January 2006 - 6673 Pure P3 - Mark scheme



Comparison of key skills specifications 2000/2002 with 2004 standardsX015461July 2004Issue 1

| | | |

|Question Number|Scheme |Marks |

| | [pic]+….) | |

|1 (a) | | |

| |=[pic] | |

| | |M1 (corr bin coeffs) |

| |May use McLaurin f(0)=1 and [pic] to obtain 1st two terms 1 + x |M1 (powers of –2x) |

| |Differentiates two further times and uses formula with correct factorials to give | |

| |[pic] |A1, A1 |

| | |(4) |

|Alternative |[pic]. So series is [pic] | |

| | |M1 A1 |

| | |M1 |

| |Uses f(2) = 0 to give 16 – 4 + 2a + b = 0 | |

| | |A1 |

| |Uses f(-1) = 6 to give -2 – 1 +-a + b = 6 |(4) |

| | | |

|(b) |Solves simultaneous equations to give a = -7, and b = 2 |M1A1 ft |

| | |(2) |

| | | |

|2 |Uses circle equation | |

| |[pic] |M1 A1 |

| |Multiplies out to give [pic]and thus [pic] (*) | |

| | |M1 A1 |

| |Or states equation of circle is [pic]has centre (-g, -f) and so | |

| |g = - 4 and f = - 3 |M1 A1 A1 |

| |Uses [pic]to give [pic], i.e. c = 20 |(7) |

|3 (a) |[pic] | |

| | |M1 |

| |y = 2x meets the circle when [pic][pic] |A1 |

| |[pic] | |

| |Solves and substitutes to obtain x = 2 and y = 4. Coordinates are (2, 4) | |

| |Or Implicit differentiation attempt ,[pic] |A1 |

| |Uses y = 2x and [pic]=2 to give 10x – 20 = 0. |(3) |

|Alternative |Thus x =2 and y =4 | |

| | | |

| | | |

| | |M1 A1 |

| | | |

| | |A1 |

| | |(3) |

|(b) | |M1 |

| | | |

| | |A1 |

| | |M1 A1 |

| | |(4) |

| | | |

| | |M1 A1 |

| | | |

| | | |

| | |M1 A1 |

| | |(4) |

| | | |

|Question Number|Scheme |Marks |

| | | |

|4.(a) |[pic] |M1 A1 |

| |[pic] | |

| | |M1 A1 |

| | |(4) |

| |[pic] | |

|(b) |= [pic] | |

| |= [pic] |M1 A1 |

| |= [pic] | |

| | | |

| | | |

| | | |

| |[pic], so A = - 1 |A1 |

| | | |

| |Uses 18 = B ( 3– 2x) + C ( 3+2x ) and attempts to find B and C |M1 A1 |

| | |(5) |

| |B = 3 and C = 3 | |

| |Or | |

| |Uses [pic] = A([pic])+ B ( 3– 2x) + C ( 3+2x ) and attempts to find A, B and C | |

|5. (a) | | |

| |A = -1, B = 3 and C = 3 |B1 |

| | | |

| | | |

| | |M1 |

| |Obtains [pic] | |

| |Substitutes limits and subtracts to give 2[pic] |A1 A1 |

| | |(4) |

| |= -2 +3ln5 or –2 +ln125 | |

| | |M1 |

| | | |

| | |A1, A1, A1 |

| | |(4) |

| | | |

| | | |

|(b) | | |

| | |M1 A1 |

| | | |

| | | |

| | |M1 A1ft |

| | | |

| | | |

| | |A1 |

| | |(5) |

| | | |

| | | |

| | | |

| | | |

|Question Number|Scheme |Marks |

| | | |

| | | |

|6 (a) |[pic]; rate of decrease/negative sign; k constant of proportionality/positive constant | |

| | |B1 |

| | |(1) |

| |[pic] | |

| |[pic] | |

| |[pic] | |

|(b) | |M1 |

| | | |

| |At t = 0 [pic], [pic] |M1 |

| |and at t = 4 [pic], [pic], | |

| |[pic]and [pic] , [pic] |A1 |

| | |(3) |

|(c) | | |

| | | |

| |Solves [pic]to give [pic] so p = 3 |B1 |

| |Solves [pic]to give [pic] so q = 5 | |

| | |M1 |

| |[pic]so unit vector is [pic]6i +6j +3k) | |

| | |M1, A1 |

| |[pic] | |

|7 (a) | |(4) |

| |[pic] | |

| | |M1 A1 |

|(b) |Write down two of [pic] |M1 A1 |

| |Solve to obtain [pic]or [pic] |(4) |

| |Obtain coordinates ( 5, 3, 5) | |

|(c) | |M1 A1 |

| | |(2) |

| | | |

| | |M1 A1 |

| | | |

| | |A1 |

|(d) | |(3) |

| | | |

| | |B1 B1 |

| | | |

| | |M1 A1 |

| | | |

| | |A1 |

| | |(5) |

| | | |

|Question Number|Scheme |Marks |

| | | |

| | | |

|8(a) |[pic] therefore [pic] | |

| |When x = 0, t =[pic] |M1 A1 |

| |Gradient is [pic] | |

| |Line equation is [pic] |B1 |

| | | |

| |Area beneath curve is [pic] | |

| |=[pic] |M1 |

| |[pic] | |

| |Uses limits 0 and [pic] to give [pic] |M1 A1 |

| |Area of triangle beneath tangent is [pic]= [pic] |(6) |

| |Thus required area is [pic]- [pic]= [pic] | |

|(b) | | |

| | |M1 |

| | | |

| |The integration of the product of two sines is worth 3 marks (lines 2 and 3 of scheme to part (b)) |M1 |

| |If they use parts | |

| |[pic] | |

| |8I = cost sin3t – 3 cos3t sint |M1 A1 |

| | | |

| | | |

| | |M1 A1 |

| | | |

| | | |

| | |M1 A1 |

| | | |

| | | |

| | |A1 |

| | |(9) |

| | | |

|N.B. | | |

| | | |

| | | |

| | | |

| | | |

| | |M1 |

| | | |

| | | |

| | |M1 A1 |

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download