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Unformatted text preview: CSET ® ® California Subject Examinations for Teachers ® TEST GUIDE MATHEMATICS SUBTEST I Sample Questions and Responses and Scoring Information Copyright © 2004 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.™ CS-TG-QR110X-01 Sample Test Questions for CSET: Mathematics Subtest I Below is a set of multiple-choice questions and constructed-response questions that are similar to the questions you will see on Subtest I of CSET: Mathematics. The term "enhanced" is used in this test guide to identify complex multiple-choice 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 I. (ENHANCED) 1. Which of the following statements refutes the claim that GLR(3), the set of 3 × 3 invertible matrices over the real numbers, is a field? A. There exist elements A and B of GLR(3) such that AB ≠ BA. There exist elements A and B of GLR(3) such that det(AB) = det(A)det(B). If A is an element of GLR(3), then there exists a matrix A–1 such that A–1A = I. If A is an element of GLR(3), then there exists a matrix A such that det(A) ≠ 1. B. C. D. 2. Which of the following sets is an ordered field? A. B. C. D. the complex numbers the rational numbers the integers the natural numbers California Subject Examinations for Teachers Test Guide 1 Mathematics Subtest I 3. Use the graph below to answer the question that follows. y 10 9 8 7 6 5 4 3 2 1 –10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 1 –1 –2 –3 –4 –5 –6 –7 –8 –9 –10 2 3 4 5 6 7 8 9 10 x Which of the following systems of inequalities represents the shaded region above? A. B. C. D. 3y – 2x ≥ 12 2 y + 3x > 18 3y – 2x ≤ 12 2 y + 3x > 18 2x – 3y > –12 3 x + 2y ≤ 18 2x – 3y < –12 3 x + 2y ≤ 18 2 California Subject Examinations for Teachers Test Guide Mathematics Subtest I (ENHANCED) 4. If f(x) = –2x 2 + 8x + 16, then which of the following is the absolute value of the difference between the zeros of f(x)? A. B. C. D. 4 4i 43 46 (ENHANCED) 5. Which of the following are the imaginary parts of the roots of iz 2 + (2 + i)z + 1? A. –1 ± 3 2 –2 ± 3 2 1± 3 2 2± 3 2 B. C. D. California Subject Examinations for Teachers Test Guide 3 Mathematics Subtest I 6. Use the graph of a polynomial function below to answer the question that follows. y 9 8 7 6 5 4 3 2 1 –9 –8 –7 – 6 –5 –4 –3 –2 –1 1 –1 –2 –3 –4 –5 –6 –7 –8 –9 2 3 4 5 6 7 8 9 y = p(x) x Which of the following statements about p(x) must be true? A. B. C. D. p(x) has at least one complex root p(x) is divisible by (x – 2) p(x) is an odd function p(x) is divisible by x 2 – 6x + 9 7. Which of the following is a point of intersection of the function y = 2 x and its inverse function? A. B. C. D. ⎛ 1 , 1⎞ ⎝4 ⎠ (1, 2) (4, 4) (8, 4 2) 4 California Subject Examinations for Teachers Test Guide Mathematics Subtest I (ENHANCED) 8. Which of the following is the solution of the equation d in terms of L? 2(n – 1) = 2 – p2 L A. L=± d+1 2 p 2 + ⎛ 2n ⎞ ⎝ ⎠ 2 d p 2 + ⎛2n + 1⎞ B. L=± ⎝ ⎝ ⎝ ⎠ C. L=± 2 d p 2 + ⎛2(n – 1)⎞ ⎠ D. L=± 2 d2 p 2 + ⎛ 2(n 2 – 1)⎞ ⎠ (ENHANCED) 9. Line B passes through the points (–7, –6) and (8, 14). What is the x-intercept of the line that is perpendicular to line B at its y-intercept? A. B. C. D. 25 6 13 3 40 9 9 2 California Subject Examinations for Teachers Test Guide 5 Mathematics Subtest I 10. If f(x) = e5x + 6 2 and g(f(x)) = x, then which of the following is equivalent to g(x)? Bn (2x – 6) 5 2x – 6 5 Bn (2x – 6) e5 Bn 2x – 6 5 A. B. C. D. 11. Given any two vectors = and = such that ⏐= ⏐ = ⏐= ⏐ = 1, a b a b which of the following statements about the inner product, = • =, must be true? ab A. B. C. D. =•==1 ab –1 ≤ = • = ≤ 1 ab 1≤=•=≤ 2 ab 1≤=•=≤2 ab 6 California Subject Examinations for Teachers Test Guide Mathematics Subtest I (ENHANCED) 12. Use the matrix below to answer the question that follows. ⎡ 1 2 3⎤ ⎢–2 –3 2⎥ ⎣ 1 2 1⎦ Which of the following matrices has the same determinant as the matrix above? A. ⎡–2 –3 2⎤ ⎢ 1 2 3⎥ ⎣ 1 2 1⎦ ⎡ 1 2 3⎤ ⎢–2 –3 2⎥ ⎣ 2 4 4⎦ ⎡ 2 4 6⎤ ⎢–2 –3 2⎥ ⎣ 1 2 1⎦ ⎡ 2 1 3⎤ ⎢–3 –2 2⎥ ⎣ 2 1 1⎦ B. C. D. 13. In order to identify all the prime numbers less than 200, a person writes the numbers from 1 to 200, and eliminates all the multiples of 2, then all the multiples of 3. To complete this task, the person will have to eliminate the multiples of which additional numbers? A. B. C. D. 5, 7, 9, 11 7, 9, 11, 13 5, 7, 11, 13 7, 11, 13, 17 California Subject Examinations for Teachers Test Guide 7 Mathematics Subtest I 14. If x, y, and z are nonnegative integers, what is the total number of factors of 2x 3y 5z? A. B. C. D. (2 + 3 + 5)(x + y + z) xyz (x + 1)(y + 1)(z + 1) x 2y 3z 5 15. Use the informal proof below to answer the question that follows. Let x be a positive integer that is not a perfect square. y Assume that x = z , and that y and z are relatively prime y 2 y2 integers. It follows that x = ⎛z⎞ = z 2. Since x is an integer, ⎝⎠ z2 is equal to 1, and x is the square of y. But this is not possible. Which of the following is shown by this informal proof? A. B. C. D. The square root of any integer that is not a perfect square is an algebraic number. The square of any rational number is also a rational number. The square root of any positive integer that is not a perfect square is irrational. The square of any real number is also a real number. 8 California Subject Examinations for Teachers Test Guide Mathematics Subtest I 16. Complete the exercise that follows. Show that the subset of complex numbers of the form a + bi with a and b rational numbers satisfies the axioms of a field under the operations of addition and multiplication of complex numbers. California Subject Examinations for Teachers Test Guide 9 Mathematics Subtest I 17. Complete the exercise that follows. A cubic function of the form f(x) = x 3 + c, where c is a real number, has one zero at x = 1 + i 3. • • Find the cubic function; and sketch the graph of the function and label any intercepts. 10 California Subject Examinations for Teachers Test Guide Mathematics Subtest I 18. Complete the exercise that follows. If vectors = = (a1, a2) and = = (b1, b2) are perpendicular, A = ⎛ a b cos θ ⎝ sin θ –sin θ⎞ , and if we cos θ⎠ identify vectors with column matrices in the usual manner, then show that the vectors A= and a A= are perpendicular for all values of θ. b California Subject Examinations for Teachers Test Guide 11 Mathematics Subtest I 19. Complete the exercise that follows. Use the principle of mathematical induction to prove the following statement. 1 1 1 n + 2 • 3 + . . . + n (n + 1 ) = n + 1 1•2 12 California Subject Examinations for Teachers Test Guide Sample Written Response Sheets for CSET: Mathematics Subtest I For questions 16–19, examinees 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 limited to the lined space available on the response sheet. A sample of the response sheet is provided below. California Subject Examinations for Teachers Test Guide 13 Mathematics Subtest I 14 California Subject Examinations for Teachers Test Guide Annotated Responses to Sample Multiple-Choice Questions for CSET: Mathematics Subtest I Algebra 1. Correct Response: A. (SMR Code: 1.1) A field is a commutative ring with identity in which every nonzero element has an inverse. The set of 3 × 3 invertible matrices is not a commutative ring because there exist matrices A and B such that AB ≠ BA. 2. Correct Response: B. (SMR Code: 1.1) Of the sets given in the response choices, only the complex numbers and the rational numbers are fields. Since the complex numbers are not ordered, response choice B is the correct response. 3. Correct Response: D. (SMR Code: 1.2) The equation of the solid line is y = – 2x + 9, or 2y + 3x = 18 in standard form. Substituting a point in the shaded region [(0, 6), for example] shows that the inequality is 2y + 3x ≤ 18. The dotted line is given by 3y – 2x = 12. Substituting (0, 6) gives the inequality 3y – 2x > 12 (the dotted line indicates that points on the boundary line are not solutions to the inequality). This inequality can be rewritten as –3y + 2x < – 12, so response choice D is correct. 4. Correct Response: C. (SMR Code: 1.2) To find the zeros of f(x), set the function equal to zero and solve for x by using the quadratic formula: –2x2 + 8x + 16 = 0 ⇒ x = x= – 8 ± 192 . Simplifying –4 192 gives x = – 8 ± 82 – 4(–2)(16) ⇒ 2(–2) 3 –8 ± 8 3 = 2 ± 2 3. The absolute value of the difference –4 between the zeros is ⏐( 2 + 2 3) – ( 2 – 2 3) ⏐, which simplifies to ⏐4 3⏐ or 4 3. 5. Correct Response: D. (SMR Code: 1.2) Finding the roots of iz2 + (2 + i)z + 1 involves setting the expression equal to 0 and solving for z. The quadratic formula gives z = –(2 + i) ± (2 + i)2 – 4(i)(1) ⇒ 2(i) –2 – i ± 3 1 i3 1⎛ 3⎞ z= ⇒ z = i – 2 ± 2 . Since the roots of the equation are – 2 + ⎜1 ± 2 ⎟i, their imaginary 2i ⎝ ⎠ 2± 3 parts are 2 . 6. Correct Response: D. (SMR Code: 1.2) The polynomial p(x) has x-intercepts at 3 and –2, so (x – 3) and (x + 2) are factors. In fact, x = 3 is a double root, since the function is tangent to the x-axis at x = 3. This means that (x – 3)2, or (x2 – 6x + 9), is a factor of p(x), so p(x) is divisible by x2 – 6x + 9. California Subject Examinations for Teachers Test Guide 15 Mathematics Subtest I 7. Correct Response: C. (SMR Code: 1.3) The inverse of y = 2 x is x = 2 y, which can be rewritten as x2 y = 4 . However, domain restrictions limit this inverse to only positive x-values, so the inverse of y = 2 x x2 x4 x2 is y = 4 , x ≥ 0. To find the intersection of these two functions, solve 2 x = 4 for x values ≥ 0. 4x = 16 ⇒ x4 – 64x = 0 ⇒ x(x3 – 64) = 0. Hence x = 0 or x = 4. Since y = 2 x, (0, 0) and (4, 4) are points where the two curves intersect. Therefore C is the correct response. 8. Correct Response: C. (SMR Code: 1.3) To solve for L, first isolate L2 – p2 by multiplying L2 – p 2 both sides of the equation by 2(n – 1) . This gives 2 d ⇒ L2 = p2 + ⎛2(n – 1)⎞ . ⎝ ⎠ 2 d d L2 – p2 = 2(n – 1) . Then L2 – p2 = ⎛2(n – 1)⎞ ⎝ ⎠ Solving for L requires taking the square root of both sides, so L=± 9. 2 d p2 + ⎛2(n – 1)⎞ . ⎝ ⎠ –6 – 14 4 Correct Response: C. (SMR Code: 1.3) The slope of line B is –7 – 8 , or 3 . Substituting one of the 10 4 4 10 given points into y = 3 x + b gives an equation of y = 3 x + 3 , so the y-intercept of B is ⎛0, 3 ⎞. Therefore, ⎝ ⎠ a line perpendicular to B at its y-intercept has an equation of y = – 4 x + 3 . To find the x-intercept of this line, substitute 0 for y and solve for x. The x-intercept is 9 . 40 3 10 10. Correct Response: A. (SMR Code: 1.3) If g(f(x)) = x, then f(x) and g(x) are inverses of each other. If f(x) = e5x + 6 e5g(x) + 6 5g(x) , then x = 2 2 . This equation can be rewritten as 2x – 6 = e , and taking the natural log of both sides yields Bn(2x – 6) = Bn e5g(x) ⇒ 5g(x)(Bn e) = Bn(2x – 6). Then, since Bn e = 1, 1 g(x) = 5 Bn(2x – 6). 11. = = Correct Response: B. (SMR Code: 1.4) The dot product of vectors a and b can be written as == = = = = a • b = ⏐ a ⏐ ⏐ b ⏐ cos θ, where θ is the angle between the two vectors. Given that a and b are unit == == vectors, a • b = cos θ. Since –1 ≤ cos θ ≤ 1, –1 ≤ a • b ≤ 1. 12. Correct Response: B. (SMR Code: 1.4) The matrix in each response choice is obtained from the given matrix by an elementary row operation. The matrix in response choice B is obtained from the given matrix by adding row 1 to row 3. This elementary row operation does not change the value of the determinant of the matrix. 16 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Number Theory 13. Correct Response: C. (SMR Code: 3.1) To identify all prime numbers less than 200, it is only necessary to eliminate all multiples of each prime number less than (or equal to) 200. This is because no composite number less than 200 will have a factor greater than 200 without having another factor less than 200. Since 200 ≈ 14.14, the person would need to check multiples of 2, 3, 5, 7, 11 and 13. 14. Correct Response: C. (SMR Code: 3.1) Each factor of 2x3y5z is the product of between 0 and x twos, between 0 and y threes, and between 0 and z fives. Since this yields (x + 1) possible products of twos, (y + 1) possible products of threes and (z + 1) possible products of fives, there are (x + 1)(y + 1)(z + 1) factors. 15. Correct Response: C. (SMR Code: 3.1) The proof begins by assuming that the square root of a positive integer that is not a perfect square is rational, by rewriting it as a fraction. This assertion leads to a contradiction, which shows that the initial assumption is not correct. California Subject Examinations for Teachers Test Guide 17 Examples of Responses to Sample Constructed-Response Questions for CSET: Mathematics Subtest I Algebra Question #16 (Score Point 4 Response) Since the complex numbers are a field, they obey the commutative and associative properties of addition and multiplication, and the distributive property of multiplication over addition. ∴ numbers of the form a + bi, with a and b rational numbers, obey these properties. Hence, we need only show that numbers of this form are closed under addition, multiplication, and inverses. For example, adding or multiplying 2 complex numbers of this form results in a complex number with real and imaginary parts that are rational. Let a + bi and c + di be two complex numbers with a, b, c and d all rational. Closed under addition: (a + bi) + (c + di) = (a + c) + (b + d)i Since the sum of two rational numbers is a rational number, a + c and b + d are both rational numbers. ∴ the sum of two complex numbers with rational real and imaginary parts (a, c & b, d) is a complex number with rational real and imaginary parts (i.e., a + c and b + d, resp.). continued on next page 18 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Question #16 (Score Point 4 Response) continued Additive identity: 0 + 0i is the additive identity, since (a + bi) + (0 + 0i) = (a + 0) + (b + 0)i = a + bi and 0 is a rational number. Additive inverse: For any a + bi, –a – bi is its additive inverse. => a + bi + (–a – bi) = (a + –a) + (b + –b)i = 0 + 0i. Since a and b are rational, –a and –b are rational. Closed under multiplication: (a + bi )(c + di) = (ac – bd) + (ad + bc)i Since the product, difference, and sum of two rational numbers are each a rational number, ac – bd and ad + bc are both rational numbers. ∴ product of two complex numbers with rational real and imaginary parts (i.e., a, c & b, d) is a complex number with rational real and imaginary parts (i.e., ac – bd & ad + bc, resp.). Multiplicative identity: 1 = 1 + 0i is the multiplicative identity, since (a + bi) (1 + 0i) = a + bi + a • 0i – b • 0 = a + bi + 0i – 0 = (a – 0) + (b + 0)i = a + bi Both 0 and 1 are rational numbers. Multiplicative inverse: The multiplicative inverse can be found as follows: (a + bi) (a – bi) = a2 + abi – abi + b2 = a2 + b2. continued on next page California Subject Examinations for Teachers Test Guide 19 Mathematics Subtest I Question #16 (Score Point 4 Response) continued Since the square of a rational number is a rational, a2 and b2 are both rationals, and so a2 + b2 is a rational number. If a and b ≠ 0, then (a + bi) (a – bi) a2 + b2 = 1. Hence the multiplicative inverse of a + bi is a2 + b2 , which can be written as a – bi a +b 2 a 2 – a + b2 i . 2 b Since the square, sum, and division of two rational numbers are each a rational number, a +b 2 a 2 and a + b2 2 b are both rational numbers. Therefore, the set of numbers of the form a + bi, with a and b rational, satisfies the axioms of a field under addition and multiplication. 20 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Question #16 (Score Point 3 Response) Let z = a + bi and w = c + di be two complex numbers with a, b, c and d rational. Closure: (a + bi) + (c + di) = (a + c) + (b + d)i Since the sum of two rationals is a rational, a + c and b + d are both rationals. (a + bi) (c + di) = (ac – bd) + (ad + bc)i Since the product, difference, and sum of two rationals are each rational, ac – bd and ad + bc are both rationals. ∴ the sum and the product of two complex numbers with rational real and imaginary parts is a complex number with rational real and imaginary parts. Identity: (a + bi) + (0 + 0i) = (a + 0) + (b + 0)i = a + bi So, 0 + 0i is the additive identity of a + bi. (a + bi) (1 + 0i) = a + bi + a • 0i – b • 0 = a + bi + 0i – 0 = (a – 0) + (b + 0)i = a + bi So, 1 + 0i = 1 is the multiplicative identity, where 0 and 1 are rational numbers. Hence the subset of numbers of the form z = a + bi, with a and b rational, satisifes the axioms of a field. California Subject Examinations for Teachers Test Guide 21 Mathematics Subtest I Question #16 (Score Point 2 Response) (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) – (c + di) = (a – c) + (b – d)i (a + bi) (c + di) = (ac – bd) + (ad + bc)i So adding, subtracting, and multiplying complex numbers of the form a + bi results in a complex number of the same form. Question #16 (Score Point 1 Response) Field Properties are: Commutative, Associative, and Distributive. Commutative: (3 + 2i) + (4 + 5i) = (4 + 5i) + (3 + 2i) (3 + 2i)(4 + 5i) = (4 + 5i)(3 + 2i) Associative: [(3 + 2i) + (4 + 5i)] + (2 + 3i) = (3 + 2i) + [(4 + 5i) + (2 + 3i)] [(3 + 2i)(4 + 5i)] (2 + 3i) = (3 + 2i) [(4 + 5i)(2 + 3i)] Distributive: [(3 + 2i) + (4 + 5i)] (2 + 3i) = (3 + 2i) (2 + 3i) + (4 + 5i)(2 + 3i) 22 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Question #17 (Score Point 4 Response) If f(x) = x3 + c, where c is a real number, and f(x) has a zero at x = 1 + i 3, then its conjugate 1 – i 3 must also be a root. Thus x – 1 + i 3 ( ) and x – (1 – i 3 ) both are factors of f(x) = x3 + c, so their product is a quadratic factor of f(x) = x3 + c. ⎡x – 1 + i 3 ⎤ ⎡x – 1 – i 3 ⎤ = ⎡x – 1 – i 3 ⎤ ⎡x – 1 + i 3 ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ = x2 - x + xi 3 – x + 1 – i 3 – xi 3 + i 3 – 3i2 = x2 - 2x + 4 is a quadratic factor of f(x) = x3 + c Let a = 3 ( ) ( ) c, so f(x) = x3 + a3. Now factor x3 + a3 as the sum of 2 cubes: f(x) = x3 + a3 = (x + a)(x2 - ax + a2) = (x + a)(x2 - 2x + 4), since x2 - 2 x + 4 is a factor of x3 + c. Thus a = 2 ⇒ f(x) = x3 + a3 = x3 + 23 = x3 + 8 x 0 -2 f(x) 8 0 continued on next page California Subject Examinations for Teachers Test Guide 23 Mathematics Subtest I Question #17 (Score Point 4 Response) continued The graph of f(x) = x3 + 8 is the graph of f(x) = x3 shifted up 8 units: y (0, 8) x Intercepts are (0, 8) and (-2, 0). (-2, 0) Question #17 (Score Point 3 Response) ⎛ ⎞3 ⎝1 + i 3⎠ + c = 0 ⎛ ⎞3 ⎝1 + i 3⎠ = –c ⎛ ⎝1 + ⎛ 2 ⎝1 + 2i 3 + i • i 3⎠ ⎝1 -c i 3⎠= + 3 ⎠⎝-c + i 3⎠= 1 ⎞2 ⎛ ⎞ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎝-2 + 2i 3⎠⎝1 + i 3⎠ = -c -2 – 2i 3 + 2i 3 + 2 • 3 • i2 = -c c=8 f(x) = x3 + 8 continued on next page 24 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Question #17 (Score Point 3 Response) continued y (0, 8) x Intercepts are (0, 8) and (-2, 0). (-2, 0) Question #17 (Score Point 2 Response) Since one root is 1 + i 3, there must be another root 1 – i 3. ⎛ ⎞⎛ ⎞ ⎝1 + i 3 ⎠⎝1 – i 3 ⎠ + c = 0 1 – i 3 + i 3 – 3i2 = -c 4 = -c c = -4 f(x) = x3 - 4 x 0 –1 2 –2 1 x3 – 4 0–4 –1 – 4 1–4 y –4 –5 4 –12 –3 8–4 –8 – 4 (0, -4) California Subject Examinations for Teachers Test Guide 25 Mathematics Subtest I Question #17 (Score Point 1 Response) x3 + c is a cubic function so it should have 3 roots. Since one root is imaginary, there must be 2 real roots. ⎡ ⎢x – ( 1 + ⎣ 3i) ⎥(x – a)(x – b) = 0 ⎤ ⎦ simplify? (–b, 0) (a, 0) (0, 1 + Ö 3 i) 26 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Question #18 (Score Point 4 Response) == = The dot product a • b = a1 b1 + a2b2 = 0 if and only if the vectors a = (a1 , a2) and = = = b = (b1 , b2) are perpendicular. Want to show that A a • B b = 0. = ⎛ cos θ -sin θ ⎞ ⎟ Aa = ⎜ ⎜ sin θ cos θ ⎟ ⎝ ⎠ = ⎛ cos θ -sin θ ⎞ ⎟ Ab = ⎜ ⎜ sin θ cos θ ⎟ ⎝ ⎠ ⎡ a 1 cos θ - a2 sin θ ⎤ ⎡ a1 ⎤ ⎥ ⎢ ⎥=⎢ ⎢ a 1 sin θ + a2 cos θ ⎥ ⎢ a2 ⎥ ⎣⎦ ⎣ ⎦ ⎡b1 ⎤ ⎡ b 1 cos θ - b2 sin θ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ b2 ⎥ ⎢ b 1 sin θ + b2 cos θ ⎥ ⎣⎦ ⎣ ⎦ = = A a • A b = (a1 cos q – a2 sin q, a1 sin q + a2 cos q) • (b1 cos q – b2 sin q, b1 sin q + b2 cos q) = a1 b1 cos2 q – a1 b2 cos q sin q – a2b1 cos q sin q + a2b2 sin2 q + a1 b1 sin2 q + a1 b2 cos q sin q + a2b1 cos q sin q + a2b2 cos2 q = a1 b1 cos2 q + a1 b1 sin2 q + a2b2 sin2 q + a2b2 cos2 q = a1 b1 (cos2 q + sin2 q) + a2b2 (cos2 q + sin2 q) = = = = Since cos2 q + sin2 q = 1, A a • A b = a1 b1 + a2b2 = 0 (since a and b are perpendicular). = = Hence the vectors A a and A b are perpendicular for all q. California Subject Examinations for Teachers Test Guide 27 Mathematics Subtest I Question #18 (Score Point 3 Response) = = == a • b = 0 since a and b are perpendicular. = ⎛ cos θ -sin θ ⎞ ⎟ Aa = ⎜ ⎜ sin θ cos θ ⎟ ⎝ ⎠ = ⎛ cos θ -sin θ ⎞ ⎟ Ab = ⎜ ⎜ sin θ cos θ ⎟ ⎝ ⎠ = = A a • Ab ⎡ a 1 cos θ - a2 sin θ ⎤ ⎡ a1 ⎤ ⎥ ⎢ ⎥=⎢ ⎢ a 1 sin θ + a2 cos θ ⎥ ⎢ a2 ⎥ ⎣⎦ ⎣ ⎦ ⎡b1 ⎤ ⎡ b 1 cos θ - b2 sin θ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ b2 ⎥ ⎢ b 1 sin θ + b2 cos θ ⎥ ⎣⎦ ⎣ ⎦ = (a1 cos q – a2 sin q, a1 sin q + a2 cos q) • (b1 cos q – b2 sin q, b1 sin q – b2 cos q) = a1 b1 cos2 q – a1 b2 cos q sin q – a2b1 cos q sin q + a2b2 sin2 q + a1 b1 sin2 q + a1 b2 cos q sin q + a2b1 cos q sin q + a2b2 cos2 q = a1 b1 cos2 q + a1 b1 sin2 q + a2b2 sin2 q + a2b2 cos2 q = a1 b1 (sin2 q + cos2 q) + a2b2 (sin2 q + cos2 q) =0 = = Therefore A a and A b are perpendicular. 28 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Question #18 (Score Point 2 Response) = = = = If A a • A b = 0, then A a and A b are perpendicular. = ⎛ cos θ -sin θ ⎞ ⎟ Aa = ⎜ ⎜ sin θ cos θ ⎟ ⎝ ⎠ = ⎛ cos θ -sin θ ⎞ ⎟ Ab = ⎜ ⎜ sin θ cos θ ⎟ ⎝ ⎠ = = A a • Ab ⎡ a 1 cos θ - a2 sin θ ⎤ ⎡ a1 ⎤ ⎥ ⎢ ⎥=⎢ ⎢ a 1 sin θ + a2 cos θ ⎥ ⎢ a2 ⎥ ⎣⎦ ⎣ ⎦ ⎡b1 ⎤ ⎡ b 1 cos θ - b2 sin θ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ b2 ⎥ ⎢ b 1 sin θ + b2 cos θ ⎥ ⎣⎦ ⎣ ⎦ = (a1 cos q – a2 sin q, a1 sin q + a2 cos q) • (b1 cos q – b2 sin q, b1 sin q – b2 cos q) = a1 b1 cos2 q + a2b2 sin2 q + a1 b1 sin2 q + a2b2 cos2 q = cos2 q (a1 b1 + a2b2) + sin2 q (a2b2 + a1 b1) = = So A a • A b = 0 if a1 b1 + a2b2 = 0 Question #18 (Score Point 1 Response) If two vectors are perpendicular, then a • b = 0 ⎡cos q –sin q⎤ ⎡ a 1 ⎤ ⎢ ⎥⎢⎥ Aa = ⎢ ⎥a ⎣ sin q cos q ⎦ ⎣ 2⎦ ⎡cos q –sin q⎤ ⎡ b1 ⎤ ⎢ ⎥⎢⎥ Bb = ⎢ ⎥b ⎣ sin q cos q ⎦ ⎣ 2⎦ Aa • Bb = 0. California Subject Examinations for Teachers Test Guide 29 Mathematics Subtest I Number Theory Question #19 (Score Point 4 Response) Initial Step: show the statement is true for n = 1. left-hand side: 1 1 1 1 1 + +… + = = 1 •2 2 •3 n(n + 1) 1(1 + 1) 2 right-hand side: n+1 n = 1 1+1 = 1 2 ∴ the statement is true for n = 1. Inductive Step: assume the statement is true for n = k, and then prove that it is true for n = k + 1. So assume 1 1•2 + 1 2•3 +…+ 1 1•2 k(k + 1) + 1 2•3 1 = k+1 k (1) 1 Now want to show that Adding 1 1•2 1 (k + 1)(k + 2) +…+ (k + 1)(k + 2) = k+2 k+1 to each side of equation (1) gives: 1 1 + 1 2•3 +…+ k(k + 1) + (k + 1)(k + 2) = = k+1 k + (k + 1)(k + 2) 1 (k + 1)(k + 2) (k + 1)(k + 2) (k + 1)(k + 2) (k + 2) continued on next page k(k + 2) + 1 k2 + 2 k + 1 = = (k + 1)(k + 1) (k + 1) = 30 California Subject Examinations for Teachers Test Guide Mathematics Subtest I Question #19 (Score Point 4 Response) continued So the statement is true for n = k + 1. Therefore, since both the initial and inductive steps have been completed, by induction, the statement is true for all natural numbers n. Question #19 (Score Point 3 Response) Assume the statement is true for n = k, then show that it is true for n = k + 1. For n = k: 1 1•2 + 1 2•3 +…+ k(k + 1) 1 = k+1 k Show true for n = k + 1, i.e. 1 1•2 1 1•2 + 1 2•3 1 2•3 +…+ (k + 1) 1 = (k + 1)(k + 2) (k + 2) 1 + 1 = + +…+ k(k + 1) (k + 1)(k + 2) k+1 = k + (k + 1)(k + 2) 1 (k + 1)(k + 2) (k + 1)(k + 2) (k + 1)(k + 2) (k + 2) (k + 1) (k + 1)(k + 1) k(k + 2) + 1 k2 + 2 k + 1 = = = Therefore, by induction, the statement is true for all n. California Subject Examinations for Teachers Test Guide 31 Mathematics Subtest I Question #19 (Score Point 2 Response) Show the statement is true for n = 1: left-hand side: 1 1 1 1 1 + + …+ = = 1•2 2•3 n(n + 1) 1(1 + 1) 2 right-hand side: ∴ true for n = 1. n+1 n = 1+1 1 = 2 1 Assume the statement is true for n = k: 1•2 1 + 2•3 1 +…+ k(k + 1) 1 = k+1 k Show true for n = k + 1: (k + 1) 1 1 1 + ++ = 1•2 2•3 (k + 1)(k + 2) (k + 2) ∴ by induction, the proof is complete. Question #19 (Score Point 1 Response) 1•2 1 + 2•3 1 1 +…+ 1 n(n + 1) 1 = n+1 n For n = 1: For n = 2: For n = k: 2 2 3 = 2 2 3 = = k k+1 k+1 k Therefore, by induction, the statement is true. 32 California Subject Examinations for Teachers Test Guide Scoring Information for CSET: Mathematics Subtest I 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 I 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 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 34). 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 I The following performance characteristics will guide the scoring of responses to the constructed-response questions on CSET: Mathematics Subtest I. PURPOSE SUBJECT MATTER KNOWLEDGE SUPPORT DEPTH AND BREADTH OF UNDERSTANDING The extent 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. California Subject Examinations for Teachers Test Guide 33 Mathematics Subtest I Scoring Scale for CSET: Mathematics Subtest I Scores will be assigned to each response to the constructed-response questions on CSET: Mathematics Subtest I 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 high-quality, 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 34 California Subject Examinations for Teachers Test Guide ...
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This note was uploaded on 03/05/2009 for the course CS 5 taught by Professor Prof during the Spring '09 term at UCSB.

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