JANUARY 1999 FINAL (answers below)

MQ5 home   1995 final

STUYVESANT HIGH SCHOOL                                                              MATHEMATICS DEPARTMENT
Jinx Perullo, Principal                                                                                  Danny Jaye, Chairman

MQ5 Final Examination January, 1999

PART I: ANSWER ALL QUESTIONS IN PART I ON THE NCS SCANTRON FORM (3 points each)

1. If n is an integer and , the expression i 4n + 3 is equivalent to:
    A. i     B. - i     C. 1 + i     D. 1 - i     E. 1 + i3

2. The number of values of q in the interval 0° £ q < 360° for which sin2q = 1/9 is:
    A. 0     B. 1     C. 2      D. 3     E. 4

3. Expressed in terms of a single function,           1           is equivalent to:     
                                                                 tan2q + 1
    A. cos2q     B. cot2q     C. csc2q     D. sec2q     E. sin2q

5. If the point with coordinates (4, -7) lies on the terminal side of q , cscq equals:

   

6. The sum (S) and product (P) of the roots of 4x2 - 6x + 5 = 0 are:
    A. S = 5/4, P = 3/2     B. S = - 5/4, P = 3/2     C. S = 3/2, P = 5/4     D. S = 3/2, P = - 5/4     E. S = - 3/2, P = 5/4

7. The expression        1           +        1           is equivalent to:
                             1 + sin q        1 - sin q
    A. 2cos2q     B. 2csc2q     C. 2sec2q     D. 2tan2q     E. 2 - cos2q

8. If a circle with a central angle of 4.5 radians intercepts 15 inches of arc, the length of its radius, in inches, is: 
    

9. The solution for x in | x + 3 | - 2 > 5 is:
    A. x < - 4 or x > 10      B. x < - 10 or x > 4      C. x > 4     D. - 4 < x < 10     E. - 10 < x < 4

10. The solution set for cos q - sec q = 0 for q such that 0° £ q < 360° is:
    A. {0° }     B. {90° }     C. {180° }      D. {0° , 180° }     E. {90° , 270° }99-11.gif (2791 bytes)

 
       The measure of AB, in units, is:
       A. 5      B. 6     C. 8     D. 10      E. 12

12. Expressed in simplest form, is: 

13. The expression (3x-1 - 3x-2) / (1 - x-2) is equivalent to:
    A.    3         B.    3        C.    3x         D.    3x        E. 399-14.gif (2592 bytes)
         x + 1           x - 1            x + 1           x - 1


      The measure of AP, in units, is: A. 6 B. 9 C. 12 D. 15 E. 18

15. If tan q = 2/3 and sec q < 0, sin q equals:
          

16. The set of all values of q in the interval 0° £ q < 360° that satisfy sin2q = - sin q is:
    A. {270°}    B. {90°, 270°}    C. {180° , 270°}    D. {0°, 180°, 270°}     E. {90°, 180°, 270°}
17. The expression is equivalent to:
     

18. The set of all the values of x for which is a real number is:
    A. {x < 3}    B. {x > 3}    C. {x ³ 3}    D. {x < - 3 or x > 3}    E. {x £ - 3 or x ³ 3}

19. When factored completely, 6x2 - 10xy + 21x - 35y is equivalent to:
    A. (3x + 7y)(2x - 5)    B. (3x - 7y)(2x + 5)   C. (3x - 5y)(2x + 7)    D. (3x + 5y)(2x - 7)   E. (3x + 5y)(2x - 7y)

20.     3                1           equals:
       x - y         2y - 2x
    A.      7           B.     5          C.      5y - 3x            D.      7         E.       5     
         2(x - y)         2(x - y)           2(x - y)(y - x)         2(y - x)         2(y - x)


PART II: Answer TWO of the following questions and SHOW ALL WORK (20 each)


     june94.gif (2807 bytes)

      f. Solve for x over the set of real numbers and check:         [6]


2a. Express as a single fraction in lowest terms with the denominator in
      factored form.   [10]
2b. Solve the equation tan2q + secq - 1 = 0 for q in the interval 0 £ q < 2p   [10]

3a. Solve for x over the set of real numbers: [10]
3b. Prove that the equation is an identity:    [10]

PART ONE SOLUTIONS:  1. B    2. E    3. A    4. C     5. E    6. C    7. C    8. B    9. B
10. D    11. A    12. A    13. A     14. D    15. B    16. D     17. B    18. E    19. C     20. A
PART TWO SOLUTIONS: 
1. Let x = m arc AB, 2x = m arc BC, 2x = m arc AC. Then x + 2x + 2x = 360, x = m arc AB = 72
    2x = m arc BC = m arc AC = 144. Since m< DAC = 48, m arc CD = 2(48) = 96
    a. m arc AB = 72
   b. m arc AD = m arc AC - m arc CD = 144 - 96 = 48
    c. m<CPD = (1/2)(m arc CD - m arc AD) = (1/2)(96 - 48) = 24
    d. m<CED = (1/2)(m arc CD + m arc  AB) = (1/2)(96 + 72) = 84
    e. m<ADP = (1/2)m arc AD = (1/2)(48) = 24
    f. reject x = 1, therefore {5}
2a. 2/[(x - 3)(x - 1)]
2b. {0, 2p/3, 4p/3}
3a. Reject  -2, therefore {0}
3b. [1/(1 + cosq)]·[(1 - cosq)/(1 - cosq)] =  (1 - cosq)/(1 - cos2q) = (1 - cosq)/sin2q =
       1/sin2q - cosq/sin2q  = csc2q - (1/sinq)(cosq/sinq) =  csc2q - cscqcotq


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