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Mathematics St. Bonaventure College and High School Form 4, Quiz 4 (Ch1&2: Quadratic Equation I & II) 1 Name: Class: ( ) Date: Mark: /29 Time: 40min, Full Marks: 29 SECTION A: Structure Questions 1. HKCEE MATH 1988 Q4

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Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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Name:    Class:                                                                  (                          )  Date:    Mark:                                                                            /29  

 Time:  40min,  Full  Marks:  29  SECTION  A:  Structure  Questions  1. HKCEE  MATH  1988  Q4  

       

Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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2. HKCEE  A.MATH  1998  Q2  

   

Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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3. HKCEE  MATH  1993  Q6  

   

Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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SECTION B: Multiple Choices (Correct answer 2M, Wrong answer -1M) Each question carry one CORRECT answer only. 4. HKCEE  MATH  2008  II  Q7  

   5. HKCEE  MATH  2007  II  Q4  

Let x be the smaller one of two consecutive integers. If the sum of the squares of the two integers is less than three times the product of the two integers by 1, then  

A. x2 + (x + 1)2 = 3x(x + 1) – 1 B. x2 + (x + 1)2 = 3x(x + 1) + 1 C. 3[x2 + (x + 1)2] = x(x + 1) - 1 D. 3[x2 + (x + 1)2] = x(x + 1) + 1

6. HKCEE  MATH  2006  II  Q8  

 7. HKCEE  MATH  2006  II  Q9  

 8. HKCEE  MATH  2005  II  Q5  

 

Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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9. HKCEE  MATH  2005  II  Q7  

     

Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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Solution:    1. (a)  9𝑥! − 𝑘 + 1 𝑥 + 1 = 0… ∗  ℎ𝑎𝑠  𝑒𝑞𝑢𝑎𝑙  𝑟𝑜𝑜𝑡, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒,∆= 0      …1M  

(𝑘 + 1)! − 4 9 1 = 0      …1M  𝑘! + 2𝑘 + 1− 36 = 0  𝑘! + 2𝑘 − 35 = 0  𝑘 − 5 𝑘 + 7 = 0    𝑘 = 5    𝑜𝑟    𝑘 = −7      …1A      (b)  If  k  takes  the  negative  value  obtained  in  part  (a),  that  means  take  k=-­‐7,  then  we  have  9𝑥! − −7+ 1 𝑥 + 1 = 0… ∗      …1M  9𝑥! − −6 𝑥 + 1 = 0  9𝑥! + 6𝑥 + 1 = 0  (3𝑥 + 1)! = 0      …1M  3𝑥 + 1 = 0  𝑥 = − !

!      …1A  

Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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Mathematics  St.  Bonaventure  College  and  High  School  

Form  4,  Quiz  4  (Ch1&2:  Quadratic  Equation  I  &  II)    

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 4. C  5. A  6. D  7. A  8. B  (85)  9. D  (47)