10 The matrix \(\left( \begin{array} { l l } 1 & 1
1 & 0 \end{array} \right)\) is denoted by \(\mathbf { M }\).
- Evaluate \(\mathbf { M } ^ { 3 }\).
The Fibonacci series \(F _ { 0 } , F _ { 1 } , F _ { 2 } , F _ { 3 } , \ldots\) is defined by
\(F _ { n + 1 } = F _ { n } + F _ { n - 1 }\) for \(n \geq 1 , F _ { 0 } = 0 , F _ { 1 } = 1\). - Prove by mathematical induction that \(\boldsymbol { M } ^ { n } = \left( \begin{array} { c c } F _ { n + 1 } & F _ { n }
F _ { n } & F _ { n - 1 } \end{array} \right)\) for \(n \geq 1\). - Use the result of part (b) to find an expression for \(F _ { n + 1 } F _ { n - 1 } - F _ { n } { } ^ { 2 }\) in terms of \(n\), for \(n \geq 1\).
\section*{ADDITIONAL ANSWER SPACE}
If additional space is required, you should use the following lined page(s). The question number(s) must be clearly shown.
| 1(a) | | \(\left[ \ln \left\{ x + \sqrt { x ^ { 2 } - 9 } \right\} \right] _ { 3 } ^ { 5 }\) \(= \ln ( 5 + 4 ) - \ln ( 3 ) = \ln ( 9 / 3 )\) | | \(= \ln 3\) |
| | | Correct indefinite integral | | Substitute and simplify to single logarithm ln 3 only |
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| 1(b) | \(\begin{gathered} { \left[ \frac { 1 } { 2 } \sinh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } \text { or } \left[ \frac { 1 } { 2 } \cosh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } } |
| \text { or } \left[ \frac { 1 } { 8 } \left( e ^ { 2 x } + e ^ { - 2 x } \right) \right] _ { 0 } ^ { \ln 3 } |
| = 1 / 2 \left[ 1 / 2 \left( \mathrm { e } ^ { \ln 3 } - \mathrm { e } ^ { - \ln 3 } \right) \right] ^ { 2 } - 0 \text { or equivalent } |
| = \frac { 1 } { 2 } \left[ \frac { 1 } { 2 } \left( 3 - \frac { 1 } { 3 } \right) \right] ^ { 2 } = \frac { 8 } { 9 } \end{gathered}\) | | | Correct method (e.g. factors can be wrong) | | Correct indefinite integral | | E.g. 25/9 - 1 | | Final answer 8/9, completely correct |
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| 2(a) | \(\begin{gathered} f ( 0 ) = \pi / 4 |
| f ^ { \prime } ( x ) = \frac { 1 } { 1 + ( x + 1 ) ^ { 2 } } |
| f ^ { \prime } ( 0 ) = \frac { 1 } { 2 } |
| f ^ { \prime \prime } ( x ) = - \frac { 2 ( x + 1 ) } { \left( 1 + ( x + 1 ) ^ { 2 } \right) ^ { 2 } } |
| f ^ { \prime \prime } ( 0 ) = - \frac { 2 } { 4 } = - \frac { 1 } { 2 } \end{gathered}\) | | | Correct f(0) | | First derivative | | Substitute \(0 , \mathrm { ft }\) on their \(f ^ { \prime } ( x )\) | | Second derivative (need not be expanded) | | Substitute 0 |
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| 2(b) | \(\begin{gathered} \frac { \pi } { 4 } + \frac { 1 } { 2 } x + \frac { 1 } { 2 } \left( - \frac { 1 } { 2 } \right) x ^ { 2 } |
| \frac { \pi } { 4 } + \frac { 1 } { 2 } x - \frac { 1 } { 4 } x ^ { 2 } \end{gathered}\) | | | Substitute into Maclaurin formula | | CAO |
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| 3(a) | \(\frac { 1 } { 6 a + 1 } \left( \begin{array} { c c c } 3 a | - 3 | 1 |
| - a | 1 | 2 a |
| 1 | 6 | - 2 \end{array} \right)\) | | | Obtain cofactors | | All cofactors correct and in right place | | Divide by determinant | | Determinant correct ( \(- 6 a - 1\) ) | | Completely correct (expect to see all signs reversed) |
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| 3(b) | \(- \frac { 1 } { 6 }\) | A1ft | ft on their determinant |
| 4(a) | \includegraphics[max width=\textwidth, alt={}]{6280d53b-3c1c-4dc9-a96b-2c58f2a7bf51-22_199_391_282_420} | | Give B1 if: just one obvious flaw in sketch, or if part of curve for negative \(r\) given, e.g. |
| 4(b) | \(\frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 2 } - \cos \theta \right) ^ { 2 } d \theta\) \(\begin{gathered} = \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 4 } - \cos \theta + \cos ^ { 2 } \theta \right) d \theta |
| = \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 3 } { 4 } - \cos \theta + \frac { 1 } { 2 } \cos 2 \theta \right) d \theta |
| = \frac { 1 } { 2 } \left[ \frac { 3 } { 4 } \theta - \sin \theta + \frac { 1 } { 2 } \sin 2 \theta \right] _ { \pi / 3 } ^ { 5 \pi / 3 } |
| = \frac { 1 } { 2 } \pi + \frac { 3 } { 8 } \sqrt { 3 } \end{gathered}\) | | | Correct expression for area | | Multiply out and use \(\cos 2 \theta\) | | Correct integrand (ignore leading 1/2) | | Correct indefinite integral (ignore leading 1/2) | | Final answer, this or clear exact equivalent only |
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| 5(a) | 2 | A1 | |
| 5(b) | | Roots are \(\alpha ( 1 + 1 / 2 ) , \beta ( 1 + 1 / 2 ) , \chi \left( 1 + \frac { 1 } { 2 } \right)\) | | Hence put \(y = 3 x / 2\) \(x = 2 y / 3\) | | \(3 ( 2 y / 3 ) ^ { 3 } + 2 ( 2 y / 3 ) ^ { 2 } - 7 ( 2 y / 3 ) - 6 = 0\) | | \(4 y ^ { 3 } + 4 y ^ { 2 } - 21 y - 27 = 0\) |
| | | Use value of \(\alpha \beta \gamma\) in \(\alpha ( 1 + 1 / \alpha \beta \gamma )\) etc Correct new variable, ft on their \(\alpha \beta \gamma\) Inverse function, ft Substitute inverse function | | Complete equation including 0, any letter, any rational multiple |
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| or | \(\begin{aligned} | \Sigma \alpha = - 2 / 3 , \Sigma \alpha \beta = - 7 / 3 , \alpha \beta \gamma = 2 |
| \qquad \begin{array} { l } ( \alpha + \beta + \gamma ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ( = - 1 ) |
| ( \alpha \beta + \beta \gamma + \gamma \alpha ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 2 } \left( = - \frac { 21 } { 4 } \right) \end{array} |
| \qquad \alpha \beta \gamma \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 3 } \left( = \frac { 27 } { 4 } \right) |
| \text { Hence equation is } x ^ { 3 } + x ^ { 2 } - \frac { 21 } { 4 } x - \frac { 27 } { 4 } = 0 \end{aligned}\) | | | All three correct | | Correct formula for \(\Sigma \alpha ^ { \prime }\) and substitute | | Correct formula for \(\Sigma \alpha ^ { \prime } \beta ^ { \prime }\) and substitute | | Correct formula for \(\alpha ^ { \prime } \beta ^ { \prime } \gamma ^ { \prime }\) and substitute | | Correct final equation including 0, any letter, any rational multiple |
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\captionsetup{labelformat=empty}
\caption{Further Mathematics Core (Pure) Mark Scheme (total 84)}
| 6(a) | | Circle, centre \(( 3 , - 1 )\) | | Radius \(\sqrt { } 2\) indicated | | Interior of circle indicated |
| | | Allow +/- errors | | Needs not to include origin | | Any clear indication of interior |
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| 6(b) | | Point of contact \(( P )\) between circle and tangent through \(O\) clearly identified | | Form appropriate right-angled \(\Delta\) | | \(O C = \sqrt { 10 } , P C = \sqrt { 2 } \Rightarrow O P = 2 \sqrt { 2 }\) and \(\angle P O C = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)\) | | \(\angle x O C = \tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)\) | | \(\angle x O P = \tan ^ { - 1 } \left( \frac { \frac { 1 } { 2 } - \frac { 1 } { 3 } } { 1 + \frac { 1 } { 2 } \cdot \frac { 1 } { 3 } } \right) = \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)\) or \(n = 7\) |
| | | Can be implied by correct working | | Can be implied by correct working | | Use exact method for a difference of angles Either \(\tan ^ { - 1 } ( 1 / 7 )\) or \(n = 7\). | | If exactly 1/7 from calculator, give M0A0 |
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| or | | Point of contact ( \(P\) ) between circle and tangent through \(O\) clearly identified | | \(y = m x\) meets \(( x - 3 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 2\) | | \(( x - 3 ) ^ { 2 } + ( m x + 1 ) ^ { 2 } = 2\) has double roots \(\left( m ^ { 2 } + 1 \right) x ^ { 2 } + 2 ( m - 3 ) x + 8 = 0\) has double roots | | \(4 ( m - 3 ) ^ { 2 } = 4.8 \left( m ^ { 2 } + 1 \right)\) | | \(7 m ^ { 2 } + 6 m - 1 = 0\), so \(m = \frac { 1 } { 7 }\) or - 1 | | But \(m\) is not - 1 as this gives minimum, not maximum, argument |
| | | Can be implied by correct working, provided final answer does not include - 1 | | Line and circle equations used | | Discriminant set to zero | | Obtain \(\tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)\) or \(n = 7\) | | Reject - 1 , with reason. Not enough to say " \(m = \frac { 1 } { 7 }\) or - 1 so \(n = 7\) " |
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| or | | Point of contact \(( P )\) between circle and tangent through \(O\) clearly identified | | Appropriate right-angled \(\Delta\) with centre \(C\) | | \(O C = \sqrt { 10 } , P C = \sqrt { 2 } \Rightarrow O P = 2 \sqrt { 2 }\) | | Circles, centre \(O , r = 2 \sqrt { 2 }\), \ | \(C , r = \sqrt { 2 }\) | | \(x ^ { 2 } + ( 3 x - 8 ) ^ { 2 } = 8,5 x ^ { 2 } - 24 x + 28 = 0\) | | \(\Rightarrow P = \left( 2 \frac { 4 } { 5 } , \frac { 2 } { 5 } \right)\) | | \(\theta = \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)\) or \(n = 7\) |
| | | Can be implied by correct working | | Can be implied by correct working | | \(x ^ { 2 } + y ^ { 2 } = 8\) and \(( x - 3 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 2\) | | Eliminate \(y\), solve quadratic in \(x\) | | Correct \(P\), both cords. allow decimals | | Either \(\tan ^ { - 1 } \left( \frac { 1 } { 7 } \right)\) or \(n = 7\) |
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| 7(a) | | Distance is \(\frac { | \mathbf { b } \mathbf { . n } - \mathbf { p } | } { | \mathbf { n } | }\) \(\mathbf { b } = ( 9,5,2 ) , \mathbf { n } = ( 3,2 , - 4 )\) | | Scalar product is 29 | | Distance is \(32 / \sqrt { } 29\) |
| | | Use correct formula with correct \(\mathbf { b }\) and \(\mathbf { n }\) | | Correct scalar product | | Answer (accept 5.94 (3s.f.)) |
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| 7(b) | \(\begin{aligned} | ( 3 \mathbf { i } + \mathbf { k } ) \times ( \mathbf { i } - \mathbf { j } - 2 \mathbf { k } ) |
| = \mathbf { i } + 7 \mathbf { j } - 3 \mathbf { k } \end{aligned}\) \(\begin{gathered} x + 7 y - 3 z = 0 |
| \left( \begin{array} { c } 1 |
| 7 |
| - 3 \end{array} \right) \cdot \left( \begin{array} { c } 3 |
| 2 |
| - 4 \end{array} \right) \end{gathered}\) \(\begin{gathered} \cos \theta = \frac { 29 } { \sqrt { 59 } \sqrt { 29 } } |
| 134.5 ^ { \circ } \text { or } 2.35 \text { radians } \end{gathered}\) | | | Find perpendicular vector | | Any multiple of this | | Cartesian equation of second plane | | Scalar product of normal vectors of 2 planes | | Correct scalar product and moduli (ft their vectors) | | Correct obtuse angle (awrt \(135 ^ { \circ }\) or 2.35 rad ) |
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| 8(a) | \(\begin{gathered} \frac { A } { x } + \frac { B x + C } { 4 x ^ { 2 } + 1 } |
| A \left( 4 x ^ { 2 } + 1 \right) + ( B x + C ) x \equiv 8 x ^ { 2 } - x + 2 |
| A = 2 , B = 0 , C = - 1 |
| \frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } \end{gathered}\) | | | Attempt at partial fractions | | Set up identity | | Compare coefficients or substitute x -values | | All correct values | | Final answer |
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| 8(b) | IF \(\exp \left( \int 2 / x \mathrm {~d} x \right) = x ^ { 2 }\) \(\begin{gathered} y x ^ { 2 } = \int \frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } d x |
| y x ^ { 2 } = 2 \ln x - \frac { 1 } { 2 } \tan ^ { - 1 } ( 2 x ) + c |
| y = \frac { 4 \ln x - \tan ^ { - 1 } ( 2 x ) + c ^ { \prime } } { 2 x ^ { 2 } } \end{gathered}\) | | | Method for IF | | \(x ^ { 2 }\) | | Multiply through by IF and spot link to (a) | | Use \(\tan ^ { - 1 }\) | | \(2 \ln x\) and \(+ c\) | | Final answer, completely correct, any equivalent form, any correct way of writing constant (e.g. c or 2c) |
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| 9(a) | \(\begin{aligned} | ( c + \mathrm { is } ) ^ { 5 } = c ^ { 5 } + 5 \mathrm { is } ^ { 4 } - 10 s ^ { 2 } c ^ { 2 } - 10 \mathrm { is } ^ { 3 } c ^ { 2 } + 5 s ^ { 4 } c + \mathrm { is } ^ { 5 } |
| \cos 5 \theta = \cos ^ { 5 } \theta - 10 \sin ^ { 2 } \theta \cos ^ { 3 } \theta + 5 \sin ^ { 4 } \theta \cos \theta |
| = c ^ { 5 } - 10 \left( 1 - c ^ { 2 } \right) c ^ { 3 } + 5 \left( 1 - c ^ { 2 } \right) ^ { 2 } c |
| = 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta \quad \text { AG } \end{aligned}\) | | | Use de Moivre (can ignore im parts if clear) | | Correct equating of real parts | | Use \(\sin ^ { 2 } \theta = 1 - \cos ^ { 2 } \theta\) | | Correctly obtain given answer |
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| 9(b) | | Using \(\cos 5 \theta = 0\) gives \(16 c ^ { 5 } - 20 c ^ { 4 } + 5 c = 0\) \(c \neq 0 \text { so } 16 c ^ { 4 } - 20 c ^ { 2 } + 5 = 0\) | | \(c ^ { 2 } = ( 20 \pm \sqrt { } 80 ) / 32\) \(c = \pm \sqrt { \frac { 5 \pm \sqrt { 5 } } { 8 } }\) | | \(\cos \pi / 10 > 0\) so first \(\pm\) is + | | \(\cos \pi / 10 > \cos ( 3 \pi / 10 )\) so second \(\pm\) is + |
| | | Use \(\theta = \pi / 10\) and \(\cos ( \pi / 2 ) = 0\) | | Justify cancellation of \(c\) | | Solve quadratic in \(c ^ { 2 }\) by exact method | | Correct expression for \(c\), any combination of \(\pm\) and + signs | | Justify outer + | | Justify inner + |
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| 10(a) | \(\left( \begin{array} { l l } 3 | 2 |
| 2 | 1 \end{array} \right)\) | | If B0, give B1 if \(\mathbf { M } ^ { 2 } = \left( \begin{array} { l l } 2 | 1 |
| 1 | 1 \end{array} \right)\) seen |
| 10(b) | | Base case: \(n = 1 , \mathbf { M } = \left( \begin{array} { l l } 1 | 1 | | 1 | 0 \end{array} \right) = \left( \begin{array} { l l } F _ { 2 } | F _ { 1 } | | F _ { 1 } | F _ { 0 } \end{array} \right)\) | | Inductive step: assume \(\mathbf { M } ^ { k } = \left( \begin{array} { c c } F _ { k + 1 } | F _ { k } | | F _ { k } | F _ { k - 1 } \end{array} \right)\) | | \(\Rightarrow \mathbf { M } ^ { k + 1 } = \left( \begin{array} { c c } F _ { k + 1 } | F _ { k } | | F _ { k } | F _ { k - 1 } \end{array} \right) \left( \begin{array} { l l } 1 | 1 | | 1 | 0 \end{array} \right)\) \(\begin{aligned} | = \left( \begin{array} { c c } F _ { k + 1 } + F _ { k } | F _ { k } + F _ { k - 1 } | | F _ { k } + F _ { k - 1 } | F _ { k } \end{array} \right) | | \left( \begin{array} { c c } F _ { k + 2 } | F _ { k + 1 } | | F _ { k + 1 } | F _ { k } \end{array} \right) \text { as required } \end{aligned}\) | | Base case and inductive step both true, so statement true for all \(n \in \mathbb { N }\) by mathematical induction |
| | | Fully correct | | Correct "assume" statement, allow \(n\) | | Consider \(\mathbf { M } \times\) hypothesised \(\mathbf { M } ^ { k }\) | | Correctly show this result | | Award only if all 4 other marks gained |
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| 10(c) | | Det \(\left( \mathbf { M } ^ { n } \right) = F _ { n + 1 } F _ { n - 1 } - F _ { n } { } ^ { 2 }\) | | \(\operatorname { Det } \left( \mathbf { M } ^ { n } \right) = ( \operatorname { Det } \mathbf { M } ) ^ { n }\) \(= ( - 1 ) ^ { n }\) |
| | | Use determinant of \(\mathbf { M } ^ { n }\) | | Relate \(\operatorname { det } \mathbf { M } ^ { \boldsymbol { n } }\) to \(\operatorname { det } \mathbf { M }\) | | Correct answer |
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