1



Fanling Lutheran Secondary School

F.4 Mathematics

Holiday Study Material (5/2 – 14/2)

[pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic][pic]

Revision: Chapter 1 – 6.1

Chapter 1: Number system

1. Convert [pic] into a fraction. (3 marks)

( 2. Let z =[pic], where k is a real number.

(a) Express z in the form a + bi. (2 marks)

(b) If the real part and the imaginary part of z are equal, find the value of k. (3 marks)

Chapter 2: Equations of Straight Lines

3. In each of the following, find the equation of the straight line L in the general form.

(Leave the radical sign ‘√’ in the answers if necessary.)

(a) (3 marks) (b) (2 marks)

4. The slope of the straight line L1 is (5. The straight line L2 passes through A((2 , 4) and L2 ( L1.

(a) Find the equation of L2 in the general form. (3 marks)

(b) If L2 passes through B(k , 6), find the equation of the vertical line passing through B. (2 marks)

5. L1: y = 4x ( 2, L2: 8x ( 2y + 3 = 0 and L3: ax ( 3y + b = 0 are three straight lines.

[pic] (a) Do L1 and L2 intersect? Explain your answer. (3 marks)

(b) If L1 and L3 have infinitely many points of intersection, find the values of a and b. (3 marks)

Chapter 3: Quadratic Equations in One Unknown

6. Solve the following quadratic equations by any algebraic method.

(Leave the radical sign ‘√’ in the answers if necessary.)

(a) x2 + 9x = 0 (2 marks) (b) x2 + 4x ( 21 = 0 (3 marks)

(c) 49 ( 4x2 = 0 (3 marks) (d) 3x2 ( x ( 7 = 0 (3 marks)

7. The quadratic equation (k + 1)x2 ( 8x + 2 = 0 has real root(s), where k ( (1.

(a) Find the range of values of k. (3 marks)

(b) If k takes the maximum value in (a), form a quadratic equation in x with roots k and k ( 2.

(2 marks)

8. The quadratic equation 9x2 + mx = 5 ( m has one double real root.

(a) Find the values of m. (3 marks)

(b) If m takes the minimum value in (a), solve the quadratic equation. (2 marks)

(9. If ( and ( are the roots of the equation x2 ( 7x ( 4 = 0, find the value of (2 ( 3(( + (2. (3 marks)

Chapter 4: Basic Knowledge of Functions

10. Let f(x) ’ x2 + 3ax – 5a, where a is a constant. If f(–4) ’ 67, find the value of a. (2 marks)

11. Let f(x) ’ x2 + 6x + k, where k is a constant. If f(–5) ( 5f(1) = 20, find the value of k. (3 marks)

12. Let f(x + 2) ’ x2 + 5x ( 4.

(a) Find f(x). (3 marks)

(b) Find [pic]. (2 marks)

Chapter 5: Quadratic Functions

13. The y-intercept and the equation of the axis of symmetry of the graph of the function

y = (x2 + (k + 2)x + 2k ( 3 are 9 and x = 4 respectively.

(a) Find the value of k. (2 marks)

(b) Find the maximum or minimum value of the function. (2 marks)

(14. The minimum value of the quadratic function y = a(x ( 6)2 + b is (18. The graph of the function passes through the origin.

(a) Find the values of a and b. (3 marks)

(b) Find the coordinates of the vertex of the graph of the function. (1 mark)

( 15. The maximum value of the function y = (2 ( x)(x + 8) + 4 ( k is 1. Find the value of k. (4 marks)

Chapter 6.1: Division of Polynomials

16. Let f(x) = x3 ( 3x2 ( 18x + k, where k is a constant. When f(x) is divided by x ( 2, the quotient is

x2 + bx + c, where b and c are constants. It is given that f(2) = 0.

(a) Find the values of b and c. (4 marks)

(b) Solve the equation f(x) = 0. (3 marks)

Pre-learning: Chapter 6.2 – 6.3

Chapter 6.2: Remainder Theorem

|Remainder Theorem |

|(a) When a polynomial f(x) is divided by x – a, the remainder is f(a). |

|(b) When a polynomial f(x) is divided by mx – n, the remainder is[pic]. |

|Instant Example 1 |Instant Practice 1 |

|Find the remainder when f(x) = 2x2 + 4x – 1 is divided by x – 2. |Find the remainder when f(x) = 4x2 – x + 3 is divided by x + 1. |

|Remainder = f(2) |Remainder = f( ) |

|= 2(2)2 + 4(2) – 1 |= 4( )2 – ( ) + 3 |

|= 8 + 8 – 1 |= |

|= 15 |= |

Use the Remainder Theorem to find the remainder in each of the following.

1. (x3 – 7x – 2) ÷ (x – 3) 2. (5x3 + 3x2 – x + 1) ÷ (x + 2) (Ex 6B: 2–13

|Instant Example 2 |Instant Practice 2 |

|When x2 + kx + 2 is divided by x – 3, the remainder |When x2 + kx – 5 is divided by x + 2, the remainder |

|is 8. Find the value of k. |is 11. Find the value of k. |

|Let f(x) = x2 + kx + 2. |Let f(x) = . |

|f(3) = 8 |f( ) = |

|32 + k(3) + 2 = 8 | |

|11 + 3k = 8 | |

|3k = –3 | |

|k = –1 | |

3. When kx2 + 4x – 2 is divided by x + 3, the 4. When x2 – 2x + 3k is divided by x + 4, the

remainder is 13. Find the value of k. remainder is 9. Find the value of k.

Chapter 6.3: Factor Theorem

|Factor Theorem |

|(a) If f(x) is a polynomial and f(a) = 0, then x – a is a factor of f(x). |

|Conversely, if x – a is a factor of the polynomial f(x), then f(a) = 0. |

|(b) If f(x) is a polynomial and[pic]= 0, then mx – n is a factor of f(x). |

|Conversely, if mx – n is a factor of the polynomial f(x), then[pic]= 0. |

|Instant Example 4 |Instant Practice 5 |

|Let f(x) = 2x3 + 3x2 – 5. Use the Factor Theorem to |Let f(x) = 3x3 + x2 – x + 1. Use the Factor Theorem to determine whether x + 1 |

|determine whether x – 1 is a factor of f(x). |is a factor of f(x). |

|f(1) = 2(1)3 + 3(1)2 – 5 ( Check whether f(1) = 0. |f( ) = 3( )3 + ( )2 – ( ) + 1 |

|= 0 |= ( ) |

|∴ x – 1 is a factor of f(x). |∴ |

In each of the following, determine whether the expression in brackets is a factor of the polynomial f(x).

3. f(x) = x3 – 2x2 – 3x + 6 [x – 2] 4. f(x) = x3 – 4x2 + x + 2 [x – 3]

5. Let f(x) = x3 + 4x2 + x – 6. 6. Let f(x) = x3 + 3x2 – 4.

(a) Show that x + 2 is a factor of f(x). (a) Show that x – 1 is a factor of f(x).

(b) Factorize f(x). (b) Factorize f(x). (Ex 6C: 8–11

|Factorization of Polynomials with Integral Coefficients |

|Factorize f(x) = ax3 + bx2 + cx + d, where a, b, c and d are integers. |

|Step 1: Use all the factors of a and d to list all the possible linear factors of f(x). |

|Step 2: Use the Factor Theorem to find one of the factors mx + n of f(x). |

|Step 3: Use long division to find the quotient Q(x) of f(x) ( (mx + n). Hence, f(x) = (mx + n)Q(x). |

|Step 4: Further factorize f(x) if possible. |

|Instant Example 6 |Instant Practice 7 |

|Factorize f(x) = x3 + 5x2 + 7x + 3. |Factorize f(x) = x3 + 2x2 – 7x + 4. |

|[pic] ( Step 1 |[pic] ( Step 1 |

|f(–1) = (–1)3 + 5(–1)2 + 7(–1) + 3 ( Step 2 |f( ) = ___________________________ ( Step 2 |

|= 0 |= 0 |

|∴ x + 1 is a factor of f(x). |∴ _________ is a factor of f(x). |

|Using long division, ( Step 3 |Using long division, ( Step 3 |

| | |

|) |) |

| | |

| | |

| | |

| | |

| | |

|f(x) = (x + 1)(x2 + 4x + 3) |f(x) = ( )( ) |

|= (x + 1)(x + 1)(x + 3) ( Step 4 |= ( Step 4 |

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

Factorize the following polynomials.

7. f(x) = x3 + 6x2 + 3x – 10 8. f(x) = x3 + 2x2 – 9x – 18

Suggested assignment 6.2 – 6.3

Ex 6B (5, 13, 16, 20, 24, 29, 31)

Ex 6C (6a, 7b, 9, 13, 15, 20, 35, 39)

-----------------------

L: slope =[pic]

7

x

y

0

(3 , 2)

x

y

O

L

60(

If x – a = 0, then x = a.

If mx – n = 0, then x =[pic].

x + 1 = 0

x = ___

If x – a = 0, then x = a.

If mx – n = 0, then x =[pic].

The possible linear factors of f(x) are x ( 1 and x ( 3.

x2 + 4x + 3

x + 1 x3 + 5x2 + 7x + 3

x3 + x2

4x2 + 7x

4x2 + 4x

3x + 3

3x + 3

The possible linear factors of f(x) are

.

The possible linear factors of f(x) are

.

The possible linear factors of f(x) are

x ± 1, x ± 2, x ± 5 and x ± 10.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download