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

Morning Fog

Jennifer Cudmore Prof. L. Gertsma English Composition 1 6 September, 2012 â€Å"Morning Fog† I often wonder how many opportunities I’ve let slip by throughout my life. Countless colorful sunsets, too many moons rising into the night sky to even count, or even something as simple as a glance in the mirror at my own reflection. Far too many times I have been too busy or too tired to stop and notice what I’m missing. On one particular morning a few days ago, I awoke to the piercing sound of my alarm blaring in my ear.Even though the piece of technology had merely sprouted legs of its own to bury itself under my pillow, I could still hear it as clear as a bull horn through my sleepiness. I managed to groggily press the correct button on the flat piece of glass that was the surface of my cell phone. There’s two buttons to choose from and if I’m not careful, I would find myself pressing the button that would allow me to drift off to the land of odd happen ings, to unicorns and fairy dust, and to the place where time seems to stand still.Often times, when this unfortunate mishap has occurred, I would wake in such frenzy that I could feel the labored breaths as they attempted to escape from my lungs. My heart was beating in such a fashion as to erupt straight through my chest. Luckily this was not one of those days and I began to untangle myself from the covers that so gently held me throughout the night. The temperature change was abrupt and waves began to rush over my skin like lake water lapping the rocks of a shoreline. My long, cold fingers reached through the darkness toward the switch on the wall.My eyes quickly clenched shut as if anticipating the searing pain that was about to commence once the switch was turned. With a loud click, electricity rushed the filament of the crystal globe and exceptionally bright light stretched through every corner of my room that was just as dark as dirty oil a moment ago. Making the unbearable a ttempt to adjust to the sudden change from darkness to light, my eyes began to blink repeatedly and tears formed at the corners of each eye. With each blink, it became apparent very quickly that contacts would not be an option today.I could almost hear the pleading of each eye screaming out to me saying, â€Å"Please go back to bed! We’re not ready yet! † With the moon still hanging high outside my window, I turn to my closet. I remember thinking to myself that the moon looked particularly bright this morning and perhaps that was the first sign that I was looking a little closer at what I was doing than I usually did. I quickly chose my outfit for the day and began down the hallway towards the steps. The potent aroma of coffee hit my senses as my feet landed on the fourth step from the top of the narrow stairwell.My â€Å"lifeline† is what I officially dubbed this miracle liquid for no matter how many hours of sleep I had shorted myself the night before, coffee had always helped me push through. I made my way down the remainder of the steps and into the coffee scented kitchen where I find a mug to place sugar and creamer in. I managed to pour a cupful without spilling even a drop of the precious, blistering hot liquid. Even with the first few sips of the coffee within the cup, I start to feel my body slowly awaken from the sudden surge of caffeine and I push forward.With a quick glance at the clock on the wall, I make the decision that there is no time for a piece of toast or bowl of cereal. Almost as if hearing my thoughts, an animal like growling emanated from my stomach in an attempt to change my mind. The plea was quickly ignored and I walked past the refrigerator door. A mental note formed in my mind of yet another missed opportunity, as I walked past my father who was so intently focused on the plate of breakfast sitting in front of him.With cup in hand I enter the bathroom, where I finish getting ready for the day. With a quick glan ce around, as if taking a mental inventory of what would be needed, a hairbrush, toothbrush and toothpaste quickly make their way to the countertop, lining up like patrons at a checkout in a supermarket. It wasn’t until my hair was done and teeth were brushed that I made the connection. How many years had I gotten myself up? How many times had I dressed myself and how many cups of coffee had it taken to get me going in the morning?I had come to the shocking realization that in the past 15 minutes of my 32 years I managed to get myself ready in the presence of 3 different mirrors in 3 separate places of my house but not once did I take the time to gaze into one. I had looked long enough to guarantee that every hair was in its place and that every pearly white tooth got its own little one on one with the toothbrush but not once did I truly look at the reflection that was quietly standing there staring back at me.I stopped and forced myself to truly look and realized that the re flection was the same with the exception of a few extra lines, or that the shine of my hair wasn’t quite as luminescent as it once was. However, when I actually looked into my eyes it came quite apparent that the many years of broken hearts, lost loves, and life experiences had changed them in such a way that it was almost unrecognizable. There was pain there, yet an even stronger wisdom irradiated brighter than the moon or the sun combined. Sometimes, all it takes is a little time to really appreciate who you are and what you have accomplished in your life.

Rituals in anthropology essays

Rituals in anthropology essays Rituals are a significant part of our society and the way we live our lives. They are important because they give us a sense of security and loyalty to a group. Rituals are also a main aspect of religion, and studying the differences among cultures helps us gain a greater understanding of how rituals influence our lives. We perform rituals to demonstrate our values and beliefs, to pray, and many times to reinforce unity within a group. Certain rituals become so fixed in our everyday routine, we dont realize we are performing them. Anthropologists believe that ritual has existed since the very beginning of our time. Some believe that it is ritual that has kept us here so long, and without this regular practice, we would not have the security or solidarity needed to survive. Invariably, there are things in life that we cannot control, no matter how hard we try. Rituals are a way of increasing our command while creating a comfort zone. The articles that are going to be discussed deal wi th ritual in ancient societies, the way they relate to modern rituals and the presence of rituals in a cultural activity such as baseball. The first article Rituals of Death: Capital Punishment and Human Sacrifice is a unique comparison of ancient Aztec rituals of human sacrifice in Mexico and capital punishment in America. Human sacrifice was a part of the offerings for the gods, whose hearts and blood were considered the supreme gift. Similarly, capital punishment can be seen has a modern day form of human sacrifice. In this article Elizabeth Purdum compares these two rituals with an analysis of each according to the sequential order in which the rituals are performed. She begins by giving a brief explanation of who is put to death. The Aztecs sacrificed an estimated total of 20,000 to 250,000 people annually. Most of the human sacrifices were male war captives from other tribes, and sometimes children sold to priests by poor....

11 GI Bill An In-Depth Analysis

Research Essay Sample on Post 9/11 GI Bill An In-Depth Analysis Introduction In the year 2008, Post 9/11 veterans’ educational assistance act was made law by the congress. This measure was responsible for the amendment of title 38 of the US constitution. This was done so as to include the third chapter which deals with extension of educational benefits for veterans who have been in military service since September 11 2001. It is these education benefits that are commonly referred to as the post 911 GI Bill. Currently people are calling it the new GI BILL. The social condition that the bill seeks to address is emotional and secondary illiteracy among war veterans all over the country. This is done through paying of college fees and providing welfare assistance to veterans in a model similar to that of the original GI Bill that came into force immediately after the Second World War (Budahn, 2011).. The New GI Bill is attempting to address these problems. For veterans who have been in military service for more than three years since 2001, the act provides that they receive 100% funding for a four year undergraduate program. The veterans are also able to transfer the benefits to their spouses and children on condition that they serve for more than ten years. This act was proposed by Senator Jim Webb in 2007. The act was affected on august 2009 with some parts of it earmarked for change. For one to be eligible for the benefits they must be in active military service, they must have attained high school certification. The officers must opt for university or college education (Scoot, 2009). It is the department of veteran affairs that is charged with managing the veterans. For us to talk about the roles, and functions of the department of veteran affairs, we have to understand how it operates and the challenges it is facing. The VA as it is commonly referred to is facing a growing challenge in its operating environment. There has been a significant increase in claims and services per patient while legislative and national security policies are full of uncertainties. By understanding these obstacles and their effects on VA, we will be able to analyze the roles and functions of the VA department. A big challenge to the VA is the changing veteran population. Due to old age, the veterans and their families is developing complex needs with expectations that the VA should be able to provide for them. The Vietnam veterans constitute a significant percentage of veterans with age related complications like prostate cancer and diabetes. This has increased their demand for better health care services. The VA is also expected to provide benefits and services to the families of these ageing veterans (Whitney, 2007). Disability compensation is the area that has been profoundly affected in recent years. This is due to change in nature of wounds inflicted during combat. Most disabilities experienced by veterans’ today e.g.| are more complex and require advanced treatment. TODAY, THE VA PROVIDES THE FOLLOWING SERVICES: Provision of high quality health services to war veterans The VA has maintained the distinction of being the largest integrated health care system in North America. It has grown from 54 hospitals in 1931 to 153 fully furnished hospitals today. There are more than seven hundred outpatient clinics that are community based. Currently, the VA runs two hundred and sixty vet centers across the country. The medical facilities offer a wide range of services ranging from medical to rehabilitation. The VA also provides telemedicine which is intended to increase efficiency in service provision (Gaytan, 2011). Provision of benefits and compensation to veterans Another function of VA is to provide compensation and welfare benefits to the war veterans. These pension and compensation cover about four million veterans. Honoring veterans The VA is also charged with the task of honoring the veterans through establishment of unique cemeteries reserved for the veterans. The agency also undertakes maintenance of the cemeteries and other national shrines. In the year 2010, the agency has maintained more than three million gravesites and one hundred and sixty four properties related to the veterans(Alford, 2010). Eliminating Veteran homelessness Currently, there are about 107000 homeless veterans in the country. The VA plans to reduce this number to zero by the year 2014 through the provision of home support funds. Enabling full delivery of 21st century benefits and services The VA is committed to reducing paperless claims related to disability by the year 2014. It also strives to ensure veterans process their claims in a period shorter than 125 days (Gaytan, 2011). Improvement of veteran mental health 97% of veterans are suffering from alcohol abuse; this is one of VA’s goals to reduce alcohol consumption among veterans. The department ensures that the veterans receive eight sessions of psychotherapy (Alford, 2010). Doing research that enhance long term well being of veterans The department continues with scientific researches that are geared toward service improving and portfolio balance in the NRAC (Bertoni, 2011). Improve health care services while keeping costs low The department is charged with the responsibility of saving cost incurred by the veteran’s health requirements thereby increasing their benefits. It also ensures that the cases of mistaken payments are eliminated. GI Bill analysis and framework In its strategic planning framework, the department of veteran affairs has split the GI policy implementation strategy into four main components. These components are responsible for better discharge of services by the VA. They are the ones that define the functions and roles of the VA. These components include; four strategic goals which are crosscutting, integrated objectives’, integrated strategies and radical initiatives. The first component is not limited to one specific goal. It is charged with the task of providing a common set of premises on which initiatives and operation strategies are based. The objectives are meant to help the VA to build strategies on which the department’s goals of service provision can be achieved. Integrated objectives are courses of action that are meant to realize VA’s vision and objectives in implementing the GI policy (Bergmann, Duggan 2007). Veterans have derived a lot of benefits from the services offered by this agency. First, psychotherapy sessions have been successful in changing their drinking habits while ensuring their mental well being. The department’s collaboration with NRAC is a crucial step in understanding specific needs of veterans. The adoption of seamless interactivity between departments has enabled the veterans to acquire all information without having to move from one office to another (Budahn, 2011). Elimination of homelessness is another benefit that veterans are enjoying under the VA. The intention of this plan is to ensure that each veteran has a house and their families are well catered for. The use of SCIP process has eliminated wastage through unification of the budget affairs. This has increased the departments efficiency in dealing with veteran affairs. Reducing hiring cycle through human capital management has ensured that veterans and their families continue to enjoy their benefits without unnecessary delays. Previously it took more than 125 days to process claims. This has significantly reduced because the veterans can now enjoy the full 21st benefits and services (Bertoni, 2011). The central government through the department of defense has been financing these benefits and services since the adoption of the act. The money is sourced from the taxpayer’s kitty. Historical analysis of the GI Bill After the Second World War, the government passed into law Title 38 which was supposed to cater for all the men and women who had retired from the military so that they could live a comfortable life. In this act, the veterans were to enjoy full benefits of military personnel including free medical care and social services. Although it did not cover the education and spouses’ part, this act went a long way in ensuring that the veterans had a decent life after service. This was the only way of dealing with the problems afflicting the veterans. Supplemental appropriations act of 2008 was adopted through the act of congress in order to introduce modifications to the previous bill. After adopting the act in 2008, the state amended the third part of US code so that retiring military officers who served in the army from 2001 would be able to enjoy more benefits and services. It also opened eligibility to members of the National Guard. The law has also reduced the housing allowance for online learners which has enabled the service members and their spouses to get annual stipend of thousand dollars. The act has also removed the state to state tuition fees for the servicemen who decide to enroll in state colleges. In addition to these benefits, the bill was also modified to set 17000 dollars as the cap for veterans who may desire to attend private colleges (Gaytan, 2011) Historical ineffectiveness of the first GI Bill The original GI bill also provided for college education for veterans although there were some limitations. The ability of the veteran to transfer benefits to the spouse or children was not provided for in the original bill. These two acts were initiated by decorated war veterans turned politicians who felt that it was necessary to safeguard the future of the servicemen and their families. Incorporation of Historical lessons in the new program In his proposal for improvement of the act, Senator Jim Webb who is a decorated Vietnam veteran wanted the war veterans to be accorded better living standards due to the sacrifices they make for the sake of the country. Other prominent people who sponsored the bill are representative Bobby Scott and senator Olympia Snowe. The democrats and a few republicans support the passage of the supplemental bill after a bipartisan deal. The veteran education assistance bill was then passed on June 19 2008 by majority (Bertoni, 2011). The controversy generated has led the bill to face opposition from some politicians. Issues of compensation during the buy up option were a key concern for the opponents of this bill. The bill does not offer any provision on how the contribution is supposed to be incorporated into it when it is fully enforced. The VA has asserted that some service members will not be refunded their contribution of six hundred dollars. The bill faced opposition from some officials from the department of defense, President Bush and senator McCain believed it would impact negatively on military retention which is, critical to the country’s defense (Scoot, 2009). Effective approach to veteran affairs is essential in building the morale of the disciplined forces. It also ensures that no serviceman will be abandoned at his hour of need by the state. The ineffectiveness of the first bill meant that veterans would continue to be locked out of certain aspects of their lives. By denying them a chance to study on subsidized fees, the bill was discriminating against these important members of the American society. The other weakness that promoted ineffectiveness of the first bill was the exclusion of children and spouses from enjoying the veteran’s benefits (Alford, 2010). This meant that the veterans would still suffer since they are the ones expected to provide for their children. Historical records show that most veterans commit suicide due to stress and psychological problems. By offering eight sessions of psychoanalysis to the veterans as one of the benefits, the bill has eliminated cases of suicides. It is also necessary to note that the policy eliminated the stress related to academic achievements because the veterans have been given a chance to pursue higher education, which will enable them to get part-time employment.

Parts Per Million Definition

Parts Per Million Definition Parts per million (ppm) is a commonly used unit of concentration for small values. One part per million is one part of solute per one million parts solvent  or 10-6. Parts per million and other parts per notations (e.g., parts per billion or parts per trillion) are dimensionless quantities with no units. Preferred methods for expressing parts per million include  Ã‚ µV/V (microvolume per volume),  Ã‚ µL/L (microliters per liter), mg/kg (milligram per kilogram),  Ã‚ µmol/mol (micromole per mole), and  µm/m (micrometer per meter). The parts per notation is used to describe dilute solutions in chemistry and engineering, but its meaning is ambiguous and it is not part of the SI system of measurement. The reason the system is ambiguous is because the concentration depends on the original unit fraction that is used. For example, comparing one milliliter of a sample to a million milliliters is different from comparing one mole to a million moles or one gram to one million grams. Sources Milton R. Beychok (2005). Air Dispersion Modeling Conversions and Formulas. Fundamentals of Stack Gas Dispersion (4th ed.). Milton R. Beychok. ISBN 0964458802.Schwartz and Warneck (1995). Units for use in atmospheric chemistry (PDF). Pure Appl. Chem. 67: 1377–1406. doi:10.1351/pac199567081377

International Legal and Ethical Issues in Business IP Week 4 Essay

International Legal and Ethical Issues in Business IP Week 4 - Essay Example Competition law and the antitrust laws are examples of such regulations. The laws are set of rules and regulations designed to enhance competitive environment in the business environment. This paper evaluates an example of a merger between two major telecommunication companies. Some issues arise due to generic competition. Generic competition stems from producers who do not incur costs in research before they launch a product to market. It results in the original manufacture imposing some restrictions to protect their brands. This paper will address legal barriers in introducing a new product to market and possible dilemmas. Key words: competition, legal, mergers, antitrust and law. International Legal and Ethical Issues in Business Introduction The antitrust laws were put in place by the federal and state governments in United States to regulate businesses. The laws ensure that companies do not become too big and they do not fix their prices. The law also ensures that there is perfe ct competition in market so that the consumer welfare is maintained. The federal governments are also mandated in reviewing potential mergers to attempt to prevent market concentration. The antitrust laws apply to businesses and individuals. The laws were enacted to stop businesses that go too large from blocking competition and abusing their power (Baker, 2004). The antitrust law is aimed at ensuring perfect competition. ... It is estimated that pharmaceutical companies spend an average of $800-1 billion and between eight and sixteen years to research a new drug (Crandal & Clifford, 2003). Research need to be conducted to ensure that the drug introduced to the market can compete perfectly. The drug should also meet the target population needs; it should be in a position to solve their problems. As a result, an extensive research should be conducted to make sure that the brand conforms to set standard and market needs. Due to this, high cost is incurred. Legal Barriers to Market Entry Legal requirements have to be followed by the drug manufacturers to ensure that they enter the market with legal approval. There are legal barriers that control the entry of drugs to the market in the United States. In 2003, President Bush signed into law the Medicare Prescription Drug, Improvement and Modernization Act of Approval. The act has had major impact on the entry of generic pharmaceutical drugs to the market. The 2003 act contains three rules that control the entry of the drugs to the market. The act allows a minimum of one 30- month stay per generic application, clarifies the types of patents that must not be submitted to Food and Drug administration for listing in the orange book, revises the information required to be submitted on patents, and consolidates all patent information on declaring forms to ensure that the submissions are more informative and precise. In addition to the 30 month stay per application, the FDA tightened control on the types of patent claims submitted by the innovator company. The law ensures that the innovator drug companies would no longer be able to submit patents claiming packaging, metabolites and

Eddington and Everyday Experience Essay Example | Topics and Well Written Essays - 1500 words

Eddington and Everyday Experience - Essay Example Often though it is something that is lacking a foundational source, the idea, 'it is there because it is' often falls into place, especially in a philosophical sense. For example Arthur Eddington's interpretation of the existence of the world is a key interpretation of this type of thinking. In his following statement it is obvious how he utilizes the areas of philosophy to try and make sense out of the universe surrounding him. "The world, which spontaneously appears around me when I open my eyes," is "a strange compound of external nature, mental imagery, and inherited prejudice". Factual knowledge is not as simple or self-evident as it so often seems to be. Thus, the process of learning cannot be taken for granted. Several different theories of the learning process have been established in Western philosophy and the memes of culture. One of these is Skepticism. Skepticism itself questions everything and places doubt where none should exist. The main message it relays to mankind is one of a negative nature, claiming that man will never reach a heightened sense of knowledge about any certain issue in life or the world in particular (Hooker 1996). So, it is found within the realm of skepticism nothing is for certain and the foundational sustenance of the utilization of epistemologies themselves finds doubt in anything and everything. Although skepticism is admonished by those who don't follow the theorization in behind it, it still shows some crude evidence as to why some doubt so much, in even the simplest of things. The reason for these doubts within this the ory is due to the imperfections of the human mind, which can possibly include: faultiness in reasoning and judgment, poor memory, limited accessibility to an object of scrutiny, a lack in the accuracy of the senses, the possibility of mistaking illusions (such as dreams) for reality, and the possibility of misinformation. The issue then, with skepticism is that it finds fault with everything, even those that take simple, common sense notions where the majority of people would find to be, "self evident". Although scientific people are typically described as taking the sceptical view of a new idea that seems "wild", the scientific method is most accurately rooted in the philosophy of Empiricism. To Empiricists, the senses are indeed highly accurate and, moreover, they are our foremost tool for acquiring knowledge about life, the universe, and everything. In stating "which spontaneously appears around me when I open my eyes", Arthur Eddington indicates that our environment is perceived by our senses. There exists an outside world to which we are only connected through our senses and construct a mental image of it. There exists a disparity between what is perceived to be and what reality essentially is. Even in an ordinary view of the world, it is sometimes dubious if we can rely entirely on our sensory data. Some simple examples are illusions, in which we can not trust our senses since they mislead us to draw odd conclusion. Most Empiricists, however, recognize the existence of a prior i truths, which are those of mathematics and logic. In Eddington's claim of "inherited prejudice", the interpretation of it can go in several directions. Prejudice is neither knowledge nor belief, although it is rooted in the latter. The word prejudice comes from the Latin word