California Subject Examinations for Teachers

California Subject aminations for Teachers TST GUID MTHMTICS SUBTST III Sample Questions and Responses and Scoring Information Copyright 04 Pearson ducation, Inc. or its affiliate(s). ll rights reserved. valuation Systems, Pearson, P.O. Bo 6, mherst, M 0004 California Subject aminations for Teachers, CST, and the CST logo are trademarks of the Commission on Teacher Credentialing and Pearson ducation, Inc. or its affiliate(s). Pearson and its logo are trademarks, in the U.S. and/or other countries, of Pearson ducation, Inc. or its affiliate(s). CS-TG-QRX-0

6 6 = where 0BSample Test Questions for CST: 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 CST: Mathematics. Please note that, as on the actual test form, approimately one third of the multiple-choice questions in this test guide are more comple questions that require minutes each to complete. 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 CST:. Note: In CST: Mathematics subtests, log represents the base-0 logarithm of.. What are the solutions to tan + π 6 π. 6 5π B. π C. π D. 6 5π, π, 4π, 7π, 0 π? California Subject aminations for Teachers Test Guide

. 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. t 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?. (t) = sin(0πt) πt 5 B. (t) = sin C. (t) = cos(0πt) πt 5 D. (t) = cos California Subject aminations for Teachers Test Guide

?. 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?. 4r tan θ B. r cos θ sin θ C. 4r D. 4r tan θ cos θ sin θ 4. Which of the following epressions is equal to sin arccos. 9 ± 9 B. C. D. 9 + 9 + California Subject aminations for Teachers Test Guide

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

'(c) '(c 6. Use the graph below to answer the question that follows. Let f() be the function whose graph is shown above. Which of the following statements best describes why f() is discontinuous at = c?. f(c) does not eist B. lim f(c) f() c C. lim does not eist f() c D. f f + ) California Subject aminations for Teachers Test Guide 5

equal 7. Let f() and g() be functions such that lim f() c and lim g() c eist. If lim [f() + g()] = 6 and lim [f() g()] =, what is c c lim f()? c. 7 7 B. 5 C. 5 D. 8. For which of the following values of k does sin + sin cos lim k sin cos? π. B. π π C. D. π 6 California Subject aminations for Teachers Test Guide

for '(c) by 9. Given the function g() = 9 4, define g(4) so that g() is continuous at = 4.. g(4) = 0 B. g(4) = 6 C. g(4) = 5 8 D. g(4) = 0. 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?. 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) = N. C. f() is differentiable on the interval (a, b). D. There eists a number c in (a, b) such that f = 0.. n economist needs to approimate the function f() = a line tangent to f() at =. Which of the following lines should be used?. y = B. y = C. y = D. y = 4 California Subject aminations for Teachers Test Guide 7

6 7 sin(). Determine lim. 0. B. 0 C. 6 D.. Given f() =, for what value of does the instantaneous rate of change of f() equal the average rate of change of f() on the interval 4?. = 4 B. = 7 C. = D. = 8 California Subject aminations for Teachers Test Guide

ln and and and 4. curve in the -y plane is defined by + y + y =. For what values of is the line tangent to the curve horizontal?. = and = B. = = C. = = + D. = 4 = 4 5. 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?. 0 5 B. 5 ln 0 C. 5 ln 00 D. 00 ln 5 California Subject aminations for Teachers Test Guide 9

6. Use the graph below to answer the question that follows. The height and width of each of the above n rectangles is given by f( i ) and respectively, where i n. n, If the sum of the areas of the n rectangles is given by n i = n f( i ) = 08n 8n 7, then by the use of Riemann 6n sums f() d is equal to which of the following? 0. 4.5 B..5 C. 8 D. 08 0 California Subject aminations for Teachers Test Guide

5 5 ln = 7. valuate ( + ) 4 d.. 6 B. 8 C. 0 4 D. 0 0 8. The acceleration of a particle traveling in one dimension d along the -ais at time t is given by dt 6t. t time t =, the particle's position is at = 0 and its velocity is 0. What is the position of the particle when t = 4?. B. 9 C. 44 D. 55 9. Which of the following represents the area under the curve of 4 the function h() = + on the interval [0, ]? 5. 6 5 B. 6 C. 4 ln 4 4 D. 4 California Subject aminations for Teachers Test Guide

and about 0. What is the volume of the solid formed by revolving the region bounded by y = y = the -ais? 8π. 4π B. 8π C. 5 6π D. 9. What is the sum of the series given by n = ( ) n e n n?. + e e B. C. e e D.. For what values of does the infinite series ( ) + ( ) ( ) + converge?. > B. < C. < < D. < or > California Subject aminations for Teachers Test Guide

. For what value(s) of does the series n = ( ) n n 4 n converge?. = B. < 7 C. < 7 D. < < 7 4. What is the Taylor series representation for f() = 0 generated at = 0?. n! ()n n = 0 B. ln 0 n () n n = 0 C. (ln 0) n n! () n n = 0 (log0 n) D. n! () n n = California Subject aminations for Teachers Test Guide

BConstructed-Response ssignments Read each assignment carefully before you begin your response. Think about how you will organize your response. n erasable notebooklet will be provided at the test center for you to make notes, write an outline, or otherwise prepare your response. For the eamination, your final responses to each constructed-response assignment must be either: ) typed into the on-screen response bo, ) written on a response sheet and scanned using the scanner provided at your workstation, or ) provided using both the on-screen response bo (for typed tet) and a response sheet (for calculations or drawings) that you will scan using the scanner provided at your workstation. 4 California Subject aminations for Teachers Test Guide

5. 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() + California Subject aminations for Teachers Test Guide 5

6. 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. 6 California Subject aminations for Teachers Test Guide

g() = = f() and g() f() =. Hence, f(). and = is is 6 = or = 6 = f() 6 or f() = 6 f() or f() for Bnnotated Responses to Sample Multiple-Choice Questions for CST: 5BCalculus. Correct Response: D. (SMR Code: 5.) tan + π 6, + π 6 arctan + π 6 π π 4π π π π 4π π 7π 0 π, since tan tan either + + =.. Correct Response: C. (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 0π radians per second, i.e., θ = 0πt (t) = cos (0πt). The initial -coordinate is, so when t = 0, (t) should equal. Hence the equation is (t) = cos(0πt).. 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. 4π, 4. Correct Response:. (SMR Code: 5.) The arccos the inverse cosine of an angle θ in a right triangle, so arccos θ. Therefore, the cos θ = the side adjacent to the angle has length, while the hypotenuse of the triangle has length. By the Pythagorean theorem, the length of the third side of the triangle is 9. Thus, the sine of θ = sin arccos the ratio of the length of the opposite side over the hypotenuse or sin θ = 9. 5. Correct Response: D. (SMR Code: 5.) By DeMoivre's Theorem, z = [cos ( 75 ) + i sin ( 75 )] = cos 5 + i sin 5. Since the argument of z is 5, response choice D is the correct response. 6. Correct Response: B. (SMR Code: 5.) Note from the graph of y = f() that California Subject aminations for Teachers Test Guide 7 lim c + = and lim c f() = ; thus lim =. lso note from the graph of y = f() that f(c) =. Since lim f(c), c c f() is discontinuous at = c. 7. Correct Response: B. (SMR Code: 5.) From the properties of limits, lim[f() + g()] = 6 lim f() + lim g() = 6 and lim = 6 lim Likewise, c c [f() g()] = lim c c c lim f() lim g() c c = and lim = lim f(). Since lim g() = lim g(), then lim = 6 lim c c c c c c 7 lim f() + lim f() = 6 + lim = 7 lim f() = c c c c. c

= or =. = = = '() = yields = 8. Correct Response: B. (SMR Code: 5.) Begin by simplifying the epression: sin + sin cos lim k sin cos = sin ( + cos ) lim k sin cos + cos lim k cos = + cos k cos k. To find the value of k that + cos makes this limit equal to, solve: k cos + cos k = cos k cos k + cos k + = 0 k (cos k + )(cos k + ) = 0 cos k =. The angle with a cosine of is π radians, so k = π. 9. Correct Response: C. (SMR Code: 5.) The function g() is continuous at = 4 if and only if lim g() = g(4). Since g(4) is undefined, define g(4) = ( + 4)( 4) lim 4 4 lim 9 4 ( + )( 4) ( + 4) (4 + 4) 6 lim 4 ( + ) 5 (4 + ) 0. Correct Response: B. (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).. Correct Response: B. (SMR Code: 5.) If f() =, then f =, so the slope of the line tangent to f() at is ( ). 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 =.. Correct Response:. (SMR Code: 5.) Since substituting = 0 into sin () 0 0, L'Hôpital's rule can be used to calculate this limit. Consecutive applications of L'Hôpital's rule (until substituting = 0 does not result in an indeterminate form) gives: lim 0 sin () = lim 0 cos () = sin cos lim lim 0 6 0 6. Since cos 0 =, the limit is 6. 8 California Subject aminations for Teachers Test Guide

( M = = = 5 = + 5 = = = = = 5 6 = = =.. = f(). Correct Response: D. (SMR Code: 5.) The average rate of change is given by. Over the f(4) f() interval 4, the average rate of change = 4 change is given by the first derivative of f() =, or f 4 4 6. The instantaneous rate of '() =. The average rate of change equals the instantaneous rate of change over the given interval when = or = ± 7. Hence, response D is the correct response. dy 4. Correct Response: B. (SMR Code: 5.) The slope of a tangent line is given by d. Horizontal dy d lines have a slope of 0; therefore, 0. By implicit differentiation, d d ( + y + y d ) = d () d d d ) + d (y) + d d (y d dy dy dy )= d () + y + y 0. Therefore, d d d y + y Setting this equal to dy y zero results in the following: 0 = d + y y = 0 y =. Substituting this value for y into + y + y = gives + ( ) + ( ) = or = ±. dm 5. Correct Response:. (SMR Code: 5.) This situation is modeled by the differential equation km dt where M is the mass and t represents time. This is a separable differential equation and can be rewritten dm as k dt. Integrating each side of the equation gives ln(m) = kt + c. Using M = 500 grams when t = 0 gives c = ln(500). Using M = 00 grams when t = 0 gives ln(00) = 0k + ln(500) k = ln 0 5 6. Correct Response: C. (SMR Code: 5.4) Calculating the area under a curve from = 0 to = can be done either by finding f() d or 0 Since lim n n i = n f( i ) 08n lim n 8n 7 6n n n by finding its equivalent, lim 8 08 lim n 6 n 7 n 08 8, f() d = 8. 0 i = n f( i ). 7. 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, and ( + ) 4 d = 0 uv v du = ( + ) 5 0 0 California Subject aminations for Teachers Test Guide 9 5 d ( + )5 0 ( + ) 6 = 0 4 0.

π and = = d = and. 0 n 8. Correct Response: D. (SMR Code: 5.4) Since (t) is the position of the particle with respect to time, v(t) = d dt dv a(t) = dt. Hence, v(t) = ( 6t ) dt = t t + C. To the value of C, note that v() = 0 so that 0 = () () + C C =. (t) = ( t ) t dt = t t t + C. To find lso, find the value of C, use the fact that () = 0 so that 0 = () () + C C =. The position when t = 4 is found by substituting 4 into (t) = t t t + or (4) = (4) (4) 4 + = 55. 9. Correct Response: D. (SMR Code: 5.4) Finding the area under the curve of h() involves 4 computing the integral + d. Using substitution, where u = + and du = d, 0 4 4 + d 8 4 8 4 u du ln u = ln 4. 0 0. Correct Response:. (SMR Code: 5.4) By the washer (disk) method, the volume is given by V = π a b ( f() g() ) d, where f() is the radius of the outer circle and g() is the radius of the inner circle of a washer of thickness d. Note that the radius of the outer circle is y = and the radius of the inner circle is given by y = To find the values of a and b it is necessary to find the points where 0 4 = ( ) or 4 = 0 = 0 and = 4. Hence, the volume of this region is given by π [ ] 4 = 8π.. Correct Response:. (SMR Code: 5.5) Writing the first few terms of the series results in n = a = ( ) n e n n = e 0 e + e e 4 +.... This is a geometric series of the form e r =. The sum of an infinite geometric series is given by found by e +. e a r. n = 0 e, where Thus, the sum of the series is. Correct Response: C. (SMR Code: 5.5) The series ( ) + ( ) ( ) +... is a geometric series of the form a + ar + ar +... + ar k +..., where a = and r = ( ). geometric series of this form converges if r <. Therefore, this series converges if ( ) <, i.e., <, which means < <. Thus, < <. 0 California Subject aminations for Teachers Test Guide

'() ''() '''() ( and ( f '(0) ''(0) '''(0),. Correct Response: C. (SMR Code: 5.5) By the ratio test, if lim n a n+ a n <, the series converges. ( ) If the limit equals one or does not eist, then the test is inconclusive. For the series n n 4 n ( ) a n = n n 4 n lim n ( ) ( ) a n+ = n+ (n + ) 4. Therefore, n+ n 4 n (n + ) 4 n+ = lim n n 4n + 4 lim n = a = n a n+ lim n ( ) n+ (n + ) 4 n+ n = n 4 n ( ) n = 4. By the ratio test, the series converges when 4 < or for < < 7. Note that the limit equals for = and = 7, which is inconclusive. It is therefore necessary to check these values in the given series for convergence. When = 7, ( ) = 4 and the resulting harmonic series diverges. When =, the alternating series + + 4... results. By the alternating series test, this series converges since its terms are alternating and their absolute values decrease to zero. Thus, the interval of for which the given series converges is < 7. 4. Correct Response: C. (SMR Code: 5.5) The Taylor series for f( a) has the form n = 0 f (n) (a) n! a) n, where f (n) is the nth derivative of f() and a is the base point around which the series is generated. In this case, f() = 0 and a = 0. Taking derivatives and evaluating them for = 0 results in the values below. f() = 0 f(0) = f = 0 ln 0 = ln 0 f = 0 ln 0 f = ln 0 f = 0 ln 0 f = ln 0 Therefore, n = 0 f (n) (a) ln 0 n! a) n = + ln 0 + ln 0 + ln 0 +... + n n +....!! n! California Subject aminations for Teachers Test Guide

Bamples of Strong Responses to Sample Constructed-Response Questions for CST: 6BCalculus Question #5 (Score Point 4 Response) y y = cos() + y = sin 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() + () From the identity sin + cos =, sin = - cos (4) Substituting equation (4) into () gives cos = cos() + cos = cos 0 = cos + cos continued on net page California Subject aminations for Teachers Test Guide

+ + + + ( π, Question #5 (Score Point 4 Response) continued Factor: 0 = cos ( + cos ) => cos = 0 or + cos = 0, for -π < < π cos = => = π, π, π π, or = π, π Substitute -values into equation () to get y-values: y = cos π = y = cos π = y = cos π = y = cos π = y = cos( π) + = 0 y = cos(π) + = 0 Coordinates of points of intersection are π,, π,, π, π,,, 0), and (π, 0). California Subject aminations for Teachers Test Guide

h (f( Question #6 (Score Point 4 Response) The derivative of F at, F (), is defined by: F () = lim h 0 F( + h) F(). h Using F() = f(t) dt, F () = lim a h 0 + h a f ( t) dt - f ( t) dt a. h Since f ( t ) dt = ( ) a a f t dt, F () = lim h 0 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 () = lim h 0 + h f ( t) dt. h The mean value theorem for integrals says that for some [, th], + h f ( t) dt = f( ) h. So F () = lim h 0 ) h) = lim h 0 f( ). Since + h, approaches as h approaches 0. Thus, f( ) approaches f() as h approaches 0, by the continuity of f. Therefore, F () = lim h 0 f( ) = f(). 4 California Subject aminations for Teachers Test Guide

4BScoring Information for CST: 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 two constructed-response questions in Subtest III of CST: Mathematics. ach 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). ach response will be assigned a score based on an approved scoring scale (see page 6). Your performance on the subtest will be evaluated against a standard determined by the Commission on Teacher Credentialing based on professional judgments and recommendations of California educators. 7BPerformance Characteristics for CST: The following performance characteristics will guide the scoring of responses to the constructed-response questions on CST:. PURPOS SUBJCT MTTR KNOWLDG SUPPORT DPTH ND BRDTH OF UNDRSTNDING The etent to which the response addresses the constructed-response assignment's charge in relation to relevant CST subject matter requirements. The application of accurate subject matter knowledge as described in the relevant CST subject matter requirements. The appropriateness and quality of the supporting evidence in relation to relevant CST subject matter requirements. The degree to which the response demonstrates understanding of the relevant CST subject matter requirements. California Subject aminations for Teachers Test Guide 5

8BScoring Scale for CST: Scores will be assigned to each response to the constructed-response questions on CST: Mathematics Subtest III according to the following scoring scale. SCOR POINT 4 U B SCOR POINT DSCRIPTION The "4" response reflects a thorough command of the relevant knowledge and skills as defined in the subject matter requirements for CST: 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 "" response reflects a general command of the relevant knowledge and skills as defined in the subject matter requirements for CST: 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 limited command of the relevant knowledge and skills as defined in the subject matter requirements for CST: 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 eamples. The response reflects a limited 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 CST: 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 nglish, or does not contain a sufficient amount of original work to score. The "B" (Blank) is assigned to a response that is blank. 6 California Subject aminations for Teachers Test Guide