5 Let \(\mathrm { f } ( x ) = \sin ^ { - 1 } ( x )\).
- Determine \(f ^ { \prime \prime } ( x )\).
- Determine the first two non-zero terms of the Maclaurin expansion for \(\mathrm { f } ( x )\).
- By considering the first two non-zero terms of the Maclaurin expansion for \(\mathrm { f } ( \mathrm { x } )\), find an approximation to \(\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x\). Give your answer correct to 6 decimal places.
- By writing \(\mathrm { f } ( x )\) as \(\sin ^ { - 1 } ( x ) \times 1\), determine the value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x\). Give your answer in exact form.
\section*{Total Marks for Question Set 6: 37}
\section*{Mark scheme}
\section*{Marking Instructions}
a An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not always be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed.
b The following types of marks are available.
\section*{M}
A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified.
A method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words "Determine" or "Show that", or some other indication that the method must be given explicitly.
\section*{A}
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded.
\section*{B}
Mark for a correct result or statement independent of Method marks.
\section*{E}
A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result.
Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.
c When a part of a question has two or more 'method' steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation 'dep*' is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.
d The abbreviation FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only - differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, what is acceptable will be detailed in the mark scheme. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be 'follow through'.
e We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.
- When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.
- When a value is not given in the paper accept any answer that agrees with the correct value to \(\mathbf { 3 ~ s } . \mathbf { f }\). unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range.
Follow through should be used so that only one mark in any question is lost for each distinct accuracy error.
Candidates using a value of \(9.80,9.81\) or 10 for \(g\) should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric.
f Rules for replaced work and multiple attempts:
- If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.
- If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.
- if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately.
For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate's data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors.
If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate's own working is not a misread but an accuracy error.
h If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold "In this question you must show detailed reasoning", or the command words "Show" or "Determine". Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero.
\begin{table}[h]
\captionsetup{labelformat=empty}
\caption{Abbreviations}
| Abbreviations used in the mark scheme | Meaning |
| dep* | Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark |
| cao | Correct answer only |
| ое | Or equivalent |
| rot | Rounded or truncated |
| soi | Seen or implied |
| www | Without wrong working |
| AG | Answer given |
| awrt | Anything which rounds to |
| BC | By Calculator |
| DR | This question included the instruction: In this question you must show detailed reasoning. |
| Question | Answer | Marks | AO | Guidance |
| 1 | (a) | DR \(\begin{aligned} | u = x ^ { 2 } |
| 3 ( \sqrt { u } ) ^ { 3 } - 2 ( \sqrt { u } ) ^ { 2 } - 5 \sqrt { u } - 4 ( = 0 ) \end{aligned}\) \(3 u \sqrt { u } - 5 \sqrt { u } = 2 u + 4 \Rightarrow u ( 3 u - 5 ) ^ { 2 } = ( 2 u + 4 ) ^ { 2 }\) \(\begin{aligned} | u \left( 9 u ^ { 2 } - 30 u + 25 \right) = 4 u ^ { 2 } + 16 u + 16 = > |
| 9 u ^ { 3 } - 34 u ^ { 2 } + 9 u - 16 = 0 \end{aligned}\) | | | | Correct substitution chosen | | Oe Attempting to make substitution | | Rearranging and squaring bs to remove the square root(s) Rearranging to answer |
| | or preparation for substitution by removing odd powers. \(\operatorname { eg } x ^ { 2 } \left( 3 x ^ { 2 } - 5 \right) ^ { 2 } = \left( 2 x ^ { 2 } + 4 \right) ^ { 2 } \ldots\) | | ...and then substituting \(u ( 3 u - 5 ) ^ { 2 } = ( 2 u + 4 ) ^ { 2 }\) | | Equation can be in \(x\) |
|
| | | Alternative method | | DR \(\alpha ^ { 2 } \beta ^ { 2 } \gamma ^ { 2 } = ( \alpha \beta \gamma ) ^ { 2 } = \left( - \frac { - 4 } { 3 } \right) ^ { 2 } = \frac { 16 } { 9 }\) | | \(\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }\) \(= ( \alpha \beta + \beta \gamma + \gamma \alpha ) ^ { 2 } - 2 \alpha \beta \gamma ( \alpha + \beta + \gamma )\) \(\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = ( \alpha + \beta + \gamma ) ^ { 2 } - 2 ( \alpha \beta + \beta \gamma + \gamma \alpha )\) | | \(u ^ { 3 } - \left( \left( \frac { 2 } { 3 } \right) ^ { 2 } - 2 \times \frac { - 5 } { 3 } \right) u ^ { 2 } + \left( \left( \frac { - 5 } { 3 } \right) ^ { 2 } - 2 \times \frac { 4 } { 3 } \times \frac { 2 } { 3 } \right) u - \frac { 16 } { 9 }\) \(= u ^ { 3 } - \frac { 34 } { 9 } u ^ { 2 } + u - \frac { 16 } { 9 } = 0 \Rightarrow 9 u ^ { 3 } - 34 u ^ { 2 } + 9 u - 16 = 0\) |
| | | | Writing the expression in terms of standard symmetrical forms | | Writing the expression in terms of standard symmetrical forms Substituting in and rearranging to answer |
| | Must include one intermediate step \(\mathrm { NB } \sum \alpha = \frac { 2 } { 3 } , \sum \alpha \beta = - \frac { 5 } { 3 } , \alpha \beta \gamma = \frac { 4 } { 3 }\) | | Condone without factorisation of "2" | | \(\mathrm { NB } \sum \alpha ^ { 2 } = \frac { 34 } { 9 } , \sum \alpha ^ { 2 } \beta ^ { 2 } = 1\) |
|
| | | [4] | | | |
| (b) | | DR \(\frac { \sum \alpha ^ { 2 } \beta ^ { 2 } } { \alpha \beta \gamma } = \frac { \left( \frac { 9 } { 9 } \right) } { \left( \frac { 4 } { 3 } \right) } \text { or } \frac { 1 } { \left( \frac { 4 } { 3 } \right) }\) | | \(= \frac { 3 } { 4 }\) |
| | | Their \(\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }\) from part (a) over \(\pm \frac { 4 } { 3 }\) | Strict ft |
| | | | | | |
| Question | Answer | Marks | AO | Guidance |
| 2 | (a) | \(\begin{aligned} | \text { DR } |
| ( r + 2 ) ( r - 1 ) \end{aligned}\) \(\begin{aligned} | \frac { A } { r - 1 } + \frac { B } { r + 2 } |
| A = 1 , B = - 1 |
| = \end{aligned}\) \(\begin{array} { c c c c c c } \frac { 1 } { 4 } | - | \frac { 1 } { 7 } | \cdots | - | \frac { 1 } { n - 1 } |
| \frac { 1 } { 5 } | - | \frac { 1 } { 8 } | \frac { 1 } { n - 3 } | - | \frac { 1 } { n } |
| \frac { 1 } { 6 } | - | \frac { 1 } { 9 } | \frac { 1 } { n - 2 } | - | \frac { 1 } { n + 1 } |
| \frac { 1 } { 7 } | - | \cdots | \frac { 1 } { n - 1 } | - | \frac { 1 } { n + 2 } \end{array}\) \(\begin{aligned} | = \frac { 1 } { 4 } + \frac { 1 } { 5 } + \frac { 1 } { 6 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } |
| = \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } \end{aligned}\) | A1 | 1.1 | Correct factorisation of denominator soi Correct form for partial fractions | | M1 can be ft from any \(A , B\) having opposite signs | | For M1, condone omission of \(\frac { 1 } { 7 }\) or \(- \frac { 1 } { n - 1 }\) |
|
| | Alternative Method \(\begin{aligned} | \therefore \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 5 } ^ { n } \frac { 1 } { r + 2 } |
| = \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 8 } ^ { n + 3 } \frac { 1 } { r - 1 } |
| = \sum _ { r = 5 } ^ { 7 } \frac { 1 } { r - 1 } - \sum _ { r = n + 1 } ^ { n + 3 } \frac { 1 } { r - 1 } \end{aligned}\) | М1 | | Using partial fractions, separating into two sums, re-indexing so that the summands have identical form and cancelling central terms. | Might see start and end terms explicitly. eg \(\sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 8 } ^ { n + 3 } \frac { 1 } { r - 1 }\) |
| Question | Answer | Marks | AO | Guidance |
| \multirow{2}{*}{} | \multirow{2}{*}{} | \(\begin{aligned} | \therefore \sum _ { r = 5 } ^ { n } \frac { 3 } { r ^ { 2 } + r - 2 } = |
| = \frac { 1 } { 4 } + \frac { 1 } { 5 } + \frac { 1 } { 6 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } |
| = \frac { 37 } { 60 } - \frac { 1 } { n } - \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 } \end{aligned}\) | A1 | | AG. | | Might see formal substitution of index. eg | | Let \(R = r + 3 \Rightarrow r + 2 = R - 1\) \(\begin{aligned} | \therefore \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { r = 5 } ^ { n } \frac { 1 } { r + 2 } | | = \sum _ { r = 5 } ^ { n } \frac { 1 } { r - 1 } - \sum _ { R = 8 } ^ { n + 3 } \frac { 1 } { R - 1 } \end{aligned}\) |
|
| | | [5] | | | |
| (b) | \(= \frac { 37 } { 60 }\) or awrt 0.617 | | 2.2a | | |
| 3 | (a) | \(\begin{aligned} | \ln ( 1 + \sin \theta ) = 0 \Rightarrow 1 + \sin \theta = 1 \Rightarrow \sin \theta = 0 |
| \text { so } \alpha = 0 \text { and } \beta = \pi \end{aligned}\) | | | | |
| (b) | \(\begin{aligned} | A = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } ( \ln ( 1 + \sin \theta ) ) ^ { 2 } \mathrm {~d} \theta |
| = 0.4162 ( 4 \mathrm { sf } ) \text { cao } \end{aligned}\) | | | Correct formula for area with \(r\) correctly substituted and their limits. Must be unambiguous but can be implied by correct answer/later work BC | Incorrect formula = M0A0 Condone missing \(\mathrm { d } \theta\) |
| (c) | | \(\theta = \frac { \pi } { 2 } \Rightarrow r = \ln 2 = 0.6931 ( 4 \mathrm { sf } )\) which would be the diameter, \(D\), of the circle | | But \(A = 0.4162 ( 4 \mathrm { sf } ) = > D = 0.7280 ( 4 \mathrm { sf } )\) or \(R = 0.3640 ( 4 \mathrm { sf } )\) so the curve is not circular |
| | | | or radius \(R = 0.3466 ( 4 \mathrm { sf } )\) condone correct \(R\) or \(D\) without reasoning | | or \(R = 0.3466 ( 4 \mathrm { sf } ) (\) or \(D = 0.6931 ) \Rightarrow A = 0.3773 ( 4 \mathrm { sf } )\) which is not \(0.4162 ( 4 \mathrm { sf } )\) |
| | It must be clear that the \(r\) value would be the diameter of the circle; the calculation alone is insufficient for M1. | | M1 can be implied by area given as \(\pi \left( \frac { \ln 2 } { 2 } \right) ^ { 2 }\) | | Explanation must include comparison of \(R\) 's, \(D\) 's or \(A\) 's and conclusion . Allow correct working to 3 sf . |
|
| Question | Answer | Marks | AO | Guidance |
| 4 | (a) | DR | | | | |
| | \(\begin{aligned} | r ^ { 2 } = ( - 4 ) ^ { 2 } + ( \sqrt { 48 } ) ^ { 2 } \quad \text { or } ( r \cos \theta = - 4 \text { and } |
| r \sin \theta = \sqrt { 48 } ) \text { or } \tan \theta = - \sqrt { 3 } \text { oe } |
| r = 8 \left( \mathrm { ie } z = 8 \mathrm { e } ^ { \mathrm { i } \theta } \right) \quad \theta = 2 \pi / 3 \left( \mathrm { ie } z = r \mathrm { e } ^ { \mathrm { i } 2 \pi / 3 } \right) |
| \sqrt [ 3 ] { 8 } \text { or } 2 |
| \frac { 2 \pi } { 9 } \text { soi } |
| \frac { 2 \pi } { 3 } + 2 \pi k \text { for } k = 1 \text { and } 2 \text { oe seen } |
| 2 \mathrm { e } ^ { \frac { 2 } { 9 } \pi i } , 2 \mathrm { e } ^ { \frac { 8 } { 9 } \pi i } \text { and } 2 \mathrm { e } ^ { - \frac { 4 } { 9 } \pi i } \end{aligned}\) | A1 | 2.1 | Correct use of relevant formula(e). Some working must be seen. | Correct answer with no working: M0A0 or eg \(\theta = 8 \pi / 3\) |
| | | B1ft | 2.1 | Argument of (principal) cube root is one third of their argument | |
| | | M1 | 2.2a | Considering further arguments at angular distance \(2 \pi\) | |
| | | A1 | 1.1 | or eg \(2 \mathrm { e } ^ { \frac { 2 } { 9 } \pi i } , 2 \mathrm { e } ^ { \frac { 8 } { 9 } \pi i }\) and \(2 \mathrm { e } ^ { \frac { 14 } { 9 } \pi i }\) | Must be in exponential form, not just \(r =\) and \(\theta =\). Do not condone any missing i's. |
| (b) | | DR | | The cube roots form an equilateral triangle which has (3) lines of symmetry, (one) through each vertex \(\theta = \frac { 2 \pi } { 9 } , \theta = \frac { 8 \pi } { 9 } \text { and } \theta = - \frac { 4 \pi } { 9 } \text { soi }\) |
| B1 | | | for one | | for all three without extras |
| | ft their angles if \(2 \pi / 3\) apart. | | If valid alternatives, must come from clear explanation/diagram |
|
| Question | Answer | Marks | AO | Guidance |
| 5 | (a) | (i) | \(\begin{aligned} | \mathrm { f } ^ { \prime } ( x ) = \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } } \text { from the formula book } |
| \text { so } \mathrm { f } ^ { \prime \prime } ( x ) = - \frac { 1 } { 2 } \cdot \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \cdot ( - 2 x ) |
| = \frac { x } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \end{aligned}\) | М1 | 1.1 | Formula from the Formula Booklet and attempt differentiation | To within sign error |
| (a) | (ii) | \(\begin{aligned} | f ( 0 ) = 0 , f ^ { \prime } ( 0 ) = 1 \text { and } f ^ { \prime \prime } ( 0 ) = 0 |
| f ^ { \prime \prime \prime } ( x ) = \frac { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } - x \cdot \frac { 3 } { 2 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \cdot ( - 2 x ) } { \left( 1 - x ^ { 2 } \right) ^ { 3 } } |
| \text { so } f ^ { \prime \prime \prime } ( 0 ) = 1 \text { and } f ( x ) = x + \frac { 1 } { 6 } x ^ { 3 } + \ldots \end{aligned}\) | | | | or \(a _ { 0 } = 0 , a _ { 1 } = 1\) and \(a _ { 2 } = 0\) | | Differentiate and simplify far enough to be able to justify value 1 | | Condone 3! In place of 6 |
| | Ignore sign error in \(\mathrm { f } ^ { \prime \prime } ( x )\) | | Either full derivative or "zero term" denoted as such | | Not BC. If M0 then SC1 for correct expansion |
|
| (a) | (iii) | \(\begin{aligned} \int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) \mathrm { d } x \approx \int _ { 0 } ^ { \frac { 1 } { 2 } } x | + \frac { 1 } { 6 } x ^ { 3 } \mathrm {~d} x |
| = | 0.127604167 \ldots |
| = | 0.127604 \text { to } 6 \mathrm { dp } \end{aligned}\) | | | | Integral of their 2 term cubic with limits | | Could be BC |
| |
| (b) | | \(\begin{aligned} | \int 1 \times \sin ^ { - 1 } x \mathrm {~d} x = x \sin ^ { - 1 } x - \int \frac { x } { \sqrt { 1 - x ^ { 2 } } } \mathrm {~d} x |
| = x \sin ^ { - 1 } x + \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } ( + \mathrm { c } ) |
| \int _ { 0 } ^ { \frac { 1 } { 2 } } \mathrm { f } ( x ) = \frac { \pi } { 12 } + \frac { \sqrt { 3 } } { 2 } - 1 \end{aligned}\) | | | Attempt integration by parts | ignore limits. Formula for parts must be correct |