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A-Level · Further Pure Mathematics (Further Calculus)

FP.CP4 Further Calculus

EDEXCEL 9FM0 · Calculator allowed · about 155 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question concerns the differentiation of inverse trigonometric functions.

(a)Given that y = arcsin(4x), find dy/dx, stating the values of x for which the derivative is defined.(2)

(b)Given that y = arctan(3x - 1), find dy/dx, and hence find the exact value of dy/dx when x = 1.(3)

(c)Show that d/dx[arccos(1 - 2x)] = 1/√x - x², for 0 < x < 1.(3)

(Total for Question 1 is 8 marks)

This question concerns inverse hyperbolic functions.

(a)Given that y = arsinh(x), so that sinh(y) = x, prove that dy/dx = 1/√1+x².(4)

(b)Hence write down dy/dx when y = arsinh(2x).(2)

(c)Show that arsinh(2x) = ln(2x + √4x²+1). Hence find the exact value of arsinh(4) as a single natural logarithm, and evaluate it correct to 3 significant figures.(4)

(Total for Question 2 is 10 marks)

This question concerns standard integrals leading to inverse trigonometric and inverse hyperbolic functions.

(a)Find the exact value of the integral: integral of 1/√9-x² dx.(3)

(b)Find the integral: integral of 1/√x²+16 dx.(3)

(c)Hence evaluate integral from x=0 to x=4 of 1/√25-x² dx, giving your answer to 3 significant figures.(4)

(Total for Question 3 is 10 marks)

This question concerns integration of a quadratic under a square root by completing the square.

(a)Show that 4x² + 8x + 13 = 4(x+1)² + 9.(2)

(b)Hence find integral of 1/√4x²+8x+13 dx, giving your answer in the form k*arsinh(f(x)) + c.(4)

(c)Hence evaluate integral from x=0 to x=1 of 1/√4x²+8x+13 dx, giving your answer to 3 significant figures.(4)

(Total for Question 4 is 10 marks)

Let I_n = integral from x=0 to x=1 of xⁿ * e^x dx, for n ≥ 0.

(a)Using integration by parts, show that, for n ≥ 1, I_n = e - n*I_(n-1).(4)

(b)Write down the exact value of I₀.(2)

(c)Hence find the exact value of I₃, giving your answer in the form a - b*e where a and b are integers.(4)

(Total for Question 5 is 10 marks)

Let f(x) = 1/√4-x² for 0 ≤ x ≤ √2.

(a)State the formula for the mean value of a function f(x) over the interval [a, b].(1)

(b)Find the mean value of f(x) over the interval [0, √2], giving your answer as an exact expression involving π and a surd.(4)

(c)Determine, showing your working, whether f(1) exceeds the mean value found in part (b).(3)

(Total for Question 6 is 8 marks)

A curve C has equation y = (2/3) x^3/2, for 0 ≤ x ≤ 3.

(a)Find dy/dx, and show that 1 + (dy/dx)² = 1 + x.(3)

(b)State the formula used to find the arc length of a curve y=f(x) between x=a and x=b.(1)

(c)Hence find the exact length of the arc of C between x=0 and x=3, giving your answer as a fraction in its simplest form.(4)

(Total for Question 7 is 8 marks)

A curve is defined parametrically by x = 3t², y = 2t³, for 0 ≤ t ≤ 2.

(a)Find dx/dt and dy/dt, and show that (dx/dt)² + (dy/dt)² = 36t²(1+t²).(3)

(b)State the formula for the arc length of a curve defined parametrically for α ≤ t ≤ β, and use it, together with the fact that t ≥ 0 on this curve, to show that the arc length is given by integral from t=0 to t=2 of 6t*√1+t² dt.(2)

(c)Hence show that the exact arc length of the curve for 0 ≤ t ≤ 2 is 10*√5 - 2, and find this length to 3 significant figures.(5)

(Total for Question 8 is 10 marks)

The curve y = √x, for 1 ≤ x ≤ 4, is rotated through 2*π radians about the x-axis to form a surface of revolution.

(a)Find dy/dx, and show that 1 + (dy/dx)² = (4x+1)/(4x).(3)

(b)Using the formula S = 2*π*integral of y*√1+(dy/dx)² dx for a surface of revolution about the x-axis, show that S = π*integral from x=1 to x=4 of √4x+1 dx.(2)

(c)Hence find the exact surface area generated, giving your answer in the form (π/6)(a*√a - b*√b), and evaluate it to 3 significant figures.(5)

(Total for Question 9 is 10 marks)

This question concerns improper integrals.

(a)Show that integral from x=1 to x=R of 1/x² dx = 1 - 1/R. Hence determine whether integral from x=1 to infinity of 1/x² dx converges, and if so find its value.(3)

(b)Determine whether integral from x=1 to infinity of 1/√x dx converges or diverges, justifying your answer by considering the limit as R tends to infinity.(4)

(c)The integral from x=0 to x=1 of 1/√x dx is improper due to a singularity at x=0. Show that it converges and find its exact value.(4)

(Total for Question 10 is 11 marks)

Let I_n = integral from x=0 to x=π/2 of sinⁿ(x) dx, for n ≥ 0.

(a)Using integration by parts on sinⁿ(x) = sin(x)*sin^n-1(x), show that, for n ≥ 2, I_n = ((n-1)/n) * I_(n-2).(5)

(b)State the exact values of I₀ and I₁.(2)

(c)Hence find the exact value of I₅, giving your answer as a fraction in its simplest form, and state its value correct to 3 significant figures.(4)

(Total for Question 11 is 11 marks)

A structural engineer in Bristol is modelling a section of a suspension cable using the catenary curve y = 5*cosh(x/5), where x and y are measured in metres, for -4 ≤ x ≤ 4.

(a)Find dy/dx, and show that 1 + (dy/dx)² = cosh²(x/5).(2)

(b)State the arc length formula, and use the result from part (a) to show that the length of the cable between x = -4 and x = 4 is given by integral from x=-4 to x=4 of cosh(x/5) dx.(2)

(c)Hence find the length of this section of cable, giving your answer in metres correct to 3 significant figures.(4)

(Total for Question 12 is 8 marks)

Mark scheme · FP.CP4 Further Calculus

Question 1

(a) M1 uses d/dx[arcsin(u)] = u'/√1-u² with u = 4x
(a) A1 dy/dx = 4/√1-16x², oe, valid for -1/4 < x < 1/4
(a) Answer: dy/dx = 4/√1 - 16x², for -1/4 < x < 1/4
(b) M1 uses d/dx[arctan(u)] = u'/(1+u²) with u = 3x-1
(b) A1 dy/dx = 3/(1+(3x-1)²), oe
(b) A1 substitutes x=1 to obtain dy/dx = 3/5, cao
(b) Answer: dy/dx = 3/(1+(3x-1)²); at x = 1, dy/dx = 3/5
(c) M1 uses d/dx[arccos(u)] = -u'/√1-u² with u = 1-2x, so du/dx = -2
(c) M1 1-(1-2x)² simplified to 4x - 4x² = 4x(1-x)
(c) A1 correctly simplifies 2/√4x(1-x) to 1/√x-x², cso
(c) Answer: d/dx[arccos(1-2x)] = 1/√x-x² (printed result)

Question 2

(a) M1 differentiates sinh(y) = x implicitly to get cosh(y) dy/dx = 1
(a) M1 uses cosh²(y) - sinh²(y) = 1 to write cosh(y) = √1+sinh²(y)
(a) M1 substitutes sinh(y) = x to get cosh(y) = √1+x²
(a) A1 dy/dx = 1/√1+x², cso (printed result)
(a) Answer: dy/dx = 1/√1+x² (printed result)
(b) M1 applies the chain rule with u = 2x, du/dx = 2
(b) A1 dy/dx = 2/√1+4x², oe
(b) Answer: dy/dx = 2/√1+4x²
(c) M1 sets sinh(y) = 2x, writes (e^y - e^-y)/2 = 2x and forms e^{2y} - 4x e^y - 1 = 0
(c) M1 solves the quadratic in e^y, taking the positive root e^y = 2x + √4x²+1
(c) A1 arsinh(2x) = ln(2x + √4x²+1), cso (printed result)
(c) A1 substitutes x=2 to get arsinh(4) = ln(4+√17), awrt 2.09
(c) Answer: arsinh(2x) = ln(2x+√4x²+1) (printed result); arsinh(4) = ln(4+√17) = awrt 2.09

Question 3

(a) M1 recognises the standard form integral 1/√a²-x² dx = arcsin(x/a) + c with a=3
(a) A1 arcsin(x/3) seen
(a) A1 + c included, cao
(a) Answer: arcsin(x/3) + c
(b) M1 recognises the standard form integral 1/√x²+a² dx = arsinh(x/a) + c with a=4
(b) A1 arsinh(x/4) seen
(b) A1 + c included, oe (accept equivalent log form ln(x+√x²+16)+k)
(b) Answer: arsinh(x/4) + c
(c) M1 writes the integral as [arcsin(x/5)] with limits 0 to 4
(c) M1 substitutes limits to get arcsin(4/5) - arcsin(0)
(c) A1 exact value arcsin(4/5) (or arcsin(0.8)), oe
(c) A1 awrt 0.927 (radians)
(c) Answer: arcsin(4/5) = awrt 0.927

Question 4

(a) M1 writes 4x²+8x+13 = 4(x²+2x)+13 = 4[(x+1)² - 1] + 13
(a) A1 simplifies to 4(x+1)² + 9, cso (printed result)
(a) Answer: 4x²+8x+13 = 4(x+1)²+9 (printed result)
(b) M1 writes √4x²+8x+13 = √4(x+1)²+9 = 2sqrt((x+1)²+9/4)
(b) M1 uses standard form 1/√u²+a² du = arsinh(u/a)+c with u=x+1, a=3/2
(b) A1 (1/2) arsinh(2(x+1)/3) seen, oe
(b) A1 + c included, cao
(b) Answer: (1/2) arsinh(2(x+1)/3) + c
(c) M1 substitutes limits into (1/2)arsinh(2(x+1)/3), giving (1/2)[arsinh(4/3) - arsinh(2/3)]
(c) M1 uses arsinh(4/3) = ln(3) (since 4/3+5/3=3) or equivalent numeric evaluation
(c) A1 arsinh(2/3) evaluated correctly, awrt 0.625
(c) A1 final answer awrt 0.237
(c) Answer: awrt 0.237

Question 5

(a) M1 chooses u=xⁿ, dv/dx = e^x, so du/dx = n*x^n-1, v = e^x
(a) M1 applies integration by parts: I_n = [xⁿ e^x]_0¹ - n*integral of x^n-1 e^x dx
(a) A1 evaluates boundary term [xⁿ e^x]_0¹ = e - 0 = e
(a) A1 I_n = e - n*I_(n-1), cso (printed result)
(a) Answer: I_n = e - n*I_(n-1) (printed result)
(b) M1 states I₀ = integral from 0 to 1 of e^x dx = [e^x]_0¹
(b) A1 I₀ = e - 1, cao
(b) Answer: I₀ = e - 1
(c) M1 uses I₁ = e - 1*I₀ = e-(e-1) = 1
(c) M1 uses I₂ = e - 2*I₁ = e-2
(c) A1 uses I₃ = e - 3*I₂ = e-3(e-2) = 6-2e
(c) A1 I₃ = 6 - 2e, cao (awrt 0.563)
(c) Answer: I₃ = 6 - 2e (awrt 0.563)

Question 6

(a) B1 mean value = (1/(b-a)) * integral from a to b of f(x) dx, oe
(a) Answer: Mean value = (1/(b-a)) integral from a to b of f(x) dx
(b) M1 writes mean value = (1/(√2-0)) * integral from 0 to √2 of 1/√4-x² dx
(b) M1 integrates to arcsin(x/2), and substitutes limits to get arcsin(√2/2) - arcsin(0)
(b) A1 recognises arcsin(√2/2) = π/4, giving integral = π/4
(b) A1 mean value = (π*√2)/8, oe cao
(b) Answer: Mean value = (π*√2)/8 (awrt 0.555)
(c) M1 evaluates f(1) = 1/√4-1 = 1/√3, awrt 0.577
(c) A1 compares with mean value awrt 0.555 from part (b)
(c) A1 correct conclusion that f(1) exceeds the mean value, ft from parts (b)
(c) Answer: f(1) = 1/√3 = awrt 0.577, which exceeds the mean value of awrt 0.555, so f(1) exceeds the mean value.

Question 7

(a) M1 differentiates to get dy/dx = (2/3)*(3/2)*x^1/2
(a) A1 dy/dx = √x, oe
(a) A1 1 + (dy/dx)² = 1 + (√x)² = 1+x, cso (printed result)
(a) Answer: dy/dx = √x; 1+(dy/dx)² = 1+x (printed result)
(b) B1 arc length = integral from a to b of √1+(dy/dx)² dx, oe
(b) Answer: Arc length = integral from a to b of √1+(dy/dx)² dx
(c) M1 sets up arc length = integral from 0 to 3 of √1+x dx
(c) M1 integrates to (2/3)(1+x)^3/2
(c) A1 substitutes limits to get (2/3)(4^3/2 - 1^3/2) = (2/3)(8-1)
(c) A1 arc length = 14/3, cao
(c) Answer: Arc length = 14/3 (awrt 4.67)

Question 8

(a) M1 differentiates to get dx/dt = 6t, dy/dt = 6t²
(a) M1 squares and adds: 36t² + 36t⁴
(a) A1 factorises to 36t²(1+t²), cso (printed result)
(a) Answer: dx/dt = 6t, dy/dt = 6t²; (dx/dt)²+(dy/dt)² = 36t²(1+t²) (printed result)
(b) B1 states arc length = integral from α to β of √(dx/dt)²+(dy/dt)² dt
(b) B1 uses t≥0 to take the positive square root, √36t²(1+t²) = 6t*√1+t²
(b) Answer: Arc length = integral from 0 to 2 of 6t*√1+t² dt
(c) M1 uses substitution u=1+t², du=2t dt to rewrite the integral as 3*integral of √u du
(c) M1 integrates to get 2u^3/2
(c) A1 changes limits (t=0 to u=1, t=2 to u=5) and substitutes to get 2(5^3/2 - 1)
(c) A1 simplifies to 10*√5 - 2, cso (printed result)
(c) A1 evaluates to awrt 20.4
(c) Answer: Arc length = 10*√5 - 2 (printed result), awrt 20.4

Question 9

(a) M1 differentiates to get dy/dx = 1/(2*√x)
(a) M1 squares to get (dy/dx)² = 1/(4x)
(a) A1 1+1/(4x) = (4x+1)/(4x), cso (printed result)
(a) Answer: dy/dx = 1/(2sqrt(x)); 1+(dy/dx)² = (4x+1)/(4x) (printed result)
(b) M1 substitutes y=√x and the result from part (a) into the surface area formula
(b) A1 simplifies √x*√(4x+1)/(4x) = √4x+1/2 to reach S = π*integral of √4x+1 dx, cso (printed result)
(b) Answer: S = π*integral from 1 to 4 of √4x+1 dx (printed result)
(c) M1 integrates √4x+1 to get (1/6)(4x+1)^3/2
(c) M1 substitutes upper limit x=4: (1/6)(17)^3/2 = (1/6)(17*√17)
(c) A1 substitutes lower limit x=1: (1/6)(5)^3/2 = (1/6)(5*√5)
(c) A1 S = (π/6)(17*√17 - 5*√5), cso
(c) A1 evaluates to awrt 30.8
(c) Answer: S = (π/6)(17*√17 - 5*√5) (awrt 30.8)

Question 10

(a) M1 integrates to get [-1/x]_1^R = -1/R - (-1) = 1-1/R (printed result)
(a) M1 considers the limit as R tends to infinity, noting 1/R tends to 0
(a) A1 concludes the integral converges to 1, cso
(a) Answer: Integral converges to 1
(b) M1 integrates to get integral from 1 to R of x^-1/2 dx = [2*√x]_1^R
(b) M1 evaluates to 2*√R - 2
(b) A1 considers limit as R tends to infinity and notes 2*√R tends to infinity
(b) A1 correct conclusion that the integral diverges, with justification
(b) Answer: Integral diverges (2*√R-2 tends to infinity as R tends to infinity)
(c) M1 considers integral from x=epsilon to x=1 of x^-1/2 dx = [2*√x]_epsilon¹
(c) M1 evaluates to 2 - 2*√epsilon
(c) A1 takes the limit as epsilon tends to 0+ and notes 2*√epsilon tends to 0
(c) A1 concludes the integral converges to 2, cso
(c) Answer: Integral converges to 2

Question 11

(a) M1 chooses u=sin^n-1(x), dv/dx=sin(x), so du/dx=(n-1)sin^n-2(x)cos(x), v=-cos(x)
(a) M1 applies parts and evaluates boundary term [-cos(x)sin^n-1(x)]_0^π/2 = 0
(a) M1 uses cos²(x) = 1-sin²(x) to write (n-1)*integral of (1-sin²(x))sin^n-2(x) dx
(a) M1 expands to I_n = (n-1)(I_(n-2) - I_n) and rearranges to collect I_n terms: n*I_n = (n-1)*I_(n-2)
(a) A1 I_n = ((n-1)/n)*I_(n-2), cso (printed result)
(a) Answer: I_n = ((n-1)/n) * I_(n-2) (printed result)
(b) B1 I₀ = π/2
(b) B1 I₁ = 1
(b) Answer: I₀ = π/2, I₁ = 1
(c) M1 uses I₃ = (2/3)*I₁ = 2/3
(c) M1 uses I₅ = (4/5)*I₃
(c) A1 I₅ = 8/15, cao
(c) A1 awrt 0.533
(c) Answer: I₅ = 8/15 (awrt 0.533)

Question 12

(a) M1 differentiates to get dy/dx = sinh(x/5)
(a) A1 uses cosh²(x/5) - sinh²(x/5) = 1 to show 1+sinh²(x/5) = cosh²(x/5), cso (printed result)
(a) Answer: dy/dx = sinh(x/5); 1+(dy/dx)² = cosh²(x/5) (printed result)
(b) B1 states arc length = integral from a to b of √1+(dy/dx)² dx
(b) B1 uses cosh(x/5) > 0 to take the positive square root, √cosh²(x/5) = cosh(x/5)
(b) Answer: Arc length = integral from -4 to 4 of cosh(x/5) dx
(c) M1 integrates to get [5*sinh(x/5)]_(-4)⁴
(c) M1 uses sinh being an odd function to write 5[sinh(4/5)-sinh(-4/5)] = 10*sinh(0.8)
(c) A1 exact form 10*sinh(0.8) (or equivalent using exponentials), cao
(c) A1 evaluates to awrt 8.88 metres
(c) Answer: Length = 10*sinh(0.8) metres, awrt 8.88 m

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