FP3 June 2017 Q3
3.
| Scheme | Marks |
|---|---|
| \(\cosh 2x \equiv 2\cosh^2 x - 1\) Note that exponentials must be used in (a) | |
| Way 1: rhs \(= 2\cosh^2 x - 1 = 2\left(\dfrac{e^x + e^{-x}}{2}\right)^2 - 1\) | M1 |
| \(= 2\left(\dfrac{e^{2x} + 2 + e^{-2x}}{4}\right) - 1\) | dM1 |
| \(= \dfrac{e^{2x} + e^{-2x}}{2} + 1 - 1\) | |
| \(= \dfrac{e^{2x} + e^{-2x}}{2} = \cosh 2x =\) lhs* | A1* |
| (3) |
Notes
M1: Substitutes the correct exponential form into the rhs
dM1: Squares correctly to obtain an expression in \(e^{2x}\) and \(e^{-2x}\). Dependent on the previous mark.
A1*: Complete proof with no errors
(a) Way 2
| Scheme | Marks |
|---|---|
| lhs \(= \cosh 2x = \dfrac{e^{2x} + e^{-2x}}{2}\) | M1 |
| \(= 2\left(\dfrac{\left(e^x + e^{-x}\right)^2 - 2}{4}\right)\) | dM1 |
| \(2\left(\dfrac{e^x + e^{-x}}{2}\right)^2 - 1 = 2\cosh^2 x - 1 =\) rhs* | A1* |
M1: Substitutes the correct exponential form
dM1: Completes the square correctly to obtain an expression in \(e^x\) and \(e^{-x}\) Dependent on the previous mark.
A1*: Complete proof with no errors
| Scheme | Marks |
|---|---|
| Way 1: \(29\cosh x - 3\left(2\cosh^2 x - 1\right) = 38\) | M1 |
| \(6\cosh^2 x - 29\cosh x + 35 = 0 \Rightarrow \cosh x = \ldots\) | M1 |
| \(\cosh x = \dfrac{7}{3}\) or \(\cosh x = \dfrac{5}{2}\) | A1 |
| \(\cosh x = \alpha \Rightarrow x = \ln\left(\alpha + \sqrt{\alpha^2 - 1}\right)\) or \(\cosh x = \alpha \Rightarrow x = \ln\left(\alpha - \sqrt{\alpha^2 - 1}\right)\) or \(\dfrac{e^x + e^{-x}}{2} = \alpha \Rightarrow x = \ldots\) | M1 |
| \(x = \ln\left(\dfrac{7}{3} \pm \sqrt{\dfrac{40}{9}}\right)\) and \(x = \ln\left(\dfrac{5}{2} \pm \sqrt{\dfrac{21}{4}}\right)\) Or equivalent exact forms e.g. \(x = \ln\dfrac{7 \pm 2\sqrt{10}}{3}\) and \(x = \ln\dfrac{5 \pm \sqrt{21}}{2}\) \(x = \pm\ln\left(\dfrac{7 + 2\sqrt{10}}{3}\right)\) and \(x = \pm\ln\left(\dfrac{5 + \sqrt{21}}{2}\right)\) \(x = \ln\left(7 \pm 2\sqrt{10}\right) - \ln 3\) and \(x = \ln\left(5 \pm \sqrt{21}\right) - \ln 2\) | A1A1 |
| (6) | |
| (9 marks) |
Notes
M1: Substitutes the result from part (a)
M1: Forms a 3-term quadratic and attempt to solve for \(\cosh x\). You can apply the General Principles for solving a 3TQ if necessary.
A1: Both correct (or equivalent values)
M1: Uses the correct ln form for arcosh to find at least one value for \(x\) for \(\alpha > 1\) or uses the correct exponential form for cosh and solves the resulting 3TQ in \(e^x\) to find at least one value for \(x\) for \(\alpha > 1\)
A1: Any 2 of these 4 solutions. Penalise lack of brackets once where necessary, the first time it occurs and penalise lack of simplification once, the first time it occurs e.g. \(\ln\dfrac{5}{2} \pm \dfrac{\sqrt{21}}{2},\ \ln\left(\dfrac{5}{2} \pm \sqrt{\left(\dfrac{5}{2}\right)^2 - 1}\right)\)
A1: All 4 correct
Note that the decimal answers are, \(\pm 1.49\ldots,\ \pm 1.56\ldots\),
(b) Way 2
| Scheme | Marks |
|---|---|
| \(29\left(\dfrac{e^x + e^{-x}}{2}\right) - 3\left(\dfrac{e^{2x} + e^{-2x}}{2}\right) = 38\) or \(6\left(\dfrac{e^x + e^{-x}}{2}\right)^2 - 29\left(\dfrac{e^x + e^{-x}}{2}\right) + 35 = 0\) | M1 |
| \(3e^{4x} - 29e^{3x} + 76e^{2x} - 29e^x + 3 = 0\) | M1A1 |
| \(\left(3e^{2x} - 14e^x + 3\right)\left(e^{2x} - 5e^x + 1\right) = 0 \Rightarrow x = \ldots\) | M1 |
| \(x = \ln\left(\dfrac{7}{3} \pm \sqrt{\dfrac{40}{9}}\right)\) and \(x = \ln\left(\dfrac{5}{2} \pm \sqrt{\dfrac{21}{4}}\right)\) Or equivalent exact forms e.g. \(x = \ln\dfrac{7 \pm 2\sqrt{10}}{3}\) and \(x = \ln\dfrac{5 \pm \sqrt{21}}{2}\) \(x = \pm\ln\left(\dfrac{7 + 2\sqrt{10}}{3}\right)\) and \(x = \pm\ln\left(\dfrac{5 + \sqrt{21}}{2}\right)\) \(x = \ln\left(7 \pm 2\sqrt{10}\right) - \ln 3\) and \(x = \ln\left(5 \pm \sqrt{21}\right) - \ln 2\) e.g. \(\ln\dfrac{5}{2} \pm \dfrac{\sqrt{21}}{2},\ \ln\left(\dfrac{5}{2} \pm \sqrt{\left(\dfrac{5}{2}\right)^2 - 1}\right)\) | A1A1 |
M1: Substitutes the correct exponential forms
M1: Multiplies by \(e^{2x}\) or \(e^{-2x}\) to obtain a quartic in \(e^x\) or \(e^{-x}\)
A1: Correct quartic in any form (not necessarily all on one side)
M1: Solves their quartic to find at least one value for \(x\)
A1: All 4 correct