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Math 3204
Grading Standards
June 2004
Pre-Marking Appraisal
The board considered the exam fair and of the appropriate difficulty and length. It was similar to and had a parallel structure to the June 2003 exam.
The following notes were made.
- #52 had 2 solutions. Students could rewrite in transformational form or use the formula t = 2 − ba.
- #54 the board agreed that the widest x -interval to be used which would still warrant full marks would be 1.9 to 2.1.
Marking Standard and Consistency
Marker reliability was checked by obtaining a random sample of 35 papers that went through the marker panel and marks were assigned to each question on a separate sheet of paper. The 35 exams were put back into the original stack of exams and corrected again when they appeared. The two values were compared and if there were discrepancies, the chief marker would review the scoring with the individual marker.
Throughout the marking process there was statistical analysis run on item data to enhance reliability and consistency of marking.
Commentary on Responses
Students performed fair to well on most questions.
- #49 was difficult for math 3204 students.
- #52 had two solutions. Half of the responses were presented using the formula t = 2 − ab ,the other half were presented by rewriting the function in transformational form.
- #54 saw many variations of the interval that was used. The board agreed that the widest interval that would be accepted would be 1.9 to 2.1.
- #59 was a question on which students did not score well. By giving the center, this perhaps made the question more difficult. Students made a wide range of errors and the board was surprised to see such a poor performance on a relatively easy question.
- A common error in #62 was students drawing a 10cm chord. A coincidence within this question was that the circumference and the area of the sector were the same 62.83. Some students used this and arrived at the right answer.
- #63 was not attempted by many students and subsequently students scored the lowest on this question. It was decided that including an extension of the tangent line, and the radius to the point of tangency would have made this question fairer for a Math 3204 question. This question was marked on all papers but was not used in determining a student’s mark if it was to the student’s detriment (i.e., Part II was marked out of 46 or 50, whichever benefitted the individual student).
OR
−
2
2
2
h t t
h t t
h t
Vertex (4,83) Max. is 83 at t 4 seconds
Value 4 52. A cannonball is fired and its height, h , in metres, above the ground, t seconds
after being fired, is given by h t ( ) = − 5 t^2 + 40 t +3.Algebraically determine the
maximum height attained by the cannonball and the time taken to reach this height.
4s 2 2( 5)
h t t t b t a
h (4) = − 5(4)^2 + 40(4) + 3 = 83 metres
Commentary on Response
This was a fair question and the majority of students attempted this question. Two methods were commonly used to complete the question including:
(i) (^) V (^) ( − 2 ba , y )→ students found the time element ‘t’ using (^) t = − 2 b a ,then substituted to find the max. height ‘h(t)’.
(ii) completing the square →students transformed the given equation into
standard and/or transformational form.
Some students substituted values for ‘t’ until the maximum appeared. Some listed in a table, while others showed calculations for h(1), h(2), h(3), h(4), h(5), etc. In both instances, the calculator may have been used to generate the given results.
Some students used V (^) ( 2 -b a^ , 4 ac 4 a^^ − b ).This is a lot of work to find h(t), however the correct was result obtained often times.
Some students found the average of the zeros then rounded to 4 which was the correct time
Common Errors
- Students used the quadratic formula and simply found the zeros. [i.e., -0. and 8.07]
- Students found h = 2 − ab ( i e V h k. ., ( , ))then assumed ‘h’ was the height.
Common answers included t = 83 , s h = 4 m when it should be
t = 4 , s h = 83 m.
- Students found h (0) = 3 the[ y −int .]then substituted t = 3 to find maximum
height.
- Students were confused with completing the square method
y = − (^5) ( t^2 − 8 t + (^16) )+ 3 −?, many errors here including adding -16, subtracting 80, etc.
- Students divided h(t) by -1 and completed the square, but did not divide both sides by -1.
- Students found h (4) = 83 then divided by 4 to get height = 20.75m.
- Students used quadratic formula to get t = 8.075 [rejected the negative zero] and substituted this value to find h(t).
4 54. A diver jumps off a spring board. Her height h , in metres, above the water, t seconds after she jumps, is given by h ( t ) = - 4.9 t^2 +8 t + 5. Algebraically determine the approximate instantaneous rate of change in her height at 2 seconds.
t h
1.99 1.
2.01 1.
So, instantaneous rate of change is approx.
Instantaneous RoC.
2 1 2 1 1.28351 1. 2.01 1.
11.6m / s
y y x x
− − − − −
Commentary on Response
Overall this question was well done. Unfortunately, no largest interval had been specified for IROC nor had an appropriate number of decimal places been specified, thus a variety of answers were accepted. Three students solved the problem using derivatives.
Common Errors
Students made mistakes by;
- not following order of operations
- using too large a time interval for IROC
- using inadequate number of decimal places
- simply evaluating h (2) getting 1.4 and leaving it as the answer
- inverting the slope formula
- dropping negative sign
−
−
(^13)
(^23)
5 1 2
5 1
2 3
1 3
x x x x x x
3 55. Solve for x : 35 x −^1 =^3 9.
Commentary on Response
This question was well answered. Students knew the concept of equating exponents when bases are equal on each side of the equation.
Common Errors
Students made the following computational errors:
(^1 ) 2
3
3 2 3
3
x
x
( )
( ) ( ) ( )
120
120
1 2 5 1 42 2 5 1 42 120 2
log log
t
t
t
−−
− −
120 0.92430.3010 120
120
0.9243 ( 0.3010)
368.5 years
t t t
t
Value 4 57. An element has a half-life of 120 years. If its initial mass is 42 grams, algebraically determine how long it will take to decrease to 5 grams.
( )
=
=
i
1 120 (^422)
t h t
y a b
y
Commentary on Response
This question was not well done. Most students seemed to know the base equation but many had difficulty substituting the correct values into the equation.
x c
y = ab
For those who correctly substituted the values, many got stuck at the step where they are to take the log of both sides. A large number of students found the answer by inspection to be ~360 or by starting with the initial mass and using half life to approximate the answer.
Common Errors
Students made mistakes by;
- multiplying (^42) ( 12 )(i.e., (^5 42) ( 12 )^120 5 21120 ).
t (^) t = ⇒ =
- not writing 120^ t^ as a power. (i.e., 5 = (^42) ( 12 ) 120 t ).
- writing formula incorrectly (i.e., y = abx ).
- substituting values into the base formula incorrectly (i.e., (^120 12) ( 42 )^5 .).
t
Value 4 58. Hot chocolate cools exponentially over time after it is brought into a stadium and the cooling is described by the function given with temperature, T , in degrees Celsius and time, m , in minutes.
T = 45(0.95) m + 5
Sketch a labeled graph for the first 100 minutes. State the initial temperature of the hot chocolate and the air temperature in the stadium.
Initial temp. of hot Chocolate: 50 OC Air temp. of Stadium: 5OC
Commentary on Response
This question was fairly well done. Many students generated a table of values to produce the graph. In drawing the graph, many students did not draw the asymptote. Many students also made a mistake with the initial temperature, using 45 O^ C instead of 50 O^ C. Students are unaware of the fact that a labeled graph means asymptote & y-intercept. As well they were unaware of all the necessary information required for this question.
Common Errors
- Students used 45 O^ C as the initial temperature instead of 50 O^ C.
- Students did not include the asymptote on the graph. Some recognized its existence but did not draw it on the graph.
- Students did not realize that a table of values was not really required, only a y-intercept and asymptote, along with the decreasing nature of graph.
Value
4 60. Write 9 x^2 + y^2 + 18 x − 6 y + 9 = 0 in transformational form.
( ) ( )
( ) ( )
( ) ( )
( ) ( )
2 2 2 2
2 2
2 1 2 9 2 1 2 3
x x y y x x y y
x y
x y
x y
Commentary on Response
This question was well done. Most students attempted it but many had trouble completing the square especially with 9 x^2 + 18 x + .Some students made the
mistake of putting the final answer into the form (^) ( ) ( ) 1 2
a y^ −^ k^ =^ x^ − h
which is the tranformational form for a quadratic. A lot of students received 3 or 3 12 because they didn’t finish the answer, (i.e., they didn’t put the answer into the form ( ) ( ) 1 2 1 2 ⎡⎣ (^) a x − h ⎤⎦ + ⎡⎣ (^) b y − k ⎤⎦ =1. More emphasis is needed on changing from general to standard to transformational form when doing quadratics and other conic sections.
Common Errors
- Students left the answer as ( ) ( ) ( ) ( ) 2 1 2 2 1 2 x + 1 + 9 y − 3 = 1 instead of ⎡⎣ x +1 ⎤⎦ + ⎡⎣ 3 y − 3 ⎤⎦ =1.
- Students did not complete the square properly for (^9) x^2 + 18 x with most work done properly thereafter.
i.e.,
( ) ( )
( ) ( )
( ) ( )
2 2
2 2
2 2
x x y y
x x y y
x y
- Students did not complete the square properly for 9 x^2 + 18 x continued.
( ) ( )
( ) ( )
2 2
2 2
x x y y
x y
- Students added improperly:
( ) ( )
( ) ( )
2 2
2 2
x x y y
x y
- Students got correct answer but then put it into (^) ( ) ( ) 1 2 a y^ −^ k^ =^ x^ − h
i.e., correct answer: (^) ( ) ( )
2 1 2
⎡⎣ x^ +^1 ⎤⎦ +^ ⎡⎣ 3 y −^3 ⎤⎦ =^1
then continued, with: (^) ( ) ( ) 1 2 2 3 y^ −^3 =^ x +^1 or: (^) ( ) ( )
2 2
y − 3 = 3 x + 1
- Students completed the square properly but subtracted on the right side instead of adding
ie., (^ )^ (^ )
should be = -9 + 9 + 9
x + x + + y − y + = − − −
(^10 )
A B
72 °
- Students assumed the question was “to prove that ABCD is a rectangle” and not “to find the area of the rectangle”.
- Students did not plot the points on the grid but just sketched a rectangle.
- Students made calculation errors: eg : 3 2 + −( 2) 2 = 6 + 4 = 10.
- Students made errors with squaring numbers: (^) eg : (^) ( -2 (^) )^2 = − 4.
4 62. AB is a chord of circle with radius 10 cm. If m ABpis 72 O^ , find the area of the
region enclosed by AB and pAB.
1 2 Area 2 (10)(10)sin 72 47.6cm
Area (^) sector = π = ⋅ 72 2 72 2 2 360 r 360 314.16cm^ 62.83cm
−
2 2 2 Areashaded region 62.83cm 47.6cm 15.23cm
Commentary on Response
Students had difficulty in creating the diagram. Most students could not label the diagram correctly. Many students formed the area of the region by subtracting the area of sector from the area of the circle. Two methods were used, to find the area of the triangle to solve the problem: (i) Some students used the formula A (^) ∆ =^12 bc sin A; and, (ii) some students used trigonometric ratios to find the height and base and then applied the formula A (^) ∆ =^12 bh.
Common Errors
- Students used 10 cm as the chord
- Students used calculator in radian mode (i.e., Sin 72 O^ = 0.2538)
- Students used 10 cm as the base of the triangle (i.e., assumed the triangle was equilateral)
- Students used an incorrect central angle for the sector (i.e., 36o, 54o, 90o^ , 144 o).
- Students used (^2) π r or 2π r^2 to find the area of a circle.
y
- 12– 10– 8– 6– 4– 2 2 4 6 8 x
2
4
6
8
10
12
O(-2,1)
P(6,7)
−
2 1 OP 2 1
6 8 3 4
y
m =
y
x x
= −^4 mtangent 3
−
= +
= +
= − +
=
4 3
m
7 (6) b
7 8 b
15 b
y x b (^) ∴ = − (^4) +
y 3 x 15
Value 4 63. Point P(6, 7) lies on the circle, with centre (-2, 1), as shown. Determine the equation of the tangent line to the circle at P.
Commentary on Response
Many students did not attempt this question. Some students found the equation of the circle to represent the tangent line. Some students extended the tangent line across the y-axis to approximate the y-intercept.
This question was marked on all papers but was not used in determining a student’s mark if it was to the student’s detriment (i.e., Part II was marked out of 46 or 50, whichever benefitted the individual student).
Common Errors
- Students used an incorrect slope formula.
- Students extended the tangent line to find the y-intercept.
- Students substituted the centre (-2, 1) into the equation y = mx + b to find b.
- Students didn’t take the negative reciprocal of the slope of the line segment OP for the slope of the tangent line.
SELECTED - RESPONSE ITEM ANALYSIS
Item
Responses Multiple Answers or No Response
A B C D
PART II CONSTRUCTED - RESPONSE ITEM ANALYSIS
- MATHEMATICS
- Correct answer is B .1 11.9 68.3 9.8 9. % % % % %
- Correct answer is B .2 28.6 61.2 6.5 3.
- Correct answer is D .2 24.6 17.2 16.2 41.
- Correct answer is D .1 19.9 35.4 7.3 37.
- Correct answer is A .2 24.5 9.5 32.0 33.
- Correct answer is D .1 13.9 20.8 19.4 45.
- Correct answer is D .1 13.6 12.8 6.4 67.
- Correct answer is D .1 9.4 4.0 3.7 82.
- Correct answer is A .1 63.0 7.3 25.9 3.
- Correct answer is C .5 20.4 8.9 62.0 8.
- Correct answer is B .1 9.0 74.3 10.7 5.
- Correct answer is D .1 8.9 22.5 11.8 56.
- Correct answer is C .4 9.7 10.7 65.0 14.
- Correct answer is D .0 3.4 6.1 19.3 71.
- Correct answer is A .2 63.0 3.3 10.6 22.
- Correct answer is C .2 2.5 2.9 89.9 4.
- Correct answer is B .0 2.3 78.4 8.1 11.
- Correct answer is C .2 11.3 5.3 71.9 11.
- Correct answer is C .4 7.9 12.0 57.1 22.
- Correct answer is B .1 12.3 75.6 3.1 9.
- Correct answer is B .1 11.9 62.1 17.5 8.
- Correct answer is C .4 16.9 25.9 40.5 16.
- Correct answer is A .1 41.6 8.8 35.5 14.
- Correct answer is B .0 15.2 70.1 10.5 4.
- Correct answer is B .2 30.1 38.4 19.7 11.
- Correct answer is C .2 19.2 13.4 57.8 9. % % % % %
- Correct answer is B .1 7.2 70.6 7.3 14.
- Correct answer is D .5 15.1 22.4 14.6 47.
- Correct answer is C .5 12.7 10.7 58.9 17.
- Correct answer is D .2 15.9 19.7 5.6 58.
- Correct answer is C .1 24.6 9.8 58.2 7.
- Correct answer is A .3 71.3 4.4 13.5 10.
- Correct answer is D .0 15.8 4.9 2.5 76.
- Correct answer is D .1 19.7 9.7 9.9 60.
- Correct answer is C .1 12.7 16.8 61.0 9.
- Correct answer is A .1 53.7 21.7 5.6 18.
- Correct answer is C .0 8.6 9.4 73.7 8.
- Correct answer is C .1 1.8 6.1 85.4 6.
- Correct answer is A .3 54.9 19.4 16.5 8.
- Correct answer is A .1 68.4 8.4 7.6 15.
- Correct answer is A .4 57.7 30.3 9.0 2.
- Correct answer is C .3 38.2 10.8 38.7 12.
- Correct answer is D .3 11.8 11.4 35.9 40.
- Correct answer is D .6 12.5 30.8 30.6 25.
- Correct answer is D .3 4.5 21.8 15.4 57.
- Correct answer is C .3 32.0 12.2 49.0 6.
- Correct answer is D .4 8.8 20.6 23.0 47.
- Correct answer is D .4 19.7 14.9 16.5 48.
- Correct answer is D .7 33.4 22.5 22.7 20.
- Correct answer is B .7 15.4 32.4 26.5 25.
- PART II Item Completing ItemStudents Value Average Average % PerItem
- 51 3502 4 2.67 66.
- 52 3502 4 2.30 57.
- 53 3502 4 1.79 44.
- 54 3502 4 2.722 68.
- 55 3502 3 1.74 58.
- 56 3502 4 1.92 48.
- 57 3502 4 2.03 50.
- 58 3502 4 1.59 39.
- 59 3502 3 1.08 36.
- 60 3502 4 2.32 58.
- 61 3502 4 2.805 70.
- 62 3502 4 1.51 37.
- 63 3502 4 .43 10.
