Hyperbolic Functions - Worksheets, Questions and Revision

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A-Level · Core Pure Mathematics (Further Maths)

FP.CP7 Hyperbolic Functions

EDEXCEL 9FM0 · Calculator allowed · about 135 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question concerns the exponential definitions of the hyperbolic functions.

(a)State, using exponential functions, the definitions of cosh x and sinh x.(2)

(b)Given that cosh x = 2 and x > 0, use the identity cosh²(x) - sinh²(x) = 1 to find the exact value of sinh x, and hence write down the exact value of tanh x.(3)

(c)Find the exact value of e^x.(2)

(Total for Question 1 is 7 marks)

Throughout this question you must derive each result from the exponential definitions of cosh x and sinh x, showing each line of algebra clearly.

(a)Show that cosh²(x) - sinh²(x) = 1 for all real x.(3)

(b)Show that sinh 2x = 2 sinh x cosh x.(3)

(c)Hence, or otherwise, show that cosh 2x = 1 + 2 sinh²(x).(3)

(Total for Question 2 is 9 marks)

This question concerns the graphs of the hyperbolic functions.

(a)Sketch the graph of y = cosh x, marking the y-intercept and stating the coordinates and nature of any turning point.(3)

(b)Sketch the graph of y = tanh x, stating the equations of its horizontal asymptotes.(2)

(c)With reference to the graph of y = sinh x, explain why the equation sinh x = k has exactly one real solution for every real number k.(2)

(Total for Question 3 is 7 marks)

Consider the equation 2 cosh x - sinh x = 2.

(a)Show that the equation 2 cosh x - sinh x = 2 can be written as e^2x - 4e^x + 3 = 0.(4)

(b)Hence solve the equation 2 cosh x - sinh x = 2, giving your answer(s) exactly.(4)

(Total for Question 4 is 8 marks)

This question builds towards solving a hyperbolic equation using the double angle formula for cosh.

(a)Using the exponential definitions of cosh x and sinh x, show that cosh(x+y) = cosh x cosh y + sinh x sinh y.(4)

(b)Hence show that cosh 2x = 2 cosh²(x) - 1.(3)

(c)Hence, or otherwise, solve the equation cosh 2x = 5 cosh x - 4, giving each solution exactly in terms of natural logarithms.(5)

(Total for Question 5 is 12 marks)

This question is about the inverse hyperbolic function arsinh x, defined for all real x.

(a)Let y = arsinh x, so that x = sinh y. By writing this equation in exponential form, show that e^2y - 2x e^y - 1 = 0, and hence show that arsinh x = ln(x + √x²+1) for all real x.(5)

(b)Hence find, in the form ln(a + √b) where a and b are integers, the exact value of arsinh(2).(2)

(c)Solve the equation arsinh(x) = ln 3, giving your answer as an exact fraction.(3)

(Total for Question 6 is 10 marks)

This question is about differentiating hyperbolic and inverse hyperbolic functions.

(a)Differentiate y = sinh(3x² - 1) with respect to x.(2)

(b)Find dy/dx where y = ln(cosh 2x), giving your answer in terms of tanh 2x.(3)

(c)Given that y = arcosh(3x), for x > 1/3, show by implicit differentiation that dy/dx = 3/√9x²-1.(3)

(d)Find the equation of the tangent to the curve y = arsinh(x/2) at the point where x = 0, giving your answer in the form y = mx.(4)

(Total for Question 7 is 12 marks)

This question is about integrating functions involving hyperbolic functions.

(a)Find, in terms of x, the integral: integral of sinh(4x - 1) dx.(2)

(b)Find, in terms of arsinh, the integral: integral of 1/√x²+9 dx.(3)

(c)Using the substitution x = 4 sinh t, show that integral of 1/√x²+16 dx = arsinh(x/4) + c.(5)

(d)Hence evaluate exactly the integral from x = 0 to x = 4 of 1/√x²+16 dx, giving your answer as a single natural logarithm.(2)

(Total for Question 8 is 12 marks)

This question is about integration leading to the logarithmic form of artanh.

(a)Using the standard result integral of 1/(a²-x²) dx = (1/a) artanh(x/a) + c and the logarithmic form artanh(u) = (1/2) ln((1+u)/(1-u)), show that, for -3 < x < 3, integral of 1/(9-x²) dx = (1/6) ln((3+x)/(3-x)) + c.(4)

(b)Hence find the exact value of the integral from x = 0 to x = 1 of 1/(9-x²) dx, giving your answer in the form (1/6) ln k, stating the value of k.(3)

(Total for Question 9 is 7 marks)

The curve C has parametric equations x = 5 cosh t, y = 3 sinh t, for t in R.

(a)Show that the Cartesian equation of C is x²/25 - y²/9 = 1.(3)

(b)Find dy/dx in terms of t, in its simplest form.(3)

(c)The point P on C has parameter t = p. Show that the equation of the tangent to C at P can be written as (x cosh p)/5 - (y sinh p)/3 = 1.(5)

(Total for Question 10 is 11 marks)

Consider the equation 3 sinh²(x) - 5 cosh x + 3 = 0.

(a)Show that the equation 3 sinh²(x) - 5 cosh x + 3 = 0 can be written as cosh x (3 cosh x - 5) = 0.(3)

(b)Hence solve the equation 3 sinh²(x) - 5 cosh x + 3 = 0, explaining why one of the values of cosh x found must be rejected, and giving your final answer(s) as exact logarithms.(5)

(Total for Question 11 is 8 marks)

An engineer, Priya, models a cable hanging between two supports of equal height as a catenary with equation y = 20 cosh(x/20) - 20, for -15 ≤ x ≤ 15, where x and y are measured in metres and x = 0 at the lowest point of the cable.

(a)Find the sag of the cable, that is, the value of y at x = 15, giving your answer in metres to 3 significant figures.(3)

(b)Find dy/dx and hence find the gradient of the cable at x = 15, giving your answer to 3 significant figures.(3)

(c)Using the arc length formula L = integral from x=-15 to x=15 of √1 + (dy/dx)² dx, and the identity 1 + sinh²(t) = cosh²(t), show that the total length of cable between the two supports is 32.9 m, correct to 3 significant figures.(5)

(Total for Question 12 is 11 marks)

Mark scheme · FP.CP7 Hyperbolic Functions

Question 1

(a) B1 cosh x = (e^x + e^-x)/2 oe
(a) B1 sinh x = (e^x - e^-x)/2 oe
(a) Answer: cosh x = (e^x + e^-x)/2, sinh x = (e^x - e^-x)/2
(b) M1 sinh²(x) = cosh²(x) - 1 = 2² - 1 = 3
(b) A1 sinh x = √3 (positive root taken since x > 0), cao
(b) B1 tanh x = √3/2, ft from their sinh x
(b) Answer: sinh x = √3, tanh x = √3/2
(c) M1 e^x = cosh x + sinh x
(c) A1 e^x = 2 + √3, cao
(c) Answer: e^x = 2 + √3

Question 2

(a) M1 expand cosh²(x) = (e^x+e^-x)²/4 and sinh²(x) = (e^x-e^-x)²/4
(a) A1 correct expansions (e^2x+2+e^-2x)/4 and (e^2x-2+e^-2x)/4
(a) A1 subtract to obtain 4/4 = 1, cso (answer printed)
(a) Answer: cosh²(x) - sinh²(x) = 1
(b) M1 write 2 sinh x cosh x = 2 . [(e^x-e^-x)/2] . [(e^x+e^-x)/2]
(b) A1 simplify to (e^2x - e^-2x)/2
(b) A1 identify this as sinh 2x by the exponential definition, cso
(b) Answer: sinh 2x = 2 sinh x cosh x
(c) M1 write cosh 2x = (e^2x+e^-2x)/2 by definition
(c) M1 expand 2 sinh²(x) = (e^2x - 2 + e^-2x)/2
(c) A1 1 + 2sinh²(x) simplifies to (e^2x+e^-2x)/2 = cosh 2x, cso (answer printed)
(c) Answer: cosh 2x = 1 + 2 sinh²(x)

Question 3

(a) B1 correct shape: even function, symmetrical about the y-axis, y ≥ 1 for all x
(a) B1 y-intercept marked at (0, 1)
(a) B1 minimum turning point identified at (0, 1)
(a) Answer: U-shaped curve symmetrical about the y-axis, minimum at (0,1), y-intercept (0,1), no x-intercepts
(b) B1 correct S-shaped, increasing, odd-function curve through the origin
(b) B1 asymptotes y = 1 and y = -1 stated
(b) Answer: y = 1 and y = -1
(c) B1 sinh x is a strictly increasing function for all real x, with range all of R
(c) B1 hence the horizontal line y = k meets the graph exactly once for any real k
(c) Answer: sinh x is a strictly increasing bijection from R to R, so y = k meets the graph exactly once for every real k

Question 4

(a) M1 write 2cosh x - sinh x in exponential form
(a) A1 simplify to e^x + 3e^-x = 4 oe
(a) M1 multiply through by e^x
(a) A1 e^2x - 4e^x + 3 = 0, cso (answer printed)
(a) Answer: e^2x - 4e^x + 3 = 0
(b) M1 let u = e^x to form u² - 4u + 3 = 0
(b) M1 factorise (u-1)(u-3) = 0, or use the quadratic formula
(b) A1 u = 1 or u = 3, i.e. e^x = 1 or e^x = 3
(b) A1 x = 0 and x = ln 3, both required, cao
(b) Answer: x = 0 or x = ln 3

Question 5

(a) M1 write cosh x cosh y = (e^x+e^-x)(e^y+e^-y)/4 and sinh x sinh y = (e^x-e^-x)(e^y-e^-y)/4
(a) A1 expand both products correctly in terms of e^x+-y
(a) M1 add the two expansions, cancelling the e^x-y and e^-x+y terms with opposite sign
(a) A1 result simplifies to (e^x+y+e^-(x+y))/2 = cosh(x+y), cso
(a) Answer: cosh(x+y) = cosh x cosh y + sinh x sinh y
(b) M1 set y = x in the result from part (a): cosh 2x = cosh²(x) + sinh²(x)
(b) M1 substitute sinh²(x) = cosh²(x) - 1
(b) A1 cosh 2x = 2cosh²(x) - 1, cso (answer printed)
(b) Answer: cosh 2x = 2 cosh²(x) - 1
(c) M1 substitute cosh 2x = 2cosh²(x) - 1 into the equation
(c) M1 rearrange to the quadratic 2cosh²(x) - 5cosh x + 3 = 0
(c) M1 factorise (2cosh x - 3)(cosh x - 1) = 0 to give cosh x = 3/2 or cosh x = 1
(c) A1 cosh x = 1 gives x = 0
(c) A1 cosh x = 3/2 gives x = +-ln((3+√5)/2) (awrt ±0.962), cao
(c) Answer: x = 0, x = ln((3+√5)/2), x = -ln((3+√5)/2) (awrt 0, 0.962, -0.962)

Question 6

(a) M1 write x = sinh y = (e^y - e^-y)/2
(a) M1 multiply by 2e^y and rearrange to e^2y - 2x e^y - 1 = 0
(a) M1 solve this quadratic in e^y using the quadratic formula: e^y = x ± √x²+1
(a) A1 reject the negative root since e^y > 0 but x - √x²+1 < 0 for all x
(a) A1 hence y = ln(x + √x²+1), i.e. arsinh x = ln(x+√x²+1), cso
(a) Answer: arsinh x = ln(x + √x²+1)
(b) M1 substitute x = 2 into the formula from part (a)
(b) A1 arsinh(2) = ln(2 + √5), cao
(b) Answer: arsinh(2) = ln(2 + √5)
(c) M1 take sinh of both sides: x = sinh(ln 3) = (3 - 1/3)/2
(c) M1 simplify (3 - 1/3)/2 = (8/3)/2
(c) A1 x = 4/3, cao
(c) Answer: x = 4/3

Question 7

(a) M1 apply the chain rule to obtain the form k . x . cosh(3x²-1)
(a) A1 dy/dx = 6x cosh(3x²-1), cao
(a) Answer: dy/dx = 6x cosh(3x² - 1)
(b) M1 use d/dx[ln u] = u'/u with u = cosh 2x, u' = 2 sinh 2x
(b) A1 dy/dx = 2 sinh 2x / cosh 2x
(b) A1 dy/dx = 2 tanh 2x, cao
(b) Answer: dy/dx = 2 tanh 2x
(c) M1 write cosh y = 3x and differentiate implicitly: sinh y . dy/dx = 3
(c) M1 use sinh y = √cosh²(y) - 1 = √9x²-1, taking the positive root since y > 0
(c) A1 dy/dx = 3/√9x²-1, cso
(c) Answer: dy/dx = 3/√9x² - 1
(d) B1 identify the point (0, 0) since arsinh(0) = 0
(d) M1 differentiate using d/dx[arsinh(u)] = u'/√u²+1 with u = x/2, u' = 1/2
(d) A1 dy/dx at x = 0 is (1/2)/√0+1 = 1/2
(d) A1 tangent: y = (1/2)x, oe (e.g. 2y = x), cao
(d) Answer: y = (1/2)x

Question 8

(a) M1 form k cosh(4x-1)
(a) A1 (1/4)cosh(4x-1) + c, cao
(a) Answer: (1/4)cosh(4x-1) + c
(b) B1 quote the standard result integral of 1/√x²+a² dx = arsinh(x/a) + c
(b) M1 identify a² = 9, so a = 3
(b) A1 arsinh(x/3) + c, cao
(b) Answer: arsinh(x/3) + c
(c) M1 substitute x = 4 sinh t, dx = 4 cosh t dt
(c) M1 use 1 + sinh²(t) = cosh²(t) to write √x²+16 = √16sinh²(t)+16 = 4cosh t
(c) A1 integral becomes integral of (4cosh t)/(4cosh t) dt = integral of 1 dt
(c) A1 = t + c, and t = arsinh(x/4) since x/4 = sinh t
(c) A1 integral of 1/√x²+16 dx = arsinh(x/4) + c, cso (answer printed)
(c) Answer: arsinh(x/4) + c
(d) M1 substitute limits: arsinh(1) - arsinh(0)
(d) A1 = ln(1 + √2), cao
(d) Answer: ln(1 + √2)

Question 9

(a) M1 apply the standard result with a = 3: (1/3) artanh(x/3) + c
(a) M1 substitute artanh(x/3) = (1/2) ln((1+x/3)/(1-x/3))
(a) A1 simplify (1/3)(1/2)ln((1+x/3)/(1-x/3)) to (1/6) ln((3+x)/(3-x))
(a) A1 integral of 1/(9-x²) dx = (1/6) ln((3+x)/(3-x)) + c, cso (answer printed)
(a) Answer: (1/6) ln((3+x)/(3-x)) + c
(b) M1 substitute limits: (1/6)ln(4/2) - (1/6)ln(3/3)
(b) A1 simplify to (1/6) ln 2, since ln(3/3) = ln 1 = 0
(b) A1 k = 2, cao
(b) Answer: (1/6) ln 2, so k = 2

Question 10

(a) M1 write cosh t = x/5 and sinh t = y/3
(a) M1 use cosh²(t) - sinh²(t) = 1
(a) A1 x²/25 - y²/9 = 1, cso (answer printed)
(a) Answer: x²/25 - y²/9 = 1
(b) M1 differentiate: dx/dt = 5 sinh t, dy/dt = 3 cosh t
(b) M1 use dy/dx = (dy/dt) / (dx/dt)
(b) A1 dy/dx = (3/5) coth t, oe (e.g. 3cosh t / (5 sinh t)), cao
(b) Answer: dy/dx = (3/5) coth t
(c) M1 write the tangent at P(5cosh p, 3sinh p) using gradient (3/5)coth p: y - 3sinh p = (3/5)coth p (x - 5cosh p)
(c) M1 multiply throughout by 5 sinh p to clear fractions
(c) A1 obtain 5y sinh p - 15sinh²(p) = 3x cosh p - 15cosh²(p)
(c) M1 use cosh²(p) - sinh²(p) = 1 to rearrange to 3x cosh p - 5y sinh p = 15
(c) A1 divide by 15 to obtain (x cosh p)/5 - (y sinh p)/3 = 1, cso (answer printed)
(c) Answer: (x cosh p)/5 - (y sinh p)/3 = 1

Question 11

(a) M1 substitute sinh²(x) = cosh²(x) - 1
(a) A1 simplify to 3cosh²(x) - 5cosh x = 0, oe
(a) A1 factorise to cosh x (3cosh x - 5) = 0, cso (answer printed)
(a) Answer: cosh x (3 cosh x - 5) = 0
(b) M1 set each factor to zero: cosh x = 0 or cosh x = 5/3
(b) B1 reject cosh x = 0, since cosh x ≥ 1 for all real x (no real solution)
(b) M1 cosh x = 5/3 gives x = +-arcosh(5/3)
(b) A1 arcosh(5/3) = ln(5/3 + √25/9-1) = ln(5/3+4/3) = ln 3
(b) A1 x = ln 3 or x = -ln 3, cao
(b) Answer: x = ln 3 or x = -ln 3