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MATHEMATICS SUBTEST III Sample Questions and Responses and Scoring Information Copyright © 2005 by National Evaluation Systems, Inc. (NES®) “California Subject Examinations 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.™ CSTGQR112X02 Sample Test Questions for CSET: Mathematics Subtest III
Below is a set of multiplechoice questions and constructedresponse 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 complex multiplechoice questions that require 2–3 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 next 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⎛x + 6⎞ = 3 where 0 ≤ x ≤ 2π? ⎝ ⎠ A. B. C. D.
11π 5π 6 ,6 5π 2 π 3,3 π 4π 3, 3 π 7π 6, 6 π 1. California Subject Examinations for Teachers Test Guide 1 Mathematics Subtest III 2. Use the diagram below to answer the question that follows.
y r q x 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. B. C. D. 4r tan θ r 2 ⏐cos θ sin θ⏐ 4r 2 tan2θ 4r 2 ⏐cos θ sin θ⏐ 2 California Subject Examinations for Teachers Test Guide Mathematics Subtest III (ENHANCED) 3. Use the diagram below to answer the question that follows.
y P (0, 0) x Point P is on the edge of a wheel of radius of 1 that rotates counterclockwise around the origin at a speed of 15 revolutions per second. At t = 0, P has coordinates (–1, 0). Which of the following functions could be used to describe the xcoordinate of P as a function of time, t, measured in seconds? A. B. C. D. x(t) = sin(30πt) –1 x(t) = –sin⎛ 15 ⎞ – 1 ⎝⎠ x(t) = –cos⎛ 15 ⎞ ⎝⎠ x(t) = –cos(30πt)
2πt 2πt California Subject Examinations for Teachers Test Guide 3 Mathematics Subtest III 4. Given that z = cos 75° + i sin 75°, which of the following best represents z 3 in vector form? A.
imaginary B. imaginary real real C. imaginary D. imaginary real real 4 California Subject Examinations for Teachers Test Guide Mathematics Subtest III (ENHANCED) 5. For which of the following values of k does 2 lim sin x + sin x cos x equal –2? x→k sin x cos x A. B. C. D. 0
π 2 π
3π 2 6. If f(x) 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. B. C. D. The inverse of f(x) is defined on the interval [f(a), f(b)]. There exists a number c in (a, b) such that f'(c) = 0. f(x) is differentiable on the interval (a, b). There exists a number c in (a, b) such that f(c) = N. 7. An economist needs to approximate the function f(x) = x by a line tangent to f(x) at x = – 1. Which of the following lines should be used? A. B. C. D. y = –x – 1 y = –x – 2 y = – 2x – 3 y = – 3x – 4 1 California Subject Examinations for Teachers Test Guide 5 Mathematics Subtest III (ENHANCED) 8. sin(x) – x Determine lim . x→0 x3 A. B. C. D. –1
6 0
1 6 ∞ (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 100 grams are left at time t = 20, what is the value of k? A. B. C. D.
1 ⎛ 1⎞ 20 Bn ⎝5⎠ 1 ⎛1⎞ 5 Bn ⎝20⎠ 1 ⎛1⎞ 5 Bn ⎝100⎠ 1 ⎛ 1⎞ 100 Bn 5⎠ 6 California Subject Examinations for Teachers Test Guide Mathematics Subtest III 10. Use the graph below to answer the question that follows.
y y = f(x) x 0 1 The height and width of each of the above n rectangles is given by f(xi) and n , respectively, where 1 ≤ i ≤ n. If the sum of the areas of the n rectangles is given by
n 1 ∑
i=1 1 108n 2 – 81n – 27 f(xi) = , then by the use of Riemann n 6n 2
1 0 sums ⌠ f(x) dx is equal to which of the following? ⌡ A. B. C. D. 4.5 13.5 18 108 California Subject Examinations for Teachers Test Guide 7 Mathematics Subtest III (ENHANCED) 11. Evaluate ⌠ x(x + 1)4 dx. ⌡
0 1 A. B. C. D. 1 5 16 5 28 10 43 10 12. Which of the following represents the area under the curve of 4 the function h(x) = 3x + 2 on the interval [0, 2]? A. B. C. D.
5 16 15 16 4 Bn 4
4 3 Bn 4 13. For what values of x does the infinite series 1 – (x – 2) + (x – 2)2 – (x – 2)3 + … converge? A. B. C. D. x>1 x<3 1<x<3 x < 1 or x > 3 8 California Subject Examinations for Teachers Test Guide Mathematics Subtest III 14. A Babylonian tablet dating from circa 300 B.C. contains the following problem:
There are two fields whose total area is 1800 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 2 of a bushel per square yard. If the total yield is 1100 bushels, what is the size of each field?
1 2 The above problem indicates the beginning of which of the following contemporary topics of mathematics? A. B. C. D. systems of linear equations calculus quadratic equations Euclidean geometry 15. 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. B. C. D. David Hilbert Blaise Pascal René Descartes Bernhard Riemann California Subject Examinations for Teachers Test Guide 9 Mathematics Subtest III 16. Complete the exercise 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 –2π ≤ x ≤ 2π. y = sin2x y = cos(x) + 1 10 California Subject Examinations for Teachers Test Guide Mathematics Subtest III 17. Use the function below to complete the exercise that follows. x4 f(x) = 4 + x 3 • • Find the coordinates of any zeros, relative extrema, and/or inflection points of this function. Sketch the graph of this function. California Subject Examinations for Teachers Test Guide 11 Mathematics Subtest III 18. Use the Fundamental Theorem of Calculus stated below to complete the exercise that follows. Let f be continuous on a closed interval [a, b] and let x be any point in [a, b]. If F is defined by F(x) = ⌠ f(t) dt, ⌡
a x then F'(x) = f(x) at each point x in the interval [a, b]. Using the formal definition of the derivative, prove the above theorem. 12 California Subject Examinations for Teachers Test Guide Mathematics Subtest III 19. Use the information below to complete the exercise that follows. Prior to the existence of trigonometric tables, Archimedes found that 71 < π < 7 by comparing the perimeters of 96sided regular polygons inscribed in and circumscribed about a circle of radius 1. Archimedes started with 6sided polygons and found numerical values of the perimeters of these inscribed and circumscribed hexagons. He then derived a recursive formula that used the numerical values found to calculate the perimeters of 12sided polygons. This formula was equivalent to the modern halfangle formulas of trigonometry. The method was then repeated with 24sided, 48sided, and 96sided polygons.
223 22 θ O • Repeat Archimedes's method using trigonometry to find upper and lower bounds for π by determining the perimeters of a hexagon inscribed in and a circumscribed about a circle of radius 1. Express the answer using exact values. Explain (without actually doing) how the trigonometric halfangle formulas can be used to extend the result to 96sided inscribed and circumscribed polygons to obtain a more accurate approximation of π. • California Subject Examinations for Teachers Test Guide 13 Sample Written Response Sheets for CSET: Mathematics Subtest III
For questions 16–19, examinees would record their written response to each question on a twopage response sheet located in their answer document. The length of their response to each question is limited to the lined space available on the response sheet. A sample of the response sheet is provided below. 14 California Subject Examinations for Teachers Test Guide Mathematics Subtest III California Subject Examinations for Teachers Test Guide 15 Annotated Responses to Sample MultipleChoice Questions for CSET: Mathematics Subtest III
Calculus
1. Correct Response: D. (SMR Code: 5.1) tan⎛x + 6⎞ = ⎝ ⎠ for 0 ≤ x ≤ 2π, since tan 3 =
π 4π π 3, ⇒ ⎛x + 6⎞ = arctan ⎝ ⎠
π π π π 3 ⇒ ⎛ x + 6 ⎞ = 3 or 3 , ⎝ ⎠
4π π 7π π π 4π 3 and tan 3 = 3. Hence, either x + 6 = 3 or x + 6 = 3 ⇒ x = 6 or 6 . 2. Correct Response: D. (SMR Code: 5.1) 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⏐= r2⏐sin θ cos θ⏐. Thus the large rectangle has an area of 4r2⏐sin θ cos θ⏐. 3. Correct Response: D. (SMR Code: 5.1) The xcoordinate of point P on the unit circle is the cosine of the angle measured from (1, 0) to point P, θ, ⇒ x = cos θ. Since the wheel turns 15 revolutions per second and each revolution is equivalent to a rotation of 2π radians, the change in the angle of rotation is 30π radians per second, i.e., θ = 30πt ⇒ x(t) = cos (30πt). The initial xcoordinate is –1, so when t = 0, x(t) should equal –1. Hence the equation is x(t) = –cos(30πt). 4. Correct Response: D. (SMR Code: 5.1) By de Moivre's theorem, z3 = 13 [cos (3 • 75°) + i sin (3 • 75°)] = cos 225° + i sin 225°. Since the argument of z is 225°, response choice D is the correct response. 5. Correct Response: C. (SMR Code: 5.2) Begin by simplifying the expression:
2 2 2 2 lim sin x + sin x cos x = lim sin x (1 + cos x) = lim 1 + cos x = 1 + cos k. To find the value of k that x→k x→k x→k sin x cos x sin x cos x cos x cos k 2k 1 + cos makes this limit equal to –2, solve: cos k = – 2 ⇒ 1 + cos2k = – 2 cos k ⇒ cos2k + 2 cos k + 1 = 0 ⇒ (cos k + 1)(cos k + 1) = 0 ⇒ cos k = – 1. The angle with a cosine of –1 is π radians, so k = π. 6. Correct Response: D. (SMR Code: 5.2) 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. 1 1 Correct Response: B. (SMR Code: 5.3) If f(x) = x , then f'(x) = – x2 , so the slope of the line tangent to –1 f(x) at –1 is (–1)2 or –1. Then the equation of the tangent line is y = – x + b. To find b, substitute the values x = – 1 and y = f(–1) = –1 into y = – x + b, which gives b = – 2. Thus, the equation of the line tangent to f(x) is y = – x – 2. 16 California Subject Examinations for Teachers Test Guide Mathematics Subtest III 8. Correct Response: A. (SMR Code: 5.3) Since substituting x = 0 into 0 sin (x) – x yields 0 , L'Hôpital's 3 x Rule can be used to calculate this limit. Consecutive applications of L'Hôpital's rule (until substituting x = 0 does not result in an indeterminate form) gives: lim sin (x3) – x = lim cos (x) – 1 = lim – sin x = lim – cos x. Since cos 0 = 1, the limit is – 1 . x→0 x→0 x→0 x0 6 x 3x2 6x 6 9. Correct Response: A. (SMR Code: 5.3) This situation is modeled by the differential equation dt = kM where M is the mass and t represents time. This is a separable differential equation and can be rewritten as 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 = 100 grams when t = 20 gives Bn (100) = 20k + Bn (500) ⇒ k = 20 Bn⎛5⎞. ⎝⎠
1 dM dM 1 10. Correct Response: C. (SMR Code: 5.4) Calculating the area under a curve from x = 0 to x = 1 can be done either by finding ⌠ f(x) dx or by finding its equivalent, ⌡0 1 n→∞ lim ∑
i=1 n 1 n f(xi). Since lim ⎜
n→∞ ⎛ ⎜ ⎝i=1 ∑ n 81 27 ⎞ 108 – n – n2 1 108n 2 – 81n – 27 lim 108 1 ⎟ lim = n→∞ = 6 = 18, ⌠ f(x) dx = 18. n f(xi)⎟ = n→∞ ⌡0 6n 2 6 ⎠ 11. Correct Response: D. (SMR Code: 5.4) To integrate this function, use integration by parts, letting u = x and dv = (x + 1)4 dx. Then du = dx, v = 5(x + 1)5, and
1 ⌠ x(x + 1)4 dx = uv – ⌠ v du = ⌡ ⌡0
12. Correct Response: D. 1 1 5 x (x + 1)5 ⏐0 – ⌠ 5 (x + 1)5 dx = 5 – 30 (x + 1)6 ⏐0 = 10 . ⌡0 5.4) Finding the area under the curve of h(x) involves 1 1 1 32 1 1 43 (SMR Code: 2 4 computing the integral ⌠ 3x + 2 dx. Using substitution, where u = 3x + 2 and du = 3 dx, ⌡0 2 8 4 4 ⌠8 1 4 4 ⌠ dx = du = Bn u⏐2 = Bn 4. 3 ⌡2 u 3 3 ⌡0 3x + 2 13. Correct Response: C. (SMR Code: 5.5) The series 1 – (x – 2) + (x – 2)2 – (x – 2)3 + … is a geometric series of the form a + ar + ar2 + … + ark – 1 + …, where a = 1 and r = – (x – 2). A geometric series of this form converges if ⏐r⏐< 1. Therefore, this series converges if ⏐–(x – 2)⏐< 1, i.e., ⏐x – 2⏐< 1, which means –1 < x – 2 < 1. Thus, 1 < x < 3. California Subject Examinations for Teachers Test Guide 17 Mathematics Subtest III History of Mathematics
14. Correct Response: A. (SMR Code: 6.1) Let A1 and A2 represent the area of fields one and two, respectively. Then A1 + A2 = 1800 and 3A1 + 2A2 = 1100. This is a system of linear equations. 15. Correct Response: C. (SMR Code: 6.1) Omar Khayyam showed that thirddegree 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.
2 1 18 California Subject Examinations for Teachers Test Guide Examples of Responses to Sample ConstructedResponse Questions for CSET: Mathematics Subtest III
Calculus
Question #16 (Score Point 4 Response) y 2 1 –2 –3 2 – – –1 –2 y = cos(x) + 1 y = sin2 x 3 2 2 x 2 2 To find the points of intersection: solve the simultaneous set of equations (1) y = sin2x y = cos(x) + 1 (2)
To solve the set of equations: substitute equation (1) into (2) => sin2x = cos(x) + 1 (3) (4) From the identity sin2x + cos2x = 1, sin2x = 1  cos2x Substituting equation (4) into (3) gives 1 – cos2x = cos(x) + 1 –cos2x = cos x 0 = cos x + cos2x continued on next page California Subject Examinations for Teachers Test Guide 19 Mathematics Subtest III Question #16 (Score Point 4 Response) continued Factor: 0 = cos x (1 + cos x) => cos x = 0 or 1 + cos x = 0 cos x = –1 => x = – 3π 2 ,– π π 3π , , 222 or , for 2π < x < 2π x = –π, π Substitute xvalues into equation (2) to get yvalues: ⎛ 3π⎞ y = cos⎜– 2 ⎟ + 1 = 1 ⎝⎠
π y = cos⎛2⎞ + 1 = 1 ⎝⎠ y = cos⎛– 2⎞ + 1 = 1 ⎝⎠ π ⎛3π⎞ y = cos⎜ 2 ⎟ + 1 = 1 ⎝⎠
y = cos(π) + 1 = 0 y = cos(–π) + 1 = 0 Coordinates of points of intersection are ⎜– and (π, 0). ⎛ 3π ⎞ ⎛ π ⎞ ⎛π ⎞ ⎛3π ⎞ , 1⎟, ⎜– , 1⎟, ⎜ , 1⎟, ⎜ , 1⎟, (–π, 0), ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝2 ⎠ ⎝ 2 ⎠ 20 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #16 (Score Point 3 Response) y 2 1 –2 –3 2 – – y = cosx + 1 y = sin2 x 3 2 2 x 2 –1 –2 2 sin2x = cos x + 1 1  cos2x = cos x + 1 0 = cos x + cos2x 0 = cos x (1 + cos x) => cos x = 0 or 1 + cos x = 0 cos x = –1 => x = – Points: ⎜– 3π π π 3π ,– , , 2 22 2 or 1⎟, ⎜ x = –π, π ⎛ 3π ⎞ ⎛ π , 1⎟, ⎜– , ⎝ 2 ⎠⎝2 ⎞ ⎛π 1⎟, ⎜ , ⎠ ⎝2 3π ⎞ ⎛ , 1⎞, (–π, 0), and (π, 0). ⎟ ⎠ ⎝2 ⎠ California Subject Examinations for Teachers Test Guide 21 Mathematics Subtest III Question #16 (Score Point 2 Response) y 2 1 –2 –3 2 – – 2 –1 2 –2 3 2 2 x Solve the set of equations y = sin2x y = cos x + 1 sin2x = cos x + 1 From the identity sin2x + cos2x = 1, sin2x = 1 – cos2x => 1 – cos2x = cos x + 1 –cos2x = cos x cos x = –1 x = –π, π y = cos(–π) + 1 = 0 22 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #16 (Score Point 1 Response) y 2 1 x –2 –3 2 – – 2 –1 –2 2 3 2 2 sin2x = cos x + 1 cos2x + 1 = cos x + 1 cos x = 1 x= 2 π ⎛π⎞ y = cos⎜2⎟ + 1 = 1 ⎝⎠ California Subject Examinations for Teachers Test Guide 23 Mathematics Subtest III Question #17 (Score Point 4 Response) Find zeros: set f(x) = 4 + x3 = 0 => factor: x4 x4 + 4x3 = 0 x3 (x + 4) = 0 => x = 0, –4 So, the zeros are at points (0, 0) and (–4, 0). Find possible extrema: set f′(x) = x3 + 3x2 = 0 factor: x2 (x + 3) = 0 => x = 0, –3 (critical values) Find relative extrema: do first derivative test –∞ x2 x+3 f′(x) = x2 (x + 3) f(x) + – – decreasing –3 + + + increasing 0 + + + increasing ∞ Since f(x) is decreasing on (–∞, –3) and increasing on (–3, 0), f(x) has a relative minimum at the critical value x = –3. At x = –3, f(x) = 4 – 27 = –64
81 3 => relative minimum is at point ⎜–3, –64⎟. ⎝ ⎠ ⎛ 3⎞ Find possible inflection points: set f′′(x) = 3x2 + 6x = 0 factor: 3x(x + 2) = 0 => x = 0, –2 continued on next page 24 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #17 (Score Point 4 Response) continued Find inflection points: do second derivative test –∞ 3x x+2 f′′(x) = 3x(x + 2) f(x) – – + concave up –2 – + – concave down 0 + + + concave up ∞ Since f(x) changes concavity at x = –2 and 0, f(x) has inflection points at x = –2 and 0. At x = –2, f(x) = 4 – 8 = –4 At x = 0, f(x) = 0 => the inflection points are at (0, 0) and (–2, –4). y 6 3 –4 –3 –2 –1 1234 –3 –6 x California Subject Examinations for Teachers Test Guide 25 Mathematics Subtest III Question #17 (Score Point 3 Response) Find zeros: set f(x) = 4 + x3 = 0 => factor:
x4 x4 + 4x3 = 0 x3 (x + 4) = 0 => x = 0, –4 Zeros are at points (0, 0) and (–4, 0). Find possible extrema: set f′(x) = x3 + 3x2 = 0 factor: x2 (x + 3) = 0 => x = 0, –3 (critical values) Find relative extrema: do second derivative test f′′(x) = 3x2 + 6x f′′(0) = 0
f′′(–3) = 27 – 18 > 0 => test fails
=>
3 relative minimum At x = –3, f(x) = 4 – 27 = –64 81 => relative minimum is at point (–3, –64 ). 3 Find inflection points: set f′′(x) = 3x2 + 6x = 0 factor: At x = 0, f(x) = 0 At x = –2, f(x) = 4 – 8 = –4 => the inflection points are at (0, 0) and (–2, –4). 3x(x + 2) = 0 => x = 0, –2 y 6 3 –4 –3 –2 –1 1234 –3 –6 x 26 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #17 (Score Point 2 Response) f(x) = 4 + x3 f′(x) = x3 + 3x2 = 0 x
2 x4 => => –∞ x4 + 4x3 = 0 x2 (x + 3) = 0 –3 + – + + => x = 0, –3 0 + + ∞ x+3 => relative minimum at x = –3. y x California Subject Examinations for Teachers Test Guide 27 Mathematics Subtest III Question #17 (Score Point 1 Response) f(x) = 4 + x3 = 0 f′(x) = x3 + 3x2 = 0 x4 => => x4 + 4x3 = 0 x2 (x + 3) = 0 => => x+4=0 x = 0, –3 => x = –4 f′(x) < 0 for x < –3 and f′(x) > 0 for x > –3 f′′(x) = 3x2 + 6x = 0 y –4 x . 28 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #18 (Score Point 4 Response) The derivative of F at x, F′(x), is defined by: F′(x) = lim
x F (x + h) – F(x) h
. h →0 . Using F(x) =
x ∫a f(t) dt, F′(x) = lim
a h →0 ∫a x +h f (t ) dt h ∫a f (t ) dt x Since ∫a f (t ) dt = – ∫ f (t ) dt ,
x F′(x) = lim h →0 ∫x a f (t ) dt + h
a ∫a x +h f (t ) dt .
x +h Notice that ∫x f (t ) dt + ∫a x +h f (t ) dt = ∫x f (t ) dt , so F′(x) = lim h →0 ∫x x +h f (t ) dt h . The Mean Value Theorem for integrals says that for some x1 ∈ [x, xth], ∫x x +h f (t ) dt = f(x1) • h. So F′(x) = lim h (f(x1) • h) = lim f(x1). h →0 h →0 1 Since x ≤ x1 ≤ x + h, x1 approaches x as h approaches 0. Thus, f(x1) approaches f(x) as h approaches 0, by the continuity of f. Therefore, F′(x) = lim f(x1) = f(x).
h →0 California Subject Examinations for Teachers Test Guide 29 Mathematics Subtest III Question #18 (Score Point 3 Response) The definition of a derivative: F′(x) = lim F (x + h) – F (x ) h h →0 . So, F′(x) = lim h →0 ∫a ∫a ∫x x +h f (t ) dt h ∫a f (t ) dt . ∫x
a x x +h = lim f (t ) dt + h f (t ) dt h f (t ) dt h →0 x +h = lim h →0 By the Mean Value Theorem, ∫x
So x+h f(t) dt = h i f(x 1) for some x1 between x and (x + h). F′(x)= lim h →0 h i f(x1 ) h F′(x) = lim f(x1)
h →0 Since lim f(x1) = f(x),
h →0 F′(x) = f(x) 30 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #18 (Score Point 2 Response) The definition of a derivative says that: F′(x) = lim F (x + h) – F (x ) h h →0 . Then F′(x) = lim h →0 ∫a x +h f (t ) dt h f (t ) dt h
x +h ∫a f (t ) dt . x = lim = lim If 1 h h →0 ∫x
h
1 x +h h →0 ∫x f (t ) dt ∫x x +h f (t ) dt = f(x), then F′(x) = lim f(x) = f(x).
h →0 Question #18 (Score Point 1 Response) By the definition of the derivative, F′(x) = lim F′(x) = lim
⎡x h ⎢ ∫a ⎣ F (x + h)  F (x ) h h →0 , then 1 +h h →0 f (t ) dt  ∫a f (t ) dt ⎥ . ⎦ x ⎤ Clearly lim h →0 1 ⎡h ⎤ ⎢ ∫ 0 f (t ) dt ⎥ = f(x). Therefore F′(x) = f(x). h⎣ ⎦ California Subject Examinations for Teachers Test Guide 31 Mathematics Subtest III History of Mathematics
Question #19 (Score Point 4 Response) • A hexagon inscribed in a circle will have a perimeter slightly smaller than the circumference of the circle, while the hexagon 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 1
30 each hexagon. A regular hexagon 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° = 1 , so
a = sin 30°. So the perimeter of the
a 1 inscribed hexagon is 12 sin 30°. To find the value of b in the diagram, use tan 30° = 1 , so b = tan 30°. This gives the perimeter of the circumscribed hexagon to be 12 tan 30°. Then, since the circumference of the circle is 2π(1), 12 sin 30° < 2π < 12 tan 30°, so 6 sin 30°< π < 6 tan 30° ⇒ 6 ⎜2⎟ < π < 6 3 ⇒ 3 < π < 2 ⎝⎠
•
b ⎛1⎞ 3 3. If a dodecagon (12 sides) were used instead of a hexagon, the number of sides would double and each angle would be divided by two, so the inequality obtained would be 12 sin 2 < π < 12 tan 2 ⇒ 12 sin 15° < π < 12 tan 15°. Halfangle formulas can be used to obtain the sine and tangent of 15° in terms of sines, cosines, and tangents of 30°, which are known exactly. This process can be repeated for 24sided polygons (using 7.5°), 48sided polygons, and 96sided polygons. Each iteration of this process would result in a better approximation of π.
30° 30° 32 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #19 (Score Point 3 Response) • 1 1 1 inscribed hexagon circumscribed hexagon Since a regular hexagon consists of six equilateral triangles, the ⎛2 3⎞ ⎜ ⎟ = 4 3 . The inscribed circumscribed hexagon has a perimeter of 6⎜ 3⎟ ⎝ ⎠
hexagon has a perimeter of 6(1) = 6. The circle has a circumference of 2π(1), so 6 < 2π < 4 3 3<π<2 3
• A twelvesided polygon could be used in a similar manner, only the circumscribed polygon would have a perimeter of 6(2 tan 15°) = 12 tan 15°, and the inscribed polygon would have a perimeter of 6(2 sin 15°) = 12 sin 15°. Halfangle formulas could be used to find sin 15° and tan 15° using sin 30°, cos 30°, and tan 30°. This process could be repeated using 24 sides, 48 sides, and finally 96 sides, each time using half the previous angle (and the halfangle formulas). California Subject Examinations for Teachers Test Guide 33 Mathematics Subtest III Question #19 (Score Point 2 Response) • A regular hexagon consists of 6 equilateral triangles, so finding its perimeter involves using 30°  60°  90° triangles: inscribed x x circumscribed y 1
30 1
x 30 = sin 30° 1 y 1 = tan 30° 2x = 2 sin 30° 2y = 2 tan 30° Since the hexagons each contain 6 triangles, the perimeters of the inscribed and circumscribed hexagons are (respectively) 12x and 12y, or 12 sin 30° and 12 tan 30°. The circumference of the circle is 2π, so 12 sin 30° < 2π < 12 tan 30° 6 sin 30° < π < 6 tan 30° 6 ⎜ 2⎟ < π < 6 ⎝⎠ 3< π<6 3
• ⎛1⎞ 3 Using a 96sided polygon instead of a 6sided polygon would change the angle and the number of sides, but the process would be the same. 34 California Subject Examinations for Teachers Test Guide Mathematics Subtest III Question #19 (Score Point 1 Response) • The inscribed hexagon consists of six equilateral triangles with sides of length 1, so its perimeter is 6. So, the circumference of the circle is more than 6 and less than the perimeter of the circumscribed hexagon, x. ⇒ 6 < 2π < x ⇒3<π<2 So the perimter of the circumscribed hexagon, x, is more than 6.
x • A twelvesided polygon would have twice as many sides, each half as long as the hexagon's sides, so the perimeter would be the same as the hexagon's perimeter. We could do the same with a 24sided polygon, 48sided polygon, or 96sided polygon. California Subject Examinations for Teachers Test Guide 35 Scoring Information for CSET: Mathematics Subtest III
Responses to the multiplechoice questions are scored electronically. Scores are based on the number of questions answered correctly. There is no penalty for guessing. There are four constructedresponse 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— approximately 10–15 minutes. Responses to constructedresponse 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 constructedresponse questions on CSET: Mathematics Subtest III. PURPOSE SUBJECT MATTER KNOWLEDGE SUPPORT DEPTH AND BREADTH OF UNDERSTANDING The extent to which the response addresses the constructedresponse 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 Examinations for Teachers Test Guide Mathematics Subtest III Scoring Scale for CSET: Mathematics Subtest III
Scores will be assigned to each response to the constructedresponse questions on CSET: Mathematics Subtest III according to the following scoring scale. SCORE POINT 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 highquality, relevant examples. • 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 examples. • The response reflects an adequate understanding of the assignment. The "2" response reflects a limited 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 limited accurate application of relevant subject matter knowledge. • The supporting evidence is limited; there are few relevant examples. • The response reflects a limited understanding of the assignment. The "1" 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 examples. • 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. 4 3 2 1 U B California Subject Examinations for Teachers Test Guide 37 ...
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