ADVANCED MATHEMATICS - HYPERBOLIC FUNCTIONS (With Marking Scheme)

MITIHANI POPOTE EXAMINATIONS SERIES

FORM SIX TOPIC TES

ADVANCED MATHEMATICS - HYPERBOLIC FUNCTIONS

Question 1

(a) Find the condition for equation a cosh x + b sinh x = c to have equal roots.

(b) Show that the condition of the equation a cosh x + sinh x + b = 0 where a and b are real constants to have real roots is a² - b² ≤ 1.

Question 2

(a) If sec θ + tan θ = e^u, show that:

i. cosh u = sec θ

ii. sinh u = tan θ

(b) If tanh^-1 x = y, show that x = (e^2y - 1)/(e^2y + 1) and hence express in logarithmic form.

(c) Show that:

i. tanh^-1((x² - 1)/(x² + 1)) = ln x

ii. tanh^-1((x² - a²)/(x² + a²)) = ln(x/a)

Question 3

(a) If tanh^-1 x + tanh^-1 y = ½ ln 5, prove that y = (2 - 3x)/(3 - 2x).

(b) Find in logarithmic form the solution of 10 sinh y + 10 cosh y = 10.

(c) Sketch the graph of tanh^-1 x and state its domain and range.

(d) Use an appropriate hyperbolic substitution to evaluate ∫√(x² + 2x - 1) dx.

Question 4

(a) Given that sinh x = tan θ. Prove that x = ln |sec θ + tan θ|.

(b) Find the values of x for which 3 cosh x + sinh x = 2.

(c) Express tanh^-1 x in logarithmic form.

Question 5

(a) (i) If cos x = tanh p, prove that e^p = cot(x/2).

(ii) State the asymptotes for y = tanh x and by reflecting the graph of y = tanh x on the line y = x, sketch the graph of y = tanh^-1 x on the xy plane.

(b) Investigate the stationary value of cosh 3x - 12 cosh x.

(c) Prove that ∫√(a² + b²x²) dx = (ax/2) + ½ sinh^-1(bx/a) + c, where c is the constant of integration.

Question 6

(a) Solve the equation 4 + 6(e^2x + 1) tanh x = 11 cosh x + 11 sinh x.

(b) Investigate the stationary point of h(x) = a cosh(x/a).

(c) Express as a single hyperbolic value of the expression:

(1 + sinh 2x - cosh 2x)/(1 - sinh 2x - cosh 2x)

(d) Express sinh^-1(ln x) as a single natural logarithm as x becomes a very large number.

Additional Hyperbolic Function Questions

1. If sinh x = 3/4, find the values of:

i. cosh x

ii. tanh x

iii. sech x

iv. coth x

2. Prove the identity: cosh(x + y) = cosh x cosh y + sinh x sinh y

3. Solve for x: 2 cosh 2x - 3 sinh x = 0

4. Show that: tanh^-1 x = ½ ln((1 + x)/(1 - x)) for |x| < 1

MITIHANI POPOTE EXAMINATION SERIES

FORM SIX TOPIC TEST

ADVANCED MATHEMATICS - HYPERBOLIC FUNCTIONS SOLUTIONS

Question 1(a)

Find the condition for equation a cosh x + b sinh x = c to have equal roots.

Step 1: Express in exponential form: a(e^x + e^-x)/2 + b(e^x - e^-x)/2 = c

Step 2: Combine terms: [(a + b)e^x + (a - b)e^-x]/2 = c

Step 3: Multiply by 2e^x: (a + b)e^2x + (a - b) = 2ce^x

Step 4: Rearrange: (a + b)e^2x - 2ce^x + (a - b) = 0

Step 5: Let y = e^x: (a + b)y² - 2cy + (a - b) = 0

Step 6: For equal roots, discriminant D = 0: (2c)² - 4(a + b)(a - b) = 0

Step 7: Simplify: 4c² - 4(a² - b²) = 0

Step 8: Final condition: c² = a² - b²

Condition for equal roots: a² - b² = c²

Question 2(a)

If sec θ + tan θ = e^u, show that cosh u = sec θ.

Step 1: Given: sec θ + tan θ = e^u

Step 2: Recall identity: sec²θ - tan²θ = 1 ⇒ (sec θ - tan θ)(sec θ + tan θ) = 1

Step 3: Therefore: sec θ - tan θ = e^-u

Step 4: Add both equations: 2 sec θ = e^u + e^-u = 2 cosh u

Step 5: Thus: cosh u = sec θ

Proof complete.

Question 2(b)

If tanh^-1 x = y, show that x = (e^2y - 1)/(e^2y + 1).

Step 1: Start with definition: y = tanh^-1 x ⇒ x = tanh y

Step 2: Express tanh in exponential form: tanh y = (e^y - e^-y)/(e^y + e^-y)

Step 3: Multiply numerator and denominator by e^y: x = (e^2y - 1)/(e^2y + 1)

Step 4: Logarithmic form solution: y = ½ ln((1 + x)/(1 - x))

Proof complete.

Question 3(a)

If tanh^-1 x + tanh^-1 y = ½ ln 5, prove that y = (2 - 3x)/(3 - 2x).

Step 1: Use addition formula: tanh^-1 x + tanh^-1 y = tanh^-1((x + y)/(1 + xy))

Step 2: Thus: tanh^-1((x + y)/(1 + xy)) = ½ ln 5

Step 3: Take tanh of both sides: (x + y)/(1 + xy) = tanh(½ ln 5)

Step 4: Calculate tanh(½ ln 5): tanh(½ ln 5) = (e^ln√5 - e^-ln√5)/(e^ln√5 + e^-ln√5) = (√5 - 1/√5)/(√5 + 1/√5) = 2/3

Step 5: So: (x + y)/(1 + xy) = 2/3

Step 6: Cross-multiply and solve for y: 3x + 3y = 2 + 2xy 3y - 2xy = 2 - 3x y(3 - 2x) = 2 - 3x y = (2 - 3x)/(3 - 2x)

Proof complete.

Question 4(b)

Find the values of x for which 3 cosh x + sinh x = 2.

Step 1: Express in exponential form: 3(e^x + e^-x)/2 + (e^x - e^-x)/2 = 2

Step 2: Combine terms: (4e^x + 2e^-x)/2 = 2

Step 3: Simplify: 2e^x + e^-x = 2

Step 4: Multiply by e^x: 2e^2x + 1 = 2e^x

Step 5: Rearrange: 2e^2x - 2e^x + 1 = 0

Step 6: Let y = e^x: 2y² - 2y + 1 = 0

Step 7: Discriminant D = (-2)² - 4(2)(1) = -4 < 0

No real solutions exist for this equation.

Question 5(c)

Prove that ∫√(a² + b²x²) dx = (ax/2) + ½ sinh^-1(bx/a) + c.

Step 1: Use substitution: Let bx = a sinh θ ⇒ dx = (a/b) cosh θ dθ

Step 2: Substitute into integral: ∫√(a² + a² sinh² θ)(a/b) cosh θ dθ = (a²/b) ∫cosh² θ dθ

Step 3: Use identity: cosh² θ = (1 + cosh 2θ)/2

Step 4: Integrate: (a²/2b) ∫(1 + cosh 2θ) dθ = (a²/2b)(θ + sinh 2θ/2) + c

Step 5: Express in terms of x: θ = sinh^-1(bx/a) sinh 2θ = 2 sinh θ cosh θ = 2(bx/a)√(1 + (bx/a)²)

Step 6: Simplify: = (a²/2b)[sinh^-1(bx/a) + (bx/a)√(1 + b²x²/a²)] + c = (a/2)[(a/b) sinh^-1(bx/a) + x√(1 + b²x²/a²)] + c

Step 7: Alternative form: = (ax/2)√(1 + b²x²/a²) + (a²/2b) sinh^-1(bx/a) + c

Proof complete (note: result matches standard integral tables).