January 1995 MQ5 Final

JANUARY 1995 FINAL (answers below)

MQ5 home Test 12 1999 final

STUYVESANT HIGH SCHOOL MATHEMATICS DEPARTMENT
M. Kahn, Principal R. Rothenberg, A.P.S.

MQ5 Final Examination: January 1995

PART I: ANSWER ALL QUESTIONS IN THIS PART ON SCANTRON SHEET (3 points each)

1. If the sum and product of the roots of 4x² - 12x + 21 = 0 are S and P, respectively, then:
A. S = - 3, P = 21/ 4 B. S = 3, P = 21/ 4 C. S = - 12, P = 21 D. S = 12, P = 21 E. S = 21/ 4, P = 3

2. If sin q - csc q = 0 and 0° £ q < 360° , the solution set for q is:
A. {0° } B. {90° } C. {180° } D. {0° , 180° } E. {90° , 270° }

3. The reciprocal of 6 + 2i equals:
A. 3 - i B. 3 + i C. 1 - i D. 1 + i E. 3 + i
20 20 20 20 15 20 15 20 20 15

4. x + 6 + 1 equals: A. x + 7 B. 2x C. 2 D. x + 4 E. 2x
x² - 4 x + 2 x² - 4 x - 2 x - 2 x - 2 x + 2

5. If x² + 2x > 35, the solution for x is:
A. x < -7 or x > 5 B. -7 < x < 5 C. x > 5 D. x < -5 or x > 7 E. -5 < x < 7

A. 3(x - 3) B. 4 C. 3(x + 3) D. 3 E. 4
x x + 3 x x + 3 x

7. If sin q = -2/3 and tan q > 0, then cos q equals:
A. Ö5/ 3 B. 1/ 3 C. -Ö5/3 D. - 1/3 E. Ö5/5

8. If ½7 - 2x½ > 11 , the solution for x is:
A. -2 < x < 9 B. x > 2 or x < -9 C. x < -2 D. x < -2 or x > 9 E. x < -9 or x > 2

9. If tan² q = tan q and 0° £ q < 360° , the solution set for q is:
A. {0°, 45°} B. {45°} C. {45°, 315°} D. {0°, 45° , 180°, 225°} E. {0°, 45°, 225°, 270°}

10. If the terminal side of q contains point (2Ö3, -2), and 0° £ q < 360°, then the degree measure of q is:
A. 150° B. 210° C. 240° D. 300° E. 330°

11. equals: A. 2/a B. 2/b C. a + b D. -1 E. 1
2(ab + 1) 2a 2b

12. 6i⁴² + 2i^-5 equals: A. -6 - i B. 6 + i C. -6 - 2i D. 6 + 2i E. 6 - i
32 32 32

13. The real values of k for which kx² + 12x + 9 = 0 has imaginary roots are:
A. k > 4 B. k < 4 C. k ³ 4 D. k £ 4 E. k = 4

14. In a circle whose radius is 14 inches, the radian measure of a central angle which intercepts 35 inches of arc is:
A. p /5 B. 2p/5 C. 5p/2 D. 2/5 E. 5/2

15. 1 + 1 equals: A. 2csc²q B. 2sec²q C. 2 - tan²q D. 2 + tan²q E. 2cot²q
1 - cosq 1 + cosq

16. The product sin (2p/3) cot (2p/3) equals: A. Ö3/2 B. -Ö3/2 C. 1/2 D. -1/2 E. Ö3/3

17. If sec(5x + 4°) = csc(3x - 10°) , the least positive solution for x in degree measure is:
A. 6 B. 12 C. 24 D. 48 E. 357

A. Ö3 - 1/2 B. Ö3 - 1 C. Ö3/2 - 1 D. Ö3/2 - 1/2 E. 2Ö3 - 1/2

19. In a circle where chords AB and CD intersect at point E; if CE = 14, ED = 2, and EB is 3 units less than AE,
then AE equals: A. 4 B. 6 C. 7 D. 8 E. 12

20. In a circle where secant segments PAB and PDC are drawn from external point P; if PB = 6, PA = 4,
and CD = 5, then PD equals: A. 3 B. 4 C. 6 D. 8 E. 9

A. -8x⁶ B. 8ix⁶ C. 8ix⁴ D. -4x⁴ E. 4ix⁴

PART II: Answer TWO of the following questions and SHOW ALL WORK (18 each)

1. Complete 1a and 1b as instructed.

1a. Solve and check over the set of real numbers: [8, 2 points]

1b. Prove that the equation is an identity: [8 points]

2. Solve for q, to the nearest degree, in the interval 0° £ q < 360°: [18 points]
2sin²q - 3cosq - 3 = 0

wpeD.gif (4156 bytes)

PART ONE SOLUTIONS: 1. B     2. E    3 A    4. C    5. A    6. E    7. C    8. D     9. D
10. E    11. A    12. C    13. A    14. E    15. A    16. D     17. B    18. D    19. C     20. A    21. D
PART TWO SOLUTIONS:
1a. Reject x = -1, therefore {6}
1b. (tan x csc²x)/(1 + tan²x) = (tan x csc²x)/sec²x = tan x (csc²x / sec²x) = tan x (cos²x / sin²x) =
      tan x cot²x = cot x
2. {120°, 180°, 240°}
3. let x = m arc AB, 234 - x = m arc BC. Therefore, 234 - x = 2x - 54; -3x = -288; and x = 96
    Then m arc BC = 138, m arc AD = 42, m arc CD = 84 (Hint: fill in all the arc measures in diagram)
    a. m arc AB = 96
    b. m<ABD = (1/2)m arc AD = (1/2) 42 = 21
    c. m<BPC = (1/2)(m arc BC - m arc AD) = (1/2)(138 - 42) = 48
   d. m<AEB = (1/2)(m arc AB + m arc CD) = (1/2)(96 + 84) = 90
    e. m<ADP = 180 - m<CDA = 180 - (1/2)m arc ABC = 180 - (1/2)234 = 63

MQ5 home Test 12 1999 final