Further Mathematics

1. 1.

   Using exponential definitions, find sinh(ln 3), cosh(ln 3) and tanh(ln 3) exactly.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Here e^x = 3 and e^(-x) = 1/3, so both definitions involve rational numbers.
   2. Divide sinh by cosh for tanh; the common denominator cancels.

   Marking points

   • sinh(ln 3) = (3 - 1/3)/2 = 4/3.
   • cosh(ln 3) = (3 + 1/3)/2 = 5/3.
   • tanh(ln 3) = 4/5.

   Examiner tip: cosh uses a sum and sinh uses a difference of exponentials.

2. 2.

   Prove cosh^2 x - sinh^2 x = 1 directly from exponential definitions.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The two squared binomials differ only in their cross terms.
   2. Unlike the circular identity, the hyperbolic identity contains a subtraction.

   Marking points

   • Substitute (e^x + e^(-x))^2/4 - (e^x - e^(-x))^2/4.
   • The e^(2x) and e^(-2x) terms cancel, leaving 4e^x e^(-x)/4.
   • Since e^x e^(-x) = 1, the identity follows.

   Examiner tip: Expand the minus sign across the whole second square.

3. 3.

   Solve cosh x = 5/4 for real x, giving exact values.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Multiply by u only after noting u cannot be zero.
   2. Both positive exponential roots are admissible; the two x values reflect the evenness of cosh.

   Marking points

   • Let u = e^x > 0; u + 1/u = 5/2.
   • 2u^2 - 5u + 2 = 0 gives u = 2 or 1/2.
   • x = ln 2 or -ln 2.

   Examiner tip: The inverse cosh returns the non-negative branch, but the equation has two roots.

4. 4.

   Evaluate integral from 0 to ln 2 of sinh(2x) dx exactly.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The inner derivative is two, so integration introduces a factor one-half.
   2. At the lower limit cosh(0) = 1, not zero; subtract this before halving.

   Marking points

   • An antiderivative is cosh(2x)/2.
   • cosh(2 ln 2) = (4 + 1/4)/2 = 17/8.
   • The integral is (17/8 - 1)/2 = 9/16.

   Examiner tip: Do not use sinh at the endpoints after choosing a cosh antiderivative.

5. 5.

   Derive arsinh x = ln(x + sqrt(x^2 + 1)) from y = arsinh x, and hence differentiate arsinh x for real x.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. sqrt(x^2 + 1) exceeds |x|, so the logarithm argument is positive for every real x.
   2. Use cosh y > 0 with cosh^2 y = 1 + sinh^2 y to choose the positive square root in the derivative.

   Marking points

   • x = (e^y - e^(-y))/2 gives u^2 - 2xu - 1 = 0 with u = e^y.
   • Only u = x + sqrt(x^2 + 1) is positive; hence y = ln(x + sqrt(x^2 + 1)).
   • Implicit differentiation of x = sinh y gives dy/dx = 1/cosh y = 1/sqrt(1 + x^2).

   Examiner tip: Reject the negative quadratic root because e^y is strictly positive.

6. 6.

   Evaluate integral from 0 to sqrt(3) of sqrt(1 + x^2) dx using x = sinh u. Give an exact answer.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. At U, sinh U = sqrt(3) and cosh U = 2, so sinh(2U) = 4sqrt(3).
   2. Both terms of the antiderivative vanish at u = 0; no back substitution is needed if the limits are changed.

   Marking points

   • dx = cosh u du and sqrt(1 + x^2) = cosh u; upper limit U = ln(2 + sqrt(3)).
   • Integrate cosh^2 u = (cosh(2u) + 1)/2 to get sinh(2u)/4 + u/2.
   • The answer is sqrt(3) + ln(2 + sqrt(3))/2.

   Examiner tip: The Jacobian dx adds a second cosh factor.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.