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1 CSET California Subject Eaminations for Teachers TEST GUIDE MATHEMATICS SUBTEST III Sample Questions and Responses and Scoring Information Copyright 005 by National Evaluation Systems, Inc. (NES ) California Subject Eaminations for Teachers, CSET, and the CSET logo are registered trademarks of the California Commission on Teacher Credentialing and National Evaluation Systems, Inc. (NES ). NES and its logo are registered trademarks of National Evaluation Systems, Inc. CS-TG-QRX-0

2 Sample Test Questions for CSET: Mathematics Subtest III Below is a set of multiple-choice questions and constructed-response questions that are similar to the questions you will see on Subtest III of CSET: Mathematics. The term "enhanced" is used in this test guide to identify comple multiple-choice questions that require 3 minutes each to complete. Please note that enhanced mathematics questions will not be identified on the actual test form. You are encouraged to respond to the questions without looking at the responses provided in the net section. Record your responses on a sheet of paper and compare them with the provided responses. Note: The use of calculators is not allowed for CSET: Mathematics Subtest III.. What are the solutions to tan + π 6 = 3 where 0 π? A. B. C. D. π 6, 5π 6 5π 3, π 3 π 3, 4π 3 π 6, 7π 6 California Subject Eaminations for Teachers Test Guide

3 . Use the diagram below to answer the question that follows. y r q The diagram shows a rectangle inscribed in a circle of radius r. What is the area of the rectangle as a function of the angle θ? A. 4r tan θ B. r cos θ sin θ C. 4r tan θ D. 4r cos θ sin θ California Subject Eaminations for Teachers Test Guide

4 (ENHANCED) 3. Use the diagram below to answer the question that follows. y P (0, 0) Point P is on the edge of a wheel of radius of that rotates counterclockwise around the origin at a speed of 5 revolutions per second. At t = 0, P has coordinates (, 0). Which of the following functions could be used to describe the -coordinate of P as a function of time, t, measured in seconds? A. (t) = sin(30πt) B. (t) = sin C. (t) = cos πt 5 πt 5 D. (t) = cos(30πt) California Subject Eaminations for Teachers Test Guide 3

5 4. Given that z = cos 75 + i sin 75, which of the following best represents z 3 in vector form? A. B. imaginary imaginary real real C. D. imaginary imaginary real real 4 California Subject Eaminations for Teachers Test Guide

6 (ENHANCED) 5. For which of the following values of k does sin + sin cos k sin cos equal? A. 0 B. π C. π D. 3π 6. If f() is continuous on the interval [a, b], and N is any number strictly between f(a) and f(b), which of the following must be true? A. The inverse of f() is defined on the interval [f(a), f(b)]. B. There eists a number c in (a, b) such that f'(c) = 0. C. f() is differentiable on the interval (a, b). D. There eists a number c in (a, b) such that f(c) = N. 7. An economist needs to approimate the function f() = by a line tangent to f() at =. Which of the following lines should be used? A. y = B. y = C. y = 3 D. y = 3 4 California Subject Eaminations for Teachers Test Guide 5

7 (ENHANCED) 8. Determine 0 sin() 3. A. 6 B. 0 C. 6 D. (ENHANCED) 9. The rate at which the mass of a radioactive isotope changes with respect to time is directly proportional to the mass of the isotope, where k is the constant of proportionality. If 500 grams of the isotope are present at time t = 0 and 00 grams are left at time t = 0, what is the value of k? A. B. C. D. 0 Bn 5 5 Bn 0 5 Bn Bn 5 6 California Subject Eaminations for Teachers Test Guide

8 0. Use the graph below to answer the question that follows. y y = f( ) 0 The height and width of each of the above n rectangles is given by f( i ) and n, respectively, where i n. If the sum of the areas of the n rectangles is given by n n f( i ) = 08n 8n 7 6n, then by the use of Riemann i = sums f() d is equal to which of the following? 0 A. 4.5 B. 3.5 C. 8 D. 08 California Subject Eaminations for Teachers Test Guide 7

9 (ENHANCED). Evaluate ( + ) 4 d. 0 A. B. C. D Which of the following represents the area under the curve of 4 the function h() = 3 + on the interval [0, ]? A. B C. 4 Bn 4 D. 4 3 Bn 4 3. For what values of does the infinite series ( ) + ( ) ( ) 3 + converge? A. > B. < 3 C. < < 3 D. < or > 3 8 California Subject Eaminations for Teachers Test Guide

10 4. A Babylonian tablet dating from circa 300 B.C. contains the following problem: There are two fields whose total area is 800 square yards. One produces grain at the rate of 3 of a bushel per square yard while the other produces grain at the rate of of a bushel per square yard. If the total yield is 00 bushels, what is the size of each field? The above problem indicates the beginning of which of the following contemporary topics of mathematics? A. systems of linear equations B. calculus C. quadratic equations D. Euclidean geometry 5. In the eleventh century, the mathematician Omar Khayyam presented the generalized geometric solution to some cubic equations using intersecting conic sections. This concept of applying algebra to geometry, which centuries later developed into the topic of analytical geometry, is more notably attributed to which of the following mathematicians? A. David Hilbert B. Blaise Pascal C. René Descartes D. Bernhard Riemann California Subject Eaminations for Teachers Test Guide 9

11 6. Complete the eercise that follows. Graph the system of equations below on the same coordinate system and then use analytic techniques to find the coordinates of the points of intersection of the graphs for π π. y = sin y = cos() + 0 California Subject Eaminations for Teachers Test Guide

12 7. Use the function below to complete the eercise that follows. f() = Find the coordinates of any zeros, relative etrema, and/or inflection points of this function. Sketch the graph of this function. California Subject Eaminations for Teachers Test Guide

13 8. Use the Fundamental Theorem of Calculus stated below to complete the eercise that follows. Let f be continuous on a closed interval [a, b] and let be any point in [a, b]. If F is defined by F() = f(t) dt, a then F'() = f() at each point in the interval [a, b]. Using the formal definition of the derivative, prove the above theorem. California Subject Eaminations for Teachers Test Guide

14 9. Use the information below to complete the eercise that follows. Prior to the eistence of trigonometric tables, Archimedes found that 3 7 < π < 7 by comparing the perimeters of 96-sided regular polygons inscribed in and circumscribed about a circle of radius. Archimedes started with 6-sided polygons and found numerical values of the perimeters of these inscribed and circumscribed heagons. He then derived a recursive formula that used the numerical values found to calculate the perimeters of -sided polygons. This formula was equivalent to the modern half-angle formulas of trigonometry. The method was then repeated with 4-sided, 48-sided, and 96-sided polygons. θ O Repeat Archimedes's method using trigonometry to find upper and lower bounds for π by determining the perimeters of a heagon inscribed in and a circumscribed about a circle of radius. Epress the answer using eact values. Eplain (without actually doing) how the trigonometric half-angle formulas can be used to etend the result to 96-sided inscribed and circumscribed polygons to obtain a more accurate approimation of π. California Subject Eaminations for Teachers Test Guide 3

15 Sample Written Response Sheets for CSET: Mathematics Subtest III For questions 6 9, eaminees would record their written response to each question on a two-page response sheet located in their answer document. The length of their response to each question is ited to the lined space available on the response sheet. A sample of the response sheet is provided below. 4 California Subject Eaminations for Teachers Test Guide

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17 Annotated Responses to Sample Multiple-Choice Questions for CSET: Mathematics Subtest III Calculus. Correct Response: D. (SMR Code: 5.) tan + π 6 = 3, + π 6 = arctan 3 + π 6 = π 3 or 4π 3, for 0 π, since tan π 3 = 3 and tan 4π 3 = 3. Hence, either + π 6 = π 3 or + π 6 = 4π 3 = π 6 or 7π 6.. Correct Response: D. (SMR Code: 5.) To find the area of the inscribed rectangle, find the area of the rectangle in Quadrant I and multiply its area by 4. The length of the horizontal side of this rectangle is r cos θ, and its vertical side is r sin θ, so the area of the rectangle is r cos θ r sin θ = r sin θ cos θ. Thus the large rectangle has an area of 4r sin θ cos θ. 3. Correct Response: D. (SMR Code: 5.) The -coordinate of point P on the unit circle is the cosine of the angle measured from (, 0) to point P, θ, = cos θ. Since the wheel turns 5 revolutions per second and each revolution is equivalent to a rotation of π radians, the change in the angle of rotation is 30π radians per second, i.e., θ = 30πt (t) = cos (30πt). The initial -coordinate is, so when t = 0, (t) should equal. Hence the equation is (t) = cos(30πt). 4. Correct Response: D. (SMR Code: 5.) By de Moivre's theorem, z 3 = 3 [cos (3 75 ) + i sin (3 75 )] = cos 5 + i sin 5. Since the argument of z is 5, response choice D is the correct response. 5. Correct Response: C. (SMR Code: 5.) Begin by simplifying the epression: sin + sin cos k sin cos = sin ( + cos ) k sin cos makes this it equal to, solve: + cos k cos k = k + cos cos = + cos k cos k. To find the value of k that = + cos k = cos k cos k + cos k + = 0 (cos k + )(cos k + ) = 0 cos k =. The angle with a cosine of is π radians, so k = π. 6. Correct Response: D. (SMR Code: 5.) The Intermediate Value Theorem states that if a function is continuous on [a, b] and if f(a) f(b), then somewhere in the interval (a, b) the function must take on each value between f(a) and f(b). So if f(a) < N < f(b), the function must equal N at some point in (a, b). 7. Correct Response: B. (SMR Code: 5.3) If f() =, then f'() =, so the slope of the line tangent to f() at is ( ) or. Then the equation of the tangent line is y = + b. To find b, substitute the values = and y = f( ) = into y = + b, which gives b =. Thus, the equation of the line tangent to f() is y =. 6 California Subject Eaminations for Teachers Test Guide

18 sin () 8. Correct Response: A. (SMR Code: 5.3) Since substituting = 0 into 3 yields 0 0, L'Hôpital's Rule can be used to calculate this it. Consecutive applications of L'Hôpital's rule (until substituting = 0 does not result in an indeterminate form) gives: sin () 0 3 = cos () 0 3 = sin 0 6 = cos 0 6. Since cos 0 =, the it is Correct Response: A. (SMR Code: 5.3) This situation is modeled by the differential equation dm dt = km where M is the mass and t represents time. This is a separable differential equation and can be rewritten as dm M = k dt. Integrating each side of the equation gives Bn(M) = kt + c. Using M = 500 grams when t = 0 gives c = Bn(500). Using M = 00 grams when t = 0 gives Bn (00) = 0k + Bn (500) k = 0 Bn Correct Response: C. (SMR Code: 5.4) Calculating the area under a curve from = 0 to = can be done either by finding Since n n i = f() d or by finding its equivalent, 8 n f( i ) = 08n 8n 7 n 6n = 08 n 7 n n 6 0 n n = 08 6 = 8, f() d = 8. 0 i = n f( i ).. Correct Response: D. (SMR Code: 5.4) To integrate this function, use integration by parts, letting u = and dv = ( + ) 4 d. Then du = d, v = 5 ( + )5, and ( + ) 4 d = uv v du = 5 ( + )5 0 5 ( + )5 d = ( + )6 = Correct Response: D. (SMR Code: 5.4) Finding the area under the curve of h() involves computing the integral d = d. Using substitution, where u = 3 + and du = 3 d, u du = 4 3 Bn u 8 = 4 3 Bn Correct Response: C. (SMR Code: 5.5) The series ( ) + ( ) ( ) 3 + is a geometric series of the form a + ar + ar + + ar k +, where a = and r = ( ). A geometric series of this form converges if r <. Therefore, this series converges if ( ) <, i.e., <, which means < <. Thus, < < 3. California Subject Eaminations for Teachers Test Guide 7

19 History of Mathematics 4. Correct Response: A. (SMR Code: 6.) Let A and A represent the area of fields one and two, respectively. Then A + A = 800 and 3 A + A = 00. This is a system of linear equations. 5. Correct Response: C. (SMR Code: 6.) Omar Khayyam showed that third-degree polynomial equations can be solved by a suitable transformation that reduces the problem of finding roots to that of finding the points of intersection of conic sections. This method was further generalized by Descartes in the seventeenth century when he published his work on analytical geometry. 8 California Subject Eaminations for Teachers Test Guide

20 Eamples of Responses to Sample Constructed-Response Questions for CSET: Mathematics Subtest III Calculus Question #6 (Score Point 4 Response) y y = cos() + y = sin 3 3 To find the points of intersection: solve the simultaneous set of equations y = sin () y = cos() + () To solve the set of equations: substitute equation () into () => sin = cos() + (3) From the identity sin + cos =, sin = - cos (4) Substituting equation (4) into (3) gives cos = cos() + cos = cos 0 = cos + cos continued on net page California Subject Eaminations for Teachers Test Guide 9

21 Question #6 (Score Point 4 Response) continued Factor: 0 = cos ( + cos ) => cos = 0 or + cos = 0, for -π < < π cos = => = 3π, π, π 3π, or = π, π Substitute -values into equation () to get y-values: y = cos 3π + = y = cos π + = y = cos π + = y = cos 3π + = y = cos( π) + = 0 y = cos(π) + = 0 Coordinates of points of intersection are 3π,, π,, π 3π,,,, ( π, 0), and (π, 0). 0 California Subject Eaminations for Teachers Test Guide

22 Question #6 (Score Point 3 Response) y y = cos + y = sin 3 3 sin = cos + - cos = cos + 0 = cos + cos 0 = cos ( + cos ) => cos = 0 or + cos = 0 cos = => = 3π, π, π 3π, or = π, π Points: 3π,, π,, π 3π,,,, ( π, 0), and (π, 0). California Subject Eaminations for Teachers Test Guide

23 Question #6 (Score Point Response) y 3 3 Solve the set of equations y = sin y = cos + sin = cos + From the identity sin + cos =, sin = cos => cos = cos + cos = cos cos = = π, π y = cos( π) + = 0 California Subject Eaminations for Teachers Test Guide

24 Question #6 (Score Point Response) y 3 3 sin = cos + cos + = cos + cos = = π y = cos π + = California Subject Eaminations for Teachers Test Guide 3

25 Question #7 (Score Point 4 Response) Find zeros: set f() = = 0 => = 0 factor: 3 ( + 4) = 0 => = 0, 4 So, the zeros are at points (0, 0) and ( 4, 0). Find possible etrema: set f () = = 0 factor: ( + 3) = 0 => = 0, 3 (critical values) Find relative etrema: do first derivative test f () = ( + 3) + + f() decreasing increasing increasing Since f() is decreasing on (, 3) and increasing on ( 3, 0), f() has a relative minimum at the critical value = 3. At = 3, f() = = 63 4 => relative minimum is at point 3, Find possible inflection points: set f () = = 0 factor: 3( + ) = 0 => = 0, continued on net page 4 California Subject Eaminations for Teachers Test Guide

26 Question #7 (Score Point 4 Response) continued Find inflection points: do second derivative test f () = 3( + ) + + f() concave up concave down concave up Since f() changes concavity at = and 0, f() has inflection points at = and 0. At =, f() = 4 8 = 4 At = 0, f() = 0 => the inflection points are at (0, 0) and (, 4). y California Subject Eaminations for Teachers Test Guide 5

27 Question #7 (Score Point 3 Response) Find zeros: set f() = = 0 => = 0 factor: 3 ( + 4) = 0 => = 0, 4 Zeros are at points (0, 0) and ( 4, 0). Find possible etrema: set f () = = 0 factor: ( + 3) = 0 => = 0, 3 (critical values) Find relative etrema: do second derivative test f () = f (0) = 0 => test fails f ( 3) = 7 8 > 0 => relative minimum At = 3, f() = = 63 4 => relative minimum is at point ( 3, ). Find inflection points: set f () = = 0 factor: 3( + ) = 0 => = 0, At = 0, f() = 0 At =, f() = 4 8 = 4 => the inflection points are at (0, 0) and (, 4). y California Subject Eaminations for Teachers Test Guide

28 Question #7 (Score Point Response) f() = => = 0 f () = = 0 => ( + 3) = 0 => = 0, => relative minimum at = 3. y California Subject Eaminations for Teachers Test Guide 7

29 Question #7 (Score Point Response) f() = = 0 => = 0 => + 4 = 0 => = 4 f () = = 0 => ( + 3) = 0 => = 0, 3 f () < 0 for < 3 and f () > 0 for > 3 f () = = 0 y 4. 8 California Subject Eaminations for Teachers Test Guide

30 Question #8 (Score Point 4 Response) The derivative of F at, F (), is defined by: F () = F( + h) F(). h Using F() = f(t) dt, F () = a + h a f ( t) dt - f ( t) dt a. h Since f ( t ) dt = ( ) a a f t dt, F () = a + h f ( t) dt + f ( t) dt a. h Notice that a + h f ( t) dt + f ( t) dt = a + h f ( t) dt, so F () = + h f ( t) dt. h The Mean Value Theorem for integrals says that for some [, th], + h f ( t) dt = f( ) h. So F () = h (f( ) h) = f( ). Since + h, approaches as h approaches 0. Thus, f( ) approaches f() as h approaches 0, by the continuity of f. Therefore, F () = f( ) = f(). California Subject Eaminations for Teachers Test Guide 9

31 Question #8 (Score Point 3 Response) The definition of a derivative: F () = F ( + h) F ( ) h. So, F () = = = + h a + h a + h f ( t) dt - f ( t) dt a. h f ( t) dt + f ( t) dt h f ( t) dt h a By the Mean Value Theorem, +h f(t) dt = h i f( ) for some between and ( + h). So F ()= h i f( ) h F () = f( ) Since f( ) = f(), F () = f() 30 California Subject Eaminations for Teachers Test Guide

32 Question #8 (Score Point Response) The definition of a derivative says that: F () = F ( + h) F ( ) h. Then F () = = + h a + h f ( t) dt - f ( t) dt a. h f ( t) dt h = h + h f ( t) dt If h + h f ( t) dt = f(), then F () = f() = f(). Question #8 (Score Point Response) By the definition of the derivative, F () = F ( + h) - F ( ) h, then F () = h f ( t) dt - f ( t) dt a + h. a Clearly h f ( t) dt = f(). Therefore F () = f(). h 0 California Subject Eaminations for Teachers Test Guide 3

33 History of Mathematics Question #9 (Score Point 4 Response) A heagon inscribed in a circle will have a perimeter slightly smaller than the circumference of the circle, while the heagon circumscribed about the circle will have a perimeter slightly larger than the circumference of the circle. So we need to find the perimeter of b a 30 each heagon. A regular heagon is made up of 6 equilateral triangles, so q = 30. The value of a (in the diagram) can be found using the equation sin 30 = a, so a = sin 30. So the perimeter of the inscribed heagon is sin 30. To find the value of b in the diagram, use tan 30 = b, so b = tan 30. This gives the perimeter of the circumscribed heagon to be tan 30. Then, since the circumference of the circle is π(), sin 30 < π < tan 30, so 6 sin 30 < π < 6 tan 30 6 < π < < π < 3. If a dodecagon ( sides) were used instead of a heagon, the number of sides would double and each angle would be divided by two, so the inequality obtained would be sin < π < tan sin 5 < π < tan 5. Half-angle formulas can be used to obtain the sine and tangent of 5 in terms of sines, cosines, and tangents of 30, which are known eactly. This process can be repeated for 4-sided polygons (using 7.5 ), 48-sided polygons, and 96-sided polygons. Each iteration of this process would result in a better approimation of π. 3 California Subject Eaminations for Teachers Test Guide

34 Question #9 (Score Point 3 Response) circumscribed heagon inscribed heagon Since a regular heagon consists of si equilateral triangles, the circumscribed heagon has a perimeter of = 4 3. The inscribed heagon has a perimeter of 6() = 6. The circle has a circumference of π(), so 6 < π < < π < 3 A twelve-sided polygon could be used in a similar manner, only the circumscribed polygon would have a perimeter of 6( tan 5 ) = tan 5, and the inscribed polygon would have a perimeter of 6( sin 5 ) = sin 5. Half-angle formulas could be used to find sin 5 and tan 5 using sin 30, cos 30, and tan 30. This process could be repeated using 4 sides, 48 sides, and finally 96 sides, each time using half the previous angle (and the half-angle formulas). California Subject Eaminations for Teachers Test Guide 33

35 Question #9 (Score Point Response) A regular heagon consists of 6 equilateral triangles, so finding its perimeter involves using triangles: inscribed circumscribed y = sin 30 y = tan 30 = sin 30 y = tan 30 Since the heagons each contain 6 triangles, the perimeters of the inscribed and circumscribed heagons are (respectively) and y, or sin 30 and tan 30. The circumference of the circle is π, so sin 30 < π < tan 30 6 sin 30 < π < 6 tan 30 6 < π < < π < 6 3 Using a 96-sided polygon instead of a 6-sided polygon would change the angle and the number of sides, but the process would be the same. 34 California Subject Eaminations for Teachers Test Guide

36 Question #9 (Score Point Response) The inscribed heagon consists of si equilateral triangles with sides of length, so its perimeter is 6. So, the circumference of the circle is more than 6 and less than the perimeter of the circumscribed heagon,. 6 < π < 3 < π < So the perimter of the circumscribed heagon,, is more than 6. A twelve-sided polygon would have twice as many sides, each half as long as the heagon's sides, so the perimeter would be the same as the heagon's perimeter. We could do the same with a 4-sided polygon, 48-sided polygon, or 96-sided polygon. California Subject Eaminations for Teachers Test Guide 35

37 Scoring Information for CSET: Mathematics Subtest III Responses to the multiple-choice questions are scored electronically. Scores are based on the number of questions answered correctly. There is no penalty for guessing. There are four constructed-response questions in Subtest III of CSET: Mathematics. Each of these constructedresponse questions is designed so that a response can be completed within a short amount of time approimately 0 5 minutes. Responses to constructed-response questions are scored by qualified California educators using focused holistic scoring. Scorers will judge the overall effectiveness of your responses while focusing on the performance characteristics that have been identified as important for this subtest (see below). Each response will be assigned a score based on an approved scoring scale (see page 37). Your performance on the subtest will be evaluated against a standard determined by the California Commission on Teacher Credentialing based on professional judgments and recommendations of California educators. Performance Characteristics for CSET: Mathematics Subtest III The following performance characteristics will guide the scoring of responses to the constructed-response questions on CSET: Mathematics Subtest III. PURPOSE SUBJECT MATTER KNOWLEDGE SUPPORT DEPTH AND BREADTH OF UNDERSTANDING The etent to which the response addresses the constructed-response assignment's charge in relation to relevant CSET subject matter requirements. The application of accurate subject matter knowledge as described in the relevant CSET subject matter requirements. The appropriateness and quality of the supporting evidence in relation to relevant CSET subject matter requirements. The degree to which the response demonstrates understanding of the relevant CSET subject matter requirements. 36 California Subject Eaminations for Teachers Test Guide

38 Scoring Scale for CSET: Mathematics Subtest III Scores will be assigned to each response to the constructed-response questions on CSET: Mathematics Subtest III according to the following scoring scale. SCORE POINT 4 3 U B SCORE POINT DESCRIPTION The "4" response reflects a thorough command of the relevant knowledge and skills as defined in the subject matter requirements for CSET: Mathematics. The purpose of the assignment is fully achieved. There is a substantial and accurate application of relevant subject matter knowledge. The supporting evidence is sound; there are high-quality, relevant eamples. The response reflects a comprehensive understanding of the assignment. The "3" response reflects a general command of the relevant knowledge and skills as defined in the subject matter requirements for CSET: Mathematics. The purpose of the assignment is largely achieved. There is a largely accurate application of relevant subject matter knowledge. The supporting evidence is adequate; there are some acceptable, relevant eamples. The response reflects an adequate understanding of the assignment. The "" response reflects a ited command of the relevant knowledge and skills as defined in the subject matter requirements for CSET: Mathematics. The purpose of the assignment is partially achieved. There is ited accurate application of relevant subject matter knowledge. The supporting evidence is ited; there are few relevant eamples. The response reflects a ited understanding of the assignment. The "" response reflects little or no command of the relevant knowledge and skills as defined in the subject matter requirements for CSET: Mathematics. The purpose of the assignment is not achieved. There is little or no accurate application of relevant subject matter knowledge. The supporting evidence is weak; there are no or few relevant eamples. The response reflects little or no understanding of the assignment. The "U" (Unscorable) is assigned to a response that is unrelated to the assignment, illegible, primarily in a language other than English, or does not contain a sufficient amount of original work to score. The "B" (Blank) is assigned to a response that is blank. California Subject Eaminations for Teachers Test Guide 37