SPS SPS FM Pure (SPS FM Pure) 2020 February

1 In this question you must show detailed reasoning.

1. Determine the value of $$\int _ { 3 } ^ { 5 } \frac { 1 } { \sqrt { x ^ { 2 } - 9 } } d x$$ giving your answer as a single logarithm.
2. Determine the exact value of $$\int _ { 0 } ^ { \ln 3 } \sinh x \cosh x d x$$

2

1. Given that $$f ( x ) = \tan ^ { - 1 } ( x + 1 )$$ find \(f ( 0 )\) and \(f ^ { \prime } ( 0 )\), and show that \(f ^ { \prime \prime } ( 0 ) = - \frac { 1 } { 2 }\).
2. Hence find the first three terms in the Maclaurin series for \(f ( x )\)

3

1. Find the inverse of the matrix \(\left( \begin{array} { l l l } 2 & 0 & 1
   0 & 1 & a
   1 & 3 & 0 \end{array} \right)\) in terms of \(a\).
2. State the value of \(a\) for which the matrix is singular.

4 The equation of a curve in polar coordinates is \(r = \frac { 1 } { 2 } - \cos \theta\).

1. Sketch the polar graph of the curve.
2. Find the exact area enclosed by the curve between \(\theta = \frac { 1 } { 3 } \pi\) and \(\theta = \frac { 5 } { 3 } \pi\).

5 The roots of the equation \(\quad 3 x ^ { 3 } + 2 x ^ { 2 } - 7 x - 6 = 0\) are \(\alpha , \beta\) and \(\gamma\).

1. Find the value of \(\alpha \beta \gamma\).
2. Hence, by making use of a suitable substitution, or otherwise, find a cubic equation whose roots are \(\alpha \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) , \beta \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) , \gamma \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right)\).

6 Throughout this question, the complex number \(z\) satisfies \(\left| z - z _ { 0 } \right| \leq \sqrt { 2 }\), where \(z _ { 0 } = 3 - \mathrm { i }\).

1. Draw an Argand diagram to illustrate the locus of \(z\).
2. In this question you must show detailed reasoning. Show that the greatest possible argument of \(z\) can be written as \(\tan ^ { - 1 } \left( \frac { 1 } { n } \right)\), where \(n\) is a positive integer to be determined and \(\arg z \in ( - \pi , \pi ]\).

7

1. Find the shortest distance from the point (9, 5, 2) to the plane \(3 x + 2 y - 4 z = - 3\).
2. The vector equation of a second plane is given by $$\boldsymbol { r } = \lambda \left( \begin{array} { l } 3
   0
   1 \end{array} \right) + \mu \left( \begin{array} { c } 1
   - 1
   - 2 \end{array} \right)$$ By finding the Cartesian equation of the plane, calculate the obtuse angle between this plane and the plane \(3 x + 2 y - 4 z = - 3\).
   [0pt] [6]

8

1. Express \(\frac { 8 x ^ { 2 } - x + 2 } { x \left( 4 x ^ { 2 } + 1 \right) }\) in partial fractions.
2. Hence find the general solution to the differential equation $$\frac { d y } { d x } + \frac { 2 y } { x } = \frac { 8 x ^ { 2 } - x + 2 } { x ^ { 3 } \left( 4 x ^ { 2 } + 1 \right) }$$

9

1. Show that \(\cos 5 \theta = 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta\).
2. Use the result of part (a) to prove that \(\cos \frac { \pi } { 10 } = \sqrt { \frac { 5 + \sqrt { 5 } } { 8 } }\).

10 The matrix \(\left( \begin{array} { l l } 1 & 1
1 & 0 \end{array} \right)\) is denoted by \(\mathbf { M }\).

1. 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 \(n \geq 1 , F _ { 0 } = 0 , F _ { 1 } = 1\).
2. 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 \(n \geq 1\).
3. 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)	\(\left[ \ln \left\{ x + \sqrt { x ^ { 2 } - 9 } \right\} \right] _ { 3 } ^ { 5 }\) \(= \ln ( 5 + 4 ) - \ln ( 3 ) = \ln ( 9 / 3 )\) \(= \ln 3\)	B1 M1 A1 [3]	Correct indefinite integral Substitute and simplify to single logarithm ln 3 only
   1(b)	\(\begin{gathered} { \left[ \frac { 1 } { 2 } \sinh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } \text { or } \left[ \frac { 1 } { 2 } \cosh ^ { 2 } x \right] _ { 0 } ^ { \ln 3 } }
   \text { or } \left[ \frac { 1 } { 8 } \left( e ^ { 2 x } + e ^ { - 2 x } \right) \right] _ { 0 } ^ { \ln 3 }
   = 1 / 2 \left[ 1 / 2 \left( \mathrm { e } ^ { \ln 3 } - \mathrm { e } ^ { - \ln 3 } \right) \right] ^ { 2 } - 0 \text { or equivalent }
   = \frac { 1 } { 2 } \left[ \frac { 1 } { 2 } \left( 3 - \frac { 1 } { 3 } \right) \right] ^ { 2 } = \frac { 8 } { 9 } \end{gathered}\)	M1 A1 M1 A1 [4]	Correct method (e.g. factors can be wrong) Correct indefinite integral E.g. 25/9 - 1 Final answer 8/9, completely correct
   2(a)	\(\begin{gathered} f ( 0 ) = \pi / 4
   f ^ { \prime } ( x ) = \frac { 1 } { 1 + ( x + 1 ) ^ { 2 } }
   f ^ { \prime } ( 0 ) = \frac { 1 } { 2 }
   f ^ { \prime \prime } ( x ) = - \frac { 2 ( x + 1 ) } { \left( 1 + ( x + 1 ) ^ { 2 } \right) ^ { 2 } }
   f ^ { \prime \prime } ( 0 ) = - \frac { 2 } { 4 } = - \frac { 1 } { 2 } \end{gathered}\)	B1 M1 A1ft M1 A1 [5]	Correct f(0) First derivative Substitute \(0 , \mathrm { ft }\) on their \(f ^ { \prime } ( x )\) Second derivative (need not be expanded) Substitute 0
   2(b)	\(\begin{gathered} \frac { \pi } { 4 } + \frac { 1 } { 2 } x + \frac { 1 } { 2 } \left( - \frac { 1 } { 2 } \right) x ^ { 2 }
   \frac { \pi } { 4 } + \frac { 1 } { 2 } x - \frac { 1 } { 4 } x ^ { 2 } \end{gathered}\)	M1 A1	Substitute into Maclaurin formula CAO
   3(a)	\(\frac { 1 } { 6 a + 1 } \left( \begin{array} { c c c } 3 a	- 3	1
   - a	1	2 a
   1	6	- 2 \end{array} \right)\)	M1 A1 M1 A1 A1 [5]	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)
   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}	B2 [2]	Give B1 if: just one obvious flaw in sketch, or if part of curve for negative \(r\) given, e.g.
   4(b)	\(\frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 2 } - \cos \theta \right) ^ { 2 } d \theta\) \(\begin{gathered} = \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 1 } { 4 } - \cos \theta + \cos ^ { 2 } \theta \right) d \theta
   = \frac { 1 } { 2 } \int _ { \pi / 3 } ^ { 5 \pi / 3 } \left( \frac { 3 } { 4 } - \cos \theta + \frac { 1 } { 2 } \cos 2 \theta \right) d \theta
   = \frac { 1 } { 2 } \left[ \frac { 3 } { 4 } \theta - \sin \theta + \frac { 1 } { 2 } \sin 2 \theta \right] _ { \pi / 3 } ^ { 5 \pi / 3 }
   = \frac { 1 } { 2 } \pi + \frac { 3 } { 8 } \sqrt { 3 } \end{gathered}\)	M1 M1 A1 A1 A1 [5]	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
   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 \(y = 3 x / 2\) \(x = 2 y / 3\) \(3 ( 2 y / 3 ) ^ { 3 } + 2 ( 2 y / 3 ) ^ { 2 } - 7 ( 2 y / 3 ) - 6 = 0\) \(4 y ^ { 3 } + 4 y ^ { 2 } - 21 y - 27 = 0\)	M1 A1ft M1 M1 A1 [5]	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
   or	\(\begin{aligned}	\Sigma \alpha = - 2 / 3 , \Sigma \alpha \beta = - 7 / 3 , \alpha \beta \gamma = 2
   	\qquad \begin{array} { l } ( \alpha + \beta + \gamma ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ( = - 1 )
   ( \alpha \beta + \beta \gamma + \gamma \alpha ) \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 2 } \left( = - \frac { 21 } { 4 } \right) \end{array}
   	\qquad \alpha \beta \gamma \left( 1 + \frac { 1 } { \alpha \beta \gamma } \right) ^ { 3 } \left( = \frac { 27 } { 4 } \right)
   	\text { Hence equation is } x ^ { 3 } + x ^ { 2 } - \frac { 21 } { 4 } x - \frac { 27 } { 4 } = 0 \end{aligned}\)	B1 M1 M1 M1 A1	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

   \begin{table}[h]
   \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	M1 B1 A1	Allow +/- errors Needs not to include origin Any clear indication of interior
   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 \(n = 7\)	B1 M1 M1 A1 B1 M1 A1 [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
   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	B1 M1 M1 A1 M1 A1 A1	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 \(n = 7\) "
   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 \(n = 7\)	B1 M1 B1 M1 M1 A1 A1	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 \(n = 7\)

   \end{table}

   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 \(32 / \sqrt { } 29\)	M2 A1 A1 [4]	Use correct formula with correct \(\mathbf { b }\) and \(\mathbf { n }\) Correct scalar product Answer (accept 5.94 (3s.f.))
   7(b)	\(\begin{aligned}	( 3 \mathbf { i } + \mathbf { k } ) \times ( \mathbf { i } - \mathbf { j } - 2 \mathbf { k } )
   	= \mathbf { i } + 7 \mathbf { j } - 3 \mathbf { k } \end{aligned}\) \(\begin{gathered} x + 7 y - 3 z = 0
   \left( \begin{array} { c } 1
   7
   - 3 \end{array} \right) \cdot \left( \begin{array} { c } 3
   2
   - 4 \end{array} \right) \end{gathered}\) \(\begin{gathered} \cos \theta = \frac { 29 } { \sqrt { 59 } \sqrt { 29 } }
   134.5 ^ { \circ } \text { or } 2.35 \text { radians } \end{gathered}\)	M1 A1 A1 M1 A1ft A1 [6]	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 )
   8(a)	\(\begin{gathered} \frac { A } { x } + \frac { B x + C } { 4 x ^ { 2 } + 1 }
   A \left( 4 x ^ { 2 } + 1 \right) + ( B x + C ) x \equiv 8 x ^ { 2 } - x + 2
   A = 2 , B = 0 , C = - 1
   \frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } \end{gathered}\)	M1 M1 M1 A1 A1 [5]	Attempt at partial fractions Set up identity Compare coefficients or substitute x -values All correct values Final answer
   8(b)	IF \(\exp \left( \int 2 / x \mathrm {~d} x \right) = x ^ { 2 }\) \(\begin{gathered} y x ^ { 2 } = \int \frac { 2 } { x } - \frac { 1 } { 4 x ^ { 2 } + 1 } d x
   y x ^ { 2 } = 2 \ln x - \frac { 1 } { 2 } \tan ^ { - 1 } ( 2 x ) + c
   y = \frac { 4 \ln x - \tan ^ { - 1 } ( 2 x ) + c ^ { \prime } } { 2 x ^ { 2 } } \end{gathered}\)	M1 A1 M1 M1 A1 A1 [6]	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)

   \section*{Further Mathematics Core (Pure) Mark Scheme (total 84)}

   9(a)	\(\begin{aligned}	( c + \mathrm { is } ) ^ { 5 } = c ^ { 5 } + 5 \mathrm { is } ^ { 4 } - 10 s ^ { 2 } c ^ { 2 } - 10 \mathrm { is } ^ { 3 } c ^ { 2 } + 5 s ^ { 4 } c + \mathrm { is } ^ { 5 }
   	\cos 5 \theta = \cos ^ { 5 } \theta - 10 \sin ^ { 2 } \theta \cos ^ { 3 } \theta + 5 \sin ^ { 4 } \theta \cos \theta
   	= c ^ { 5 } - 10 \left( 1 - c ^ { 2 } \right) c ^ { 3 } + 5 \left( 1 - c ^ { 2 } \right) ^ { 2 } c
   	= 16 \cos ^ { 5 } \theta - 20 \cos ^ { 3 } \theta + 5 \cos \theta \quad \text { AG } \end{aligned}\)	M1 A1 M1 A1 [4]	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
   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 +	M1 B1 M1 A1 A1 A1 [6]	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 +
   10(a)	\(\left( \begin{array} { l l } 3	2
   2	1 \end{array} \right)\)	B2 [2]	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	B1 B1 M1 A1 A1 [5]	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
   10(c)	Det \(\left( \mathbf { M } ^ { n } \right) = F _ { n + 1 } F _ { n - 1 } - F _ { n } { } ^ { 2 }\) \(\operatorname { Det } \left( \mathbf { M } ^ { n } \right) = ( \operatorname { Det } \mathbf { M } ) ^ { n }\) \(= ( - 1 ) ^ { n }\)	M1 M1 A1 [3]	Use determinant of \(\mathbf { M } ^ { n }\) Relate \(\operatorname { det } \mathbf { M } ^ { \boldsymbol { n } }\) to \(\operatorname { det } \mathbf { M }\) Correct answer