Human Resource Functions Of The Human Resources Department

Do you remember when you first decided to start a small business. The vision was clear and the excitement drove us from imagination to plan to reality. But, before we knew it, we went from happily wading in the waters of our areas of expertise into the deep and sometimes turbulent waves of the unknown; Human Resources. Human Resource functions include: Payroll, Employment Tax, Recruitment, Hiring, Employee Relations, Termination, Regulatory Compliance and Training to name a few. Each of these functions demanded specific skill sets and experience. They also began to expend valuable time and resources. The reality is that when reviewing a successful business plan there are two major areas to consider; Revenue streams and cost centers.†¦show more content†¦It is definitely money well spent. Be wary of firms that will provide you with free audits since you usually get what you pay for. The audit results should tell you exactly where you stand and specifically provide a HR roadmap to get and keep you on track. Next, ask the most important question: Do YOU have the expertise, time and interest to actually focus on those areas and do them well? Finally, consider the risks and rewards of alternative Human Resource solutions. After twenty five years of providing human resource solutions to hundreds of companies from small start ups to fortune 100 companies, my advice to many of those firms, especially small and start up businesses is to maximize their resources by at least partially outsourcing Human Resources. By outsourcing HR, companies get to focus on what they do best and avoid the hassles and dangers of managing potentially high risk areas. Remember, as an entrepreneur you want to maximize productivity, increase revenue and control costs. Outsourcing Human Resources and other administrative tasks allow management and staff to focus on core business functions. Most importantly, companies receive high end HR services for less than the cost of an administrative staff position. MINIMIZING RISK A critical component of any small business plan is minimizing risk. When reviewing potential Risk Management issues consider this; Regulatory Non-compliance and Discriminatory practices make up the majority

The Plight Of Low Income Individuals And Their Families

Saul Alvarez English 101 The Plight of Low Income Individuals and Their Families According to the PRB article â€Å"U.S. Low-Income Working Families Increasing†, 42.7% of people live below the poverty line, with 32.1% of them living 200% below the poverty line; this group is categorized as the low-income group.(PRB) 59% of low income working families have one or more minority parents. Living in a low-income community limits one’s standard of living through academic gaps, the selection of food available, an ever-present fear of victimization , and a lack of resources available to both children and adults. The differences in academic gaps between children and teenagers from low-income neighborhoods and affluent neighborhoods present themselves in the schools the students attend. A study by the Department of Education showed that students in low-income areas did not have access to the same rigorous courses, and when they did, the resources available to the low-income students was lacking compared to the resources the students in high income neighborhoods had access to. In the low-income areas black and Hispanic students were twice as likely to have teachers with less experience, with one or two years in the profession, compared to schools in affluent neighborhoods with well-to-do white students. Another thing increasing the academic gap is that only 22 percent of local districts reported offering pre-kindergarten or other early learning programs for low-income children.Show MoreRelatedDescription Of The Business Opportunity836 Words   |  4 Pagesin 2012. Among the reasons identified by the large population of uninsured individuals is low income. The low income implies that they have little disposable income, and therefore they spend a little on food and other daily survival necessities (Bailey, 2012). Sixty one percent of the respondents stated that they would seek insurance services if a company offered them at affordable costs that are relative to their income. Only 1.5% of the uninsured population perceived that they did not require insuranceRead MoreThe Leader Of The Field Of Healthcare1271 Words   |  6 Pagescharacteristics of leaders have been endless over a period and around the world. Leaders are individuals with vision and he/her view the future different from the commonly accepted view. A leader can influence change and communicate his/her vision to other staff and gain the support, and acceptance in order to implementing the vision. Leaders have established values which offer a basis for vision and passion to attain individual and organizational g oals. Influence is vital to attaining change and to be groundedRead MoreA Study Of The Development, Structure, And Functioning Of Human Society909 Words   |  4 Pagestheories in sociology today, says that society is a unified whole that functions because of the contributions of its separate structures: family, education, politics, and the economy (social institutions). Fortunately, only with the use of two social institutions from this theory and through culture can we understand William Julius Wilson’s article the â€Å"Economic Plight of Inner-City Black Males† from More than Just Race: Being Black and Poor in the Inner City where he addresses the question, why are soRead MoreMinimum Wage Does Not Impact The Poverty Rate600 Words   |  3 Pagesworkers below the poverty rate and reducing the opportunity for jobs. Protection Since the years after 1938, the Fair Labor Act has enabled workers to maintain a minimum level of income. The forty-hour workweek has provided a limit to employers but also in good economic times the ability to work multiple jobs to offset low-skilled workers living expenses. Today, single mothers, elderly and uneducated are the protected society. Consequently the protection comes at a cost, the minimum wage is also affectingRead MoreIdeal Healthcare System1680 Words   |  7 Pageshealthcare system, the number of persons who lack health insurance approaches 47 million. Lack of health insurance has been associated with limited or no access to comprehensive medical services, worse health outcomes, financial catastrophe for many families, and financial challenges for many service providers. The U.S. Census Bureau conducts an annual social and economic supplemental survey each March- the widely cited Current Population and economic survey (CPS) - that asks respondents about theirRead MoreWhat Does It Mean For A Minority?1584 Words   |  7 PagesWithin most societies there are divisions between the different groups that is composed of, but what is ironic about this is that a society essentially is a unit of individuals who coexist for mutual benefit. Within the constructs of these divisions there are many individuals who feel that they are inherently disadvantaged because they are a part of the minority of their respective society. In terms of their economic stature it can be easily argued that they are disadvantaged because; those of theRead MoreEconomic Inequality And Political Inequality1647 Words   |  7 PagesEconomic inequality, also known as income inequality, is the interval between the rich and the poor. Economic inequality refers to how the total wealth in the United States is distributed among people in a social class. It is needed and it is important but due to the major gap difference, it affects the Democratic Party and in addition, it also af fects Americans because they do not understand the actual wealth distribution. It is a major issue in the United States because it affects other economicRead MoreEconomic Poverty970 Words   |  4 Pagesattention to the plight of the poor when there isnt a natural disaster to put them in the headlines. Poor children in the United States are more likely to be White than Black or Latino and are more likely to live in a rural or suburban area than in an inner city (Poverty and Schooling in the U.S., 2004). There is also an economic case for reducing child poverty and any other form of it. When children grow up in poverty, they are somewhat more likely than non-poor children to have low earnings as adultsRead MoreAddressing the Challenges Faced by Young Parents1814 Words   |  7 PagesParents†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦..†¦...2 1. Education†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.†¦3 2. Stress†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦...3 3. Finances†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦..4 4. Health†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦..4 IV. Addressing the Challenges faced by Young Parents†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦4 1. Support from Family and Friends†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦...5 2. Adequate Medical and Psychological Care†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦5 3. Role of the Government†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.6 V. Conclusion†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦.†¦6 VI. References†¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦....7 Introduction In addition to the challengesRead MoreEssay on Poverty and Personal Choice1378 Words   |  6 Pagesno simple solution to resolve the plight of these often forgotten citizens. Most of us associate poor as being in a class below the poverty line. In fact there are many levels of poverty ranging from those with nothing, to those with enough to survive but too little to move up. I believe many of our nation’s poor are so by their own doing. I will share observations and personal experiences to support the argument that being poor often is a result of individual choice. One needs merely inspiration

Edexcel Maths Fp2 Paper Free Essays

Paper Reference(s) 6667 Edexcel GCE Further Pure Mathematics FP1 Advanced Level Specimen Paper Time: 1 hour 30 minutes Materials required for examination Answer Book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Items included with question papers Nil Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus candidates may NOT use calculators such as the Texas Instruments TI-89, TI-92, Casio CFX-9970G, Hewlett Packard HP 48G. Instructions to Candidates In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP1), the paper reference (6667), your surname, initials and signature. We will write a custom essay sample on Edexcel Maths Fp2 Paper or any similar topic only for you Order Now When a calculator is used, the answer should be given to an appropriate degree of accuracy. Information for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. Full marks may be obtained for answers to ALL questions. This paper has eight questions. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit. This publication may only be reproduced in accordance with London Qualifications Limited copyright policy. Edexcel Foundation is a registered charity.  ©2003 London Qualifications Limited 1. Prove that a (r r =1 n 2 – r -1 = ) 1 (n – 2)n(n + 2) . 3 (5) 2. 1 f ( x ) = ln x – 1 – . x (a) Show that the root a of the equation f(x) = 0 lies in the interval 3 lt; a lt; 4 . (2) (b) Taking 3. 6 as your starting value, apply the Newton-Raphson procedure once to f(x) to obtain a second approximation to a. Give your answer to 4 decimal places. (5) 3. Find the set of values of x for which 1 x gt; . x -3 x -2 (7) 4. f ( x ) ? 2 x 3 – 5 x 2 + px – 5, p I ?. The equation f (x) = 0 has (1 – 2i) as a root. Solve the equation and determine the value of p. (7) 5. (a) Obtain the general solution of the differential equation dS – 0. 1S = t. dt (6) (b) The differential equation in part (a) is used to model the assets, ? S million, of a bank t years after it was set up. Given that the initial assets of the bank were ? 200 million, use your answer to part (a) to estimate, to the nearest ? illion, the assets of the bank 10 years after it was set up. (4) 2 6. The curve C has polar equation r 2 = a 2 cos 2q , -p p ? q ? . 4 4 (a) Sketch the curve C. (2) (b) Find the polar coordinates of the points where tangents to C are parallel to the initial line. (6) (c) Find the area of the region bounded by C. (4) 7. Given that z = -3 + 4i and zw = -14 + 2i, find (a) w in the form p + iq where p and q are real, (4) (b) the modulus of z and the argument of z in radians to 2 decimal places (4) (c) the values of the real constants m and n such that mz + nzw = -10 – 20i . (5) 3 Turn over 8. (a) Given that x = e t , show that (i) y dy = e -t , dx dt 2 dy o d2 y – 2t ? d y c 2 – ?. =e c 2 dt ? dx o e dt (ii) (5) (b) Use you answers to part (a) to show that the substitution x = e t transforms the differential equation d2 y dy x 2 2 – 2x + 2y = x3 dx dx into d2 y dy – 3 + 2 y = e 3t . 2 dt dt (3) (c) Hence find the general solution of x2 d2 y dy – 2x + 2y = x3. 2 dx dx (6) END 4 Paper Reference(s) 6668 Edexcel GCE Further Pure Mathematics FP2 Advanced Level Specimen Paper Time: 1 hour 30 minutes Materials required for examination Answer Book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Items included with question papers Nil Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus candidates may NOT use calculators such as the Texas Instruments TI 89, TI 92, Casio CFX-9970G, Hewlett Packard HP 48G. Instructions to Candidates In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP2), the paper reference (6668), your surname, initials and signature. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Information for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. Full marks may be obtained for answers to ALL questions. This paper has eight questions. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit. This publication may only be reproduced in accordance with London Qualifications Limited copyright policy. Edexcel Foundation is a registered charity.  ©2003 London Qualifications Limited 1. The displacement x of a particle from a fixed point O at time t is given by x = sinh t. 4 At time T the displacement x = . 3 (a) Find cosh T . (2) (b) Hence find e T and T. (3) 2. Given that y = arcsin x prove that (a) dy = dx (1 – x ) 2 1 , (3) (b) (1 – x 2 ) d2 y dy -x = 0. 2 dx dx (4) Figure 1 3. y P(x, y) s A y O x Figure 1 shows the curve C with equation y = cosh x. The tangent at P makes an angle y with the x-axis and the arc length from A(0, 1) to P(x, y) is s. (a) Show that s = sinh x. (3) (a) By considering the gradient of the tangent at P show that the intrinsic equation of C is s = tan y. 2) (c) Find the radius of curvature r at the point where y = p . 4 (3) S 4. I n = o x n sin x dx. p 2 0 (a) Show that for n ? 2 ?p o I n = nc ? e 2o n -1 – n(n – 1)I n – 2 . (4) (4) (b) Hence obtain I 3 , giving your answers in terms of p. 5. (a) Find ? v(x2 + 4) dx. (7) The curve C has equation y 2 – x 2 = 4. (b) Use your answer to part (a) to fi nd the area of the finite region bounded by C, the positive x-axis, the positive y-axis and the line x = 2, giving your answer in the form p + ln q where p and q are constants to be found. (4) Figure 2 6. y O 2pa x The parametric equations of the curve C shown in Fig. are x = a(t – sin t ), y = a(1 – cos t ), 0 ? t ? 2p . (a) Find, by using integration, the length of C. (6) The curve C is rotated through 2p about Ox. (b) Find the surface area of the solid generated. (5) 7 7. (a) Using the definitions of sinh x and cosh x in terms of exponential functions, express tanh x in terms of e x and e – x . (1) (b) Sketch the graph of y = tanh x. (2) 1 ? 1 + x o lnc ?. 2 e1 – x o (c) Prove that artanh x = (4) (d) Hence obtain d (artanh x) and use integration by parts to show that dx o artanh x dx = x artanh x + 1 ln 1 – x 2 + constant. 2 ( ) (5) 8. The hyperbola C has equation x2 y2 = 1. a2 b2 (a) Show that an equation of the normal to C at P(a sec q , b tan q ) is by + ax sin q = a 2 + b 2 tan q . (6) ( ) The normal at P cuts the coordinate axes at A and B. The mid-point of AB is M. (b) Find, in cartesian form, an equation of the locus of M as q varies. (7) END U Paper Reference(s) 6669 Edexcel GCE Further Pure Mathematics FP3 Advanced Level Specimen Paper Time: 1 hour 30 minutes Materials required for examination Answer Book (AB16) Graph Paper (ASG2) Mathematical Formulae (Lilac) Items included with question papers Nil Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus candidates may NOT use calculators such as the Texas Instruments TI 89, TI 92, Casio CFX 9970G, Hewlett Packard HP 48G. Instructions to Candidates In the boxes on the answer book, write the name of the examining body (Edexcel), your centre number, candidate number, the unit title (Further Pure Mathematics FP3), the paper reference (6669), your surname, initials and signature. When a calculator is used, the answer should be given to an appropriate degree of accuracy. Information for Candidates A booklet ‘Mathematical Formulae and Statistical Tables’ is provided. Full marks may be obtained for answers to ALL questions. This paper has eight questions. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the Examiner. Answers without working may gain no credit. This publication may only be reproduced in accordance with London Qualifications Limited copyright policy. Edexcel Foundation is a registered charity.  ©2003 London Qualifications Limited 1. y = x 2 – y, y = 1 at x = 0 . dx y – y0 ? dy o Use the approximation c ?  » 1 with a step length of 0. 1 to estimate the values of y h e dx o 0 at x = 0. 1 and x = 0. 2, giving your answers to 2 significant figures. (6) 2. (a) Show that the transformation w= z -i z +1 maps the circle z = 1 in the z-plane to the line w – 1 = w + i in the w-plane. (4) The region z ? 1 in the z-plane is mapped to the region R in the w-plane. (b) Shade the region R on an Argand diagram. (2) 3. Prove by induction that, all integers n, n ? 1 , ar gt; 2 n r =1 n 1 2 . (7) 4. dy d2 y dy +y = x, y = 0, = 2 at x = 1. 2 dx dx dx Find a series solution of the differential equation in ascending powers of (x – 1) up to and including the term in (x – 1)3. (7) 5. ? 7 6o A=c c 6 2? . ? e o (a) Find the eigenvalues of A. (4) (a) Obtain the corresponding normalised eigenvectors. (6) NM 6. The points A, B, C, and D have position vectors a = 2i + k , b = i + 3j, c = i + 3 j + 2k , d = 4 j + k respectively. (a) Find AB ? AC and hence find the area of triangle ABC. (7) (b) Find the volume of the tetrahedron ABCD. (2) (c) Find the perpendicular distance of D from the plane containing A, B and C. (3) 7. ? 1 x – 1o c ? 5 A( x) = c 3 0 2 ? , x ? 2 c1 1 0 ? e o (a) Calculate the inverse of A(x). (8) ? 1 3 – 1o c ? B = c3 0 2 ? . c1 1 0 ? e o ? po c ? The image of the vector c q ? when transformed by B is cr? e o (b) Find the values of p, q and r. (4) ? 2o c ? c 3? . c 4? e o 11 8. (a) Given that z = e iq , show that zp + 1 = 2 cos pq , zp where p is a positive integer. (2) (b) Given that cos 4 q = A cos 4q + B cos 2q + C , find the values of the constants A, B and C. (7) The region R bounded by the curve with equation y = cos 2 x, rotated through 2p about the x-axis. (c) Find the volume of the solid generated. (6) p p ? x ? , and the x-axis is 2 2 END NO EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 1. Scheme Marks M1 B1 a (r r =1 n 2 – r -1 = a r2 – a r – a1 r =1 r =1 r =1 ) n n n ? n o c a1 = n ? e r =1 o = = = n (n + 1)(2n + 1) – ? 1 on(n + 1) – n c ? 6 e 2o n 2n 2 – 8 6 [ ] M1 A1 A1 (5) (5 marks) 1 n(n – 2 )(n + 2 ) 3 2. (a) f ( x) = ln x – 1 – 1 x f (3) = ln 3 – 1 – 1 = -0. 2347 3 f (4) = ln 4 – 1 – 1 = 0. 1363 4 f (3) and f (4) are of opposite sign and so f ( x ) has root in (3, 4) (b) x 0 = 3. 6 f ? (x ) = 1 1 + x x2 M1 A1 (2) M1 A1 f ? (3. 6 ) = 0. 354 381 f (3. 6) = 0. 003 156 04 Root  » 3. – f (3. 6) f ? (3. 6) M1 A1 ft A1 (5) (7 marks)  » 3. 5911 13 EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 3. Scheme x x x 2 – 3x + 3 1 1 gt; ? gt;0 ? gt;0 x-3 x-2 x-3 x-2 (x – 3)(x – 2 ) Marks M1 A1 B1 B1 Numerator always positi ve Critical points of denominator x = 2, x = 3 x lt; 2 : den = (- ve)(- ve) = + ve 2 lt; x lt; 3 : den = (- ve)(+ ve) = – ve 3 lt; x : den = (+ ve)(+ ve) = + ve M1 A1 A1 (7) (7 marks) Set of values x lt; 2 and x gt; 3 {x : x lt; 2} E {x : x gt; 3} 4. If 1 – 2i is a root, then so is 1 + 2i B1 M1 A1 M1 A1 ft A1 A1 (7) x – 1 + 2i )(x – 1 – 2i ) are factors of f(x) so x 2 – 2 x + 5 is a factor of f (x) f ( x ) = x 2 – 2 x + 5 (2 x – 1) Third root is 1 2 ( ) and p = 12 (7 marks) 5. (a) dS – (0. 1)S = t dt – ( 0. 1)dt Integrating factor e o = e -(0. 1)t M1 d Se – (0. 1)t = te – (0. 1)t dt Se – (0. 1)t = o te – (0. 1)t dt = -10te – (0. 1)t – 100e – (0. 1)t + C [ ] A1 A1 M1 A1 A1 (6) S = Ce (0. 1)t – 10t – 100 (b) S = 200 at t = 0 ? 200 = C – 100 i. e. C = 300 S = 300e (0. 1)t – 10t – 100 M1 A1 At t = 10, S = 300e – 100 – 100 = 615. 484 55 M1 A1 ft (4) (10 marks) Assets ? 615 million NQ EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 6. (a) l Scheme Marks q B1 (Shape) B1 (Labels) (2) (b) Tangent parallel to initial line when y = r sin q is stationary Consider therefore d 2 a cos 2q sin 2 q dq ( ) M1 A1 = -2 sin 2q sin 2 q + cos 2q (2 sin q cos q ) =0 2 sin q [cos 2q cos q – sin 2q sin q ] = 0 sin q ? 0 ? cos 3q = 0 ? q = p -p or 6 6 M1 A1 o ? ? o ? 1 p o? 1 -p Coordinates of the points c c a, ? c a, ? c 6 6 oe 2 e 2 A1 A1 (6) 1 o4 2 1 2o4 (c) Area = o r dq = a o cos 2q dq 2 o -p 2 o -p 4 4 p p M1 A1 a2 a2 1 2 e sin 2q u = a e = [1 – (- 1)] = 2 e 2 u -4p 4 2 u p 4 M1 A1 (4) (12 marks) 15 EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 7. (a) z = -3 + 4i, zw = -14 + 2i Scheme Marks w= = = – 14 + 2i (- 14 + 2i )(- 3 – 4i ) = (- 3 + 4i )(- 3 – 4i ) – 3 + 4i M1 A1 A1 A1 M1 A1 M1 A1 M1 A1 A1 M1 A1 (5) (13 marks) (4) (42 + 8) + i(- 6 + 56) 9 + 16 50 + 50i = 2 + 2i 25 (4) (b) z = (3 2 + 42 = 5 4 = 2. 21 3 ) arg z = p – arctan (c) Equating real and imaginary parts 3m + 14n = 10, 4m + 2n = -20 Solving to obtain m = -6, n = 2 NS EDEXCEL FURTHER PURE MATHEMATICS FP1 (6667) SPECIMEN PAPER MARK SCHEME Question number 8. (a)(i) x = et , dy dy dy dt = = e -t dt dx dt dx Scheme Marks M1 A1 ? dx t o c =e ? e dt o (ii) d 2 y dt d e – t dy u e = dt u dx 2 dx dt e u e M1 e dy d2 yu = e – t e – e -t + e -t 2 u dt dt u e e d 2 y dy u = e – 2t e 2 – u dt u e dt (b) x2 2t A1 A1 (5) d2 y dy – 2x + 2y = x3 2 dx dx – 2t e e e d 2 y dy u t – t dy + 2 y = e 3t e 2 – u, – 2e e dt u dt e dt M1 A1, A1 (3) d2 y dy – 3 + 2 y = e 3t 2 dt dt (c) Auxiliary equation m 2 – 3m + 2 = 0 (m – 1)(m – 2) = 0 Complementary function y = Ae t + Be 2t e 3t 1 Particular integral = 2 = e 3t 3 – (3 ? 3) + 2 2 General solution y = Ae t + Be 2t + 1 e 3t 2 = Ax + Bx 2 + 1 x 3 2 M1 A1 M1 A1 M1 A1 ft 6) (14 marks) 17 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 1. cosh 2 T = 1 + sinh 2 T = 1 + 16 25 = 9 9 Scheme Marks M1 A1 (2) M1 A1 A1 ft (3) cosh T =  ± 5 5 = since cosh T gt; 1 3 3 4 5 + =3 3 3 e T = cosh T + sinh T = Hence T = ln 3 2. (5 mar ks) (a) y = arcsin x ? sin y = x M1 cos y dy =1 dx dy 1 1 = = dx cos y 1- x2 M1 A1 (3) (b) d2 y dx 2 = – 1 1- x2 2 ( ) -3 2 (- 2 x ) M1 A1 = x 1- x2 ( ) -3 2 (1 – x ) 2 d2 y dy -x = 1 – x2 x 1 – x2 2 dx dx ( )( ) -3 2 – x 1- ( 1 2 -2 x ) =0 M1 A1 (4) (7 marks) NU EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 3. Scheme x 0 Marks (a) s=o e ? dy o 2 u 2 e1 + c ? u dx e e dx o u u e dy = sinh x dx 1 y = cosh x, x B1 s = o 1 + sinh 2 x 2 dx 0 [ ] 1 = o cosh x dx = sinh x 0 x M1 A1 (3) (b) Gradient of tangent dy = tan y = sinh x = s dx s = tan y M1 A1 M1 A1 A1 (2) (c) r= ds = sec2 y dy At y = p , r = sec2 p = 2 4 4 (3) (8 marks) 19 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 4. Scheme I n = o x n sin x dx = x n (- cos x ) p 2 0 Marks (a) [ ] p 2 0 – o 2 nx n -1 (- cos x )dx 0 p M1 A1 i i = 0 + ni x n -1 sin x i i [ -o 0 p 2 p 2 0 = n (p ) 2 [ n -1 – (n – 1)I n -2 n -1 ] u i (n – 1)x n- 2 sin x dxy i ? A1 So I n = n(p ) 2 2 – n(n – 1)I n -2 A1 (4) (b) ?p o I 3 = 3c ? – 3. 2 I 1 e2o I 1 = o x sin x dx = [x(- cos x )] + o cos x dx 0 p 2 0 p 2 p 2 0 M1 = [sin x ] = 1 0 p 2 A1 3p ? p o I 3 = (3)c ? – 6 = -6 4 e 2o 2 2 M1 A1 (4) (8 marks) OM EDEXCEL FURTHER PU RE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 5. Scheme x = 2 sinh t Marks B1 (a) (x 2 + 4 = 4 sinh 2 t + 4 ) ( 2 ) 1 2 = 2 cosh t dx = 2 cosh t dt I =o (x + 4 dx = 4 o cosh 2 t dt ) M1 A1 = 2 o (cosh 2t + 1) dt = sinh 2t + 2t + c M1 A1 M1 A1 ft (7) = 1 x 2 (x 2 2 ? xo + 4 + 2arsinh c ? + c e 2o 2 0 ) (b) Area = o y dx = o 0 (x ) 2 + 4 dx 2 ) M1 e1 =e x e2 = 2 ( xu u e x + 4 u + e 2arsinh u 2u0 u0 e 2 2 1 2 2 8 + 2arsinh (1) 2] = 2 2 + ln 3 + 2 A1 2 + 2 ln[1 + ( 2 ) M1 A1 (4) (11 marks) 21 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 6. Scheme 2p 0 Marks (a) s=o e e x + y u dt e u e u  · 2 1  · u2 2 dy  · dx  · = x = a (1 – cos t ); = y = a sin t dt dt s=o 2p 0 M1 A1; A1 2p 0 a (1 – cos t ) + sin 2 t 2 dt = a o 2 p ? 2 sin c 0 2p [ ] 1 [2 – 2 cos t ]2 dt M1 A1, A1 ft (6) 1 = 2a o e ? t ou to ? t , = -4a ecosc ? u = 8a e 2o e e 2 ou 0 1 o2 (b) s = 2p o = 2p o 2p 0 ? yc x + y ? dt c ? e o 1 22 2p  · 2  · 2 2p 0 a 2 (1 – cos t ) 2 dt M1 A1 M1 3 = 8pa 2 o 0 2p 0 ?to sin 3 c ? dt e 2o = 8pa 2 o 2 e t 2 ? t ou e1 – cos c 2 ? u sin 2 dt e ou e 2p 64pa 2 t 2 e 3 t u = 8pa e – 2 cos + cos u = 2 3 2u0 3 e A1 A1 ft (5) (11 mar ks) OO EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 7. Scheme tanh x = sinh x e x – e – x = cosh x e x + e – x B1 Marks (1) (a) (b) 1 y 0 x -1 B1 B1 (2) (c) artanhx = z ? tanh z = x e z – e-z e z + e -z =x M1 A1 e z – e-z = x e z + e-z ( ) 1 – x )e z = (1 + x )e – z e2z = z= 1+ x 1- x 1 ? 1 + x o lnc ? = artanh x 2 e1- x o M1 A1 M1 A1 1 x dx (4) (d) dz 1 ? 1 1 o 1 = c + ? = dx 2 e 1 + x 1 – x o 1 – x 2 o artanh x dx = (x artanh x ) – o 1 – x = (x artanh x ) + 2 M1 A1 A1 (5) 1 ln 1 – x 2 + constant 2 ( ) (10 marks) 23 EDEXCEL FURTHER PURE MATHEMATICS FP2 (6668) SPECIMEN PAPER MARK SCHEME Question Number 8. Scheme x2 y2 =1 a2 b2 2 x 2 y dy =0 a 2 b 2 dx Marks (a) M1 A1 M1 A1 dy 2 x b 2 b 2 a sec q b = 2 = 2 = dx a 2 y a b tan q a sin q Gradient of normal is then a sin q b a Equation of normal: ( y – b tan q ) = – sin q (x – a sec q ) b x si n q + by = a 2 + b 2 tan q (b) M: A normal cuts x = 0 at y = B normal cuts y = 0 at x = ( ) M1 A1 (6) (a 2 + b2 tan q b ) M1 A1 (a = ( ) a2 + b2 tan q a sin q + b2 a cos q 2 ) A1 e a2 + b2 u a2 + b2 sec q , tan q u Hence M is e 2b e 2a u Eliminating q sec 2 q = 1 + tan 2 q 2 2 ( ) M1 M1 e 2aX u e 2bY u =1+ e 2 e u u ea2 + b2 u ea + b2 u A1 2 4a 2 X 2 – 4b 2Y 2 = a 2 + b 2 [ ] A1 (7) (15 marks) OQ EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number 1. Scheme Marks ? dy o x 0 = 0, y 0 = 1, c ? = 0 – 1 = -1 e dx o 0 ? dy o y1 – y 0 = hc ? ? y1 = 1 + (0. 1)(- 1) = 0. e dx o 0 ? dy o x1 = 0. 1, y1 = 0. 9, c ? e dx o 1 ? dy o y 2 = y1 + hc ? e dx o 1 = (0. 1) – 0. 9 2 B1 M1 A1 ft A1 = -0. 89 = 0. 9 + (0. 1)(- 0. 89) = 0. 811  » 0. 81 z -i ? w( z + 1) = ( z – i ) z +1 M1 A1 (6) (6 marks) 2. (a) w= z (w – 1) = -i – w z= -i-w w -1 -i-w =1 w -1 M1 A1 z =1? i. e. w – 1 = w + i (b) z ? 1? w + i ? w -1 M1 A1 (4) B1 (line) B1 (shading) (2) (6 marks) OR qiea=liEe EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number 3. Scheme For n = 1, LHS =1, RHS = So result is true for n = 1 Assume true for n = k. Then k +1 r =1 Marks 1 2 M1 A1 r gt; 2 k2 + k +1 = = 1 2 1 k + 2k + 1 + 2 2 1 (k + 1)2 + 1 2 2 1 M1 A1 ( ) M1 A1 A1 (7) (7 marks) If true for k, true for k+1 So true for all positive integral n d2 y dy dy +y = x, y = 0, = 2 at x = 1 2 dx dx dx d2 y = 0 +1=1 dx 2 Differentiating with respect to x d 3 y ? dy o d2 y + c ? + y 2 =1 dx 3 e dx o dx 2 4. B1 M1 A1 d3 y dx 3 = -(2) + 0 + 1 = -3 2 A1 x =1 By Taylor’s Theorem y = 0 + 2(x – 1) + = 2(x – 1) + 1 1 2 3 1( x – 1) + (- 3)(x – 1) 3! 2! M1 A1 A1 (7) (7 marks) 1 (x – 1)2 – 1 (x – 1)3 2 2 OS EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number 5. Scheme A – lI = 0 Marks (a) (7 – l ) 6 6 =0 (2 – l ) M1 A1 (7 – l )(2 – l ) – 36 = 0 l2 – 9l + 14 – 36 = 0 l2 – 9l – 22 = 0 (l – 11)(l + 2) = 0 ? l1 = -2, l2 = 11 (b) l = -2 Eigenvector obtained from M1 A1 (4) 6 o ? x1 o ? 0 o ? 7 – (- 2) c ? c ? =c ? c 6 2 – (- 2)? c y 1 ? c 0 ? e oe o e o 3Ãâ€"1 + 2 y1 = 0 ? 2o 1 ? 2o c ? e. g. c ? normalised c – 3? c ? 13 e – 3o e o M1 A1 M1 A1 ft ? – 4 6 o ? x2 o ? 0o c ? c ? =c ? l = 11 c ? c ? c ? e 6 – 9o e y2 o e 0o – 2 x2 + 3 y 2 = 0 ? 3o 1 ? 3o c ? e. g. c ? normalised c 2? c ? 13 e 2 o e o A1 A1 ft (6) (10 marks) 27 EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number 6. (a) AB = (- 1, 3, – 1) ; AC = (- 1, 3, 1) . i j k Scheme Marks M1 A1 AB ? AC = – 1 3 – 1 -1 3 1 = i (3 + 3) + j (1 + 1) + k (- 3 + 3) = 6i + 2 j M1 A1 A1 Area of D ABC = = 1 AB ? AC 2 1 36 + 4 = 10 square units 2 = = = 1 AD . AB ? AC 6 M1 A1 ft (7) (b) Volume of tetrahedron ( ) M1 A1 (2) 1 – 12 + 8 6 2 cubic units 3 ? ?  ® ? ? ® (c) Unit vector in direction AB ? AC i. e. perpendicular to plane containing A, B, and C is 1 n= (6i + 2 j) = 1 (3i + j) 10 40 M1 p = n ? AD = 1 10 (3i + j) ? (- 2i + 4 j) = 1 2 -6+4 = units. 10 10 M1 A1 (3) (12 marks) OU EDEXCEL FURTHER MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number Scheme ? 1 x – 1o c ? A( x ) = c 3 0 2 ? c1 1 0 ? e o 3 o ? – 2 2 c ? Cofactors c – 1 1 x – 1? c 2 x – 5 – 3x ? e o Determinant = 2 x – 3 – 2 = 2 x – 5 ? – 2 1 c A (x ) = c 2 2x – 5 c e 3 -1 Marks 7. (a) M1 A1 A1 A1 M1 A1 M1 A1 (8) -1 1 (x – 1) 2x o ? -5 ? – 3x ? o (b) ? 2o ? po ? – 2 – 1 6 o ? 2o c ? 1c c ? ?c ? -1 1 – 5? c 3? c q ? = B c 3? = c 2 c 4? 1 c 3 cr? 2 – 9? c 4? e o e o e oe o M1 A1 ft M1 A1 = (17, – 13, – 24 ) (4) (12 marks) 29 EDEXCEL FURTHER PURE MATHEMATICS FP3 (6669) SPECIMEN PAPER MARK SCHEME Question Number Scheme zp + Marks 8. (a) 1 1 = e ipq + ipq p z e = e ipq + e -ipq = 2 cos pq ( ) M1 A1 (2) (b) By De Moivre if z = e iq zp + 1 = 2 cos pq zp 4 1o ? 4 p = 1 : (2 cos q ) = c z + ? zo e M1 A1 M1 A1 1 1 1 1 = z 4 + 4 z 3 . + 6 z 2 2 + 4 z. 3 + 4 z z z z 1 o ? 1 o ? = c z 4 + 4 ? + 4c z 2 + 2 ? + 6 z o e z o e = 2 cos 4q + 8 cos 2q + 6 M1 A1 3 8 cos 4 q = 1 cos 4q + 1 cos 2q + 8 2 A1 ft (7) (c) V =p o p 2 p 2 p 2 p 2 y dx = p o 2 p 2 p 2 cos 4 x dx =p o 3o 1 ? 1 c cos 4q + cos 2q + ? dq 8o 2 e8 p M1 A1 ft 1 3 u 2 e1 = p e sin 4q + sin 2q + q u 4 8 u-p e 32 2 M1 A1 ft 3 = p2 8 M1 A1 (6) (15 marks) PM How to cite Edexcel Maths Fp2 Paper, Papers