The SAT Subject Tests Answer Eplanations TO PRACTICE QUESTIONS FROM THE SAT SUBJECT TESTS STUDENT GUIDE Mathematics Level & Visit sat.org/stpractice to get more practice and stud tips for the Subject Test in Mathematics Level and.
SAT Subject Tests in Mathematics Level and Level This document gives detailed answer eplanations to mathematics practice questions from The SAT Subject Tests Student Guide. B reviewing them, ou ll get to know the tpes of questions on the test and learn our strengths and weaknesses. Estimated difficult level is based on a 5 scale, with the easiest and 5 the most difficult. For more about the SAT Subject Tests, go to satsubjecttests.org. SAT Subject Test in Mathematics Level. Difficult: Choice (B) is correct. The cost of the equipment is $50, and each package of 0 blank CDs costs $5.90. The total cost for the band to produce 0 CDs, that is, package, is 50 + 5.90 = 55.90 dollars, and the total cost to produce 0 CDs, that is, packages, is 50 + 5.90() = 50 +.80 = 6.80 dollars. One wa to determine the correct epression is to find the slope-intercept form of the equation of the line that passes through the two points given b the ordered pairs (0, 55.90) and (0, 6.80). The line has slope Alternativel, one can appl the interior angle sum theorem to pentagon ABECD. Because line AD is a 6.80 55.90 = 0.59 and -intercept 50, which is 0 0 the cost, in dollars, for the band to produce 0 CDs after purchasing the equipment. The equation of this line is = 0.59n + 50, where n is the number of CDs used. Alternativel, note that the total cost to produce 0 CDs is 50 + 5.90 = 50 + 0.59(0) dollars, and the cost to produce 0 CDs is 50 + 5.90() = 50 + 0.59(0) dollars, so, in general, if n is a multiple of 0, then the total cost to produce n CDs is 50 + 5.90. Difficult: ( 0n ) = 50 + 0.59n E n + 8 = 0.5. Solving for n gives n + 8 = ( 0.5) = 0.5, and so n = 0.5 8 = 8.5. D One wa to determine in terms of and z is to find the measure of each of the angles of EDC in terms of,, and z and then appl the triangle sum theorem. Because CED is supplementar to BEC and the measure of BEC is given to be,,the measure of CED is (80 ). Because line BD is a transversal to the parallel lines AB and CD, the alternate interior angles BDC and SAT Subject Tests in Mathematics Level and Level Choice (C) is correct. One wa to determine the value of n is to create and solve an algebraic equation. The phrase a number n is increased b 8 is represented b the epression n + 8, and the cube root of that result is equal to 0.5, so C z A transversal to the parallel lines AB and CD, it follows that BAD and ADC are supplementar; that is, the sum of the measures of these two angles is 80. The measure of BEC, interior to polgon ABECD, is (60 ). The measure of EBA is given to be, and the measure of DCE is given to be z. Therefore, the sum of the measures of the interior angles of pentagon ABECD is 80 + (60 ) + + z. The interior angle sum theorem applied to pentagon ABECD gives the equation 80 + (60 ) + + z = 540, which can be solved for to arrive at = + z.. Difficult: dollars. Choice (A) is correct. Let E be the point of intersection of lines AC and BD. B DBA are of equal measure. Thus, the measure of BDC is, which is also the measure of EDC.. The measure of DCE is given to be z. Therefore, the sum of the angle measures of EDC, in degrees, is (80 ) + + z. The triangle sum theorem applied to EDC gives the equation (80 ) + + z = 80, which can be solved for to arrive at = + z. Alternativel, one can invert the operations that were done to n. Appl the inverse of each operation, in the reverse order: First cube 0.5 to get 0.5 and then decrease this value b 8 to find that n = 0.5 8 = 8.5. 4. Difficult: Choice (A) is correct. To determine the value of b, appl the fact that two comple numbers are equal if and onl if the real and pure imaginar parts are equal. Because (a + b ) + 5i = 9 + ai, this gives the two equations a + b = 9 and 5i = ai;, that is, a + b = 9 and 5 = a. Therefore, 5 + b = 9 and b = 4.
5. Difficult: Choice (C) is correct. One wa to determine all values of for which 4 is first to rewrite the inequalit in an equivalent form that compares a factored epression to 0 and then reason about the arithmetic sign of the product of the factors. The inequalit 4 is equivalent to 4 + 0, which in turn is equivalent to + 6 0. Because + 6 factors as ( )( + 6) = ( )( + )( ), the original inequalit is equivalent to ( )( + )( ) 0 or ( + )( ) 0. To solve ( + )( ) 0, notice that this inequalit is satisfied b a value of precisel when either =, = or the product of the factors ( + ) and ( ) is negative; this last condition is true for < <, as shown in the table below: < < < > ( + ) Negative Positive Positive ( ) Negative Negative Positive ( + ) ( ) Positive Negative Positive Therefore, all values of for which 4 are described b the etended inequalit. Alternativel, one can use a graphing calculator. Graph the two equations = 4 and =. 0 5 0 5 O 5 0 = 4 = 5 0 The values of for which 4 are the same as the values of for which the parabolic graph of = 4 lies above or intersects the line =. Therefore, all values of for which 4 are described b the etended inequalit. 6. Difficult: 4 Choice (D) is correct. To determine for which of the five age groups in the options the projected percent increase in population from 000 to 050 is greatest, estimate the ratio of the projection for the ear 050 to the population figure for 000 for each age group. Each of the ratios described is equal to plus the decimal corresponding to the projected percent increase in population from 000 to 050, so the greatest ratio will correspond to the greatest projected percent increase in population. For the 0 9 age group, the ratio is less than 60 40 =.5, because the 050 projected population is less than 60 million, and the 000 population is greater than 40 million. Thus, the projected percent increase in population from 000 to 050 for this age group is less than 50%. For the 40 49 age group, the ratio is less than 50 =. 5, 40 because the 050 projected population is less than 50 million, and the 000 population is greater than 40 million. Thus, the projected percent increase in population from 000 to 050 for this age group is less than 5%. For the 50 59 age group, the ratio is less than 50 = 6. 6, 0 because the 050 projected population is less than 50 million, and the 000 population is greater than 0 million. The projected percent increase in population from 000 to 050 for this age group is less than 66%. For the 60 69 ear age group, the ratio is approimatel 4 million to 0 million, or as a decimal,.05, corresponding to an approimate 05% increase in population for this age group. For the 70 79 ear age group, the ratio is approimatel million to 6 million, or as a decimal, approimatel.94, corresponding to an approimate 94% increase in population for this age group. Therefore, the 60 69 ear age group has the greatest projected percent increase in population from 000 to 050 among the given options. 7. Difficult: Choice (D) is correct. One wa to determine which equation must be true is to appl the laws of logarithms to the given equation. It is given that = log c a, so c = c log c a must be true. One of the laws of logarithms states that log c c a = a must be true. Therefore, c = a must be true. Alternativel, one can recall the definition of logarithms. The logarithm of a to the base c, represented b the smbol log c a, is the eponent to which the base c must be raised to obtain a. In other words, = log c a, means precisel that c = a. 8. Difficult: Choice (E) is correct. Statement I is false: The function f has the real line as its domain. The function g has as its domain all values of ecept, because the value of the epression 9 is undefined when is substituted for. This means that the graph of f contains a point with -coordinate equal to, but the graph of g contains no such point. SAT Subject Tests in Mathematics Level and Level
Statement II is true: For an not equal to, the epressions + and 9 give the same value, because for, the epression 9 ( + )( ) = = +. It follows that for, the graphs of f and g contain the same point with that -coordinate. Statement III is true: Because statement II is true, for ever number not equal to, the point (, + ) is on both the graph of f and the graph of g. Because there are infinitel man numbers not equal to, the graphs have an infinite number of points in common. 9. Difficult: 4 Choice (D) is correct. Because line is perpendicular to the line segment with endpoints (, 0) and (0, ), the slope of line must be the negative reciprocal of the slope of the line segment. The line segment has slope 0 ( ) =, so the slope of line equals =. 0 0. Difficult: Choice (B) is correct. It is given that students have sampled all three kinds of cand bars and that 5 students have sampled eactl two kinds, so a total of + 5 = 8 students have sampled more than one kind of cand bar. Because 0 students have sampled one or more of the three kinds of cand bars, the number of students that have sampled onl one kind is 0 8 =.. Difficult: Choice (B) is correct. To determine the length of side BC, appl the definition of the tangent of an angle θ in a right triangle: the tangent of θ is equal to the length of the side opposite angle θ divided b the length of the side adjacent BC BC to angle θ. Thus, in this case, tan = =, which AC 0 gives BC = 0tan. Use a calculator, set to degree mode, to compute that 0tan 40... Difficult: 4 Choice (D) is correct. One wa to determine the maimum height the ball reaches is to rewrite the quadratic epression that defines the function h b completing the square: ( ) + ( ) + 6 + 46 + 5 = 6 t t t t 5 8 = + ( ) ( ) 6 t t 8 6 6 = (( ) ( ) ) 6 6 t 6 + 5 = 6( ( t ) ( ) + 6 + 5 6 6 It is not necessar to simplif an further, as the maimum height must correspond to t =, which is the onl value 6 5 of t that makes the term 6( t 6 ) nonnegative. B substitution, h 6 5 8 065 6 6 ( ) = ( ) + =.. Therefore, to the nearest foot, the maimum height the ball reaches is 8 feet. Alternativel, one can use a graphing calculator to determine the maimum value of the function h. Because h( 0) = 5, h( ) = 6+ 46+ 5 = 5, h( ) = 64 ( ) + 46( ) + 5 =, and h( ) = 69 ( ) + 46() + 5 =, the maimum value of h must occur for some t-value between and. Set the window so that the independent variable goes from 0 to and the dependent variable goes from 0 to 50 to view the verte of the parabola. Upon tracing the graph, the maimum value of h is slightl greater than 8. Therefore, to the nearest foot, the maimum height the ball reaches is 8 feet.. Difficult: 4 Choice (A) is correct. One wa to determine the volume of the solid is to determine the length, width w, and height h of the solid, in centimeters, and then appl the formula V = wh to compute the volume. Let, w, and h represent the length, width, and height, in centimeters, respectivel, of the solid. The area of the front face of the solid is h = 4 square centimeters, the area of the side face is wh = 8 square centimeters, and the area of the bottom face is w = square centimeters. Elimination of h b using the first two equations gives h wh = 4 8, which simplifies to =, or = w. Substitution of w for w in the third equation gives (w)w =, or w =, so w = (because onl positive values of w make sense as measurements of the length of an edge of a rectangular solid). Substitution of for w in the equation wh = 8 gives h = 8, and substitution of 8 for h in the equation h = 4 gives 8 = 4, so =. Therefore, the volume V of the solid, in cubic centimeters, is V = ()()( 8) = 4. Alternativel, one can recognize that the square of the volume of a rectangular solid is the product of the areas of the front, side, and bottom faces of the solid. That is, squaring both sides of the formula V = wh gives V = whwh = ( w)( h )( wh). Therefore, in this case, V = ()( 4)( 8) = 576, so V = 576 = 4. Note that it is not necessar to solve for the values of, w, and h. 4. Difficult: 4 Choice (C) is correct. The area of the shaded region can be found b subtracting the area of rectangle ABCD from the area of the circle. To determine the area of the circle, first find the radius r and then compute the area πr. Because rectangle ABCD is inscribed in the circle, ABC is an inscribed right angle, and thus AC is a diameter of the circle. Appling the Pthagorean theorem 4 SAT Subject Tests in Mathematics Level and Level
to right triangle ABC, one finds the length of side AC is 5 + = 69 =. Thus, the radius of the circle is, and the area of the circle is π ( = 69 π ). The area 4 of rectangle ABCD is 5 = 60 and therefore, the area of the shaded region is 69 π 60 7.7. 4 5. Difficult: Choice (E) is correct. To determine how man real numbers k satisf f (k) = is to determine how man solutions 4 the equation 9 + 4 = has, which in turn is to determine how man solutions the equation 4 9 + = 0has. It is not difficult to see, using 4 the Rational Roots Theorem, that 9 + has no factor of the form k for an whole or rational number value of k, but there are real number values of k such that k is a factor. This problem must be solved using a nonalgebraic method. One wa to determine how man numbers k satisf f( k ) = is to eamine the graph of the function 4 f ( ) = 9 + 4. Use a graphing calculator to 4 graph = 9 + 4 for on a suitabl large interval to see all intersections of the graph with the line = and then count the number of points of intersection. 5 50 O 50 There are at least four such points: A first point with -coordinate between and, a second point with -coordinate between and 0, a third point with -coordinate between 0 and, and a fourth point with -coordinate between 4 and 5. The fact that 4 9 + 4 is a polnomial of degree 4 means that there can be at most four such points. Therefore, there are four values of k for which f (k) =. 5 intervals for which the values of f (k) = at the endpoints have different signs and appl the Intermediate Value Theorem to each of those intervals. Create a table of values 4 for = 9 + 4 for whole number values of between 5 and 5, inclusive, and count the number of intervals of length for which the value of f is greater than for one of the endpoints and less than for the other endpoint. f() 5 779 4 08 85 8 0 4 7 40 77 4 76 5 9 There are four such intervals: [, ], [, 0], [0, ], and [4, 5]. B the Intermediate Value Theorem, for each of these four intervals, there is a value k in that interval such that f (k) =. This shows that there are at least four such values 4 of k, and the fact that 9 + 4 is a polnomial of degree 4 means that there can be at most four such values of k. Therefore, there are four values of k for which f (k) =. 6. Difficult: 5 Choice (C) is correct. To determine the value, to the nearest hundred dollars, of the automobile when t = 4, find the equation of the least-squares regression line and then evaluate that equation at t = 4. A calculator can be used to find the least-squares regression line. The specific steps to be followed depend on the model of calculator, but the can be summarized as follows. Enter the statistics mode, edit the list of ordered pairs to include onl the four points given in the table, and perform the linear regression. The coefficients will be, approimatel, 5, for the -intercept and,49 for the slope, so the regression line is given b the equation =, 49t + 5,. When t = 4, one gets =, 49( 4) + 5, = 5, 66. Therefore, based on the least-squares linear regression, the value, to the nearest hundred dollars, of the automobile when t = 4 is $5,600. Alternativel, one can eamine a table of values of the 4 function f ( ) = 9 + 4 and then identif SAT Subject Tests in Mathematics Level and Level 5
SAT Subject Test in Mathematics Level 7. Difficult: Choice (D) is correct. The distance d between the points with coordinates (,, z ) and (,, z ) is given b the distance formula: d = ( ) + ( ) + ( z z ). Therefore, the distance between the points with coordinates (, 6, 7) and (,, 4) is:, which simplifies to 8 9...5.0.5.0.5 O,000,000,000 4,000 5,000 6,000 7,000 8,000 9,000 0,000 8. Difficult: Choice (E) is correct. One wa to determine the value that f () approaches as gets infinitel larger is to rewrite the definition of the function to use onl negative powers of and then reason about the behavior of negative powers of as gets infinitel larger. Because the question is onl concerned with what happens to + as gets infinitel larger, one can assume that is positive. For 0, the epression + ( + ) is equivalent to the epression = +. As ( ) gets infinitel larger, the epression approaches the + value 0, so as gets infinitel larger, the epression approaches the value + 0 =. Thus, as gets infinitel 0 larger, f () approaches. Alternativel, one can use a graphing calculator to estimate the height of the horizontal asmptote for the function + f( ) =. Graph the function + = on an interval with large -values, sa, from = 00 to =,000..5.0.5.0.5 O 00 00 00 400 500 600 700 800 900,000 B eamining the graph, the -values all seem ver close to.5. Graph the function again, from, sa, =,000 to = 0,000. The -values var even less from.5. In fact, to the scale of the coordinate plane shown, the graph of the function + f( ) = is nearl indistinguishable from the asmptotic line =.5. This suggests that as gets infinitel larger, f () approaches.5, that is,. Note: The algebraic method is preferable, as it provides a proof that guarantees that the value f () approaches is. Although the graphical method worked in this case, it does not provide a complete justification; for eample, the graphical method does not ensure that the graph resembles a horizontal line for ver large -values such as 0 00 0 0. 9. Difficult: 4 Choice (C) is correct. According to the model, the world s population in Januar 995 was 5.(.0) 5 and in Januar 996 was 5.(.0) 6. Therefore, according to the model, the population growth from Januar 995 to Januar 996, in billions, was 5.(.0) 6 5.(.0) 5, or equivalentl, 5 9 5.(. 0) (. 00)( 0 ) 7, 000, 000. 0. Difficult: 4 Choice (C) is correct. First, note that the angle of the parallelogram with verte (, 5) is one of the two larger angles of the parallelogram: Looking at the graph of the parallelogram in the -plane makes this apparent. Alternativel, the sides of the angle of the parallelogram with verte (, 5) are a horizontal line segment with endpoints (, 5) and (6, 5), and a line segment of positive slope with endpoints (, ) and (, 5) that intersects the horizontal line segment at its left endpoint (, 5), so the angle must measure more than 90. Because the sum of the measures of the four angles of a parallelogram equals 60, the angle with verte (, 5) must be one of the larger angles. One wa to determine the measure of the angle of the parallelogram with verte (, 5) is to appl the Law of Cosines to the triangle with vertices (, ), (, 5), and (6, 5). The length of the two sides of the angle with verte (, 5) are ( ) + ( 5 ) = 7 and ( 6) + ( 5 5) = ; the length of the side 6 SAT Subject Tests in Mathematics Level and Level
opposite the angle is ( 6 ) + ( 5 ) = 4. Let θ represent the angle with verte (, 5) and appl the Law of Cosines: A + B ABcos θ = C, so C ( A + B ) ( 7 + 9) 6 cosθ = = = = AB ( 7)( ) 6 7 7. Therefore, the measure of one of the larger angles of the parallelogram is arccos ( ) 04. 0 7. Another wa to determine the measure of the angle of the parallelogram with verte (, 5) is to consider the triangle (, ), (, 5), and (, ). The measure of the angle of this triangle with verte (, 5) is 90 less than the measure of the angle of the parallelogram with verte (, 5). The angle of the triangle has opposite side of length = and adjacent side of length 5 = 4, so the measure of this angle is arctan. Therefore, the measure 4 of the angle of the parallelogram with verte (, 5) is 90 + arctan 04. 0. 4 Yet another wa to determine the measure of the angle of the parallelogram with verte (, 5) is to use trigonometric relationships to find the measure of one of the smaller angles and then use the fact that each pair of a larger and smaller angle is a pair of supplementar angles. Consider the angle of the parallelogram with verte (, ); this angle coincides with the angle at verte (, ) of the right triangle with vertices at (, ), (, 5), and (, ), with opposite side of length 5 = 4 and adjacent side of length =, so the measure of this angle is arctan4. This angle, together with the angle of the parallelogram with verte (, 5), form a pair of interior angles on the same side of a transversal that intersects parallel lines, so the sum of the measures of the pair of angles equals 80. Therefore, the measure of the angle of the parallelogram with verte (, 5) is 80 arctan4 04. 0.. Difficult: 4 Choice (E) is correct. To determine the numerical value of the fourth term, first determine the value of t and then appl the common difference. Because t, 5t, and 6t + are the first three terms of an arithmetic sequence, it must be true that (6t + ) (5t ) = (5t ) t, that is, t + = t. Solving t + = t for t gives t =. Substituting for t in the epressions of the three first terms of the sequence, one sees that the are 4, 9, and 4, respectivel. The common difference between consecutive terms for this arithmetic sequence is 5 = 4 9 = 9 4, and therefore, the fourth term is 4 + 5 = 9.. Difficult: Choice (A) is correct. To determine the height of the clinder, first epress the diameter of the clinder in terms of the height and then epress the height in terms of the volume of the clinder. The volume of a right circular clinder is given b V = πr h, where r is the radius of the circular base of the clinder and h is the height of the clinder. Because the diameter and height are equal, h =. r Thus, r = h. Substitute the epression h for r in the volume formula π to eliminate r : V = π( h) h = h. Solving for h gives 4 h = 4 V. Because the volume of the clinder is, the π height of the clinder is h = 8 7.. π. Difficult: Choice (E) is correct. One wa to determine the value of sin( π θ) is to appl the sine of difference of two angles identit: sin( π θ) = sinπcosθ cosπsinθ. Because sinπ = 0 and cosπ =, the identit gives sin( π θ) = 0(cos θ) ( )(sin θ ) = sinθ. Therefore, sin(π θ) = 057.. Another wa to determine the value of sin( π θ) is to appl the supplementar angle trigonometric identit for the sine: sin( π θ) = sinθ. Therefore, sin( π θ) = 057.. 4. Difficult: 4 Choice (A) is correct. One wa to determine the probabilit that neither person selected will have brown ees is to count both the number of was to choose two people at random from the people who do not have brown ees and the number of was to choose two people at random from all 0 people and then compute the ratio of those two numbers. Because 60% of the 0 people have brown ees, there are 0.60(0) = 6 people with brown ees, and 0 6 = 4 people who do not have brown ees. The number of was of choosing two people, neither of whom has brown ees, is 4 ( ) = 6 : There are 4 was to choose a first person and was to choose a second person, but there are was in which that same pair of people could be chosen. Similarl, the number of was of choosing two people at random from the 0 people is 0 ( 9 ) = 45. Therefore, the probabilit that neither of the two people selected has brown ees is 6 0.. 45 Another wa to determine the probabilit that neither person selected will have brown ees is to multipl the probabilit of choosing one of the people who does not have brown ees at random from the 0 people times the probabilit of choosing one of the people who does not have brown ees at random from the 9 remaining people after one of the people who does not have brown ees has been chosen. Because 60% of the 0 people have brown ees, the probabilit of choosing of the people who does not have brown ees at random from the 0 people is 0.60 = 0.40. SAT Subject Tests in Mathematics Level and Level 7
If one of the people who does not have brown ees has been chosen, there remain people who do not have brown ees out of a total of 9 people; the probabilit of choosing one of the people who does not have brown ees at random from the 9 people is. Therefore, if two 9 people are to be selected from the group at random, the probabilit that neither person selected will have brown ees is 40( 9 ) 0. 0.. 5. Difficult: Choice (A) is correct. B the Factor Theorem, is a factor of + k + 8 onl when is a root of + k + 8, that is, ( ) + k ( ) + ( ) 8 = 0, which simplifies to 4k + 4 = 0. Therefore, k = 6. Alternativel, one can perform the division of + k + 8 b and then find a value for k so that the remainder of the division is 0. approimatel,. for the constant and.4 for the base, which indicates that the eponential equation =.(.4) is the result of performing the eponential regression. If the calculator reports a correlation, it should be a number that is ver close to, which indicates that the data ver closel matches the eponential equation. Therefore, of the given models, =.(.4) best fits the data. Alternativel, without using a calculator that has a statistics mode, one can reason about the data given in the table. The data indicates that as increases, increases; thus, options A and B cannot be candidates for such a relationship. Evaluating options C, D, and E at = 9.8 shows that option D is the one that gives a value of that is closest to 0.. In the same wa, evaluating options C, D, and E at each of the other given data points shows that option D is a better model for that one data point than either option C or option E. Therefore, =.(.4) is the best of the given models for the data. 8. Difficult: 4 Choice (D) is correct. Statement I is false: Because C =.0F + 9.6, high values of F are associated with low values of C, which indicates that there is a negative correlation between C and F. Statement II is true: When 0% of calories are from fat, F = 0 and the predicted percent of calories from carbohdrates is C =.0(0) + 9.6 7. Because the remainder is 4k + 4, the value of k must satisf 4k + 4 = 0. Therefore, k = 6. 6. Difficult: 4 Choice (E) is correct. One wa to determine the value of f (.) 5 is to solve the equation f ( ) = 5. for. Because f ( ) = +, start with the equation + =. 5, and cube both sides to get + = (.) 5 = 75.. Isolate to get =. 75 and appl the cube root to both sides of the equation to get =. 75.. Another wa to determine the value of f (.) 5 is to find a formula for f and then evaluate at.5. Let = + and solve for : cubing both sides gives = +, so, =, and =. Therefore, f ( ) = and f (.) 5 = ( 5.) = 75... 7. Difficult: 4 Choice (D) is correct. One wa to determine which of the equations best models the data in the table is to use a calculator that has a statistics mode to compute an eponential regression for the data. The specific steps to be followed depend on the model of calculator but can be summarized as follows: Enter the statistics mode, edit the list of ordered pairs to include onl the four points given in the table, and perform an eponential regression. The coefficients are, Statement III is true: Because the slope of the regression line is.0, as F increases b, C increases b.0() =.0; that is, C decreases b.0. 9. Difficult: Choice (B) is correct. One wa to determine the slope of the line is to compute two points on the line and then use the slope formula. For eample, letting t = 0 gives the point (5, 7) on the line, and letting t = gives the point (6, 8) on the line. Therefore, the slope of the line is equal to 8 7 6 5 = =. Alternativel, one can epress in terms of. Because = 5 + t, = 7 + t, and 7+ t = ( 5+ t) +, it follows that = +. Therefore, the slope of the line is. 0. Difficult: Choice (D) is correct. The range of the function defined b f( ) = + is the set of -values such that = + for some -value. One wa to determine the range of the function defined b f( ) = + is to solve the equation = + for and then determine which -values correspond to at least one -value. To solve = + for, first subtract from both sides to get = and then take the reciprocal 8 SAT Subject Tests in Mathematics Level and Level
of both sides to get =. The equation = shows that for an -value other than, there is an -value such that = +, and that there is no such -value for =. Therefore, the range of the function defined b f( ) = + is all real numbers ecept. Alternativel, one can reason about the possible values of the term. The epression can take on an value ecept 0, so the epression + can take on an value ecept. Therefore, the range of the function defined b f( ) = + is all real numbers ecept.. Difficult: 4 Choice (A) is correct. To determine how man more dalight hours the da with the greatest number of hours of dalight has than Ma, find the maimum number of dalight hours possible for an da and then subtract from that the number of dalight hours for Ma. To find the greatest number of dalight hours possible for an da, notice that the epression 5 7 π + sin ( t 65 ) is maimized when sin π ( t 65 ) =, which corresponds to π π 65 t =, so t = = 9. 5. However, for this problem, 65 4 t must be a whole number, as it represents a count of das after March. From the shape of the graph of the sine function, either t = 9 or t = 9 corresponds to the da with the greatest number of hours of dalight, and because 5 7 + 65 9 5 7 sin( π ( ) ) > + π sin( 65 ( 9) ), the epression 5 7 π + sin ( t 65 ) is maimized when t = 9 das after March. (It is not required to find the da on which the greatest number of hours of dalight occurs, but it is 0 + 0 + + 0 das after March, that is, June 0.) Because Ma is 0 + 0 + = 4 das after March, the number of hours of dalight for Ma is 5 7 + π sin( 65 ( 4) ). ( ( )) ( ( + ( ) )) Therefore, the da with the greatest number of hours of dalight has 5 7 + sin π 65 ( 9) 5 more dalight hours than Ma. 7 sin π 65 4 0.8. Difficult: Choice (C) is correct. A correct matri representation must have eactl three entries, each of which represents the total income, in dollars, for one of the three das. The total income for Da is given b the arithmetic epression 99 0 + 99 6 + 99 9, which is the 0 single entr of the matri product 99 99 99 6 ; 9 in the same wa, the total income for Da is given b 99 8 + 99 5 + 99, the single entr of 8 99 99 99 5 ; and the total income for Da is given b 99 + 99 8 + 99 0, the single entr of 99 99 99 8 0. Therefore, the matri representation 0 8 99 99 99 6 5 8 gives the total income, 9 0 in dollars, received from the sale of the cameras for each of the three das. Although it is not necessar to compute the matri product in order to answer 0 8 the question correctl, 99 99 99 6 5 8 9 0 equals 0, 845 6, 066 4, 879. 07 The College Board. College Board, SAT, and the acorn logo are registered trademarks of the College Board. SAT Subject Tests is a trademark owned b the College Board. 00576-0 SAT Subject Tests in Mathematics Level and Level 9