms in a polynomial will multiply with all the terms of the other polynomial, and the products will be added. A polynomial can be divided by a polynomial of same degree or of lower degree. Division ends when the remainder has its degree lower than divisor.a is the zero of (x).

3

Addition and Subtraction of polynomials

4

Multiplication of polynomials

5

Division of polynomials P(x) = Q(x)D(x) + R(x)

6

1. If a bi is a zero, then a bi is also a zero of the polynomial. 2. If a b is a zero, then a b is also a zero of the polynomial. If remainder is 0, then x a is a factor of (x).

9

Factor Theorem: If x a is a factor of P (x) then P(a) = 0.

10

Completing the Square for a Quadratic Polynomial ax 2 bx c a ( x k )2 h

Remainder

8

( x) | (ax 2 bx c)Q( x) Ax BAx B ; Divisor+ ax 2 bx c

Use it to find remainder if by quadratic equation.

If (x-a) and (x- b) are the factors of P(x) then (x a)(x b) also factors of P(x) can be expressed as x2 (a+b)x + ab e.g. 2x2 4x + 5 = 2(x 1)2 + 3 2 22 3x2 + 4x + 6 = 3(x + )2 + 3 3 4

7

Remainder Theorem When a polynomial P (x) is divided by ( x a ) , the remainder is P(a).

Zeroes For polynomial P (x) , if a is a zero then (a) ! 0 .

(x) and x a is a factor of

(x) is divided

11

ax 2 bx c ! 0 ,

Quadratic formula 12 Properties of roots :b 2 4ac " 0 --b 2 4ac ! 0 --b 2 4ac 0 --b2 4ac u 0 ---

x!

b s b 2 4ac 2a

To find the roots of quadratic equation. (must know how to prove)

two roots are real and distinct two roots are real and equal two roots are complex real roots

b 2 4 ac is the discriminant of quadratic

equation ax 2 bx c = 0

13

Inequalities x a, x " a,

a x

x

a.

e.g x 5 4, 4

a or x " a

x5

4.

3 x 7 " 5, 3 x 7

5 or 3 x 7 " 5

For f ( x) " g ( x) or f ( x) ! g ( x) ! 14

f ( x)

g ( x) , use

f ( x)2 g ( x)2

and

Sometimes graphical method is preferred.

before squaring both sides.

Let

( x) R | Q ( x) , ( x) ( x) and is the remainder.

here Q ( x ) is the quotient

b) If the degree of (x) < degree of ( x) = + + ( x) 15

If E and F are the roots of the quadratic equation ax2 + bx + c = 0, then sum of the roots = E + F = b a c product of the roots = E F = a

16

x3 y3 = (x y)(x2 + xy + y2 ) x3 + y3 = (x + y)(x2 xy + y2 ) Important formula to solve simultaneous equations.

P( x) be the fractional function. D ( x) a) If the degree of (x) u degree of (x) by (x) . Express

(x), divide

Express fractional functions in partial fractions determine first whether it is a proper fraction or improper fraction. Make sure D(x) is expressed as linear factors( repeating linear factors) or quadratic factor(which is irreducible).

(x), + Some important identities: 1. E 2 F 2 ! (E F ) 2 2EF 2. (E F )2 ! (E F ) 2 4EF 3. E F ! (E F ) 2 4EF 4. E 3 F 3 ! (E F )((E F )2 3EF ) 5. E 3 F 3 ! (E F )((E F ) 2 EF )

5

GUIDED EXAMPLES1. When the polynomial 2x 3 + ax2 + 3x + b is divided by (x+1) and (2x 1) respectively the remainder are - 13 and 5. Find the value of a and b. SOLUTION: Let f(x) = 2x3 + ax2 + 3x + b f(-1) = 2(-1)3 + a(-1)2 + 3(-1) + b = -13 a + b =-5 f( ) = 2( )3 + a( )2 + 3( ) + b = 5 a + 2b = -2 from 1 and 2 2. a = -8 , b = 3

----- 1 ----- 2

Show that x 1 is a factor of P(x), where P(x) = 6x3 + 11x2 5x 12. Hence find the other factors of P(x). SOLUTION: P(1) = 6(1)3 + 11(1)2 5(1) 12 =0 Since P(1) = 0, hence x- 1 is a factor of the polynomial P(x) P(x) = 6x 3 + 11x2 5x 12 = (x 1)(6x2 + 17x + 12) = (x 1)(3x + 4)(2x + 3) Other factors of the polynomial p(x) are 3x + 4 and 2x + 3 6 x 2 17 x 12 x - 1 6 x 3 11x 2 5 x 12 6x3 6x2 17x2 -5x 17x2 -17x 12x 12 12x 12 0

3.

Show that 2x 1 is a factor of the expression 6x3 13x2 + x + 2.Hence find the set values of x such that 6x3 13x2+ x + 2 > 0. SOLUTION: Let p(x) = 6x3 13x2 + x + 2 p( ) = 6( )3 13 ( )2 + ( ) + 2 =0 Since p( ) = 0, hence 2x- 1 is a factor of the polynomial p(x) p(x) = 6x3 13x2 + x + 2 = (2x 1)(3x2 5x 2) = (2x -1)(3x + 1)(x 2) (2x -1)(3x + 1)(x 2) > 0 Try to obtain the quotient by long division

-1/3

1/2

2

{x : x IR , - 1/3 < x < 1/2 } { x : x IR, x > 2}

6

4.

If m and n are the roots of the equation 2x2 + 3x 1 = 0, find the value of m3 + n3. Obtain the 1 1 quadratic equation whose roots are 2 and 2 . m n SOLUTION: Sum of the roots , m + n = 3 1 ; product of the roots = 2 2

m3 + n3 = ( m + n)( (m + n)2 3mn) 3 3 1 = [ ( )2 3( ) 2 2 2 45 =81 1 2 2 m n 1 1 x m2 n2 1 3 ( ) 2 2( ) n m 2 = 13 2 = = 2 1 2 (mn) ( ) 2 1 = =4 1 2 ( ) 22 2

Quadratic equation with the roots 5.

1 1 and 2 is x2 -13x + 4 = 0. 2 m n

1 ., express x2 + 1/x2 in terms of y. Hence, solve the equation x x4 - 2x3 - 6x2 - 2x + 1 = 0.

If y = x

SOLUTION:x2

1 1 ! x 2 x2 x ! y2 2

2

4 ! x x2

1 x

4x 1 ! 0 x ! ! ! 4 s 4 s 2 4 s 2 2 3 3 16 4 2 12

2 1 x x2 2x 6 2 ! 0 x x 2

1 1 x 2 x 2 2 2 x 6 ! 0 x x 1 1 x { 0, x 2 2 2 x 6 ! 0 x x y2 2 2y 6 ! 0 y2 2 y 8 ! 0

! 2 s or 2 ! x x2

y - 4 ( y

2) ! 0

1 x 2x 1 ! 0 x ! 1, o x ! 1 equation are 1, 1, 2 s 3.

( x 1 )( x 1 ) ! 0

The

roots

7

the

6.

12x 2 5 x 11 Decompose in partial fractions. ( x 1) 2 ( x 2) SOLUTION:

12 x 2 5 x 1 A B C ! 2 2 ( x 1) ( x 2) x 1 ( x 1) x2 12 x 2 5 x 11 ! A( x 1)( x 2) B( x 2) C ( x 1)2 When x ! 1, 6 ! B(3), B ! 2. When x ! 2, 27 ! C (3)2 , Comparing coefficient of x , 12 ! A C A!9 12 x 2 5 x 1 9 2 3 ! 2 2 ( x 1) ( x 2) x 1 ( x 1) x2 7. Find the set of values of x which satisfy each of the following inequality SOLUTION:x 1 1 " 3 x 3x 3 x "0 3x 2x 3 "0 3x 3 x or x " 0 2 3 The solution set is ( g , ) (0, g ). 2 x 1 1 " x 32

C ! 3.

3 2

0

+

_

+

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SMART PRACTICE1. 2. Find the values of the constants m, n, p and q such that x4 + px3 + 5x2 + x + q = (x2 + 4)(x2 x + n) + m(x + 1). Find the quotient and the remainder when a) 3x5 + 5x4 4x3 + 7x + 3 is divided by x + 2 b) x4 + x3 x2 1 is divided by x + 1 c) 8x4 + 6x2 3x + 1 is divided by 2x2 x + 2 When the polynomial x 3 + px2 + qx + 5 is divided by x2 x 2 the remainder is 2x + 7. Find the value of p and q. The expression x7 - ax3 + b can be divided exactly by (x-1) and has remainder 8 when it is divided by (x-2). Find the values of a and b and the remainder when it is divided by (x+2). Show that x + 3 is a factor of the polynomial, P(x) = x 4 9x2 + 4x + 12. Hence solve the equation P(x) = 0. Given that when f(x) = 6x3 + ax2 + bx + 4 is divided by (x + 1) and (x 3) the remainders are -15 and 49 respectively. Find the values of a and b. Hence find all the factor of f(x). Given that f(x) = 2x3 + 5x2 8x 15, find f(-3), hence determine the factors of f(x). Show that 2 i is a zero of the polynomial P(x)= x 3 3x2 + x + 5. Hence find all the roots of the equation P(x) = 0. A polynomial p(x) = ax 3 + bx2 + cx + d where a, b, c and d are real constants. If two of the roots of the equation p(x) = 0 are 1 2i and 2 and p(-1) = -8, find the values of a, b, c and d.

3. 4 5. 6. 7. 8. 9.

10. One of the zeros of the expression 2x3 + ax2 + 4x + b is 2. The expression has remainder 9 when it is divided by x + 1. Find the values of a and b and hence, solve the equation 2x3 + ax2 + 4x + b = 0. 11. Find all the roots of the equation f(x) = 8x 3 9x2 + 6x 5. Find the set values of x such that f(x) > 0. 12. Show that x3 is a factor of f(x)= x4 +2x3 + x2 - 28x 60. Factorise f(x) completely. Find the set values of x such that f(x) > 0. 13. Given that f(x) = x3 + 5x2 + 8x + 6, show that -3 is a zero f(x). Find all the root of the equation f(x) = 0. Hence find the set values of x such that f(x) > 0. 14. Write 3x2 + 12x + 15 in the form a(x + k)2 + h, where a, k and h are integers to be found. Find the minimum value of 3x2 + 12x + 15 and state the value of x for which this minimum occurs. Hence write down the maximum value of the function1 . 3x 12 x 152

15. Find the set of values for k if k2 x2 + 2(2k 5)x + 8 > 0 for all real values of x. 16. Find the set of values of m such that the equation x2 + (3m 7)x + (2m + 6) = 0 has real root. 9

17. Find the set of values of m such that the equation mx2 + 2(m - 3)x + 2(m + 1) = 0 has distinct real roots. 18. If a and b are roots of the equation x2 + mx + n = 0, where m and n are non-zero constants, find both the roots of the equation nx2 + (2n m2 )x + n = 0 in terms of a and b. 19. The roots of the equation x2 + 5x + 5 = 0 are E and F . F 2 E 2 Determine a quadratic equation whose roots are and . 2 E 2 F 20. By using the substitution p = x 1 , show that the equation 2x4 + x3 6x2 + x + 2 = 0, can be x reduces to 2p2 + p 10 = 0. Hence solve 2x4 + x3 6x2 + x + 2 = 0.

21. Show that 2x -1 is a factor of the polynomial p(x) = 8x4 4x3 -2x2 + 7x 3. Factorize p(x) completely. Find all the roots of the equation p(x) = 0. Hence deduce the real roots of the equation a) x4 x3 x2 + 7x 6 = 0 b) 8x8 4x6 2x4 + 7x2 - 3 = 0 22. Express 23. Express15 - 20x (1 2x)(2 - x) 25 x 2 31 ( x 3) ( x 4)2

in partial fractions. in partial fractions.

24. (x + 1) and (x 2) are factors of the polynomial P(x) = x3 + px2 + qx - 6. Determine the values of p and q, hence find the other factor of P(x). 2 Express 2 x 5 x 13 in partial fractions. 25. Find the set of values of x which satisfy each of the following inequalities: a)1 2 x 1 x3

2 d) 3 x e 2

x

26. Solve the following simultaneous equations a) x2 + 2xy = 3 y2 xy = 4 b) x2 2xy + 8y2 = 8 3xy 2y2 = 4

(x)

b) 12 u x 2x

c) . x2 + x + 1 < x + 2 < x 2 6x + 12 f) -16 < x3 4x2 + 4x 16 0. [6]

2. If m and n are the roots of the equation 2x2 + 3x 1 = 0, find the value of m3 + n3. Obtain the 1 1 quadratic equation whose roots are 2 and 2 . [6] m n

3.

Express

12x 2 5 x 11 in partial fractions.[4] ( x 1) 2 ( x 2)

20

4. Given that when f(x) = 6x3 + ax2 + bx + 4 is divided by (x + 1) and (x 3) the remainders are -15 and 49 respectively. Find the values of a and b. Hence find all the factors of f(x).[8]

5.

1 , show that the equation 2x4 + x3 6x2 + x + 2 = 0, can be x reduces to 2p2 + p 10 = 0. Hence solve the equation 2x4 + x3 6x2 + x + 2 = 0. [9]

By using the substitution p = x

6.

1 2 1 2 1 2 Given that M = 1 2 3 and N = 13 5 7 show that MN = kI. 3 2 8 1 4 5 Hence solve the following simultaneous equation 2x + y + z = 1 ; x - 2y - 3z = 1 ; 3x + 2y + 4z = 5. [8]

21

7.

1 4 If A = 2 1 , B =

0 1 3 2 , C =

3 4 21 19 , find the matrix X such that AXB = C. [5]

8.

a b Determine all the possible matrices for A such that A = b a and A(I A) = 0. [6]

9.

1 1 2 Matrix A is given by A = 0 2 2 1 1 3 Find A2 6A + 11I. Show that A(A2 6A + 11I) = 6I and deduce A-1. [6] A factory produces three types of nuts, kacang kuda, kacang botak and kacang parang. The profit from 1 kg of kacang kuda , 1 kg of kacang botak and 2 of kg kacang parang is RM 9. The profit from 1 kg of kacang botak and 1 kg of kacang parang is RM 3. The profit from 1 kg kacang botak and 3 kg of kacang parang is equal to the profit from 1 kg of kacang kuda. If x, y and z represent the profit from 1 kg of kacang kuda, 1 kg of kacang botak and 1 kg of kacang parang respectively. Write a matrix equation to represent the above information. Hence determine the profit for 1 kg of kacang kuda, 1 kg of kacang botak and 1 kg of kacang parang . [7]

22

ANSWER: 1. SOLUTION: Let p(x) = 6x3 13x2 + x + 2 p( ) = 6( )3 13 ( )2 + ( ) + 2 =0 Since p( ) = 0, hence 2x- 1 is a factor of the polynomial p(x) p(x) = 6x3 13x2 + x + 2 = (2x 1)(3x2 5x 2) = (2x -1)(3x + 1)(x 2) (2x -1)(3x + 1)(x 2) > 0 B1 A1

M1A1

Try to obtain the quotient by long division-1/3 2

1/2

{x : x IR , - 1/3 < x < } { x : x IR, x > 2} 2. SOLUTION: Sum of the roots , m + n = 3 1 ; product of the roots = 2 2

A1A1

B1

m3 + n3 = ( m + n)( (m + n)2 3mn) 3 3 1 = [ ( )2 3( ) 2 2 2 45 =8n2 m2 1 1 = 2 = (mn ) 2 m2 n 1 1 x = m 2 n2 1 3 ( ) 2 2( ) 2 = 13 2 1 2 ( ) 2

M1 A1

B1

1 =4 1 2 ( ) 21 1 and 2 is x2 13x + 4 = 0 2 m n

B1

Quadratic equation with the roots

A1

23

3 .

12 x 2 5 x 1 ( x 1) 2 ( x 2) When x ! 1,

!

A B C x 1 ( x 1) 2 x 2 B ! 2. C ! 3.

B1 B1 M1

12 x 2 5 x 11 ! A( x 1)( x 2) B ( x 2) C ( x 1) 2 6 ! B (3), When x ! 2, 27 ! C ( 3) 2 , Comparing coefficien t of x 2 , 12 ! A C A!9 12 x 2 5 x 1 ( x 1) 2 ( x 2) 9 2 3 x 1 ( x 1) 2 x 2

!

A1

4 f ( x ) = 6 ( -1)3 + a ( -1)2 + b ( -1 ) + 4 = -15 . a - b = -13 f(3 ) = 6 ( 3)3 + a ( 3)2 + b( 3) + 4 = 49 3a + b a= -13@ f(x) = 6x3 -13x2 + 4 By try and error method

.. M 1 . M 1 . A1

= b =0

-39

f ( 2) = 6 (2)3 -13(2)2 + 4 = 0 Hence x 2 is a factor of f(x) f(x) = 6x 13x + 4 = (x 2 )( 6x - x -2 )2 3 2

B1 + A16x2 x 2 x 2 6x 13x 2 4

.. M 1

= ( x-2 )(3x-2)( 2x+ 1)

.. A1

6x3 12 x2 ____________ - x2 + 4 - x2 + 2x _________ -2x + 4 -2x + 4 _______ 0

3

The factors

of f(x) are x 2 , 3x- 2 and 2x + 1 24

A1

5 .

x2 ( 2x2 + x -6 +

1 2 ) = 0 x x2

x2 ( 2( x 2 +

1 1 ) + (x+ ) - 6 ) =0 2 x x

B1

x{ 0 2( p2 2 ) + p 6 = 0 2 2p + p -10 = 0 ( 2p + 5 )( p 2 ) = 0 p = when

B1 M1 M1 A1

5 or 2

p=2

5 1 !x 2 x

2= x+

1 x M1

2x2 + 5x + 2 = 0 M1 ( 2x + 1 )( x+ 2 ) = 0

x 2 - 2x + 1 = 0 ( x -1)(x 1 ) = 0

x= -

1 , x= -2 2

A1

x=1, x=1 .

A1

Hence, the solutions are

1 x = - , x ! 2, x ! 1, x ! 1 2

6 SOLUTION: . 4 5 1 2 7 5 9 0 0 4 13 8 2 26 24 2 10 3 1 14 15 = 0 9 0 = 9 = 6 26 32 6 10 4 3 14 20 0 0 9 2/9 1/ 9 2/9 1 1 13 / 9 5 / 9 7 / 9 Therefore, M = N = 9 8 / 9 1/ 9 5/9 2x + y + z = 1 x 2y 3z = 1 3x + 2y + 4z = 5

1 0 0 0 1 0 = 9 I 0 0 1

B1M1A1

A1

25

1 x 1 2 1 1 2 3 y ! 1 3 2 4 z 5 x 1 y ! M 1 1 z 5

x 1 M y ! 1 z 5

M1

2/9 1/ 9 1 x 2/9 y ! 13 / 9 5 / 9 7 / 9 1 = z 8 / 9 1/ 9 5 / 9 5

1 3 2

M1A1 A1

x = 1, y = 3, z = 2 7 . A1 ! 1 1 4 1 8 2 1 1 1 4 ! 9 2 1 B 1 ! 2 1 1 0 (3) 3 0 1 2 1 ! 3 3 0

B1

B1

AXB ! C

A1 AXB B 1 ! A1CB 1

A1 A X BB 1 ! A1CB 1 IXI ! A1CB 1 X !

B1

1 1 4 3 4 1 2 1 9 2 1 21 19 3 3 0

M1

1 45 72 2 1 27 27 27 3 0 1 54 81 ! 27 27 27 2 3 ! 1 1 A1 !

81 72 A1 for 27 27

26

8.1 0 a b IA! 0 1 b a 1 a b ! b 1 a AI A ! 0 a b 1 a b 0 0 b a b 1 a ! 0 0 a (1 a ) b 2 ab b (1 a ) 0 0 ! B1 b(1 a ) ab b 2 a (1 a ) 0 0

a a 2 b2 ! 0 b 2 a a 2 ! 0.

;

ab b ab ! 0 B1 either 2

b ab ab ! 0 b 2ab ! 0 b(1 2a ) ! 0 b ! 0 or a ! When b ! 0, 0 a a2 ! 0 a (1 a ) ! 0 a ! 0 or When a ! M1 a ! 1. 1 2 M1

1 , 2 1 1 b2 ! 0 2 4 1 b2 ! 4 1 b!s 2 b ! 0, a ! 0 or b ! 0 , a ! 1 or a !1 2 , 1 2

1 0 0 1 0 2 , 1 , @ A! 0 0 0 1 2

1 1 1 1 or a ! , b ! , b! 2 2 2 2 1 1 2 2 . A1 A1 all 4 correct, 1 1 2 2 1 mark if 2 correct.

27

9. . A2

1 1 2 1 1 2 = 0 2 2 0 2 2 1 1 3 1 1 3 1 5 10 2 6 10 4 4 9

=

B 1

1 5 10 1 1 2 0 0 1 0 A - 6A + 11 I = 2 6 10 60 2 2 11 1 0 0 4 4 9 1 1 3 0 12

=

4 1 2 2 5 2 2 2 2

. B 1

1 1 2 4 1 2 A ( A -6A + 11 I ) = 0 2 2 2 5 2 1 1 3 2 2 2 2

0 0 6 = 6 0 0 0 0 6

.. B 1

= 6I A ( A2 - 6A + 11 I ) A-1 A (A2 - 6A + 11 I ) I (A2 - 6A + 11 I ) A-1 = = 6 I = = 6 A-1 I 6 A -1.

B1

1 (A2 - 6A + 11 I ) 6 4 1 2 1 2 5 2 6 2 2 2 2 1 1 3 6 3 1 5 1 3 3 6 1 1 1 3 3 3

=

M 1

=

.. A 1

28

x + y + 2z = 9 y + z y + 3z = 3 = x

x + y + 2z =

9 6 0 B1 + BI either 2

0x + 2y + 2z -x + y + 3z

= =

1 1 2 x 9 6 0 2 2 y ! 8 1 1 3 z

..

M1

AB =

C A-1 C B1

(A-1 A )B = IB

= A-1C 4 1 2 9 6 2 5 2 0 2 2 2

B =

1 6

M1

=

5 2 1

A1

x=5 , y=2

z=1

either = RM 5.00 = RM 2.00 A1

Hence, the profit from 1 kg of kacang kuda 1 kg of kacang botak

1 kg of kacang parang = RM 1.00

29

ANSWERS FOR SMART PRACTICES PAPER 1: POLYNOMIALS 5 5 16. g, ? , g 1.m ! 5, n ! 1,2.

p ! 1, q ! 9.

17. 18. 19. 20. 21.

(a ) 3 x 4 x 3 2 x 2 4 x 1; 5. (b) x 3 x 1;3. 4. 5. 6. 7. 8. 9.

20 x 2 90 x 81 ! 0

2

(c) 4 x 2 2 x; 7 x 1 p ! 0, q ! 1 a ! 17, b ! 16 ; 24 3, 1, 2, 2

( 2 x 1)( x 1)( 4 x 2 4 x 3);(a) (b)

1,2,s

f (3) ! 0, x 3, 2 x 2 x 5 1,

22.

2 i, 2 i

23. 24

2 2 4 2 5 1 2 x 2 x (2 x) 2 4 2 1 x 3 ( x 3) 2 x 4 (x 1)(x - 2)(x - 3) ; 1 5 5 2( x 1) x 2 2( x 3)(c)

a ! 1, b ! 0, c ! 1, d ! 1010.

p ! 2, q ! 5; 2, 5.72, 1.2225.

a ! 13, b ! 28;11. 12.

1,

1 s 159i 162

(f) 0, 2 (2, 4) f ( x) ! ( x 3)( x 2)( x 3 x 10), (g,2) (3, g) 3 4 3 3 4 3 26. (a) 13. 3, 1 s i ; (-3, g) 3 , 3 , 3 , - 3 14. 1 1 (b) 1 3, , - 3, - , 2, 1, 2 3x 2 3, a ! 3, k ! 2, h ! 3 ; 3 , x ! 2; 2 2 3

5 (a) 2, 3, g 2 (d) ? 3, 1A ? , 3A 1

(b) ? 1, 1A\ _ a 0

(e)

30

9 _ : m I ,m m a b , b a

1a _ : m I , m " 9a m

x!

1 1 s 2i , 1, 2 2

1, 1A

- g,-5A 2,

1 ? , g 1

, - 3, 1 3, 1 ,

2,

1

MATRICES1. 2. 3. 4. m = -2 , n = 4 a=0 , b = 1 , c = 1 , d =1 2

7 0 0 1 2 2 1 0 0 7 , 7 5 3 3 0 0 7 4 8 1 6 4 2 1 2 4 6 10 2 2 2 2 3 1 3 1 3 1 9 2 9 5 9 4 9 1 9 2 9 1 b) k = - ,1 2

5.

6.

7.

a) k = -10

8.

a : a IR, x { 3, x { 4 2 3 1 1

9.

10 . x=0 , y =6 1 11. 2 1 2 1 2 1 2

12.

0 0 3 2 1 1 -1 AB = 1 0 , B = 9 5 2 , x=2,y=1,z= -1 0 0 0 1 5 3 1

0 0 1 13. AB = 1 0 x =2 , y = -1 , z = 3 0 0 0 1

31

14.

2 1 1 3 6 3 4 1 2 9 1 1 2 x 1 5 1 -1 6 2 5 2 , A = 3 6 3 , 0 2 2 y ! , 1 1 3 z 0 2 2 2 1 1 1 3 3 3 x=5, y=2 , z =1 190 0 0 AB = 0 190 0 , 0 190 0 114 190 114 A-1 = 190 152 190

15.

19 190 19 190 57 190

24 190 14 190 52 190

1 2 1 x 0 1810 20 12 6 y ! 1710 19 19 0 z

, x = 6650 , y = 10450 , z = 14250

32

MATHEMATICS T

Bertujuan membantu calon STPM yang gagal atau berprestasi lulus sebahagian ke arah meningkatkan pencapaian agar mencapai tahap LULUS PENUH Digubal oleh tenaga pengajar STPM berpengalaman di Johor Darul Takzim

33

PANEL PENGGUBAL

1. BOOPATHY RAMASAMY SMK TUN HABAB, KOTA TINGGI 2. LIM POH THAI SMK DATO PENGGAWA BARAT, PONTIAN 3. KOH AH LEH SMK DATO BENTARA LUAR, BATU PAHAT 4. GOH KIM HENG SMK SERI MERSING, MERSING 5. LEE KIM SING SEKOLAH TINGGI SEGAMAT, SEGAMAT

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DATA DESCRIPTIONCHECKLIST:AT THE END OF THIS TOPIC STUDENTS SHOULD BE ABLE TO:1. 2. 3. 4. 5. 6. 7. 8. Understand discrete, continuous, ungrouped, and grouped data Construct and interpret stemplots, boxplots, histogram and cumulative frequency curves. Derive and use the formula

(xi !1

n

i

x ) 2 ! xi2 n ( x ) 2i !1

n

Calculate and interpret the mode, median and mean. Estimate graphically measures of location. Calculate and interpret the range, semi -interquartile range and standard deviation. Estimate graphically measures of dispersion. Understand the symmetry and skewness in a data distribution.

35

SMART TIPSNo 1x = n

Formulae/Terms

Point To Note Mean of ungrouped data

xi !1

i

nn

2x=

fxi !1 n

i i

Mean of ungrouped and group data (x i = class mid-point for the ith class)

fi !1

i

3

yi =

xi k ; h

k ! assumed mean h ! scaling factor

Formula for coding method

4

xk xk 1 n if ! integer(k) 2 2 n M = xk+1 if ! decimal(k. .) 2 M= f F C. M=L+ 2 f ---C--

Median for ungrouped data

5

Median for grouped data F cumulative frequency of pre-class

frequency

M 6Q new ! Q x k Q ne ! h Q x Q new ! h Q x k

Boundaries Linear transformation for mean(can use for median and mode) MOLnew = h MOLold + k Ungrouped data Grouped data x mid-point U/L upper/lower boundaries.

7 8

Range = x max - xmin Range = xHC - xLC = UBHC LBLC

36

9 dB Mode = L + C dB d A

Mode for grouped data

frequency

dA mode Boundaries Quartiles for ungrouped data.

dB

10

n 3n is an integer = k, & is an integer = h 4 4 1 1 Q 1 = ( xk xk 1 ) & Q 3 = ( xh xh 1 ) . 2 2 n 3n is not an integer(k. ..) & is not an integer If 4 4 Q 1 = x k 1 & Q3 = x h 1 .

If

11

C Q 1= L B + 3 f B C Q 3 =L B+ 4 f Q3 B

f 4 fQ1

Quartiles for grouped data 3 f 4

12 13

Interquartile Range = Q 3 Q 1 Semi-Interquartile Range =n

14W2 =

( x x)i i !1

2

n

15W =2

xn

2

i

xi = n

2

!

f4 Q1 All types of data. All types of data. Q2

Q3 Q1 2

Basic definition for variance

xn

2

i

x

2

Variance for ungrouped data.

37

16W2

W

2

f ( x x ) ( basic definition ) ! f f x f x ! f f i 2 2 i i i i i i ne

2

Variance for ungrouped & grouped data. or

18 X = 19 s =

n x1 m x2 mean of combined data nm

x ynm2

2

2

X22

=

n( s12 x1 ) m(s 2 x 2 ) 2 X2 nm

standard deviation of combined data

20 21

( x W , x W ), Type Of Distributions Symmetrical Distribution Positively Skewed Distribution Negatively skewed Distribution

22

MOL = mean MOL = median

MOD = SD MOD = SIQR

23

Range of outlier (Q1 1.5(IQR) , Q3 + 1.5(IQR) )

#

"

17

W 2 ne ! h 2 W x

2

&

W

!h Wx

Linear combination for variance & standard deviation.

Combined data: Data 1 consists of n datums with the mean and standard deviation x1 and s1 respectively. Whereas Data 2 consists of m datums with the mean and standard deviation x2 and s2 respectively.

One standard deviation from the mean.

Can identify by using : a) MOL b) Histogram c) Box plot d) Frequency curve e) PCS(Pearson coefficient of skewness) 3( mean - median ) PCS = s tan dard deviation -3 < PCS < 3 Symmetrical distribution Skewed distribution (must know reason) The values out this range is known as outliers.

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GUIDED EXAMPLES1. The total scores given The Times for Welsh and Scottish Rugby matches on 4 February 1995 were: 34 21 23 26 24 39 56 22 28 25 63 83 42 39 24 23 32 61 44 19 31 24 24 60 45 39 34 50 46 53 48 33 43 a) Construct a stem-and-leaf diagram to display the data above. [SOLUTION] 1 9 2 1 2 3 3 4 4 4 4 5 6 8 3 1 2 3 4 4 9 9 9 4 2 3 4 5 6 8 5 0 3 6 6 0 1 3 7 8 3 Key: 3 2 is 32 b) Find the median and semi-interquartile range SOLUTION n = 33 n = 16.5 (decimal) 2 n = 8.25 (decimal) 4 3n = 24.75 (decimal) 4 Semi-interquartile range

1 1

median = x17 = 34 Q1 Q3 = x9 = 24 = x25 = 46

= 2 ( Q3 Q1 ) = 2 ( 46 24) = 11.

c) Find the mean and standard deviation. Hence find numbers of score that lies one standard deviation from the mean. SOLUTION mean,x =

x ! 1258n 332

= 38.12 / 38.1

x2 = 55300

55300 1258 = 222.531 33 33 standard deviation, s = 222.531 = 14.92 / 14.9 The range of one standard deviation from the mean is ( x s, x + s ) = ( 23.20, 53.04) Therefore, the number of scores that lie one standard deviation from the mean = 23. s2 =

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2 .

The heights to the nearest cm, of 80 soldiers are summarized in the table below. Height (in cm) 150 154 155 159 160 164 165 169 170 174 175 179 180 184 185 189 Numbers of soldiers 3 7 10 15 25 12 6 2

a) Calculate the mean and standard deviation for the grouped data, giving your answer correct to one decimal place. [SOLUTION] midpoint x 152 157 162 167 172 177 182 187 mean,x =

frequency f 3 7 10 15 25 12 6 2

s =

2

fx ! 13570 = 169.625 = 169.6 ( correct to 1 decimal place) 80 f x x = 2306860 13570 = 63.1087 80 80 2 2 2

standard deviation, s = 63.1087 = 7.9441 = 7.9 ( correct to 1 decimal place) b) Plot the cumulative requency graph or this data hence, i) estimate the median and semi-interquartile range o this distribution. [SOLUTION] x< Cumulative frequency F 149.5 0 154.5 3 159.5 10 164.5 20 169.5 35 174.5 60 179.5 72 184.5 78 189.5 80 Plot the cumulative frequency curve on a graph paper. i) From the graph: Median = 170.5 Q1 = 164.5 Q3 = 174.5 Semi-interquartile range =1 2

( Q3 Q1 ) =

1 2

(174.5 164.5)

=5

40

Cumulative frequency 80

y y

y

70

60

y

ii) 5040

y30

20

y

10

y

0

y145 150

y155 160 164.5 165 170 175 180 185 190 Height(cm)

Median=170.5Q3 = 174.5

Q1 =

140

145

150

155

160

165

170

175

180

185

190

[SOLUTION] Sketch a box plot on a graph paper. End points of whiskers : 159.5, 179.5 Box : 164.5, 174.5, 170.5 iii) Two of the soldiers are chosen at random for guard duty. Determine the probability that one, and only one, of them is taller than 179.5 cm. [SOLUTION] 8 72 72 8 x x + = 0.1823 80 79 80 79

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3

Mean and standard deviation for the weight of 200 boys are 55.8 kg and 6.5 kg respectively, whereas the mean and standard deviation for the weight of 150 girls are 57.1 kg and 5.3 kg respectively. Calculate the mean and standard deviation for weight of the whole groups. [SOLUTION] X =200 (55.8) 150 (57.1) = 56.36 200 150

s =

200(6.5 2 55.8 2 ) 150(5.3 2 57.12 ) 56.36 2 200 150

= 6.02

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SMART PRACTICE1 A random sample of 51 people were asked to record the number of miles they traveled by car in a given week. The distance, to the nearest mile, are shown below: 67, 72, 70, 58, 66, 63, 76, 77, 70, 74, 50, 44, 85, 53, 66, 44, 67, 47, 42, 41, 62, 52, 87, 57, 93, 48, 54, 74, 78, 68, 48, 86, 85, 52, 86, 81 93, 78, 60, 82, 94, 46, 56, 58, 78, 63, 52 80 43 47 72

a) Construct a stem and leaf diagram to represent these data. b) Find the median and the semi-interquartile range of this distribution c) Draw a box plot to represent these data. Identify whether there are any outliers or not. 2. The weight to the nearest kg, of 180 male members of a circus is summarized in the following table. Weight in kg 41 50 51 60 61 70 71 80 81 90 91 100 101 110 111 120 121 130 131 - 140 Number of men 2 5 11 23 37 35 28 20 15 4

a) Calculate the mean and standard deviation of the weight of the male members of the circus. b) Plot a histogram for the grouped data, hence, estimate the median and the mode of the distribution. c) Calculate the median, hence determine the different between the median in (b) and the calculated value. d) State the type of distribution for the weight of the male members of the circus. Give reason for your answer.

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3

The data shown in the stem plot below are the marks in a statistics course obtained by a group of students at a local institution of higher learning 3 4 5 6 7 8 9 1 0 1 1 0 2 1 4 2 2 3 2 3 2 3 7 6 4 9 8

key : 3/1 means 31 marks i) Find the percentage of students who obtain less than 40 marks and the percentage of students who obtain at least 80 marks. ii) Find the mean and standard deviation of the students marks. Hence find percentage of students who obtain marks within the range of one standard deviation from mean. iii) Find the median and semi-interquartile range of the students marks. iv) Construct a box plot for the above data. Comment about the distribution. 4 The table shows the lifespan, t the nearest 1000 km, of 100 tyres of a new model produced by a tyre company. Lifespan (km) Number of tyres 70 000 74 000 9 75 000 79 000 24 80 000 84 000 35 85 000 89 000 21 90 000 94 000 11 a) Calculate the mean and standard deviation for above data. b) Plot a histogram for the grouped data, and use it to estimate the median lifespan of the tyres. c) Plot a cumulative frequency curve for the data. Hence, estimate the median and semi-inter quartile range for the tyres. Hence find the percentage of the tyres where the lifespan in the range of one standard deviation from the mean. 5 Data below show heights in cm for 36 new born chicks. 31 29 30 33 34 31 32 25 37 48 41 51 60 41 32 44 37 39 31 43 50 37 35 43 28 32 30 a) Construct a stem and leaf diagram to represent these data. 28 30 37 26 41 37 35 36 37

b) Calculate the mean and standard deviation for height of the chicks. Hence find percentage of chicks whose height lying in the range of one standard deviation from mean. c) Find the median and semi-interquartile range. d) Draw a box plot to represent these data. e) Comment about the distribution of this data. f) State which of the mean and median would be a better measure of location. Justify your answer. 44

6

The weights of 25 Form three students in a school were recorded. It is found that the mean weight of the students is 47.8 kg and the standard deviation is 30.2 kg. After the data was checked again, the following three mistakes were found. Recorded weight of 14.8 159.6 77.6 student (kg) Actual weight of 41.8 59.6 55.6 students (kg) Calculate the correct the mean and standard deviation for the weights of the 25 students The following table shows the number of children in 20 families. Number of children Number of families 0 2 1 7 2 5 3 4 4 1 5 1

7

a) Calculate the mean and standard deviation of the number of children in the families. b) Find the percentage of the families with the number of children less than the mean.

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DISCRETE PROBABILITY DISTRIBUTIONCHECKLIST:AT THE END OF THIS TOPIC STUDENTS SHOULD BE ABLE TO:1. 2. 3. 4. Understand the concept of a discrete random variable. Construct a probability distribution table for a discrete random va riable. Understand the concept of mathematical expectation. Use the formulae E(aX + b) = aE(X) + b, Var(aX + b) = a 2Var(X), E(aX + bY) = aE(X) + bE(Y) and Var(aX + bY) = a2 Var(X) + b2 Var(Y). 5. 6. 7. 8. 9. Derive and use the formula E(X - Q )2 = E(X2) - Q 2. Calculate the mean and variance of a discrete random variable. Understand the binomial distribution. Use the probability function of the binomial distribution. Calculate the mean and variance fo r binomial distribution.

10. Use the binomial distribution as models for solving problems. 11. Understand the Poisson distribution. 12. Use the probability function of the Poisson distribution. 13. Calculate the mean and variance for Poisson distributi on. 14. Use the Poisson distribution as models for solving problems. 15. Use the Poisson distribution as an approximation to the binomial distribution where appropriate.

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SMART TIPSNo 1 Formulae/Terms X discrete random variable Note a) can assume only a countable finite or infinite number of values. b) result from counting. c) must assume integral values. 2n

p1 + p2 + + p n = 1 pi u 0 , i Z+

i !1

( ! xi ) = 1

To prove X is DRV or can use for problems involves DRV

3

Probability distribution of X can either list the probability individually (tabular form) or summarise them in a formula. x P(X=x) P(X=x) = 0 0.25 0.25 0 .5 1 0.125 2 0.125 3 0.5

0.125 , x ! 1 , 2 ,x!0 ,x!3

4

P(X = x)5 12 4 12 3 12

Probability Distribution Graph

y y y0 1 2 x Expectation value of X also known as mean of X Properties of expectation

5 6

E(X) =

xP( X

! x)

E(a) = a E(aX) = aE(X) E(aX s b) = aE(X) s b

7

Var (X) = E(X2 ) - (E(X))2 = E(X2) - E2(X)

Must know how to prove from basic definition (E[ X - Q ]2 ) Properties of variance

8

Var(a) = 0 Var(aX) = a2 Var(X) Var(aX s b) = a2 Var(X)

47

9 10 11

E(X s Y) = E(X) s E(Y) E(aX s bY) = aE(X) s bE(Y) Var(X s Y) = Var(X) + Var(Y) Var(aX s bY) = a 2Var(X) + b2 Var(Y) n E(X1 + X2 + + Xn) = E X i i !1 = E(X1 ) + E(X2)+ + E(Xn) = nE(X)

X and Y are any two independent random variables X and Y are any two independent random variables independent observation, X1, X2, , Xn from the same distribution as X

12

independent observation, X1, X2, , Xn n Var(X1 + X2 + + Xn) = Var X i from the same distribution as X i !1 = Var(X1)+Var(X2)++Var(Xn) = nVar(X) For a situation to be described using a binomial model: a finite number, n, trials are carried out, the trials are independent, the outcome of each trial is deemed either a success or a failure, the probability, p, of a successful outcome is the same for each trial.n P(X = x) = x C qn xpx for x = 0, 1, 2, , n

13

X b B(n , p) where n and p are the parameters of X.

14 15

X ~ B( n , p) where P(X = x) is the probability distribution of X , q = 1 - p

X ~ B( n , p) then P(X = 0) = qn and P(X = n) = pn

16

X ~ B(n , p) then E(X) = np and Var (X) = npq

17 18

np q e r e np + p

Determining the most likely value of X to occur if X ~ B(n , p) Probability distribution of Poisson Dist.

P(X = x) = e P

Px for x = 0, 1, 2, to infinity x!

19

X ~ Po( P ) then E(X) = P and Var (X) = P

20

P -1 e x e P

Determining the most likely value of X to occur if X ~ Po( P )

48

21

X ~ Po( P ) then P(X = 0) = e P and P(X = 1) = P e P P(X > n) = 1 P(X e n) = = 1 - eP ( 1 + P + P2 P3 Pn -1 ) .... 2! 3! ( n 1)!

22 n > 50, p < 0.1 ; np = P X ~ B(n , p) np g pp 0 X ~ Po( P ) Poisson distribution as an approximation for binomial distribution.

49

GUIDED EXAMPLES1. The random discrete variable X can take the values 1, 3, 5 and 9 with P(X = 1) = 0.2, P(X = 3) = 0.3 , P(X = 5) = 0.4 and P(X = 9) = 0.1. Find E(X) and Var(X). SOLUTION : X P(X=x) E(X) = 1 0.2! x)

3 0.3

5 0.4

9 0.1

x

= 1(0.1) x 3(0.3) x 5(0.4) x 9(0.1) = 4 E(2

)= x2 ( =x) = 12(0.2) + 32(0.3) +52(0.4) +92(0.1) = 21

Var( ) = E( 2) [ E( ) ]2 = 21 16 =5 2. The discrete random variable 0 ( =x) m + 0.1 a) Determine the value of m. SOLUTION: a) (m + 0.1) + (2 m + 0.2) + (3m + 0.1) = 1 6m + 0.4 = 1 m = 0.1 X 0 2 P(X=x) 0.2 0.4 b) E(X) = xP(X=x) = 0(0.2) + 2(0.4) =4(0.4) = 2.4 E(X2 ) = has the probability distribution 2 4 2m + 0.2 3m + 0.1 b) Find the expected value and variance for .

Var(X) = E(X2) [E(X)]2 = 8 2.42 = 2.24

%$(

4 0.4

x P(X=x) = 02(0.2) + 22 (0.4) + 42(0.4) = 8

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3.

A box contains a large number of pens. The probability that a pen is faulty is 0.1. How many pens would you need to select to be more than 95% certain of picking at least one faulty one? SOLUTION: Let the random variable X represents the number of faulty pen. X ~ B( n, p) P(X=x) = ncxpxqn - x p = 0.1, q = 0.9 P(X u 1)>0.95 1 P(X=0) > 0.95 1 0.9n > 0.95 0.9n < 0.05 nlog0.9 < log 0.05 n>log 0.05 log 0.9

n > 28.43 At least 29 pens must be chosen so that the probability that at least one pen is faulty is greater than 0.95. 4. The mean number of bacteria per ml of a liquid is 4. Assuming that the number of bacteria follow a Poisson distribution, find the probability that, in 1 ml of liquid, there will be a) no bacteria b) 4 bacteria c) less than 3 bacteria SOLUTION: X b Po(4)e 4 4 0 -4 a) P(X=0) = e o!

b) P( X=4 ) =

= 0.195

e 4 4 4 4!

= 0.0183 c) P (X< 3 ) = P(x=0) + P(X=1) + P(X=2) = e 4 (4 0 41 4 2 ) 2! 0! 1!

= 0.238 5.

A small bus company runs four buses with 50 seats each. Experience shows that, on average, 2% of people with tickets do not show up for the departure and the place remains free. If the company sells 205 tickets for one day, by using a suitable approximation calculate what is the probability that a) more than 200 passengers will show up? b) exactly 200 passengers will show up for the departure? SOLUTION: Let the random variable X represent the number of passengers that will not show up.P ! np ! 205 x 0.02 ! 4.1

an approximate Poisson distribution is used P(X=x) =e 4.1 4.1 x x = 0, 1, 2, 3, 4, . x!

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a) More than 200 passengers shown up mean less than 5 will not show up P(X