History And Evolution Of Sitcoms

History And Evolution Of Sitcoms In the field of entertainment, everybody could use a good laugh and this is where the comedy genre comes in. As the name states, its purpose is to bring humor and laughter to the audience. Comedies come in many formats in movies and television. In television, one of the most common genre is the situation comedy or sitcom, for short. As the name states the plot is centered on a particular situation set in a typical setting such as a home or workplace. A situation comedy features a regular cast of characters plus recurring ones who would appear in subsequent episodes as well as special guest stars. There are sitcoms that are aired performed before a live studio audience, making it similar to a theatrical play. One can tell it is live whenever a special guest star would appear as the audience would cheer enthusiastically. Another distinctive feature of the sitcom is the laugh track or what is called canned laughter which is played every time a hilarious scene unfolds. What makes sitcoms different from stand up comedy and sketch comedy is that they have a storyline and this essentially makes it a comedic drama; and as mentioned before, the setting is usually centered on family, workplace, or a group of friends as the principal characters or mainstays. Sitcoms came about when the television was introduced and this enabled audiences to return to a certain program if they like it. As a result of this (initial) trend, the performers who have key roles would become mainstays and the situations would remain the same to enable audiences to be familiar with them. Even animated shows, also adapt the sitcom format to cater to a specific audience as well, not merely children. Another feature of the sitcom is that it is often character driven and naturally, running gags often develop during a series or season. The plot of a sitcom episode is typical: It starts with the status quo where everything is normal among the characters and then, a disruption will occur, thereby affecting the usual situation and the relationships of the characters, but by the end of the episode, these issues will be settled, the situation will revert to the status quo and it is alls well that ends well until the next episode where the it will happen all over again but in a different plot. History and Evolution: Throughout its history, sitcoms over the years have evolved not only in the performers but also in the plot as well as how humor is delivered. In the early years of sitcom, the most common method of delivery is the slapstick approach of the 1950s and well into the 1960s. The slapstick is characterized by exaggerated violence where the characters appear to be hitting one another with exaggerated sound effects without getting hurt. This is common is earlier comedies such as The Three Stooges. Slapstick also features characters doing unthinkable or crazy stunts or acts or silly things in the scene to the point of making complete fools of themselves. One of the most popular sitcoms of the period was I Love Lucy starring the real-life couple of Lucille Ball and Desi Arnaz. This sitcom is one of those that features the slapstick delivery. One example is a scene in one episode where Lucy works in a chocolate factory. The chocolates are being churned out so fast that Lucy had to eat those that could not be packaged. In another episode, Lucy was mashing grapes in a winery and wrestled with another worker in the vat, making a mess out of themselves (Oppenheimer, 4-5). As for the setting, it centers on a typical couple, Lucy and Ricky (Arnaz) Ricardo. What makes the story interesting is that Lucy is not content in being a plain housewife while her husband works as a bandmaster in a club. Lucy aspires to have a career and this is the source of the humor. Besides these madcap adventures and misadventures, there is also the relationship with their neighbors Fred and Ethel who play the straight couple to their seemingly dysfunctional one to provide balance. Essentially, women would be portrayed as scatterbrained but extremely clever, men would be indignant (like Ricky), and friends or neighbors would be unwitting pawns, accomplices or villians (such as Fred and Ethel). Besides the typical family setting, early sitcoms offer different settings but with similar plots such as Sergeant Bilko, which looks at the humorous side of the military; Car 54 Where Are You? for the police officers and McHales Navy in the US Navy set during World War II to boot. The 1960s added a fantasy touch with sitcoms like I Dream of Jeannie and Bewitched. This was made possible with advancements in the realm of special effects which enabled these magic tricks to be performed. In addition, 1960s sitcoms deviated from the earlier ones in the sense that they were not filmed before live audiences but were filmed instead where the sessions are called tapings. The 1970s and 1980s saw a change in the character makeup of the sitcoms as evidenced by such shows as The Jeffersons, Different Strokes, Barney Miller and Chico and the Man where nonwhites began to share top billing as well yet the plot remains the same. This underscored by the real-world events happening the changes America has been going through by way of racial integration and that nonwhites were not regarded as equals to whites. This is what these shows intend to convey besides providing laughter (Dalton and Linder 125-126; Hamamoto 87, 90-91). In terms of delivery, studio audiences became the norm again on shows that were not filmed in several locations and several factors have to be taken into consideration such budget (Dalton and Linder 49-50). There were also changes in the composition of the characters or how they are shown in the show. It is no longer the typical family. The show would center on the kids or teens such as Facts of Life and Charles in Charge which deals prima rily with issues teenagers usually face. Besides race, feminism got into the act with shows such as Rhoda and The Mary Tyler Moore Show as women took lead roles in sitcoms as well. The 1990s saw a continued surge of similar shows such as Murphy Brown and Just Shoot Me. There were even instances when some shows would challenge morals at the time such as MarriedWith Children which features a dysfunctional family, an introduction to the genre commonly known as transgression comedy. One other noticeable feature of sitcoms starting in the 1970s was the evolution of the delivery. Slapstick was no longer used at this point in time. Rather, the producers and the performers make use of the current situations they are in, whether they are personal, affecting only them or the social situation that affects them entirely. In other words, they poke fun at their problems without having to do crazy stunts or exaggerated violence reminiscent of The Three Stooges. The 1990s saw the emergence of sitcoms in animation followed the lead of live shows such as The Simpsons which is still enjoying a following to this day, and later South Park (Dalton and Linder 270-271). This trend would go on well into the 21st century with similar shows coming out such as The Office and 30 Rock which still follow the same trend. Furthermore, sitcoms have also broken down into categories according to age groups. Disney Channel offers sitcoms for children and teenagers such as Thats so Raven, Hannah Montana and Corey in the House. Naturally, the shows centers on the younger characters with the adults in the supporting role but the plot is nonetheless similar to the mainstream sitcoms (Dalton and Linder 44-45). Analysis: By the 21st century, the trend in the entertainment industry is leaning toward reality-TV shows which feature non-actors and single-camera recordings. The reason for this increasing popularity is that the participants are not actors and the audience can easily relate to them as the cameras capture every moment of their life and they are seen in their best and their worst and there are no cuts and takes, the camera continues rolling. It is as real as it gets and all the elements or genres are there from action, drama and even comedy without the canned laughter. It is said that these shows will replace mainstream programming which would affect soap operas and sitcoms. But this does not mean sitcoms will fade into oblivion or give up without a fight. In the latter shows, for instance, The Office, 30 Rock and even The Drew Carey Show make use of an element called the pathos. Where the shows make the viewers sympathize with the characters, and relate to them on a level unlike any other show (Graham 1). Old sitcoms never had this pathos. The problems facing the characters in the show are similar problems the audience faces on a regular basis in real life (Graham 1). In addition, this pathos need not be exaggerated like in slapstick. It is shown in a plain simple way similar to what real people go through every day. The characters, especially the main ones, are depicted as the everyman or average Joe. In the case of the three sitcoms, they reflect the joys, trials and tribulations the average (American) employee goes through in their daily routine. Nothing is exaggerated and the issues they face are real since real people experience them too. Matthew G ilbert puts it nicely At their best these single hand held sitcoms are an uplifting art form, one that has come of age in the past 20 years. They can invent a unique comic lexicon, invite us to laugh at our failings, capture the brilliance of our imaginations, and satirize our culture, they can reflect, clarify, and normalize human nature(Gilbert 1). Even animated sitcoms like The Simpsons and South Park show that as well this is why such animated series also appeal to adults, breaking the notion that they are only for children. Shows like Arrested Development and The Office are amongst some of a growing number of sitcoms that look different and are produced differently from sitcoms in the past. Eric Berlin states on the changing sitcom This new genre combines oddball characters, stress on improv acting, cinematic look, expert single-camera production work, and inventive use of flashbacks, private one on one confessions with the camera and quick cut-aways. The writing is daring, smart, and resembles the everyday awkward encounters that humans experience(Berlin 3). Shows like The Office and Arrested Development have been appropriately termed by author Brett Mills as the televisual style called comedy verite(Thompson 63). What comedy verità © is doing through its distinctive televisual style is shift the source of humor in the television comedy from the constructed joke, as seen in prior sitcoms, to the observation of a comic event (Thompson 67). Thompson states The observational component of these sitcoms, w hich includes not just what they look like but also the timing of shots and the sense that at times we observe events in real time, creates a different type of engagement with the narrative. The sitcom is thus reinvigorated by a shift from the tired realm of the staged sitcom, with its three cameras, studio audience, or one-camera, coverage shooting, to an experience of observation or witness(Thompson 67). A big part of these new comedy verite style sitcoms is how they convey the observational mode that the viewer is caught in, primarily through handheld shooting and a pacing that suggests particular segments unfold in real time as if their viewer were there (Thompson 68). However these segments and scenes are increasingly taking place in intimate settings, including shots conveying a particular characters subjectivity and vulnerability, and are less marked by characters performing for the camera, a fly on the wall type of perspective (Thompson 68). This production style that is new to the Sitcom tries to convey to the viewer that they are not watching comedy but are observing the comedic acts that unfold before the ever present handheld camera. Whether the acting is improvised or carefully scripted, it looks like it just happened and thats the whole point (Thompson 71). As a mode of production, this new developed sub genre of the sitcom is addressing comedy like never before and is effective in the use of presenting what is truly funny to viewers. Furthermore, what makes sitcoms better than reality TV is, as mentioned earlier, these shows do not only try to make the audience relate to the character as far as pathos is concerned, but being a comedy, it encourages the audience to laugh off their problems as well. Furthermore, unlike reality shows, it is the characters that are laughed at, not the people themselves. Because of this, this is what will make sitcoms last longer and keep on entertaining people.

Edexcel Maths Fp2 Paper

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, com/geo-sba-cxc/" class="ilgen">candidate number, the unit title (Further Pure Mathematics FP1), the paper reference (6667), 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 provid ed. 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 < a < 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 > . 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 con stants 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 NilCandidates 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 Pac kard 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 find the area of the fin ite 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 NilCandidates 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 sho uld 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 > 2 n r =1 n 1 2 . (7) 4. dy d2 y dy +y = x, y = 0, = 2 at x = 1. 2 dx dx dxFind 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 â €“ 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 2END 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 à ¢â‚¬â€œ 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 > ? >0 ? >0 x-3 x-2 x-3 x-2 (x – 3)(x – 2 ) Marks M1 A1 B1 B1 Numerator always positive Critical points of denominator x = 2, x = 3 x < 2 : den = (- ve)(- ve) = + ve 2 < x < 3 : den = (- ve)(+ ve) = – ve 3 < x : den = (+ ve)(+ ve) = + ve M1 A1 A1 (7) (7 marks) Set of values x < 2 and x > 3 {x : x < 2} E {x : x > 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 f actors 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 2 q 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) 15EDEXCEL 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 dxSch eme 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 > 1 3 3 4 5 + =3 3 3 e T = cosh T + sinh T = Hence T = ln 3 2. (5 marks) (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 â⠂¬â€œ 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 PURE 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 + cM1 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 marks) 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 sin 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 > 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 3×1 + 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) OUEDEXCEL 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 NumberScheme 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 c os 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

Computer Motherboard (Descriptive)

Computer Motherboard Motherboard acts like the arteries of a human taking the blood all over allowing the blood to travel. It has many line-like structures that are like the veins and the arteries. These lines are the wires mounted on the motherboard that keep thousandths or even millions of connection. Motherboard is a small yet powerful device inside the computer system. The motherboard has three cards on it. First, the graphics card, it is like the eyes of the computer, it acts like the eye that saw and reflects images to the screen.Second, the sound card it is the mouth of the computer, it sends sound to the speaker, it shouts what the computer says. And the third, the modem, the nose of the computer. It sniffs messages from the wires coming from the telephone lines. If there’s a blood, there’s also the heart. The processor of the computer acts as the heart of the computer. It has a square-shaped chip that has many pins. It sits on the motherboard controlling all th e cards including the motherboard.It controls, allows, and restricts information that is being disseminated into the system like the heart. It also has a heat sink and a fan keeping the processor cool. To be able to complete the system, it also has the brain. Hard disk serves as the brain of the computer. It is rectangular in shape. Inside it, there’s a mirror-like disk where all the information are stored. The disk spin at greater speed, depending on the information /data is being written into the disk.Motherboard is also consists of chips, transistors, and capacitors. It is the other component of the motherboard that acts like the organs of the system. These chips may be small as a grain of rice and bigger as a coin. Transistor is a small, black in color and rectangular in shape. It is sometimes recognizable because of its three metal legs attached on it. Transistors is consists of thousand or even million of IC (Integrated Circuit) packed in one part. And the capacitors ar e like tanks attached in the motherboard.It varies in size; it may be small, medium or large depending on the voltage and its capacitance. Motherboard is consists of millions of chips, transistors and IC’s. It acts like the body of the computer. Without it, the computer will not work. It is like the trunk of a tree. It is an amazing thing that has millions of wires attached in a small board. It is powerful and fundamental part that the computer will not work without it. Reference: * Motherboard. Retrieved from http://www. en. wikipedia. org/wiki/motherboard

Probar Conjugation in Spanish, Translations and Examples