FP3 June 2018 Q4
4. The curve \(C\) has equation \[y = \text{arsinh}\,x + x\sqrt{x^2 + 1}, \qquad 0 \leqslant x \leqslant 1\]
| Scheme | Marks |
|---|---|
| \(y = \text{arsinh}\,x + x\sqrt{x^2 + 1},\quad 0 \leqslant x \leqslant 1\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{x^2 + 1}} + \dfrac{x^2}{\sqrt{x^2 + 1}} + \sqrt{x^2 + 1}\) | B1 B1 |
| E.g. \(= \dfrac{1 + x^2 + 1 + x^2}{\sqrt{x^2 + 1}} = \ldots\) or \(= \dfrac{1 + x^2}{\sqrt{x^2 + 1}} + \sqrt{x^2 + 1} = \ldots\) | M1 |
| \(= 2\sqrt{x^2 + 1}\) * | A1 |
| (4) |
Notes
B1: \(\dfrac{\mathrm{d}(\text{arsinh}\,x)}{\mathrm{d}x} = \dfrac{1}{\sqrt{x^2 + 1}}\)
B1: \(\dfrac{\mathrm{d}\left(x\sqrt{x^2 + 1}\right)}{\mathrm{d}x} = \dfrac{x^2}{\sqrt{x^2 + 1}} + \sqrt{x^2 + 1}\)
M1: Processes 3 terms of the form \(\dfrac{A}{\sqrt{x^2 + 1}},\ \dfrac{Bx^2}{\sqrt{x^2 + 1}}\) or \(\dfrac{Bx}{\sqrt{x^2 + 1}},\ C\sqrt{x^2 + 1}\) using correct algebra (allow sign slips only) to obtain a single term.
A1: cso Allow \(2\left(x^2 + 1\right)^{\frac{1}{2}}\)
| Scheme | Marks |
|---|---|
| \(1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + 4\left(x^2 + 1\right)\) | M1 |
| \(\displaystyle\Rightarrow (L =)\int_0^1\sqrt{5 + 4x^2}\,\mathrm{d}x\) * | A1* |
| (2) |
Notes
M1: Attempts \(\displaystyle\int\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x\) with the printed answer from part (a) (limits not needed here) but must see a step before the given answer.
A1*: Answer as printed with no errors including limits and “\(\mathrm{d}x\)” Allow \(\displaystyle\int_0^1\sqrt{4x^2 + 5}\,\mathrm{d}x\)
| Scheme | Marks |
|---|---|
| \(x = \dfrac{\sqrt{5}}{2}\sinh u \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}u} = \dfrac{\sqrt{5}}{2}\cosh u\) | |
| \(\displaystyle\Rightarrow L = \int\sqrt{5 + 5\sinh^2 u}\,\frac{\sqrt{5}}{2}\cosh u\,(\mathrm{d}u)\) | M1 |
| \(\displaystyle = \frac{5}{2}\int\cosh^2 u\,(\mathrm{d}u)\) | A1 |
| \(\displaystyle = \frac{5}{4}\int(\cosh 2u + 1)(\mathrm{d}u)\) | dM1 |
| \(= \dfrac{5}{4}\left[\dfrac{1}{2}\sinh 2u + u\right]\) | A1 |
| \(= \left[\ldots\ldots\right]_0^{\text{arsinh}\frac{2}{\sqrt{5}}\left(\text{or}\ln\sqrt{5}\right)}\) Note \(\dfrac{1}{2}\sinh\left(2\left(\text{arsinh}\dfrac{2}{\sqrt{5}}\right)\right) = \dfrac{6}{5}\) | ddM1 |
| \(= \dfrac{3}{2} + \dfrac{5}{8}\ln 5\) | A1 |
| (6) | |
| (12 marks) |
Notes
M1: Fully substitutes into \(\displaystyle\int\sqrt{4x^2 + 5}\,\mathrm{d}x\)
A1: Correct integral including the 5/2. Allow e.g. \(\displaystyle\frac{5}{2}\int\cosh u\cosh u\,(\mathrm{d}u)\)
dM1: Applies \(\cosh 2u = \pm 2\cosh^2 u \pm 1\) to an integral of the form \(k\int\cosh^2 u\,\mathrm{d}u\). Dependent on the first method mark.
A1: Correct integration: \(k(\cosh 2u + 1) \to k\left(\dfrac{1}{2}\sinh 2u + u\right)\)
ddM1: Use of correct limits or returns to \(x\) and uses 0 and 1. The use of 0 may be implied. Dependent on both method marks.
A1: Allow equivalent exact answers. E.g. \(\dfrac{3}{2} + \dfrac{5}{4}\ln\sqrt{5},\ \dfrac{3}{2} + \dfrac{5}{4}\ln\left(\dfrac{2}{\sqrt{5}} + \dfrac{3}{\sqrt{5}}\right)\)
May need to check their answer and could be implied by awrt 2.51 if their integration is correct. If the integration is incorrect and no substitution is shown, you may need to check their answer, but score M0 if the answer does not follow.
Note that the variable may change mid-solution once the substitution has been made e.g. \(u \to x\) but this should not be penalised unless there is a clear error in the solution
Note that having reached \(\displaystyle\frac{5}{2}\int\cosh^2 u\,\mathrm{d}u\), candidates may use exponentials. Score the last 4 marks in (c) as follows:
| Scheme | Marks |
|---|---|
| \(\displaystyle\frac{5}{2}\int\cosh^2 u\,\mathrm{d}u = \frac{5}{8}\int\left(e^{2u} + 2 + e^{-2u}\right)\mathrm{d}u\) | dM1 |
| \(= \dfrac{5}{8}\left[\dfrac{1}{2}e^{2u} + 2u - \dfrac{1}{2}e^{-2u}\right]\) | A1 |
| \(= \left[\ldots\ldots\right]_0^{\text{arsinh}\frac{2}{\sqrt{5}}\left(\text{or}\ln\sqrt{5}\right)}\) | ddM1 |
| \(= \dfrac{3}{2} + \dfrac{5}{8}\ln 5\) | A1 |
| (6) |
(Corrected from the printed mark scheme: it says “Score the last 4 marks in (b)”; these are the last 4 marks of part (c).)
dM1: Uses \(\cosh u = \dfrac{1}{2}\left(e^u + e^{-u}\right)\) and squares applies to an integral of the form \(k\int\cosh^2 u\,\mathrm{d}u\)
A1: Correct integration
ddM1: Use of correct limits or returns to \(x\) and uses 0 and 1. Dependent on both method marks.
A1: Allow equivalent exact answers. E.g. \(\dfrac{3}{2} + \dfrac{5}{4}\ln\sqrt{5},\ \dfrac{3}{2} + \dfrac{5}{4}\ln\left(\dfrac{2}{\sqrt{5}} + \dfrac{3}{\sqrt{5}}\right)\)
May need to check their answer and could be implied by awrt 2.51 if their integration is correct. If the integration is incorrect and no substitution is shown, you may need to check their answer, but score M0 if the answer does not follow.