Social Networks free essay sample

Also according to Lenhart and Madden, â€Å"Adults are much less likely than teens to have a profile on a social network website. About three in ten (35%) adult internet users age 18 and older have a profile on a social networking site like MySpace, Facebook or LinkedIn†(2007). How ever, teenagers are twice as likely as adults in use of these websites. Some others use it because they do not have friends in real life and they wish to make new but somehow they are afraid, and they can write anything they want to say without any fear by using social networks. Social networking technology does affect negatively on people as it causes family problems, it can be dangerous for people and it can change behavior for the worse. Social networking causes many family problems that can affect relations. As a fact, family relations are sensitive and small problems can turn to big issues if they didn’t find the right solutions for them. We will write a custom essay sample on Social Networks or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page These networks take the happy moments that people can spend with family, because children prefer to sit with their computers than family, so it makes distance between the family members. Children prefer to hide their emotions and problems from their own family to tell a stranger about it and it makes them trust strangers not relatives. Why does this happen? Social networking is turning to be the best choice for children and adults to spend time. It could be nothing more than they want some independence from their parents and freedom from parents involved. Some children want to say things to their friends who are the same age and they feel they understand them more than parents. Families are suffering with the problems websites are causing. For example, two girls committed suicide because some abuse on social networks and because of pictures that were spread on Internet from these websites. These stories was in a news website that talked about two girls one committed suicide because of sexting, and the other one because of some abuse on social networks from her friends (ABC News). Other problems we can mention are teenagers got addicted to these social networks and that is a sign of behavior changing negatively. These problems for sure increase with time and affect people who use social networks and every one who is around them. The reasons why these problems are huge is because social networks never stops producing new things that attract teens and adults and encourage them to use them more and more. Adults and teenagers invite their friends and tell them to join these networks, as they are fun. Parents are aware about these problems and of course they are trying their best to find solutions for them. Some research happened to show the opinion of parents who have children with social network profiles, â€Å"Research by YouGov showed a quarter of children aged six to 11 spend an hour a day on social networking such as Facebook and Twitter. But only one-in-10 uses the Internet daily for homework†. Extra parents of older children were distressed concerning their offspring’s grades skidding as a consequence of online distractions (mirror news, 2011). This research shows that parents are really aware and worried about social networks and the negative effects of them on their children, and supports the negative side of using social networks. These social networks can be dangerous and harmful. Teens and adults, who do not care about what will happen, they are not aware of the danger that these etworks can make, are using social networks. â€Å"Though the user’s privacy settings determine whether they can be contacted by strangers or friends, many youth may not realize or use these privacy settings, thus exposing them to online predators† (Wolak, et al, 2007). There are lots of social networks, and people from many different countries use them to find new friends and make online relations. The fact is that, in social networks teens meet wrong people who use them in wrong way. These meetings can also lead to a real life meeting and outing and here the danger increases. Teens who decide to meet new people from Internet and bring them to their real life, may not tell their parents about it, and here the problem takes a wrong direction in many cases. Basically, in social networks teens and adults are more likely to meet these wrong people and believe them. Some people pretend to be friend then divulging personal information and then they can enter others profile, they maybe send messages to other people about anything they that will damage their computers. These fraud people can send viruses using others name and spread it on Internet. Fake identities are also a big problem happens in this social networking world. Some make fake identities to get attention and it is not a problem usually, problem begins when they spread wrong information about this person that harms someone else. â€Å"Their motives could just be to have a little nameless fun but they are extra probable to be sinister, like instituting phony friendships that lead to face-to-face meetings alongside who-knows-what aftermath, or to drift invitations to mature sites†(scam Busters Organization). They can make advertisements about a famous brand and take money from people and fool them and some may believe. Teens can be fooled quickly because some of them believe everything they see, especially when they put advertisement with pictures and colors that attract teenagers and adults the most. Social networking has many negative effects on teens and adults that lead to big problems. Social networks make people change their behavior. Nowadays, children use social networks more than anything else. Dr. Gallagher believes that â€Å"young teenagers and preteens are the most vulnerable to negative consequences of social media†(2011). Teens and adults change their behavior when they feel that social networking is more important than real life relations. Parents object on their children when they use social networks all day, which leads to put anger in teens and adults so they miss behave with their parents. Even parents may not know how to react and they start shouting on their children, it can also be one of the causes of changing behavior in adults and teenagers. Children turn to be abusers and they got that violent behavior with others. As Ryan (august 26,2011) has noted, â€Å"Teens who use social networking sites are three times as likely to drink alcohol, and five times more likely to smoke tobacco than teens who dont frequent the sites. † When other people post pictures like smoking and having drugs, that effect on children who use these networks and get excited to try this even if they never tried it before. This is a big issue and a reason that makes children change to the worst. They do not respect elders anymore as Socrates pointed out â€Å"  they show disrespect for their elders and love chatter in place of exercise† (2011). Kids are producing up amid a marine of electronic media. It is vital for parents to ponder how this could encounter communal existence, progress, academics, and adjustment, and how parents can assist as a buffer from a little of the possible negative aftermath( Daiz, Y. , Evans, L. , Gallagher,R). Social networks change behavior to worst because teens like their new life and personality on these networks than the real one. People around teens and adults who use social networks can be affected. For example, children can ignore Parents as they find new friends online. These friends share the same age and think the same, so children in this age do not like their parents interfere in what they are doing. Here when children change their behavior and it becomes more violent and disrespectable for people around them. Using social networks more and more can make children lose interest in studies and that is a one face of changing behavior. According to a recent poll, 22% of teenagers log on to their favorite social media site more than 10 times a day†(council on Communications and Media). These research shows that adults and teens who use social networks more than 10 times a day, they will not have a good focus on their studies. That can cause damage to their marks in school. It will lead to their parent anger and will affect on teens and make them feel upset, and also it can cause behavior changing. Som e people may argue that social networks are good to use and they have positive things like connecting with people. They connect people and also give some skills to teenagers in their growing age. They support their argument by saying that â€Å"Talking on a social networking site may also bring a teen’s personality to the forefront, while the focus on a disability lessens â€Å"(Holmquist. J 2007). Some one also might argue that becoming cozy alongside communal networking locations might additionally assistance teens change to a globe that is intentions extra and extra alongside convoluted technologies (2007). They also say â€Å"Middle and high school students are using social media to connect with one another on homework and group projects†(Schurgin. G, Clarke. K). One seeming benefit is that youth can increase their circle of friends and even converse extra frequently alongside spread relations (Holmquist. J 2007). They could maybe show that social networks can be good enough for children to use and make them more interesting to go in and see what they are is. All that may be true; however, we must consider the harm that all these social networks cause and realize how much greater the harm is than their positive effects. Networks can connect people together and strong the relations but not for adults and children, because they are still go to school and can see their friends and connect with them in real life. It is better to talk face to face rather than facing a computer screen without knowing their emotions. Also bulling is a big problem happens on social networks as â€Å"Some research has shown that youth with disabilities are at a greater risk of being bullied, and bullying can also occur on social networking sites†(Holmquist. J 2007). In conclusion, social networking has both sides of negative and positive effects that can be good or harmful for children. Parents should be aware of these social networking activities that their children do and put limits on using it. It will surly reduce the effects of social networks on them. Parents also can take help from some child study centers like NYU child study center that could be useful for them, they will have more idea about social networking and how they work. Parents can also join their children in using these networks by being friendly with them, so children will feel more confortable talking to their mother and father. Networks can be good and bad, it depends on how we want to use it and make it more interesting or make it more scary and harmful. References Mirror News ( 9 may,2011), Parents fear social networking sites are affecting their childrens school grades.. Retrieved from http://www. mirror. co. uk/ Celizic, M. (2009). Her teen committed suicide over ‘sexting’ . Retrieved from http://today. snbc. msn. com/id/29546030/ns/today-parenting_and_family/t/her-teen-committed-suicide-over-sexting/#. T8OOPbBo0Ug Goldman, R. (2010, March 29). Teens indicted after allegedly taunting girl who hanged herself. ABC News. Retrieved from http://abcnews. go. com Marquette,S. (2011), How kids fool their parents on social networks. NBC News. Retrieved from http://www. msnbc. msn. com/ McLoughlin, C. amp; Burgess, J. (2009). Texting, sex ting and social networking among Australian youth and the need for cyber safety education. Australian Association for Research in Education International Education Research Conference 2009 Canberra. Retrieved from AARE, website http://www. aare. edu. au/09pap/mcl091427. pdf Wolak, J. , Mitchell, K. J. , Finkelhor, D. (2007). Does online harassment constitute bullying? An exploration of online harassment by known peers and online-only contacts. Journal of Adolescent Health 41: S51-S58. Daiz, Y. , Evans, L. , Gallagher,R. Anti-Social Networking: how do texting and social media affect our children?. Retrieved from http://www. aboutourkids. org/ Holmquist, J. (n. d. ). Social Networking Sites: Consider the Benefits, Concerns for Your Teenager . Retrieved from University Of Minnesota website: http://ici. umn. edu/ Schurgin. G, Clarke. K, The impact of social media on children, adolescents, and families. Retrieved from http://pediatrics. aappublications. org/ Scam Busters Organization. (n. d. ). the 5 Most Common Social Networking Scams. Retrieved from http://www. scambusters. org/socialnetworking. html

Models of Health Behavior Essays

Models of Health Behavior Essays Models of Health Behavior Paper 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