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A-Level · Further Pure Mathematics 1 and 2 (Further Calculus and Series Depth)

FP.CP14 Core Pure: Further Calculus and Series Depth

EDEXCEL 9FM0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The improper integral I = integral from x=1 to x=infinity of 4/x³ dx is defined as I = lim_{b -> infinity} integral from x=1 to x=b of 4/x³ dx.

(a)Show that integral from x=1 to x=b of 4/x³ dx = 2 - 2/b^2.(3)

(b)Hence show that I converges, and find its exact value.(2)

(Total for Question 1 is 5 marks)

The function f(x) = 1/√x-2 is undefined at x=2. The improper integral J = integral from x=2 to x=6 of 1/√x-2 dx is defined as J = lim_{a -> 2+} integral from x=a to x=6 of 1/√x-2 dx.

(a)Find integral from x=a to x=6 of 1/√x-2 dx, in terms of a.(3)

(b)Determine whether J converges, giving its exact value if it does.(3)

(Total for Question 2 is 6 marks)

The curve C has equation y = e^-x/2, for 0 ≤ x ≤ ln4. The region R is bounded by C, the x-axis, and the lines x=0 and x=ln4. R is rotated through 2*π radians about the x-axis to form a solid of revolution.

(a)Show that the volume of the solid formed is V = π * integral from x=0 to x=ln4 of e^-x dx.(2)

(b)Hence find the exact value of V, giving your answer in the form k*π, where k is a rational number to be found.(4)

(Total for Question 3 is 6 marks)

The curve C has equation y = ln x, for x ≥ 1. The region R is bounded by C, the y-axis, and the lines y=0 and y=1. R is rotated through 2*π radians about the y-axis to form a solid of revolution.

(a)Show that x² = e^2y.(2)

(b)Find the exact volume of the solid formed, giving your answer in terms of e.(5)

(c)Give the volume found in part (b) correct to 3 significant figures.(1)

(Total for Question 4 is 8 marks)

The mean value of a function f over the interval [a,b] is defined as (1/(b-a)) * integral from x=a to x=b of f(x) dx. Find the mean value of f(x) = 3*sin²(x) over the interval [0, π], using the identity sin²(x) = (1-cos(2x))/2.

(Total for Question 5 is 5 marks)

A chemical process cools a sample so that its temperature, θ degrees Celsius, t minutes after the process begins, satisfies θ = 20 + 60*e^-0.2t for t ≥ 0.

(a)Find the mean temperature of the sample over the first 10 minutes, giving your exact value and also your answer correct to 3 significant figures.(5)

(b)State one limitation of using this mean value to estimate the sample's typical temperature during the cooling process.(1)

(Total for Question 6 is 6 marks)

f(x) = ln(1+2x).

(a)Show that f'(x) = 2/(1+2x), and find f''(x) and f'''(x).(3)

(b)Hence find the Maclaurin series for f(x), up to and including the term in x^3.(3)

(c)State the range of values of x for which this expansion is valid.(1)

(Total for Question 7 is 7 marks)

You may use, without proof, the standard series e^x = 1 + x + x²/2! + x³/3! + ... and cos x = 1 - x²/2! + x⁴/4! - ...

(a)Write down the Maclaurin expansions of e^x and cos x, up to and including the term in x^3.(2)

(b)Hence show that e^x * cos x = 1 + x - (1/3)x³ + O(x⁴).(6)

(Total for Question 8 is 8 marks)

Given that 1/(r(r+2)) = (1/2)(1/r - 1/(r+2)) for r a positive integer,

(a)Verify that this identity is correct by combining the right-hand side into a single fraction.(2)

(b)Using the method of differences, show that sum_{r=1}^{n} 1/(r(r+2)) = n(3n+5)/(4(n+1)(n+2)).(6)

(c)Hence find the exact value of sum_{r=1}^{20} 1/(r(r+2)).(2)

(Total for Question 9 is 10 marks)

Let f(r) = r(r+1)(r+2)(r+3).

(a)Show that f(r) - f(r-1) = 4r(r+1)(r+2).(3)

(b)Hence, using the method of differences, show that sum_{r=1}^{n} r(r+1)(r+2) = (1/4)n(n+1)(n+2)(n+3).(5)

(c)Hence find sum_{r=1}^{15} r(r+1)(r+2).(2)

(Total for Question 10 is 10 marks)

Using the standard results sum_{r=1}^{n} r = n(n+1)/2, sum_{r=1}^{n} r² = n(n+1)(2n+1)/6 and sum_{r=1}^{n} r³ = (n(n+1)/2)²,

(a)Show that sum_{r=1}^{n} r(2r-1)(r+3) = (1/6)n(n+1)(3n²+13n-4).(6)

(b)Hence find sum_{r=1}^{12} r(2r-1)(r+3).(2)

(Total for Question 11 is 8 marks)

The curve C has equation y = 1/x, for x ≥ 1. The region R is bounded by C, the x-axis, and the line x=1, and extends without bound as x -> infinity. R is rotated through 2*π radians about the x-axis to form an infinite solid of revolution S.

(a)Show that the volume of S formed between x=1 and x=b, where b>1, is V(b) = π(1 - 1/b).(4)

(b)Hence show that, as b -> infinity, the volume of S converges to a finite limit, and state this limit.(3)

(c)It is a standard result, which you may assume without proof, that the corresponding surface area of S between x=1 and x=b is A(b) = 2*π * integral from x=1 to x=b of (1/x)*√1+1/x⁴ dx, and that A(b) -> infinity as b -> infinity. Using your results from parts (a) and (b), comment on what this implies about the volume and surface area of the infinite solid S.(2)

(d)A second, finite, solid is formed by rotating C about the x-axis between x=1 and x=10 only. Find its exact volume, giving your answer in the form k*π, where k is a rational number to be found.(3)

(Total for Question 12 is 12 marks)

f(x) = cos x.

(a)Write down the Maclaurin series for f(x), up to and including the term in x⁴, and state the range of values of x for which it is valid.(3)

(b)By substituting x=0.5 into your series from part (a), find an approximation for cos(0.5), giving your answer to 4 decimal places.(2)

(c)Given that cos(0.5) = 0.87758, correct to 5 decimal places, calculate the percentage error in the approximation found in part (b), giving your answer to 2 significant figures.(2)

(Total for Question 13 is 7 marks)

Given the identity 2*sin(1/2)*cos(r) = sin(r+1/2) - sin(r-1/2), which holds for all real r,

(a)Show that sum_{r=1}^{n} cos(r) = (sin(n+1/2) - sin(1/2)) / (2*sin(1/2)).(6)

(b)Hence find, correct to 3 significant figures, the value of sum_{r=1}^{50} cos(r), where r is measured in radians.(3)

(Total for Question 14 is 9 marks)

Mark scheme · FP.CP14 Core Pure: Further Calculus and Series Depth

Question 1

(a) M1 integrates 4x^-3 to -2x^-2 (oe)
(a) M1 substitutes the limits b and 1 correctly into -2x^-2
(a) A1 = 2 - 2/b² cso
(a) Answer: 2 - 2/b² (shown)
(b) B1 as b -> infinity, 2/b² -> 0
(b) A1 I = 2 (converges) cao
(b) Answer: I = 2

Question 2

(a) M1 integrates (x-2)^-1/2 to 2(x-2)^1/2 (oe)
(a) M1 substitutes the limits 6 and a correctly
(a) A1 = 4 - 2sqrt(a-2) oe cao
(a) Answer: 4 - 2sqrt(a-2)
(b) M1 considers the limit as a -> 2+ of 2sqrt(a-2)
(b) A1 2sqrt(a-2) -> 0 as a -> 2+
(b) A1 J = 4 (converges) cao
(b) Answer: J = 4

Question 3

(a) B1 uses V = π * integral y² dx with the correct limits 0 and ln4
(a) A1 y² = (e^-x/2)² = e^-x, giving V = π * integral from 0 to ln4 of e^-x dx cso
(a) Answer: V = π * integral from x=0 to x=ln4 of e^-x dx (shown)
(b) M1 integrates e^-x to -e^-x
(b) M1 substitutes the limits ln4 and 0 correctly
(b) A1 uses e^-ln4 = 1/4 to obtain integral = 3/4 oe
(b) A1 V = 3*π/4 cao
(b) Answer: V = 3pi/4

Question 4

(a) M1 rearranges y = ln x to x = e^y
(a) A1 squares to obtain x² = e^2y cso
(a) Answer: x² = e^2y (shown)
(b) B1 V = π * integral from y=0 to y=1 of x² dy soi
(b) M1 substitutes x² = e^2y
(b) M1 integrates e^2y to (1/2)e^2y
(b) M1 substitutes the limits y=1 and y=0 correctly
(b) A1 V = (e² - 1)*π/2 oe cao
(b) Answer: V = (e² - 1)π/2
(c) B1 V = 10.0 awrt (3 s.f.)
(c) Answer: 10.0 (awrt)

Question 5

M1 uses sin²(x) = (1-cos2x)/2 to rewrite 3sin²(x) as (3/2)(1-cos2x)
M1 integrates to obtain (3/2)(x - (1/2)sin2x) oe
M1 substitutes the limits π and 0 correctly
A1 integral from 0 to π of 3sin²(x) dx = 3*π/2 oe
A1 mean value = 3/2 cao
Answer: Mean value = 3/2

Question 6

(a) M1 integrates the constant term 20 to 20t
(a) M1 integrates 60*e^-0.2t to -300*e^-0.2t oe
(a) M1 substitutes the limits 10 and 0, and divides by 10 (the interval length)
(a) A1 exact mean = 50 - 30*e^-2 oe cao
(a) A1 mean = 45.9 awrt (3 s.f.)
(a) Answer: Mean = 50 - 30e^-2, awrt 45.9
(b) B1 gives a sensible limitation, e.g. the model θ = 20+60e^-0.2t may not hold exactly in practice (real cooling can be affected by draughts, insulation, or measurement error), so the calculated mean is only as reliable as the model
(b) Answer: Any sensible limitation, e.g. the exponential cooling model may not describe the real sample exactly, so the mean value is only an estimate based on the model.

Question 7

(a) M1 differentiates using the chain rule to obtain f'(x) = 2/(1+2x) cso
(a) A1 f''(x) = -4/(1+2x)² oe
(a) A1 f'''(x) = 16/(1+2x)³ oe
(a) Answer: f'(x) = 2/(1+2x), f''(x) = -4/(1+2x)², f'''(x) = 16/(1+2x)³
(b) M1 evaluates f(0)=0, f'(0)=2, f''(0)=-4, f'''(0)=16
(b) M1 substitutes into f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3!
(b) A1 f(x) = 2x - 2x² + (8/3)x³ oe cao
(b) Answer: ln(1+2x) = 2x - 2x² + (8/3)x³ + ...
(c) B1 -1/2 < x < 1/2 oe (allow |x| < 1/2)
(c) Answer: -1/2 < x < 1/2

Question 8

(a) B1 e^x = 1 + x + x²/2 + x³/6 oe
(a) B1 cos x = 1 - x²/2 (+ 0x³) oe
(a) Answer: e^x = 1+x+x²/2+x³/6; cos x = 1 - x²/2
(b) M1 multiplies the two series together
(b) M1 identifies the constant term as 1x1=1
(b) M1 identifies the x¹ coefficient as 1x1 + 1x0 = 1
(b) M1 identifies the x² coefficient as 1x(-1/2)+1x0+(1/2)x1 = 0
(b) A1 identifies the x³ coefficient as 1x0+1x(-1/2)+(1/2)x0+(1/6)x1 = -1/2+1/6 = -1/3
(b) A1 e^x*cos x = 1 + x - (1/3)x³ + ... cso
(b) Answer: e^x cos x = 1 + x - x³/3 + ...

Question 9

(a) M1 combines (1/2)(1/r - 1/(r+2)) over the common denominator r(r+2)
(a) A1 simplifies to 1/(r(r+2)), matching the left-hand side cso
(a) Answer: Identity verified
(b) M1 writes sum_{r=1}^n 1/(r(r+2)) = (1/2) sum_{r=1}^n (1/r - 1/(r+2))
(b) M1 lists sufficient terms of the sum (e.g. r=1,2,...,n-1,n) to show the cancellation pattern
(b) A1 identifies the surviving terms as 1 + 1/2 - 1/(n+1) - 1/(n+2)
(b) M1 combines these four terms over a common denominator
(b) A1 simplifies the numerator to 3n²+5n
(b) A1 sum = n(3n+5)/(4(n+1)(n+2)) cso
(b) Answer: sum_{r=1}^n 1/(r(r+2)) = n(3n+5)/(4(n+1)(n+2))
(c) M1 substitutes n=20 into the formula from part (b)
(c) A1 = 325/462 cao
(c) Answer: 325/462

Question 10

(a) M1 writes f(r-1) = (r-1)r(r+1)(r+2)
(a) M1 factorises f(r)-f(r-1) as r(r+1)(r+2)[(r+3)-(r-1)]
(a) A1 = 4r(r+1)(r+2) cso
(a) Answer: f(r) - f(r-1) = 4r(r+1)(r+2) (shown)
(b) M1 writes sum_{r=1}^n 4r(r+1)(r+2) = sum_{r=1}^n [f(r)-f(r-1)]
(b) M1 recognises the sum telescopes to f(n) - f(0)
(b) A1 states f(0) = 0
(b) M1 divides both sides of 4*sum r(r+1)(r+2) = f(n) by 4
(b) A1 sum_{r=1}^n r(r+1)(r+2) = (1/4)n(n+1)(n+2)(n+3) cso
(b) Answer: sum_{r=1}^n r(r+1)(r+2) = (1/4)n(n+1)(n+2)(n+3)
(c) M1 substitutes n=15 into the formula
(c) A1 = 18360 cao
(c) Answer: 18360

Question 11

(a) M1 expands r(2r-1)(r+3) = 2r³ + 5r² - 3r
(a) M1 writes the sum as 2*sum r³ + 5*sum r² - 3*sum r
(a) M1 substitutes the three standard results correctly
(a) M1 factorises out n(n+1) from the resulting expression
(a) A1 simplifies the bracket to (3n²+13n-4)/6
(a) A1 sum = n(n+1)(3n²+13n-4)/6 cso
(a) Answer: sum_{r=1}^n r(2r-1)(r+3) = n(n+1)(3n²+13n-4)/6
(b) M1 substitutes n=12 into the formula
(b) A1 = 15184 cao
(b) Answer: 15184

Question 12

(a) M1 uses V(b) = π * integral from x=1 to x=b of y² dx with y² = 1/x²
(a) M1 integrates x^-2 to -1/x
(a) M1 substitutes the limits b and 1 correctly
(a) A1 V(b) = π(1 - 1/b) cso
(a) Answer: V(b) = π(1 - 1/b) (shown)
(b) M1 considers the limit as b -> infinity of 1/b
(b) A1 1/b -> 0 as b -> infinity
(b) A1 V(b) -> π, so the volume converges to π cao
(b) Answer: The volume converges to π
(c) B1 recognises that S has a finite volume (π, from part (b)) but an infinite surface area (since A(b) -> infinity)
(c) B1 gives a correct comment, e.g. that a solid can enclose a finite volume even though its bounding surface has infinite area, so finite volume does not imply finite surface area
(c) Answer: S has finite volume π but infinite surface area; a solid can have a finite volume while its surface area is unbounded.
(d) M1 uses V = π * integral from x=1 to x=10 of (1/x)² dx
(d) M1 integrates and substitutes the limits 10 and 1 correctly
(d) A1 V = 9*π/10 cao
(d) Answer: V = 9pi/10

Question 13

(a) B1 cos x = 1 - x²/2 oe
(a) B1 + x⁴/24 oe
(a) B1 valid for all real x
(a) Answer: cos x = 1 - x²/2 + x⁴/24 + ...; valid for all real x
(b) M1 substitutes x=0.5 into 1 - x²/2 + x⁴/24
(b) A1 = 0.8776 (4 d.p.) cao
(b) Answer: 0.8776
(c) M1 calculates |approximation - true value| / true value
(c) A1 = 0.0028% awrt (2 s.f.) cao
(c) Answer: 0.0028%

Question 14

(a) M1 sums both sides of the given identity from r=1 to n, giving 2sin(1/2)*sum cos(r) = sum [sin(r+1/2) - sin(r-1/2)]
(a) M1 lists the first two and last two terms of the right-hand sum to show the telescoping pattern
(a) A1 the right-hand sum telescopes to sin(n+1/2) - sin(1/2)
(a) M1 equates 2sin(1/2)*sum cos(r) to this telescoped result
(a) B1 divides both sides by 2sin(1/2), valid since sin(1/2) is not 0
(a) A1 sum_{r=1}^n cos(r) = (sin(n+1/2) - sin(1/2))/(2sin(1/2)) cso
(a) Answer: sum_{r=1}^n cos(r) = (sin(n+1/2) - sin(1/2))/(2sin(1/2))
(b) M1 substitutes n=50 into the formula from part (a)
(b) M1 evaluates sin(50.5) and sin(0.5) correctly (in radians)
(b) A1 = -0.258 awrt (3 s.f.) cao
(b) Answer: -0.258

Core Pure: Further Calculus and Series Depth - Worksheets, Questions and Revision