HANDOUT SELF TEST SEPTEMBER 2009 FACULTY EEMCS DELFT UNIVERSITY OF TECHNOLOGY

Transcription

1 HANDOUT SELF TEST SEPTEMBER 009 FACULTY EEMCS DELFT UNIVERSITY OF TECHNOLOGY

2 Contents Introduction Tie-up Eample Test Eercises Self Test Refresher Course Important dates and schedules Eample Test References Calculus Stewart References Basisboek Wiskunde References Foundation Maths Digital Eercises Review of Algebra Formulae Sheet TU Solutions Eample Test Answers Eample Test

3 Introduction Tie-up In the first year analysis course you will have to use knowledge and mathematical skills you learned in the past. May be you do not remember it all that well and maybe you used a formulae sheet or a calculator to figure out some basic things. Taking the analysis course will be a lot easier when you have the basic knowledge and skills ready for use. Eample test You can do the eample test below in order to find out what you can do well and where there are gaps in your mathematical knowledge. We tried to cover the most important subjects. Please take the test without the use of a calculator. You can use a calculator during the analysis course, but you should be able to take this basic test without it. You can t afford to use a calculator each time one of these things pops up in the analysis course. It will take too much time. During the analysis course you can use a formulae sheet. You will find it below, but for the same reasons you should do the test without using it. Practice material Answers and solutions to the eample test are given below. There are notes on each problem and references to practice material as well. You can use these to brush up on the theory behind the troublesome problems and work some eercises. Of some formulas we say that it is useful to know them by heart. The obvious advice is, of course, to do something about that, soon. Self Test The test you will have to take at the beginning of the analysis course is similar to the eample test. For practical reasons the test contains multiple choice questions only. Again you are not allowed to use a calculator or a formulae sheet. Please write down your student number and your branch of study. If you want to able checking your answers write them down on a piece of paper before you hand in the form. The test is taken by almost every first year student of the TU, so the forms are processed automatically. The results give an over all impression, but they can also be used by teachers in order to find out where problems occur. And of course you can learn something from your possible mistakes as well. Refresher course Students of the faculties EEMCS, AS, AE and 3ME have the opportunity to take a refresher course in week 3 to 6. During four meetings they can practice under supervision of student assistants. Then there is a second opportunity to pass the test. Aforementioned faculty s state passing the test at some point in time as a necessary condition for getting an analysis grade awarded. (The test is a part of the course Caleidocoop for students Applied Mathematics.) Of course you can take analysis eams, even if you didn t pass the self test, but the grade is awarded only after you pass the self test. More information on the refresher course can be found on blackboard. The course code is wi000 (for 3ME it is wi50wbmt deel ). Important dates and schedules Dates and schedules of the refresher course and test are best found through blackboard (blackboard.tudelft.nl aforementioned codes wi000 and wi50wbmt deel ) and the registration site for eaminations TAS (tas.tudelft.nl).

4 Delft University of Technology Faculty of Electrical Engineering, Mathematics and Computer Science Mekelweg 4, Delft Eample Test Read carefully the points below This test is intended to assess your active knowledge and your command of some basic mathematical skills at this moment. Use of a calculator or table of formulas is not permitted. The test consists of multiple-choice questions. For each question eactly one of the given possibilities is correct. The test will last one hour.. One of the following statements is not true. Which one? a b c.. 3 a 5 a is equal to d. a. 5 a b. 8 a c. 3. Which number is the largest one? 8 a d. 5 a 8 a. b. 3 4 c. 4. The epression a a + a + a is equal to 4 8 d. 5 6 a. 4 a 4 a b. a a 4 c. a 4 a d. 4 a a 4 please turn over

5 5. The epression ( 7 ) ( + 7 ) is equal to a. 0 b c. 4 d How many different zeros has the polynomial f() ? a. 3 b. c. d The epression ln( e e ) ln( e) is equal to a. e b. c. 3 d If 3 ln(y) 3 + ln(8), then y is equal to a. e b. 8 e 3 3 c. 9. If f() and g() +, then f(g()) is equal to 8 3 e 3 3 d. e 3 3 a. + b. ( + ) c. ( + ) d. + ( + ) 0. Someone is asked to solve the net two equations: () ln( ) 4, () (ln()) 4 He solves these equations as follows: () ln( ) 4 ln() 4 ln() e () (ln()) 4 ln() e Which statement is true? a. Only solution () is complete c. Both solutions are complete b. Only solution () is complete d. Both solutions are incomplete.. The epression ln(e 5 e 3 ) is equal to a. b. 5 3 c. 3 + ln(e ) d. 3 ln(e ) please turn over

6 . The function f is given by f() 0 log(). The domain of f consists of all such that a. 0 < c. b. 0 < < d. ánd 0 3. The epression 7 49 log(3) is equal to a. 7 log(9) b. 3 c. 7 log( 3) d Solve the equation The equation has a. just one solution. It is true that >. b. no solutions d. two solutions 5. If h() f(g()), then h () is equal to c. just one solution. It is true that 0 < <. a. f (g()) c. f (g()) + f(g ()) b. f (g()) g () d. f (g()) g() + f(g ()) g () 6. If y , then dy d is equal to a b. 3 ( 3 + 8) c. d. 3 3 ( 3 + 8) 7. For k > 0 the integral 3k k d can be simplified as a. ln(3) b. ln( k) c. k log(3 k) d. 8. The integral ( ) 3 d is equal to 8 9 k a. 4 ln4 () b. ln(8) c. 5 6 d. 3 8 please turn over

7 9. The function f is given by f() sin(a ) + cos(a ) for some a 0. The maimum value of f() is a. c. b. d. a value dependent on a. 0. The function f, given by f() cos ( ) sin ( ), has a. period π c. period π b. period π d. a horizontal line as its graph. The derivative of f() (cos() + sin()) is a. 0 c. sin () cos () b. cos () sin () d. sin() cos(). An antiderivative of f() cos() sin() is a. b. cos () c. sin () + cos () sin () d. 4 cos () sin () end of the test

8 References Calculus Early Trancendentals James Stewart, 6 E, Thomson Brooks/Cole, ISBN Eponents/powers (problems,, 3 and 7 of the eample test) In this Handout, following these notes, there is a chapter called Review of Algebra. Under the heading of Eponents you will find a summary of the definitions and rules for working with eponents. If you had any difficulties with these problems then read that section and do some problems from the range in the Review (answers at the end). Fractions and brackets (problems, 4 and 5) These problems are about adding, subtracting and simplifying fractions. This also involves manipulating brackets and factoring numbers. See under Fractions and Factoring in the Review for a summary. Problems 7-8 and are recommended. Quadratic and higher-order equations (problems 6, 0, and 4) Quadratic equations occur very frequently; you should be able to solve them without difficulty - see problems 6-68 of the Review. In the review there is also an eplanation of the quadratic formula. You are probably used to solve inequalities with the aid of a Graphic Calculator. It is useful to be able to solve simple cases without that device, for eample using a quick sketch or a table of signs. In Appendi A of Stewart's book you will find, under Inequalities, more about this subject and in problems 3-38 plenty to practice with. In the test we did not deal with the absolute value. If you do not know what that is (anymore) then read in Appendi A the section of that name. For cubic equations there is a general `cubic formula' (known as Cardan's formula) but we do not assume that you know it. We do epect that you are able to note and perform simple steps like ``bringing outside the brackets''. Roots (problems, 3, 5, 7, and 4) Roots (square, cube, etc) are collectively called radicals (radi is the Latin word for root), you will find their theory in the Review in the section of that name. Suitable problems are Eponentials and logarithms (problems 7, 8, 0,, and 3) You can find the definitions and properties of eponential and logarithmic functions in sections.5 and.6 ( Logarithmic functions and Natural logarithms ) of Stewart's book. Suitable problems are 37-4 and 47-5 from.6. Differentiation (problems 5, 6, 9 and ) We epect you to know a few standard derivatives by heart: those of n, (also for negative and rational n), ep(), ln(), sin(), cos() and tan(). We also epect that you know the rules for differentiating sums, differences, products and quotients. Finally there is the chain rule, you must know that one too, otherwise you will keep making unnecessary mistakes. The theory can be found in sections.7 and.8 of Stewart's book; the various rules are in 3., 3., 3.3 and 3.4.In section 3.4 you will find a range of problems, 7-34, in which the rules are combined. Section.7 will eplain in more detail how to set up an equation of a tangent line at a point on a graph; see eample and in problems 5-0 of that section. Integration (problems 7, 8 and ) The Analysis course will cover advanced integration techniques etensively. These problems deal with simple functions that can be integrated using the basic rules. Here too we epect that you know how to integrate a number of standard functions n, ep(), sin(), cos() and simple combinations thereof in your head. In section 5.4 of Stewart's book you will find problems with elementary integrals, for eample 5-9, -6 and 9-3. Trigonometry (problems 9, 0, and ) Trigonometry uses a lot of formulas, called "trigonometric identities", wherein one epression is transformed into another. The tables of formulae in the front (and back) of Stewart's book (they are detachable) contain many of them and we do not epect you to know them all by heart. A few of them occur so often that it is impracticable to have to look them up each time you need them. This is especially true of the rule for sin(-), cos(-) and tan(-) and sin(π /- ) and cos(π /-). It is easy to reconstruct these formulas,

9 and also those that say what happens when you add π to (or subtract it from) the argument of the sine or cosine, by sketching the graphs of the functions or of the unit circle. The doubleangle formulas sin() and cos() are less easy to derive but they are used a lot. Our advice is to learn them by heart - if you have not done so yet. The definitions and many properties of the trigonometric functions and their graphs can be found in Appendi D of Stewart's book. It is very useful to verify simple identities using graphs or the unit circle. We strongly recommend that you have the standard values of sine and cosine for 0, π /6, π /4, π /3 and π / at your fingertips; they are needed virtually everywhere.

10 References Basisboek Wiskunde Jan van de Craats and Rob Bosch, second edition, Pearson Education, ISBN Though this book is in Dutch we recommend it as an opportunity to get acquainted with Dutch mathematical terminology (for starters: `Wiskunde' is Dutch for `Mathematics'). Parts of the book can be found at The faculties EEMCS and AS provide their students with this book. It is used in the refresher course for 3ME as well. For every item of the eample test the pages where relevant eplanations can be found are listed. Corresponding eercises are available on the adjacent pages. Item eample test Pages Basisboek Wiskunde (nd edition) 7, 3, 7 5, , ,33, , , , , 33, , , 47, , 45 45, 79 79

11 References Foundation Maths Anthony Croft and Robert Davison, fourth ed., Pearson Prentice Hall, ISBN For AE students the American book Foundation Maths is used during the refresher course. They get the book at the beginning of the year. For every item of the eample test the relevant chapter is listed below. Item sample test Chapters Foundation Maths (fourth ed.), 7 7 3, , , 0 9, 0 7, niet in dit boek, zie Stewart 6 niet in dit boek, zie Stewart 7 0, , , 9

12 Digital practice material Online learning environment Foundation Math MyMathLab is the online learning environment that comes with the book Foundation Maths ( It contains lots of eercises. AS and AE students can create an account using the code they obtained with the Basisboek Wiskunde (AS) or Foundation Maths. The courseid and a manual can be found on blackboard wi000 Course Documents. Toetsbank analyse Another set of eercises is toetsbank analyse on blackboard under courses. Its number is Online practice using Maple TA Online practice is possible using the link below. You will have to create an account. Possibly you Studentid is too long. In that case you will have to choose a shorter id. The systems gives hints and solutions. You progress is recorded and can be monitored by your teacher. Please note that the navigation of Maple TA on blackboard may not be flawless. Wizmo data bank The link directs you to a collection of practice materials for tie-up problems. You can search the bank using the table of contents of the Basisboek Wiskunde or by subject.

13 Review of Algebra

14 REVIEW OF ALGEBRA Review of Algebra Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus. Arithmetic Operations The real numbers have the following properties: a b b a ab ba a b c a b c a b c ab ac ab c a bc (Commutative Law) (Associative Law) (Distributive law) In particular, putting a in the Distributive Law, we get b c b c b c and so b c b c EXAMPLE (a) (b) (c) 3y y y t 7 t 4t 4t t If we use the Distributive Law three times, we get a b c d a b c a b d ac bc ad bd This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding the products. Schematically, we have a b c d In the case where c a and d b, we have or a b a ba ab b a b a ab b Similarly, we obtain a b a ab b

15 REVIEW OF ALGEBRA 3 EXAMPLE (a) (b) (c) Fractions To add two fractions with the same denominator, we use the Distributive Law: Thus, it is true that a b c b b a b c a c a c b b a c b a b c b But remember to avoid the following common error: a b c a b a c (For instance, take a b c to see the error.) To add two fractions with different denominators, we use a common denominator: a b c ad bc d bd We multiply such fractions as follows: a b c ac d bd In particular, it is true that a b a b a b To divide two fractions, we invert and multiply: a b c d a b d c ad bc

16 4 REVIEW OF ALGEBRA EXAMPLE 3 3 (a) 3 3 (b) (c) (d) 3 y y s t u ut s t u u s t y y y y y y y y y y y y Factoring We have used the Distributive Law to epand certain algebraic epressions. We sometimes need to reverse this process (again using the Distributive Law) by factoring an epression as a product of simpler ones. The easiest situation occurs when the epression has a common factor as follows: Epanding 3(-)3@-6 Factoring To factor a quadratic of the form b c we note that r s r s rs so we need to choose numbers r and s so that r s b and rs c. EXAMPLE 4 Factor 5 4. SOLUTION The two integers that add to give 5 and multiply to give 4 are 3 and 8. Therefore EXAMPLE 5 Factor 7 4. SOLUTION Even though the coefficient of is not, we can still look for factors of the form r and s, where rs 4. Eperimentation reveals that Some special quadratics can be factored by using Equations or (from right to left) or by using the formula for a difference of squares: 3 a b a b a b

17 REVIEW OF ALGEBRA 5 The analogous formula for a difference of cubes is 4 a 3 b 3 a b a ab b which you can verify by epanding the right side. For a sum of cubes we have 5 a 3 b 3 a b a ab b EXAMPLE 6 (a) (Equation ; a, b 3) (b) (Equation 3; a, b 5) (c) (Equation 5; a, b ) 6 EXAMPLE 7 Simplify. 8 SOLUTION Factoring numerator and denominator, we have To factor polynomials of degree 3 or more, we sometimes use the following fact. 6 The Factor Theorem If P is a polynomial and P b 0, then b is a factor of P. EXAMPLE 8 Factor SOLUTION Let P If P b 0, where b is an integer, then b is a factor of 4. Thus, the possibilities for b are,, 3, 4, 6, 8,, and 4. We find that P, P 30, P 0. By the Factor Theorem, is a factor. Instead of substituting further, we use long division as follows: Therefore Completing the Square Completing the square is a useful technique for graphing parabolas or integrating rational functions. Completing the square means rewriting a quadratic a b c

18 6 REVIEW OF ALGEBRA in the form a p q and can be accomplished by:. Factoring the number a from the terms involving.. Adding and subtracting the square of half the coefficient of. In general, we have a b c a b a c a b a a b a b a c 4a b a b c EXAMPLE 9 Rewrite by completing the square. SOLUTION The square of half the coefficient of is. Thus 4 4 ( ) 3 4 EXAMPLE Quadratic Formula By completing the square as above we can obtain the following formula for the roots of a quadratic equation. 7 The Quadratic Formula are The roots of the quadratic equation a b c 0 b sb 4ac a EXAMPLE Solve the equation SOLUTION With a 5, b 3, c 3, the quadratic formula gives the solutions 3 s s69 0 The quantity b 4ac that appears in the quadratic formula is called the discriminant. There are three possibilities:. If b 4ac 0, the equation has two real roots.. If b 4ac 0, the roots are equal. 3. If b 4ac 0, the equation has no real root. (The roots are comple.)

19 REVIEW OF ALGEBRA 7 These three cases correspond to the fact that the number of times the parabola y a b c crosses the -ais is,, or 0 (see Figure ). In case (3) the quadratic a b c can t be factored and is called irreducible. y y y FIGURE Possible graphs of ya@+b+c (a) b@-4ac>0 (b) b@-4ac0 (c) b@-4ac<0 EXAMPLE The quadratic is irreducible because its discriminant is negative: b 4ac Therefore, it is impossible to factor. The Binomial Theorem Recall the binomial epression from Equation : a b a ab b If we multiply both sides by a b and simplify, we get the binomial epansion 8 a b 3 a 3 3a b 3ab b 3 Repeating this procedure, we get a b 4 a 4 4a 3 b 6a b 4ab 3 b 4 In general, we have the following formula. 9 The Binomial Theorem If k is a positive integer, then a b k a k ka k b k k k k k 3 a k b a k 3 b 3 k k k n 3 n a k n b n kab k b k

20 8 REVIEW OF ALGEBRA EXAMPLE 3 Epand 5. SOLUTION Using the Binomial Theorem with a, b, k 5, we have Radicals The most commonly occurring radicals are square roots. The symbol s means the positive square root of. Thus sa means a and 0 Since a 0, the symbol sa makes sense only when a 0. Here are two rules for working with square roots: 0 sab sa sb a sa b sb However, there is no similar rule for the square root of a sum. In fact, you should remember to avoid the following common error: sa b sa sb (For instance, take a 9 and b 6 to see the error.) EXAMPLE 4 s8 (a) s 8 s9 3 (b) s y s sy sy Notice that s because s indicates the positive square root. (See Appendi A.) In general, if n is a positive integer, s n a means n a If n is even, then a 0 and 0. Thus s 3 8 because 3 8, but s 4 8 and s 6 8 are not defined. The following rules are valid: a s n a s n ab s n a s n n b b s n b EXAMPLE 5 s 3 4 s 3 3 s 3 3 s 3 s 3

21 REVIEW OF ALGEBRA 9 To rationalize a numerator or denominator that contains an epression such as sa sb, we multiply both the numerator and the denominator by the conjugate radical sa sb. Then we can take advantage of the formula for a difference of squares: (sa sb)(sa sb) (sa) (sb) a b s 4 EXAMPLE 6 Rationalize the numerator in the epression. SOLUTION We multiply the numerator and the denominator by the conjugate radical s 4 : s 4 s 4 s s 4 (s 4 ) (s 4 ) s 4 Eponents Let a be any positive number and let n be a positive integer. Then, by definition,. a n a a a. 3. a 0 a n a n n factors 4. a n s n a a m n s n a m (s n a) m m is any integer Laws of Eponents Let a and b be positive numbers and let r and s be any rational numbers (that is, ratios of integers). Then r s. a r a s a r s. a ab r a r b r 5. a r a s b a r a r b b 0 r a r s a rs In words, these five laws can be stated as follows:. To multiply two powers of the same number, we add the eponents.. To divide two powers of the same number, we subtract the eponents. 3. To raise a power to a new power, we multiply the eponents. 4. To raise a product to a power, we raise each factor to the power. 5. To raise a quotient to a power, we raise both numerator and denominator to the power.

22 0 REVIEW OF ALGEBRA EXAMPLE 7 (a) (b) (c) (d) y y 4 3 s4 3 s64 8 s z (e) y 3 y y y y y y y y y z 4 y y y y y y Alternative solution: 7 y 5 z 4 y y y y 4 3 (s4) Eercises A Click here for answers. 6 Epand and simplify.. 6ab 0.5ac. y y a t 4 t t t y 4 6 y 5 y t 5 t 3 8t Perform the indicated operations and simplify. 8 9b b u u. u y z 5. r s 6. s 6t a 3 ab 4 b y z a bc b ac c c 9 48 Factor the epression ab 8abc t t 9s 4. 4t t Simplify the epression

23 REVIEW OF ALGEBRA Complete the square Solve the equation Which of the quadratics are irreducible? Use the Binomial Theorem to epand the epression. 73. a b a b Simplify the radicals. s 3 s s3 s s 3 54 s sy s 3 y 8. s6a 4 b 3 8. s 5 96a 6 s 5 3a Use the Laws of Eponents to rewrite and simplify the epression a n a n a 3 b 4 y 88. a 5 b 5 y y y 3 z s 5 y (sa) s (st) 5 s 4 3 t sst 4 s Rationalize the epression. s s s5 07. s s s 09 6 State whether or not the equation is true for all values of the variable. 09. s 0. 6 a. a y y a 6 4 4a a n s 4 r n s 4 r ( s) s h s h h s sy s 4 y y 4

24 ANSWERS Answers. 3a bc. 3 y a t t y 4 y 5 y t 56t b u 3u u. b 3ab 4a z rs a b yz y 3t 6. a c b c ab 5 8c t t t 40. t 3s t 3s 4. t ( 5 ) ( 3 ) , 0 6., 4 9 s s s33 s5 67., 4 68., s 69. Irreducible 70. Not irreducible 7. Not irreducible (two real roots) 7. Irreducible a 6 6a 5 b 5a 4 b 0a 3 b 3 5a b 4 6ab 5 b 6 a 7 7a 6 b a 5 b 35a 4 b 3 35a 3 b 4 a b 5 7ab 6 b y 8. 4a bsb 8. a a n a y b y s s s 3 y 6 3 y y 9 5 z 6 t r n s 4 s s 8 3 s s sy y a 3 4 t 5 s h s h s s 3 4 s s 09. False 0. False. True. False 3. False 4. False 5. False 6. True 56 5 s3 6 8

25 FORMULAE SHEET to be used during the Calculus eaminations at the TU Delft Some trigonometric formulas. sin(α) sin α cos α. cos(α) cos α sin α cos α sin α Some limits 3. lim 4. lim 5. lim p Some Taylor series e 0 ( + a ) e a ln 0 (p > 0) p 6. e + +! ( R) 3! 7. sin 3 3! + 5 5! 7 + ( R) 7! 8. cos! + 4 4! 6 + ( R) 6! 9. ln( + ) ( + ) k + k + k(k ) +! Some integrals d tan. sin ln + C d ( tan. cos ln + π ) + C 4 d 3. + arctan + C d 4. ln + + C d 5. arcsin + C d 6. + ln( + + ) + C d 7. ln + + C + ( < ) k(k )(k ) 3 + ( < < ) 3! d + + ln( + + ) + C d + arcsin + C n n 3 n 5 π sin n d n n n 4 3 π if n is even and n 4 0 n n 3 n 5 n n n 4 4 if n is odd and n 3 5 3

26 Solutions eample test A () B. (64) ( ) 6 C D. Answer: C a a a a a a a 3. Answer: D ( ) ( ) ( ) 6 6 Answer: D a a + a + a a( + a) a( a) + ( a)( + a) ( + a)( a) a+ a + a a 4a 4 a 4 a Answer: A

27 5. ( 7) ( + 7) 7+ 7 (+ 7+ 7) Answer: D ( 8+ 6) 0 ( 4) Answer: B ln ( e e) ln ( e ) e ln ee ln e ln e ln e ln e ln e ln e 3 ln Answer: C 3 3 y 3 3 y ( ) ( ) ( ) ( ) 3 ( ) ln( y) + ln(8) ln( y ) ln8 ln 8 8 y 8e y e 3 3 y e e 3 3 e 9. Answer: D If f ( ) and g ( ) + then f ( g( )) ( + ) Answer: B

28 0. () ln( ) 4 ln( ) 4 ln( ) e e e () (ln( )) 4 ln( ) ln( ) e e.. Answer: D ln( e e ) ln( e ( e )) 3 ln( e ) + ln( e ) 3 + ln( e ) Answer: C is defined for ( ) 0 0 log( ) is defined for > 0 3. The fraction is defined for 0 log( ) 0 So the epression is defined for 0< < Answer: B 49 log(3) 49 log(3) ( log(3) ) ( ) log(3) ( log(3) ) Answer: B

29 so ( + ) + 5 and + 0 so so so and 6 6 so 4 and (negative) so Answer: C Using the chain rule: let g()u df du h ( ) f ( u) u ( ) f ( g( )) g ( ) du d Answer: B y + 8 ( + 8) Using the chain rule; dy ( + 8) 3 d 3 ( ) ( + 8) Answer: B 3k k 3k d [ ln ] k 3k ln(3 k) ln k ln ln 3 k Answer: A

30 ( ) d d Answer: D 9. One can use standard graphs. The variable a only changes the period, and not the maimum. Sketch the graphs of sin() and cos() and of their sum. Clearly the maimum is larger than and is attained on the half of the period. So the maimum is One can also use the derivative: ' f ( ) a cos( a) asin( a) f '( ) 0 implies a cos( a) a sin( a) so a π so the maimum equals 4 0. One can also use a formula f ( ) sin( a) + cos( a) sin( a) + sin( π a π sin ( a + a)cos ( a ( π a)) a a sin π cos( π) cos( π) so the maimum equals Answer: C Using the formula of the double angle: cos ( ) sin ( ) cos So the period equals π Answer: A

31 . First solution f ( ) (cos( ) + sin( )) ( sin( ) + cos( )) (cos ( ) sin ( )) cos ( ) sin ( ) Or second solution f( ) (cos( ) + sin( )) cos ( ) + sin cos + sin ( ) + sin( ) f ( ) cos( ) (cos sin ). Answer: B First solution f ( ) cos( )sin( ) sin( ) So an antiderivative is: cos( ) + c 4 4 ( sin ( )) + c sin ( ) + k Choose k0 Second solution Differentiate the given alternatives f( ) sin ( ) f ( ) sin( ) cos sincos Answer: B

32 Answers eample test Item eample test Answer C D 3 D 4 A 5 D 6 B 7 C 8 D 9 B 0 D C B 3 B 4 C 5 B 6 B 7 A 8 D 9 C 0 A B B