201408 Semester I MB0042 Managerial Eco Essay - 2227 Words

Q.No 1- Inflation is a global Phenomenon which is associated with high price causes decline in the value for money. It exists when the amount of money in the country is in excess of the physical volume of goods and services. Explain the reasons for this monetary phenomenon. Inflation is commonly understood as a situation of substantial and rapid increase in the level of prices and consequent deterioration in the value of money over period of time. It refers to the advantage rise in the general level of prices and fall in the value of money. Inflation is upward movement in the average level of prices. Cause of Inflation 1. Demand Side increase in aggregative effective demand is responsible for inflation. In this case, aggregate demand†¦show more content†¦This type of price discrimination is called perfect discrimination Discrimination of the Second Degree – In case of discrimination of second degree, the monopolist charges different prices for markets of the same commodity, but not at a maximum possible rate but at a lower rate. The monopolist will leave a certain amount of consumer’s surplus with the consumers. This is done to keep the consumers satisfied and prevent the entry of potentials rivals. This method is adopted by railway companies. Discrimination of the Third Degree – In case of discrimination of the third degree, the markets are divided into many sub-markets or Sub-groups. The price charged in each case roughly depends on the ability to pay of different subgroups in the market. This is the most common type of discrimination followed by a monopolist. Q.No-3 Define monopolistic competition and explain its characteristics It is a market structure in which a large number of small sellers sell differentiated products which are close, but not perfect substitutes for one another. Under this market, the products produced and sold are different, but they are close substitutes for one another. This leads to competition among different sellers. Thus, in this

Devil in the white City Summary Free Essays

The World’s Fair was an amazing event for our country. It represented how capable we were and how amazing our technology could become. It negated many of the stereotypes surrounding Chicago that it was only a city of animal butchery. We will write a custom essay sample on Devil in the white City Summary or any similar topic only for you Order Now It set a standard for how cities should be run and it picked up the economy with all of the work and tourism It created. But, hidden within all of this good was evil. Although pick pockets and thieves were very common, largely represented In the book Is murder. Holmes takes advantage of this situation and lures In young women who are traveling alone. Without the magnificent fair, he would have had a much more difficult time doing what he did. The good that people were creating within the fair created this room for evil. Honestly, It seems this Is a real world theme as well. Wherever good Is done, corruption Is always a possibility and often this opportunity Is taken. The sad theme that permeates this book Is that where there Is good, there will be evil. And although unfortunate, It is reality-To me, the text had one BIG theme: evil as the result of good. The World’s Fair was an amazing event for our country. It represented how capable we were and how amazing our technology could become. It negated many of the stereotypes surrounding Chicago that it was only a city of animal butchery. It set a standard for how cities should be run and it picked up the economy with all of the work and tourism it created. But, hidden within all of this good was evil. Although pick pockets and thieves were very common, largely represented in the book is ruder. Holmes takes advantage of this situation and lures in young women who are traveling alone. Without the magnificent fair, he would have had a much more difficult time doing what he did. The good that people were creating within the fair created this room for evil. Honestly, it seems this is a real world theme as well. Wherever good is done, corruption is always a possibility and often this opportunity is taken. The sad theme that permeates this book is that where there is good, there will be evil. And although unfortunate, it is reality. How to cite Devil in the white City Summary, Papers

Censorship and Pop Culture Essay Example For Students

Censorship and Pop Culture Essay One mans vulgarity is anothers lyric. Justice John M. Harlan, Cohen v. California (1971) It is probably no accident that freedom of speech is the first freedom mentioned in the First Amendment: Congress shall make no law.. .abridging the freedom of speech, or of the press, or of the people peaceably to assemble, and to petition the Government for a redress of grievances. The Constitutions framers believed that freedom of inquiry and liberty of expression were the hallmarks of a democratic society. Freedom of speech, of the press, of association, of assembly and petition this set of guarantees, protected by the First Amendment, comprises what we refer to as freedom of expression. The Supreme Court has written that this freedom is the matrix, the indispensable condition of nearly every other form of freedom. Without it, other fundamental rights, like the right to vote, would wither and die. But in spite of its preferred position in our constitutional hierarchy, the nations commitment to freedom of expression has been tested over and over again. Especially during times of national stress, like war abroad or social upheaval at home, people exercising their First Amendment rights have been censored, fined, even jailed. Those with unpopular political ideas have always borne the brunt of government repression. It was during WWI hardly ancient history that a person could be jailed just for giving out anti-war leaflets. Out of those early cases, modern First Amendment law evolved. Many struggles and many cases later, ours is the most speech-protective country in the world. (Glasser, Visions of Liberty, 1991.) Three Reasons Why Freedom of Expression Is Essential to a Free Society It is the foundation of self-fulfillment. The right to express ones thoughts and to communicate freely with others affirms the dignity and worth of each and every member of society, and allows each individual to realize his or her full human potential. Thus, freedom of expression is an end in itself and as such, deserves societys greatest protection. It is vital to the attainment and advancement of knowledge, and the search for the truth. The eminent 19th-century writer and civil libertarian, John Stuart Mill, contended that enlightened judgment is possible only if one considers all facts and ideas, from whatever source, and tests ones own conclusions against opposing views. Therefore, all points of view even those that are bad or socially harmful should be represented in societys marketplace of ideas. It is necessary to our system of self-government and gives the American people a checking function against government excess and corruption. If the American people are to be the masters of their fate and of their elected government, they must be well-informed and have access to all information, ideas and points of view. Mass ignorance is a breeding ground for oppression and tyranny. Beginning in the 1980s, religious fundamentalists and some parents groups have waged a persistent campaign to limit the variety of cultural messages available to American youth by attacking the content of some of the music industrys creative products. These attacks have taken numerous forms, including a call by the Parents Music Resource Center (PMRC) for the labeling of recordings whose themes or imagery relate to sexuality, violence, drug or alcohol use, suicide or the occult, and prosecutions of record companies and store owners for producing or selling albums that contain controversial songs. After years of pressure from the PMRC and a series of Senate hearings in 1985, the Recording Industry Association of America (RIAA) introduced, in 1990, a uniform labeling system using the logo, Parental Advisory Explicit Lyrics. The RIAA initiated this system without providing record companies with any standards, criteria or guidelines for determining which albums should be labeled. That decision is left completely up to the companies, which have chosen to label only selected rock and rap albums and not recordings of country music, opera or musical comedy that may also contain controversial material. Dissatisfied with the RIAAs labels, many would-be censors have demanded even more limits on the sale of music with controversial lyrics. As a result, legislators have introduced bills in more than 20 states in recent years that would require warning labels far more detailed .

Models of Health Behavior Essay Example

Models of Health Behavior Paper Note: If candidate uses incorrect formula: maximum 1/4 marks (for standard form) substitution into correct formula Note: If an error in subs and 4  ± ? 80 gets: and 6 states â€Å"no solution†: maximum 3/4 marks 4  ± 16 + 96 6 4  ± 112 = 6 2 ±2 7 = 3 = 2,43 or ? 1,10 112 If doesn’t conclude with â€Å"no solution†: maximum 2/4 marks 4  ± 112 or 6 decimal answer (4) Copyright reserved Mathematics/PI 3 NSC – Memorandum DBE/November 2011 OR 3x 2 ? 4 x = 8 3x 2 ? 4 x ? 8 = 0 x= = ? b  ± b ? 4ac 2a ? (? 4)  ± 2 (? 4)2 ? 4(3)(? 8) 2(3) Note: Penalise 1 mark for inaccurate rounding off to ANY number of decimal places if candidate gives decimal answers tandard form substitution into correct formula answer answer (4) = 2,43 or ? 1,10 1. 1. 3 4 x 2 + 1 ? 5x (4 x ? 1)(x ? 1) ? 0 + 0 1 4 x? OR 4 x 2 ? 5x + 1 ? 0 factors ? 0 1 + 1 4 1? ? OR ? ? ? ; ? ? [1; ? ) 4? ? 1 both critical 1 values of and 1 4 or OR ? answer (4) 1 or x ? 1 4 1 4 1 x OR 1 4 1 xNote: If candidate gives either of these correct graphical solutions but writes down the incorrect intervals or uses AND: max 3/4 marks NOTES: If a candidate gives an answer of 1 ? x ? If a candidate gives an answer of 1 then max 3/4 marks. 4 1 ? x ? 1 then max 2/4 marks. 4 1 If a candidate gives an answer of x ? nd x ? 1 then max 3/4 marks. 4 If the candidate leaves out the equality of the notation then penalty of 1 mark. 1 If a candidate gives an answer of x ? ; x ? 1 then max 3/4 marks. 4 1 If candidate gives x ? and/or x ? 1 , BREAKDOWN: max 2/4 marks. 4 If candidate gives : 0 ? 0 + + award 3/4 marks 1 1 4 Copyright reserved Please turn over Mathematics/PI 4 NSC – Memorandum DBE/November 2011 1. 2. 1 x 2 + 5 xy + 6 y 2 = 0 (x + 3 y )(x + 2 y ) = 0 x + 3y = 0 x = ? 3 y OR x = ? 3 y OR x + 2y = 0 x = ? 2 y x = ? 2 y Note: If a candidate gives x x ? = 3 or ? = 2 y y award 2/3 marks factors answers (3) Let k = y 2 x + 5 xy + 6 y 2 = 0 2 ?x? ?x? ? ? + 5? ? + 6 = 0 ? y? ? y? ? ? ? ? 2 k + 5k + 6 = 0 (k + 3)(k + 2) = 0 k = – 3 or k = – 2 x x = ? 3 or = ? 2 y y OR factors answers (3) x 2 + 5 xy + 6 y 2 = 0 x= x= ? 5 y  ± (5 y ) 2 ? 4(1)(6 y 2 ) 2(1) ? 5y  ± y2 2 ? 5y  ± y x= 2 x = ? 3 y x = ? 2 y or x x = ? 3 = ? 2 y y substitutes correctly into correct formula answers (3) OR x 2 + 5 xy + 6 y 2 = 0 ?5 x 2 + 5 xy + ? ?2 ? 5 ? y ? = ? 6 y 2 + ? ?2 ? 2 2 ? y? ? 2 5 ? 1 2 ? ? x + y? = y 2 ? 4 ? 5 1 x+ y= ± y 2 2 5 1 x=? y ± y 2 2 completing the square Copyright reservedPlease turn over Mathematics/PI x = ? 3 y x = ? 3 y x = ? y 5 NSC – Memorandum DBE/November 2011 answers (3) or x = ? 2 y OR Let k = x = ky x y x 2 + 5 xy + 6 y 2 = 0 (ky )2 + 5 y(ky ) + 6 y 2 = 0 k 2 y 2 + 5 y 2k + 6 y 2 = 0 y 2 k 2 + 5k + 6 = 0 ( (k ) 2 + 5k + 6 = 0 ) factors (k + 3)(k + 2) = 0 k = – 3 or k = – 2 x x = ? 3 or = ? 2 y y answers (3) Note: (x;y) = (0;0) is also a solution, but in this case OR x is undefined y Let y = 1 , x 2 + 5x + 6 = 0 (x + 2)(x + 3) = 0 x = ? 2 or x = ? 3 x x = ? 2 or = ? 3 y y x+ y =8 ? 3y + y = 8 ? 2y = 8 y = ? 4 x = 12 factors answers (3) x+ y =8 ? 2y + y = 8 ? y =8 y = ? 8 x = 16 1. 2. 2 OR substitution x = – 3y subs x = ? 2 y values both x values correct (5) OR 8? y = ? 3 y 8 ? y = ? 3 y 8 = ? 2 y y = ? 4 x = 12 Copyright reserved 8? y = ? 2 OR y 8 ? y = ? 2 y 8 = ? y y = ? 8 x = 16 x=8–y substitution y values both correct x values (5) Please turn over Mathematics/PI 6 NSC – Memorandum DBE/November 2011 OR x+ y =8 y =8? x x = ? 3 OR 8? x x = ? 3(8 ? x ) x = ? 24 + 3 x ? 2 x = ? 24 x = 12 y = ? 4 OR y =8? x x = ? 2 8? x x = ? 2(8 ? x ) x = ? 16 + 2 x ? x = ? 16 x = 16 y = ? 8 substitution x values correct both y values correct (5) (x + 2 y )(x + 3 y ) = 0 x+ y =8 x =8? y ( y + 8)(2 y + 8) = 0 y = ? 8 or y = ? 4 x = 16 x = 12 x =8? y ubstitution y values correct both x values correct (5) OR x = 8? y x = 8? y 2 (8 ? y ) 2 + 5(8 ? y ) y + 6 y = 0 64 ? 16 y + y 2 + 40 y ? 5 y 2 + 6 y 2 = 0 2 y 2 + 24 y + 64 = 0 y 2 + 12 y + 32 = 0 ( y + 8)( y + 4) = 0 y = ? 8 or y = ? 4 x = 16 x = 12 OR substitution factors both y values correct both x values correct (5) Copyright reserved Please turn over Mathematics/PI 7 NSC – Memorandum DBE/November 2011 OR x =8? y (8 ? y ) 2 + 5(8 ? y ) y + 6 y = 0 2 2 2 2 x = 8? y substitution 64 ? 16 y + y + 40 y ? 5 y + 6 y = 0 2 y 2 + 24 y + 64 = 0 y 2 + 12 y + 32 = 0 ? 12  ± 12 ? 4(1)(32) y= 2(1) 2 ? 12  ± 16 2 y = ? 8 or y = ? x = 16 x = 12 = Note: If a candidate uses the formula and replaces x for y and then answers are swapped: maximum 4/5 marks substitutes into correct formula both y values correct both x values correct (5) OR y =8? x x 2 + 5 x(8 ? x ) + 6(8 ? x )2 = 0 x + 40 x ? 5 x + 6 64 ? 16 x + x 2 x 2 ? 56 x + 384 = 0 x 2 ? 28 x + 192 = 0 (x ? 16)(x ? 12) = 0 x = 16 x = 12 or y = ? 8 y = ? 4 2 2 y =8? x 2 ( )= 0 substitution factors both x values correct both y values correct (5) OR y =8? x x + 5 x(8 ? x ) + 6 (8 ? x ) = 0 2 2 y =8? x substitution x + 40 x ? 5 x 2 2 + 6(64 ? 16 x + x ) = 0 2 2 x 2 ? 56 x + 384 = 0 x 2 ? 28 x + 192 = 0 x= = ? ? 28)  ± 28  ± 416 2 (? 28)2 ? 4(1)(192) 2(1) x = 12 x = 16 or y = ? 4 y = ? 8 substitutes into correct formula both x values correct both correct y values (5) [19] Please turn over Copyright reserved Mathematics/PI 8 NSC – Memorandum DBE/November 2011 QUESTION 2 2. 1. 1 x ? 4 = 32 ? x 2 x = 36 x = 18 T2 ? T1 = T3 ? T2 Note: If answer only: award 2/2 marks answer (2) a + 2d = 32 and a = 4 OR a=4 a + 2d = 32 2d = 28 d = 14 x = 14 + 4 x = 18 OR Note: If candidate writes x? 4 32 ? x only (i. e. omits equality) : 0/2 marks answer (2) substitutes correctly into arithmetic mean 4 + 32 formula i. e. 2 answers (2) T2 T3 = T1 T2 x= + 32 = 18 2 2. 1. 2 x 32 = 4 x x 2 = 128 x =  ± 128 x =  ±8 2 OR a=4 x r= 4 ? x? ar 2 = 4? ? ? 4? ? x? 32 = 4? ? ? 4? x 2 = 128 OR x =  ±11,31 OR x =  ± 2 2 7 Note: If candidate 32 x only writes 4 x (i. e. omits e quality) : 0/2 marks x 2 = 128 both answers (surd or decimal or exponential form) (3) Note: If only x = 128 then penalty 1 mark 2 2 2 ? x? 32 = 4? ? ? 4? 2 x = 128 x =  ± 128 x =  ±8 2 or x =  ±11,31 or x =  ± 2 2 7 both answers (surd or decimal or exponential form) (3) substitutes correctly into geometric mean formula i. e.  ± 4? 32 both answers (surd or decimal or exponential form) (3) Please turn over OR x =  ± 4 ? 32 =  ± 128 or x =  ±8 2 or x =  ±11,31 or x =  ± 2 2 7 Copyright reserved Mathematics/PI 9 NSC – Memorandum DBE/November 2011 2. 2 13 P = ? 3k ? 5 a = 3? 4 or +3 3? 5 =3 k =1 1? 5 +3 2? 5 + + 3 13 ? 5 = 3 ? 4 + 3 ? 3 + 3- 2 + + 38 3 ? 4 313 ? 1 = 3 ? 1 = 9841,49 Note: Correct answer only: 1/4 marks only 1 81 ( ) r =3 subs into correct formula or 9841 40 797161 or 81 81 answer (4) OR 13 P = ? 3k ? 5 k =1 1? 5 =3 + 32 ? 5 + 33 ? 5 + + 313 ? 5 2. 3 2 S n = [2a + (n ? 1)d ] + [2a + (n ? 1)d ] + + [2a + (n ? 1)d ] + [2a + (n ? 1)d ] = n[2a + (n ? 1)d ] Sn = n [2a + (n ? 1)d ] 2 S n = [a + (n ? 1)d ] + [a + (n ? 2)d ] + [a + d ] + a = 3 ? 4 + 3? 3 + 3- 2 + + 38 1 1 1 = + + + + 6561 81 27 9 40 797161 or = 9841,49 or 9841 81 81 S n = a + [a + d ] + [a + 2d ] + + [a + (n ? 2 )d ] + [a + (n ? 1)d ] Note: If the candidate rounds off and gets 9841,46 (i. e. correct to one decimal place): DO NOT penalise for the rounding off. expand the sum 13 terms in expansion answer (4) writing out Sn â€Å"reversing† Sn expressing 2Sn grouping to get 2 S n = n[2a + (n ? 1)d ] (4) writing out Sn â€Å"reversing† Sn expressing 2Sn grouping to get 2S n = n[a + a + (n ? 1)d ] (4) OR S n = a + [a + d ] + [a + 2d ] + + (Tn ? d ) + Tn S n = Tn + (Tn ? d ) + [a + d ] + a = n[a + a + (n ? 1)d ] = [2a + (n ? 1)d ] Sn = n [2a + (n ? 1)d ] 2 2 S n = a + Tn + a + Tn + a + Tn + + a + Tn Note: If a candidate uses a circular argument (eg S n +1 = S n + Tn ): max 1/4 marks (for writing out Sn) Note: If a candidate uses a specific linear sequence, then NO marks. [13] Copyright reserved Please turn over Mathematics/PI 10 NSC – Memorandum DBE/November 2011 QUESTION 3 3. 1 21; 24 Note: If candidate writes T8 = 21 T7 = 24 : award 1/2 marks 21 24 (2) 3. 2 T2 k = 3. 2 k ? 1 and so T52 = 3. 2 26 ? 1 = 100663296 Note: If candidate writes out all 52 terms and gets correct answer: award 5/5 marks . 2 k ? 1 T52 6k ? 3 T51 T2k ? 1 = 3 + 6(k ? 1) = 6k ? 3 and so T51 = 6(26) ? 3 = 153 T52 ? T51 = 100663296 ? 153 = 100663143 answer Note: If candidate used k = 52: max 2/5 Note: if candidate interchanges order i. e. does T51 ? T52 : max 4/5 marks Note: writes out all 52 terms and subtracts T51 ? T52 : max 4/5 marks (5) OR Consider sequence P: 3 ; 6 ; 12 †¦ Pn = 3. 2 n? 1 P26 = 3. 2 26 ? 1 = 100663296 Consider sequence Q: 3 ; 9 ; 15 †¦ Qn = 6n ? 3 Q26 = 6(26) ? 3 = 153 T52 ? T51 = P26 ? Q26 = 100663296 ? 153 = 100663143 Pn = 3. 2 n? 1 P26 Qn = 6n ? 3 Q26 answer (5) Copyright reserved Please turn over Mathematics/PI 1 NSC – Memorandum DBE/November 2011 3. 3 For all n ? N , n = 2k or n = 2k ? 1 for some k ? N If n = 2k : Tn = T2k = 3. 2 k ? 1 If n = 2k ? 1 : Tn = T2k ? 1 = 6k ? 3 = 3(2k ? 1) factors 3. 2 k ? 1 Note: If a candidate only illustrates divisibility by 3 with a specific finite part of the sequence, not the general term: 0/2 marks factors 3(2k ? 1) (2) In either case, Tn has a factor of 3, so is divisible by 3. OR Pn = 3. 2 n ? 1 Which is a multiple of 3 Qn = 6 n ? 3 = 3(2n ? 1) Which is also a multiple of 3 Since Tn = Q2 k ? 1 or Tn = P2 k for all n ? N , Tn is always divisible by 3 OR factors 3. 2 n ? 1 factors 3(2n ? 1) (2)The odd terms are odd multiples of 3 and the even terms are 3 times a power of 2. This means that all the terms are multiples of 3 and are therefore divisible by 3. odd multiples of 3 3 times a power of 2 (2) [9] Copyright reserved Please turn over Mathematics/PI 12 NSC – Memorandum DBE/November 2011 QUESTION 4 4. 1 The second, third, fourth and fift h terms are 1 ; – 6 ; T4 and – 14 First differences are: – 7 ; T4 + 6 ; – 14 – T4 So T4 + 6 + 7= – 14 – 2T4 – 6 T4 = – 11 d = – 11 + 6 + 7 = 2 or – 14 + 22 – 6 = 2 Note: Answer only (i. e. d = 2) with no working: 3 marks Note: Candidate gives T4 = ? 11 and d = 2 only: award 5/5 marks 7 T4 + 6 – 14 – T4 setting up equation T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) answer (5) –7 –7+d – 7 + 2d setting up equation T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) OR T2 1 -7 T3 -6 -7+d d T4 -7+2d d T5 -14 T5 ? T2 = (T5 ? T4 ) + (T4 ? T3 ) + (T3 ? T2 ) ? 15 = (? 7 + 2d ) + (? 7 + d ) + ? 7 ? 15 = ? 21 + 3d 6 = 3d d =2 Note: Candidate uses trial and error and shows this: award 5/5 marks answer (5) OR 4a + 2b + c = 1 9a + 3b + c = ? 6 5a + b = ? 7 25a + 5b + c = ? 14 16a + 2b = ? 8 10a + 2b = ? 14 6a = 6 a =1 d = 2a = 2 4a + 2b + c = 1 9a + 3b + c = ? 6 25a + 5b + c = ? 1 4 solved simultaneously answer (5) ORT1 1 – T1 T1 8 1 -7 T4+13 -6 T4+6 -20-2 T4 -14 T4 -14 T4 –7 T4 + 6 ? 14 ? T4 setting up equation answer (5) Please turn over T4 + 13 = ? 20 ? 2T4 3T4 = ? 33 T4 = ? 11 d = ? 11 + 13 d =2 Copyright reserved Mathematics/PI 13 NSC – Memorandum DBE/November 2011 OR T1 x 1–x -8+x T2 1 -7 y +13 T3 -6 y+6 20 – 2y T4 y -14 y T5 -14 –7 y+6 ? 14 ? y y + 13 = ? 20 ? 2 y 3 y = ? 33 y = ? 11 Second difference = y + 13 = ? 11 + 13 = 2 4. 2 T1 1 –6 -9 2 -7 Note: Answer only: award 2/2 marks Note: If incorrect d in 4. 1, 2/2 CA marks for T1 = d + 8 (since 1 ? T1 = ? 7 ? d ) setting up equation answer (5) method T1 = 10 (2) T1 = 10 OR =1 5a + b = ? 7 5(1) + b = ? 7 b = ? 12 a +b+c =1 4(1) + 2(? 12) + c = 1 c = 21 Tn = n 2 ? 12n + 21 T1 = (1) 2 ? 12(1) + 21 = 10 OR method T1 = 10 (2) T4 + 13 = ? 8 + T1 ? 11 + 13 = ? 8 + T1 T1 = 10 y + 13 = ? 8 + x method T1 = 10 (2) [7] OR ? 11 + 13 = ? 8 + x x = 10 Copyright res erved Please turn over Mathematics/PI 14 NSC – Memorandum DBE/November 2011 QUESTION 5 5. 1. 1 y = f (0) ? 6 = ? 1 0? 3 =1 (0 ; 1) OR ?6 ? 1 x? 3 ? 6 1= x? 3 x ? 3 = ? 6 0= x = ? 3 (? 3 ; 0) x = 0 and y = 1 5. 1. 2 Note: Mark 5. 1. 1 and 5. 1. 2 as a single question. If the intercepts are interchanged: max 3/5 marks y =1 x=0 (2) y=0 x ? 3 = ? 6 nswer (3) shape y 5. 1. 3 Note: The graph must tend towards the asymptotes in order to be awarded the shape mark (? 3; 0) (0 ; 1) 0 3 x y = ? 1 ?1 x=3 both intercepts correct horizontal asymptote vertical asymptote (4) Note: A candidate who draws only one ‘arm’ of the hyperbola loses the ‘shape’ mark i. e. max 3/4 marks 5. 1. 4 ? 3 lt; x lt; 3 OR (? 3; 3) OR ? 3 lt; x and x lt; 3 Note: if candidate writes ? 3 lt; x only: 1/2 marks Note: if candidate writes x lt; 3 only: 1/2 marks ?3 and 3 inequality OR interval notation (2) Copyright reserved Please turn over Mathematics/PI 15 NSC – Memorandum DBE/Novem ber 2011 5. 1. y= ?6 ? 1 ? 2? 3 1 = 5 1? 1 5 1 5 m= 0 ? (? 2) 2 = 5 formula substitution answer (4) OR m= = f (0) ? f (? 2) 0 ? (? 2) 1? 1 5 formula f (? 2) = 1 5 0+2 2 = 5 b lt; 0 since b lt; 0 and a lt; 0 2a y x substitution answer (4) y-intercept negative turning point on the x axis turning point on the left of the y axis maximum TP and quadratic shape 5. 2 x=? 0 (4) [19] Copyright reserved Please turn over Mathematics/PI 16 NSC – Memorandum y DBE/November 2011 QUESTION 6 f C(0 ; 4,5) g x O A B 6. 1 0 = 2x ? 8 8 = 2x 23 = 2 x x=3 A(3 ; 0) f (0) = 2 0 ? 8 = 1? 8 = ? 7 B(0 ; –7) Note: no CA marks Note: answer only: award 2/2 marks =0 answer for A x=0 answer for B (4) answer (1) 6. 2 6. 3 y = ? 8 OR y + 8 = 0 h( x ) = f ( 2 x ) + 8 = 22x ? 8 + 8 ( ) (2 2 x ? 8) answer of h( x ) = 4 x or 2 2 x (2) = 4 x or 2 2 x 6. 4 x = 4y y = log 4 x OR x = 22 y 2 y = log 2 x 1 y = log 2 x OR y = log 2 x 2 Note: answer only award 2/2 marks Note: candidate works out f -1 and gets y = l og 2 ( x + 8) award 1/2 marks log x OR y = log 4 switch x and y answer in the form y =†¦ (2) 6. 5 p ( x) = ? log 4 x OR p( x) = log 1 x 4 answer (1) OR p ( x) = log 4 OR 1 x OR 1 p( x) = ? log 2 x 2 y = ? log 2 x Copyright reserved Please turn over Mathematics/PI 17 NSC – Memorandum 5 DBE/November 2011 . 6 ? g (k ) ? ? g (k ) = g (0) + g (1) + g (2) + g (3) ? g (4) ? g (5) x = 3 is the axis of symmetry of g ? by symmetry g (2) = g (4) and g (1) = g (5) Answer = g (0) + g (3) = 4,5 + 0 = 4,5 OR k =0 k =4 3 = g (0) + g (1) + g (2) + g (3) ? g (4) ? g (5) g (2) = g (4) and g (1) = g (5) g (0) + g (3) answer (4) ? g (k ) ? ? g (k ) ? g (k ) = g (0) + g (1) + g (2) + g (3) ? g (k ) = g (4) + g (5) k =4 k =0 5 3 5 expansion k =0 3 k =4 x = 3 is the axis of symmetry of g ? by symmetry g (4) = g (2) g (5) = g (1) k =0 g (2) = g (4) and g (1) = g (5) ? g (k ) ? ? g (k ) k =4 3 5 = g (0) + g (3) = 4,5 + 0 = 4,5 OR g (0) + g (3) answer (4) ( x) = a( x ? 3) + 0 2 4,5 = a(0 ? 3) 2 + 0 4,5 = 9a 1 a= 2 1 2 g ( x) = ( x ? 3) 2 k =0 3 k =0 g ( x) = 1 (x ? 3)2 2 ? g (k ) ? ? g (k ) k =4 3 5 ? g (k ) = g (0) + g (1) + g (2) + g (3) = 4,5 + 2 + 0,5 + 0 =7 expansion Copyright reserved Please turn over Mathematics/PI 18 NSC – Memorandum DBE/November 2011 k =4 ? g (k ) = g (4) + g (5) = 0,5 + 2 = 2,5 ? g (k ) ? ? g (k ) k =4 3 5 5 k =0 = 7 ? 2,5 = 4,5 7 ? 2,5 answer (4) OR g ( x) = ax 2 + bx + c g (k ) = ak 2 + bk + c g (0) = c g (1) = a + b + c g (2) = 4a + 2b + c g (3) = 9a + 3b + c k =0 ? g (k ) = 14a + 6b + 4c 3 g (4) = 16a + 4b + c ? g (k ) = 41a + 9b + 2c 5 5 (5) = 25a + 9b + c k =4 3 k =0 ? g (k ) ? ? g (k ) = ? 27a ? 3b + 2c k =4 ? 27 a ? 3b + 2c g ( x) = a ( x ? 3)2 + 0 4,5 = a (0 ? 3) 2 + 0 4,5 = 9a 1 2 1 g ( x) = ( x ? 3)2 2 1 2 9 = x ? 3x + 2 2 a= k =0 ? g (k ) ? ? g (k ) = ? 27 a ? 3b + 2c k =4 3 5 g ( x) = 1 (x ? 3)2 2 ?9? ?1? = ? 27? ? ? 3(? 3) + 2? ? ? 2? ?2? = 4,5 answer (4) [14] Copyright reserved Please turn over Mathematics/PI 19 NSC – M emorandum DBE/November 2011 QUESTION 7 7. 1 A = P(1 ? i ) P n = P(1 ? 0,07 ) 2 1 = 0,93n 2 1 log = n log 0,93 2 1 log 2 n= log 0,93 = 9,55 years n OR A = P(1 ? i )n P = P(1 ? 0,07 )n 2 1 = 0,93 n 2 1 log 0,93 = n 2 n = 9,55 yearsP 2 subs into correct formula A= log answer (4) Note: If candidate uses incorrect formula: max 1/4 marks P for A = 2 Note: If candidate interchanges A and P A i. e. uses P = : max 2/4 marks 2 Copyright reserved Please turn over Mathematics/PI 20 NSC – Memorandum DBE/November 2011 7. 2 Radesh: A = P(1 + in ) = 8 550 Bonus = 0,05 ? 6 000 = 300 = 6 000(1 + 0,085 ? 5) A = 6 000 + 8,5% of 6000 ? 5 OR = 6000 + 510 ? 5 = 6000 + 2550 = 8 550 8 550 Received = 8 550 + 300 = R 8 850 Thandi: n A = P(1 + i ) R8 850 ? 0,08 ? = 6 000? 1 + ? 4 ? ? = R 8 915,68 20 n = 20 0,08 i= 4 answer choice made (6) 0,15 1 or or 0,0125 12 80 n = 18 n = 18 7. 3Thandis investment is bigger. Fv = initial deposit with interest + annuity ? ? ? 0,15 ? 18 ? ?1 + ? ? 1? 18 ? 12 ? ? 0,15 ? = 1 000? 1 + ? + 700? ? 0,15 12 ? ? ? ? ? ? 12 ? ? = 1 250,58 + 14 032,33 = R15 282,91 i= OR ? 0,15 ? 1 000? 1 + ? 12 ? ? ? ? 0,15 ? 18 ? ? ? 1 + ? ?1? ? 12 ? 700? ? 0,15 ? ? ? ? 12 ? ? answer (6) 18 Fv = initial deposit with interest + annuity ?18 ? ? ? ? 1 ? ?1 + 0,15 ? ? ? 18 18 ? ? 0. 15 ? 0,15 ? 12 ? ? = 1 000? 1 + ? ? + 700? 1 + 0,15 12 ? 12 ? ? ? ? ? 12 ? ? 0,15 1 or or 0,0125 12 80 n = 18 n = 18 i= 0. 15 ? ? = 1 250,58 + 11220,68? 1 + ? 12 ? ? = 1 250,58 + 14 032,33 = R15 282,91 Copyright reserved 8 ? 0,15 ? 1 000? 1 + ? 12 ? ? ?18 ? ? ? 1 ? ?1 + 0,15 ? ? ? ? 12 ? 700? 0,15 ? ? 12 ? 18 ? ? 18 0. 15 ? ? 1 + 12 ? ? ? answer (6) Please turn over Mathematics/PI 21 NSC – Memorandum DBE/November 2011 OR ? ? 0,15 ? 19 ? ? ? 1 + ? ?1? 18 ? 12 ? ? 0,15 ? Fv = 300? 1 + ? + 700? ? 0,15 12 ? ? ? ? ? ? 12 ? ? = 375,17 + 14 907,74 = R15 282,91 0,15 1 or or 0,0125 80 12 n = 19 (corresponding to 700) n = 18 (corresponding to 300) i= ? 0,15 ? 300? 1 + ? 12 ? ? ? ? 0,15 ? 19 ? ? ? 1 + ? ?1? 12 ? ? 700? ? 0,15 ? ? ? ? 12 ? ? answer (6) [16] 18 QUESTION 8 8. 1 f ? ( x ) = lim = lim f (x + h ) ? f (x ) h h;0 ? 4( x + h )2 ? 4 x 2 h h;0 ( ) Note: Incorrect notation: formula substitution expansion = lim = lim ? 4 x 2 + 2 xh + h 2 + 4 x 2 h h;0 ? 4 x 2 ? 8 xh ? 4h 2 + 4 x 2 h h;0 ( ) no lim written: penalty 2 marks lim written before equals sign: penalty 1 mark Note: A candidate who gives –8x only: 0/5 marks Note: A candidate who omits brackets in the line lim (? 8 x ? 4h ) : h ;0 ? 8 xh ? 4h 2 = lim h h;0 h(? 8 x ? 4h ) = lim h h;0 = lim (? 8 x ? 4h ) h;0 ? 8 x ? 4h answer (5) = ? 8 x NO penalty OR Copyright reserved Please turn over Mathematics/PI 22 NSC – Memorandum DBE/November 2011 f ( x ) = ? 4 x 2 f ( x + h) = ? 4( x + h) 2 = ? 4 x 2 ? xh ? 4h 2 f ( x + h) ? f ( x) = ? 8 xh ? 4h 2 ? 8 xh ? 4h 2 f ? ( x) = lim h h;0 h ( ? 8 x ? 4 h ) = lim h h;0 = lim (? 8 x ? 4h) h;0 substitution expansion formula ? 8 x ? 4h = ? 8 x 8. 2. 1 y= 3 x2 ? 2x 2 3 1 = x ? 1 ? x 2 2 2 answer (5) 3 ? 1 x 2 ? Note: Incorrect notation in 8. 2. 1 and/or 8. 2. 2: Penalise 1 mark 3 dy = ? x ? 2 ? x 2 dx 3 =? 2 ? x 2x 8. 2. 2 f ( x) = (7 x + 1) 2 3 ? 2 x 2 (3) ?x = 49 x 2 + 14 x + 1 f ? ( x) = 98 x + 14 f ? (1) = 98(1) + 14 = 112 multiplication 98 x 14 answer (4) OR f ( x) = (7 x + 1) 2 f ? ( x) = 2(7 x + 1)(7 ) By the chain rule . f ? ( x) = 98 x + 14 f ? (1) = 98(1) + 14 = 112 hain rule answer (4) [12] Copyright reserved Please turn over Mathematics/PI 23 NSC – Memorandum DBE/November 2011 QUESTION 9 9. 1 f ( x ) = ? 2 x 3 + ax 2 + bx + c f ? ( x ) = ? 6 x 2 + 2ax + b = ? 6( x ? 5)( x ? 2) = ? 6 x 2 ? 7 x + 10 ( ) = ? 6 x 2 + 42 x ? 60 2a = 42 a = 21 b = ? 60 f (5) = ? 2(5) + 21(5) ? 60(5) + c 3 2 Note: A candidate who substitutes the values of a, b and c and then checks (by substitution) that T (2; ? 9 ) and S (5;18) lie on the curve: award max 2/7 marks f ? ( x ) = ? 6 x 2 + 2ax + b ? 6( x ? 5)( x ? 2 ) b= –60 2a = 42 f (2) = ? 2(2) + 21(2) ? 60(2) + c OR ? 9 = ? 52 + c c = 43 3 2 18 = ? 5 + c c = 43 subs (5 ; 18) or (2 ; -9) c = 43 (7) Note: A candidate who substitutes the values of a, b and c into the a = 21 ; b = ? 60 ; c = 43 function i. e. gets f ( x) = ? 2 x 3 ? 21x 2 ? 60 x + 43 and then shows by substitution that T (2; ? 9 ) and S (5;18) are on the curve and works out the derivative i. e. gets f ? ( x ) = ? 6 x 2 ? 42 x ? 60 and shows (by substitution into the derivative) that the turning points are at x = 2 and x = 5 (assuming what s/he sets out to prove and proving what is given): award max 4/7 marks as follows: x = 2 from f ? ( x ) = 0 OR subs x = 2 into the derivative and gets 0 x = 5 from f ? x ) = 0 OR subs x = 5 into the derivative and gets 0 substitution of x = 2 in f and gets – 9 substitution of x = 5 in f and gets 18 f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = 0 f ? (5) = 0 6a = 126 OR f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = ? 6(2) + 2a (2) + b 0 = ? 24 + 4a + b 2 b = 24 ? 4a f ? (5) = ? 6(5) 2 + 2a (5) + b 0 = ? 150 + 10a + b 0 = ? 150 + 10a + (24 ? 4a) 0 = ? 126 + 6a 6a = 126 a = 21 b = ? 60 f (5) = ? 2(5) + 21(5) ? 60(5) + c 3 2 Note: If derivative equal to zero is not written: penalize once only b = – 60 f (2) = ? 2(2) + 21(2) ? 60(2) + c 3 2 subs (5 ; 18) or (2 ; -9) c = 43 (7) Please turn over 18 = ? 25 + c c = 43Copyright reserved a = 21 ; b = ? 60 ; c = 43 OR ? 9 = ? 52 + c c = 43 Mathematics/PI 24 NSC – Memorandum DBE/November 2011 OR f (2) = ? 9 i. e. ? 16 + 4a + 2b + c = ? 9 4a + 2b + c = 7 f (5) = 18 i. e. ? 250 + 25a + 5b + c = 18 25a + 5b + c = 268 21a + 3b = 261 f ? (5) = 0 f ? ( x ) = ? 6 x 2 + 2ax + b and f ? (2 ) = 0 OR 4a + b = 24 10a + b = 150 ? 16 + 4a + 2b + c = ? 9 and ? 250 + 25a + 5b + c = 18 f ? ( x ) = ? 6 x 2 + 2ax + b f ? (2) = 0 or f ? (5) = 0 12a + 3b = 72 9a = 189 189 9 a = 21 a= 12(21) + 3b = 72 3b = ? 180 b = ? 60 4(21) + 2(? 60 ) + c = 7 c = 43 4a + 2b + c = 7 OR 30a + 3b = 450 9a = 189 189 a= 9 a = 21 9a = 189 b = – 60 5a + 5b + c = 268 c = 43 subs (5 ; 18) or (2 ; -9) c = 43 (7) subs f ? (1) m tan = ? 24 f(1) = 2 f ? ( x) = ? 6 x + 42 x ? 60 2 OR 25(21) + 5(? 60 ) + c = 268 9. 2 f ? ( x) = ? 6 x + 42 x ? 60 2 m tan = ? 6(1) + 42(1) ? 60 2 = ? 24 3 2 f (1) = ? 2(1) + 21(1) ? 60(1) + 43 =2 Point of contact is (1 ; 2) y ? 2 = ? 24( x ? 1) y = ? 24 x + 26 9. 3 f ? ( x) = ? 6 x + 42 x ? 60 f ( x) = ? 12 x + 42 2 OR y = ? 24 x + c 2 = ? 24(1) + c c = 26 y = ? 24 x + 26 y ? 2 = ? 24( x ? 1) OR y = ? 24 x + 26 f ( x ) = ? 12 x + 42 (5) 0 = ? 12 x + 42 x= 7 2 x= 7 2 (2) OR 2+5 x= 2 Please turn over Copyright reserved Mathematics/PI 25 NSC – MemorandumDBE/November 2011 2+5 2 7 x= 2 OR ? 21 x= 3(? 2 ) 7 = 2 x= QUESTION 10 y x= 7 2 (2) x= ? 21 3(? 2 ) 7 x= 2 (2) [14] ?4 0 1 x y = f /(x) 10. 1 x-value of turning point: ? 4 +1 x= 2 3 =? 2 3 ? 3 ? ?x ;gt; ? OR ? x ? ? ? ; ? ? 2 ? 2 ? f has a local minimum at x = ? 4 because: (1; y) f ? 4 xgt;? 3 ? 3 ? OR ? ? ; ? ? 2 ? 2 ? (1) 10. 2 x=†“4 graph (3) f 1 –4 OR f ( x) lt; 0 for x lt; ? 4 , so f is decreasing for x lt; ? 4 . f / ( x) gt; 0 for ? 4 lt; x lt; 1 , so f is increasing for ? 4 lt; x lt; 1 . / i. e. –4 OR Copyright reserved ? f has a local minimum at x = ? 4 x=–4 f / ( x) lt; 0 for x lt; ? 4 f / ( x) gt; 0 for ? lt; x 0 so graph is concave up at x = – 4, so f has a local minimum at x = – 4. x=–4 gradient negative for x lt; ? 4 gradient positive for ? 4 lt; x lt; 1 (3) f ? (? 4) = 0 f (? 4) gt; 0 x=–4 (3) [4] QUESTION 11 11. 1 11. 2 V (0) = 100 ? 4(0) = 100 litres Rate in – rate out = 5 – k l / min answer (1) 5–k –4 units stated once 5 ? k = ? 4 k=9 V ? (t ) = ? 4 l / min 11. 3 5 ? k = ? 4 k = 9 l / min OR (3) (2) Note: Answer only: award 2/2 marks Volume at any time t = initial volume + incoming total – outgoing total 100 + 5t ? kt = 100 ? 4t 5t ? kt = ? 4t 9t ? kt = 0 t (9 ? k ) = 0 At 1 minute from start, t = 1, 9 â€⠀œ k = 0, so k = 9 OR 00 + 5t ? kt = 100 ? 4t k=9 (2) dV = ? 4 , the volume of water in the tank is decreasing by 4 dt litres every minute. So k is greater than 5 by 4, that is, k = 9. Since k=9 (2) [6] Copyright reserved Please turn over Mathematics/PI 27 NSC – Memorandum DBE/November 2011 QUESTION 12 Note: If the wrong inequality 50x + 25y ? 500 is used, candidate wrongly says that there are more learners than available seats. Maximum of 10 marks. 12. 1 x, y ? N x + y ? 15 50 x + 25 y ? 500 y? 8 y ? ? x + 15 OR y ? ?2 x + 20 y? 8 Note: for the inequality’s marks to be awarded, the LHS and the RHS must be correctNote: If candidate gives 50 x + 25 y = 500 : max 5/6 marks x + y ? 15 y ? 8 50 x + 25 y ? 500 (6) 12. 2 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 y x + y ? 15 50 x + 25 y ? 500 y? 8 feasible region (4) Blue buses x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Red buses 12. 3 12. 4. 1 C = 600 x + 300 y (6 ; 8) ; (7 ; 6) ; (8 ; 4) ; (9 ; 2) and (10 ; 0) NOTE: The gradient of the search line is m = ? answer 2 1 (1) 3 marks for all correct solutions 2 marks if only 3 or 4 correct solutions 1 mark if only 1 or 2 correct solutions (3) subs answer (2) 2. 4. 2 12. 5 C = 6(600) + 8(300) = R 6 000 or C = 7(600) + 6(300) = R 6 000 or C = 8(600) + 4(300) = R 6 000 or C = 9(600) + 2(300) = R 6 000 or C = 10(600) + 0(300) = R 6 000 8 red ; 4 blue answer (1) [17] TOTAL: 150 Copyright reserved Please turn over Mathematics/P1 28 NSC – Memorandum DBE/November 2011 QUESTION 12. 2 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 y Blue Buses Red Buses 1 2 3 4 5 6 7 8 x 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Copyright reserved