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HL Paper 1

Three planes have equations:

$2x - y + z = 5$

$x + 3y - z = 4$ , where $a{\text{, }}b \in \mathbb{R}$ .

$3x - 5y + az = b$

Find the set of values of $a$ and $b$ such that the three planes have no points of intersection.

Markscheme

attempt to eliminate a variable (or attempt to find det $A$ ) M1

$\left( {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ 1&3&{ - 1} \\ 3&{ - 5}&a \end{array}\,} \right|\begin{array}{*{20}{c}} 5 \\ 4 \\ b \end{array}} \right) \to \left( {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ 0&7&{ - 3} \\ 0&{ - 14}&{a + 3} \end{array}\,} \right|\begin{array}{*{20}{c}} 5 \\ 3 \\ {b - 12} \end{array}} \right)$ (or det $A = 14\left( {a - 3} \right)$ )

(or two correct equations in two variables) A1

$\to \left( {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ 0&7&{ - 3} \\ 0&{ 0}&{a - 3} \end{array}\,} \right|\begin{array}{*{20}{c}} 5 \\ 3 \\ {b - 6} \end{array}} \right)$ (or solving det $A = 0$ )

(or attempting to reduce to one variable, e.g. $\left( {a - 3} \right)z = b - 6$ ) M1

$a = 3{\text{, }}b \ne 6$ A1A1

[5 marks]

Examiners report

[N/A]

Two distinct lines, ${l_1}$ and ${l_2}$ , intersect at a point ${\text{P}}$ . In addition to ${\text{P}}$ , four distinct points are marked out on ${l_1}$ and three distinct points on ${l_2}$ . A mathematician decides to join some of these eight points to form polygons.

The line ${l_1}$ has vector equation r₁ $= \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 1 \end{array}} \right)$ , $\lambda \in \mathbb{R}$ and the line ${l_2}$ has vector equation r₂ $= \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 2 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}} 5 \\ 6 \\ 2 \end{array}} \right)$ , $\mu \in \mathbb{R}$ .

The point ${\text{P}}$ has coordinates (4, 6, 4).

The point ${\text{A}}$ has coordinates (3, 4, 3) and lies on ${l_1}$ .

The point ${\text{B}}$ has coordinates (−1, 0, 2) and lies on ${l_2}$ .

Find how many sets of four points can be selected which can form the vertices of a quadrilateral.

[2]

a.i.

Find how many sets of three points can be selected which can form the vertices of a triangle.

[4]

a.ii.

Verify that ${\text{P}}$ is the point of intersection of the two lines.

[3]

Write down the value of $\lambda$ corresponding to the point ${\text{A}}$ .

[1]

Write down $\overrightarrow {{\text{PA}}}$ and $\overrightarrow {{\text{PB}}}$ .

[2]

Let ${\text{C}}$ be the point on ${l_1}$ with coordinates (1, 0, 1) and ${\text{D}}$ be the point on ${l_2}$ with parameter $\mu = - 2$ .

Find the area of the quadrilateral ${\text{CDBA}}$ .

[8]

Markscheme

appreciation that two points distinct from ${\text{P}}$ need to be chosen from each line M1

${}^4{C_2} \times {}^3{C_2}$

=18 A1

[2 marks]

a.i.

EITHER

consider cases for triangles including ${\text{P}}$ or triangles not including ${\text{P}}$ M1

$3 \times 4 + 4 \times {}^3{C_2} + 3 \times {}^4{C_2}$ (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

consider total number of ways to select 3 points and subtract those with 3 points on the same line M1

${}^8{C_3} - {}^5{C_3} - {}^4{C_3}$ (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

56−10−4

THEN

= 42 A1

[4 marks]

a.ii.

METHOD 1

substitution of (4, 6, 4) into both equations (M1)

$\lambda = 3$ and $\mu = 1$ A1A1

(4, 6, 4) AG

METHOD 2

attempting to solve two of the three parametric equations M1

$\lambda = 3$ and $\mu = 1$ A1

check both of the above give (4, 6, 4) M1AG

Note: If they have shown the curve intersects for all three coordinates they only need to check (4,6,4) with one of " $\lambda$ " or " $\mu$ ".

[3 marks]

$\lambda = 2$ A1

[1 mark]

$\overrightarrow {{\text{PA}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ { - 2} \\ { - 1} \end{array}} \right)$ , $\overrightarrow {{\text{PB}}} = \left( {\begin{array}{*{20}{c}} { - 5} \\ { - 6} \\ { - 2} \end{array}} \right)$ A1A1

Note: Award A1A0 if both are given as coordinates.

[2 marks]

METHOD 1

area triangle ${\text{ABP}} = \frac{1}{2}\left| {\overrightarrow {{\text{PB}}} \times \overrightarrow {{\text{PA}}} } \right|$ M1

$\left( { = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} { - 5} \\ { - 6} \\ { - 2} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} { - 1} \\ { - 2} \\ { - 1} \end{array}} \right)} \right|} \right) = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} 2 \\ { - 3} \\ 4 \end{array}} \right)} \right|$ A1

$= \frac{{\sqrt {29} }}{2}$ A1

EITHER

$\overrightarrow {{\text{PC}}} = 3\overrightarrow {\,{\text{PA}}}$ , $\overrightarrow {{\text{PD}}} = 3\overrightarrow {\,{\text{PB}}}$ (M1)

area triangle ${\text{PCD}} = 9 \times$ area triangle ${\text{ABP}}$ (M1)A1

$= \frac{{9\sqrt {29} }}{2}$ A1

${\text{D}}$ has coordinates (−11, −12, −2) A1

area triangle ${\text{PCD}} = \frac{1}{2}\left| {\overrightarrow {{\text{PD}}} \times \overrightarrow {{\text{PC}}} } \right| = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} { - 15} \\ { - 18} \\ { - 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 6} \\ { - 3} \end{array}} \right)} \right|$ M1A1

Note: A1 is for the correct vectors in the correct formula.

$= \frac{{9\sqrt {29} }}{2}$ A1

THEN

area of ${\text{CDBA}} = \frac{{9\sqrt {29} }}{2} - \frac{{\sqrt {29} }}{2}$

$= 4\sqrt {29}$ A1

METHOD 2

${\text{D}}$ has coordinates (−11, −12, −2) A1

area $= \frac{1}{2}\left| {\overrightarrow {{\text{CB}}} \times \overrightarrow {{\text{CA}}} } \right| + \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{BD}}} } \right|$ M1

Note: Award M1 for use of correct formula on appropriate non-overlapping triangles.

Note: Different triangles or vectors could be used.

$\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} { - 2} \\ 0 \\ 1 \end{array}} \right)$ , $\overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ 2 \end{array}} \right)$ A1

$\overrightarrow {{\text{CB}}} \times \overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} { - 4} \\ 6 \\ { - 8} \end{array}} \right)$ A1

$\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ { - 1} \end{array}} \right)$ , $\overrightarrow {{\text{BD}}} = \left( {\begin{array}{*{20}{c}} { - 10} \\ { - 12} \\ { - 4} \end{array}} \right)$ A1

$\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{BD}}} = \left( {\begin{array}{*{20}{c}} { - 12} \\ {18} \\ { - 24} \end{array}} \right)$ A1

Note: Other vectors which might be used are $\overrightarrow {{\text{DA}}} = \left( {\begin{array}{*{20}{c}} {14} \\ {16} \\ {5} \end{array}} \right)$ , $\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} {4} \\ {4} \\ {1} \end{array}} \right)$ , $\overrightarrow {{\text{DC}}} = \left( {\begin{array}{*{20}{c}} {12} \\ {12} \\ {3} \end{array}} \right)$ .

Note: Previous A1A1A1A1 are all dependent on the first M1.

valid attempt to find a value of $\frac{1}{2}\left| {a \times b} \right|$ M1

Note: M1 independent of triangle chosen.

area $= \frac{1}{2} \times 2 \times \sqrt {29} + \frac{1}{2} \times 6 \times \sqrt {29}$

$= 4\sqrt {29}$ A1

Note: accept $\frac{1}{2} \sqrt {116} + \frac{1}{2}\sqrt {1044}$ or equivalent.

[8 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Let S be the sum of the roots found in part (a).

Find the roots of ${z^{24}} = 1$ which satisfy the condition $0 < {\text{arg}}\left( z \right) < \frac{\pi }{2}$ , expressing your answers in the form $r{e^{{\text{i}}\theta }}$ , where $r$ , $\theta \in {\mathbb{R}^ + }$ .

[5]

Show that Re S = Im S.

[4]

b.i.

By writing $\frac{\pi }{{12}}$ as $\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$ , find the value of cos $\frac{\pi }{{12}}$ in the form $\frac{{\sqrt a + \sqrt b }}{c}$ , where $a$ , $b$ and $c$ are integers to be determined.

[3]

b.ii.

Hence, or otherwise, show that S = $\frac{1}{2}\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)\left( {1 + {\text{i}}} \right)$ .

[4]

b.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\left( {r\left( {{\text{cos}}\,\theta + {\text{i}}\,{\text{sin}}\,\theta } \right)} \right)^{24}} = 1\left( {{\text{cos}}\,0 + {\text{i}}\,{\text{sin}}\,0} \right)$

use of De Moivre’s theorem (M1)

${r^{24}} = 1 \Rightarrow r = 1$ (A1)

$24\theta = 2\pi n \Rightarrow \theta = \frac{{\pi n}}{{12}}$ , $\left( {n \in \mathbb{Z}} \right)$ (A1)

$0 < {\text{arg}}\left( z \right) < \frac{\pi }{2} \Rightarrow n =$ 1, 2, 3, 4, 5

$z = {\text{e}}\frac{{\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{2\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{3\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{4\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{5\pi {\text{i}}}}{{12}}$ A2

Note: Award A1 if additional roots are given or if three correct roots are given with no incorrect (or additional) roots.

[5 marks]

Re S = ${\text{cos}}\frac{\pi }{{12}} + {\text{cos}}\frac{{2\pi }}{{12}} + {\text{cos}}\frac{{3\pi }}{{12}} + {\text{cos}}\frac{{4\pi }}{{12}} + {\text{cos}}\frac{{5\pi }}{{12}}$

Im S = ${\text{sin}}\frac{\pi }{{12}} + {\text{sin}}\frac{{2\pi }}{{12}} + {\text{sin}}\frac{{3\pi }}{{12}} + {\text{sin}}\frac{{4\pi }}{{12}} + {\text{sin}}\frac{{5\pi }}{{12}}$ A1

Note: Award A1 for both parts correct.

but ${\text{sin}}\frac{{5\pi }}{{12}} = {\text{cos}}\frac{\pi }{{12}}$ , ${\text{sin}}\frac{{4\pi }}{{12}} = {\text{cos}}\frac{{2\pi }}{{12}}$ , ${\text{sin}}\frac{{3\pi }}{{12}} = {\text{cos}}\frac{{3\pi }}{{12}}$ , ${\text{sin}}\frac{{2\pi }}{{12}} = {\text{cos}}\frac{{4\pi }}{{12}}$ and ${\text{sin}}\frac{\pi }{{12}} = {\text{cos}}\frac{{5\pi }}{{12}}$ M1A1

⇒ Re S = Im S AG

Note: Accept a geometrical method.

[4 marks]

b.i.

${\text{cos}}\frac{\pi }{{12}} = {\text{cos}}\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) = {\text{cos}}\frac{\pi }{4}{\text{cos}}\frac{\pi }{6} + {\text{sin}}\frac{\pi }{4}{\text{sin}}\frac{\pi }{6}$ M1A1

$= \frac{{\sqrt 2 }}{2}\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 2 }}{2}\frac{1}{2}$

$= \frac{{\sqrt 6 + \sqrt 2 }}{4}$ A1

[3 marks]

b.ii.

${\text{cos}}\frac{{5\pi }}{{12}} = {\text{cos}}\left( {\frac{\pi }{6} + \frac{\pi }{4}} \right) = {\text{cos}}\frac{\pi }{6}{\text{cos}}\frac{\pi }{4} - {\text{sin}}\frac{\pi }{6}{\text{sin}}\frac{\pi }{4}$ (M1)

Note: Allow alternative methods eg ${\text{cos}}\frac{{5\pi }}{{12}} = {\text{sin}}\frac{\pi }{{12}} = {\text{sin}}\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$ .

$= \frac{{\sqrt 3 }}{2}\frac{{\sqrt 2 }}{2} - \frac{1}{2}\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 - \sqrt 2 }}{4}$ (A1)

Re S = ${\text{cos}}\frac{\pi }{{12}} + {\text{cos}}\frac{{2\pi }}{{12}} + {\text{cos}}\frac{{3\pi }}{{12}} + {\text{cos}}\frac{{4\pi }}{{12}} + {\text{cos}}\frac{{5\pi }}{{12}}$

Re S = $\frac{{\sqrt 2 + \sqrt 6 }}{4} + \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 2 }}{2} + \frac{1}{2} + \frac{{\sqrt 6 - \sqrt 2 }}{4}$ A1

$= \frac{1}{2}\left( {\sqrt 6 + 1 + \sqrt 2 + \sqrt 3 } \right)$ A1

$= \frac{1}{2}\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)$

S = Re(S)(1 + i) since Re S = Im S, R1

S = $\frac{1}{2}\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)\left( {1 + {\text{i}}} \right)$ AG

[4 marks]

b.iii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

Let $z = 1 - \cos 2\theta - {\text{i}}\sin 2\theta ,{\text{ }}z \in \mathbb{C},{\text{ }}0 \leqslant \theta \leqslant \pi$ .

Solve $2\sin (x + 60^\circ ) = \cos (x + 30^\circ ),{\text{ }}0^\circ \leqslant x \leqslant 180^\circ$ .

[5]

Show that $\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}$ .

[3]

Find the modulus and argument of $z$ in terms of $\theta$ . Express each answer in its simplest form.

[9]

c.i.

Hence find the cube roots of $z$ in modulus-argument form.

[5]

c.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$2\sin (x + 60^\circ ) = \cos (x + 30^\circ )$

$2(\sin x\cos 60^\circ + \cos x\sin 60^\circ ) = \cos x\cos 30^\circ - \sin x\sin 30^\circ$ (M1)(A1)

$2\sin x \times \frac{1}{2} + 2\cos x \times \frac{{\sqrt 3 }}{2} = \cos x \times \frac{{\sqrt 3 }}{2} - \sin x \times \frac{1}{2}$ A1

$\Rightarrow \frac{3}{2}\sin x = - \frac{{\sqrt 3 }}{2}\cos x$

$\Rightarrow \tan x = - \frac{1}{{\sqrt 3 }}$ M1

$\Rightarrow x = 150^\circ$ A1

[5 marks]

EITHER

choosing two appropriate angles, for example 60° and 45° M1

$\sin 105^\circ = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ$ and

$\cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ$ (A1)

$\sin 105^\circ + \cos 105^\circ = \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}$ A1

$= \frac{1}{{\sqrt 2 }}$ AG

attempt to square the expression M1

${(\sin 105^\circ + \cos 105^\circ )^2} = {\sin ^2}105^\circ + 2\sin 105^\circ \cos 105^\circ + {\cos ^2}105^\circ$

${(\sin 105^\circ + \cos 105^\circ )^2} = 1 + \sin 210^\circ$ A1

$= \frac{1}{2}$ A1

$\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}$ AG

[3 marks]

EITHER

$z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta$

$\left| z \right| = \sqrt {{{(1 - \cos 2\theta )}^2} + {{(\sin 2\theta )}^2}}$ M1

$\left| z \right| = \sqrt {1 - 2\cos 2\theta + {{\cos }^2}2\theta + {{\sin }^2}2\theta }$ A1

$= \sqrt 2 \sqrt {(1 - \cos 2\theta )}$ A1

$= \sqrt {2(2{{\sin }^2}\theta )}$

$= 2\sin \theta$ A1

let $\arg (z) = \alpha$

$\tan \alpha = - \frac{{\sin 2\theta }}{{1 - \cos 2\theta }}$ M1

$= \frac{{ - 2\sin \theta \cos \theta }}{{2{{\sin }^2}\theta }}$ (A1)

$= - \cot \theta$ A1

$\arg (z) = \alpha = - \arctan \left( {\tan \left( {\frac{\pi }{2} - \theta } \right)} \right)$ A1

$= \theta - \frac{\pi }{2}$ A1

$z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta$

$= 2{\sin ^2}\theta - 2{\text{i}}\sin \theta \cos \theta$ M1A1

$= 2\sin \theta (\sin \theta - {\text{i}}\cos \theta )$ (A1)

$= - 2{\text{i}}\sin \theta (\cos \theta + {\text{i}}\sin \theta )$ M1A1

$= 2\sin \theta \left( {\cos \left( {\theta - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\theta - \frac{\pi }{2}} \right)} \right)$ M1A1

$\left| z \right| = 2\sin \theta$ A1

$\arg (z) = \theta - \frac{\pi }{2}$ A1

[9 marks]

c.i.

attempt to apply De Moivre’s theorem M1

${(1 - \cos 2\theta - {\text{i}}\sin 2\theta )^{\frac{1}{3}}} = {2^{\frac{1}{3}}}{(\sin \theta )^{\frac{1}{3}}}\left[ {\cos \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right)} \right]$ A1A1A1

Note: A1 for modulus, A1 for dividing argument of $z$ by 3 and A1 for $2n\pi$ .

Hence cube roots are the above expression when $n = - 1,{\text{ }}0,{\text{ }}1$ . Equivalent forms are acceptable. A1

[5 marks]

c.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

Consider the three planes

$\prod_{1} : 2 x - y + z = 4$

$\prod_{2} : x - 2 y + 3 z = 5$

$\prod_{3} : - 9 x + 3 y - 2 z = 32$

Show that the three planes do not intersect.

[4]

Verify that the point $P (1, - 2, 0)$ lies on both $\prod_{1}$ and $\prod_{2}$ .

[1]

b.i.

Find a vector equation of $L$ , the line of intersection of $\prod_{1}$ and $\prod_{2}$ .

[4]

b.ii.

Find the distance between $L$ and $\prod_{3}$ .

[6]

Markscheme

METHOD 1

attempt to eliminate a variable M1

obtain a pair of equations in two variables

EITHER

$- 3 x + z = - 3$ and A1

$- 3 x + z = 44$ A1

$- 5 x + y = - 7$ and A1

$- 5 x + y = 40$ A1

$3 x - z = 3$ and A1

$3 x - z = - \frac{79}{5}$ A1

THEN

the two lines are parallel ( $- 3 \neq 44$ or $- 7 \neq 40$ or $3 \neq - \frac{79}{5}$ ) R1

Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.

hence the three planes do not intersect AG

METHOD 2

vector product of the two normals $= (\begin{matrix} - 1 \\ - 5 \\ - 3 \end{matrix})$ (or equivalent) A1

$r = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ (or equivalent) A1

Note: Award A0 if “ $r =$ ” is missing. Subsequent marks may still be awarded.

Attempt to substitute $(1 + λ, - 2 + 5 λ, 3 λ)$ in $\prod_{3}$ M1

$- 9 (1 + λ) + 3 (- 2 + 5 λ) - 2 (3 λ) = 32$

$- 15 = 32$ , a contradiction R1

hence the three planes do not intersect AG

METHOD 3

attempt to eliminate a variable M1

$- 3 y + 5 z = 6$ A1

$- 3 y + 5 z = 100$ A1

$0 = 94$ , a contradiction R1

Note: Accept other equivalent alternatives. Accept other valid methods.
To obtain the final R1, at least the initial M1 must have been awarded.

hence the three planes do not intersect AG

[4 marks]

$\prod_{1} : 2 + 2 + 0 = 4$ and $\prod_{2} : 1 + 4 + 0 = 5$ A1

[1 mark]

b.i.

METHOD 1

attempt to find the vector product of the two normals M1

$(\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}) \times (\begin{matrix} 1 \\ - 2 \\ 3 \end{matrix})$

$= (\begin{matrix} - 1 \\ - 5 \\ - 3 \end{matrix})$ A1

$r = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ A1A1

Note: Award A1A0 if “ $r =$ ” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize lack of “ $r =$ ” only once.

METHOD 2

attempt to eliminate a variable from $\prod_{1}$ and $\prod_{2}$ M1

$3 x - z = 3$ OR $3 y - 5 z = - 6$ OR $5 x - y = 7$

Let $x = t$

substituting $x = t$ in $3 x - z = 3$ to obtain

$z = - 3 + 3 t$ and $y = 5 t - 7$ (for all three variables in parametric form) A1

$r = (\begin{matrix} 0 \\ - 7 \\ - 3 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ A1A1

Note: Award A1A0 if “ $r =$ ” is missing.
Accept any multiple of the direction vector. Accept other position vectors which satisfy both the planes $\prod_{1}$ and $\prod_{2}$ .

[4 marks]

b.ii.

METHOD 1

the line connecting $L$ and $\prod_{3}$ is given by $L_{1}$

attempt to substitute position and direction vector to form $L_{1}$ (M1)

$s = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + t (\begin{matrix} - 9 \\ 3 \\ - 2 \end{matrix})$ A1

substitute $(1 - 9 t, - 2 + 3 t, - 2 t)$ in $\prod_{3}$ M1

$- 9 (1 - 9 t) + 3 (- 2 + 3 t) - 2 (- 2 t) = 32$

$94 t = 47 \Rightarrow t = \frac{1}{2}$ A1

attempt to find distance between $(1, - 2, 0)$ and their point $(- \frac{7}{2}, - \frac{1}{2}, - 1)$ (M1)

$= |(\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + \frac{1}{2} (\begin{matrix} - 9 \\ 3 \\ - 2 \end{matrix}) - (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix})| = \frac{1}{2} \sqrt{{(- 9)}^{2} + 3^{2} + {(- 2)}^{2}}$

$= \frac{\sqrt{94}}{2}$ A1

METHOD 2

unit normal vector equation of $\prod_{3}$ is given by $\frac{(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix})}{\sqrt{81 + 9 + 4}}$ (M1)

$= \frac{32}{\sqrt{94}}$ A1

let $\prod_{4}$ be the plane parallel to $\prod_{3}$ and passing through $P$ ,
then the normal vector equation of $\prod_{4}$ is given by

$(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) = - 15$ M1

unit normal vector equation of $\prod_{4}$ is given by

$\frac{(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix})}{\sqrt{81 + 9 + 4}} = \frac{- 15}{\sqrt{94}}$ A1

distance between the planes is $\frac{32}{\sqrt{94}} - \frac{- 15}{\sqrt{94}}$ (M1)

$= \frac{47}{\sqrt{94}} (= \frac{\sqrt{94}}{2})$ A1

[6 marks]

Examiners report

Part (a) was well attempted using a variety of approaches. Most candidates were able to gain marks for part (a) through attempts to eliminate a variable with many subsequently making algebraic errors. Part (b)(i) was well done. For part (b)(ii) few successful attempts were noted, many candidates failed to use an appropriate notation "r =" while giving the vector equation of a line. Part (c) proved to be challenging for most candidates with very few correct answers seen. Many candidates did not attempt part (c).

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

In the following diagram, $\overrightarrow {{\text{OA}}}$ = a, $\overrightarrow {{\text{OB}}}$ = b. C is the midpoint of [OA] and $\overrightarrow {{\text{OF}}} = \frac{1}{6}\overrightarrow {{\text{FB}}}$ .

N17/5/MATHL/HP1/ENG/TZ0/09

It is given also that $\overrightarrow {{\text{AD}}} = \lambda \overrightarrow {{\text{AF}}}$ and $\overrightarrow {{\text{CD}}} = \mu \overrightarrow {{\text{CB}}}$ , where $\lambda ,{\text{ }}\mu \in \mathbb{R}$ .

Find, in terms of a and b $\overrightarrow {{\text{OF}}}$ .

[1]

a.i.

Find, in terms of a and b $\overrightarrow {{\text{AF}}}$ .

[2]

a.ii.

Find an expression for $\overrightarrow {{\text{OD}}}$ in terms of a, b and $\lambda$ ;

[2]

b.i.

Find an expression for $\overrightarrow {{\text{OD}}}$ in terms of a, b and $\mu$ .

[2]

b.ii.

Show that $\mu = \frac{1}{{13}}$ , and find the value of $\lambda$ .

[4]

Deduce an expression for $\overrightarrow {{\text{CD}}}$ in terms of a and b only.

[2]

Given that area $\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})$ , find the value of $k$ .

[5]

Markscheme

$\overrightarrow {{\text{OF}}} = \frac{1}{7}$ b A1

[1 mark]

a.i.

$\overrightarrow {{\text{AF}}} = \overrightarrow {{\text{OF}}} - \overrightarrow {{\text{OA}}}$ (M1)

$= \frac{1}{7}$ b – a A1

[2 marks]

a.ii.

$\overrightarrow {{\text{OD}}} =$ a $+ \lambda \left( {\frac{1}{7}b -a} \right){\text{ }}\left( { = (1 - \lambda )a + \frac{\lambda }{7}b} \right)$ M1A1

[2 marks]

b.i.

$\overrightarrow {{\text{OD}}} = \frac{1}{2}$ a $+ \mu \left( { - \frac{1}{2}a + b} \right){\text{ }}\left( { = \left( {\frac{1}{2} - \frac{\mu }{2}} \right)a + \mu b} \right)$ M1A1

[2 marks]

b.ii.

equating coefficients: M1

$\frac{\lambda }{7} = \mu ,{\text{ }}1 - \lambda = \frac{{1 - \mu }}{2}$ A1

solving simultaneously: M1

$\lambda = \frac{7}{{13}},{\text{ }}\mu = \frac{1}{{13}}$ A1AG

[4 marks]

$\overrightarrow {{\text{CD}}} = \frac{1}{{13}}\overrightarrow {{\text{CB}}}$

$= \frac{1}{{13}}\left( {b - \frac{1}{2}a} \right){\text{ }}\left( { = - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)$ M1A1

[2 marks]

METHOD 1

${\text{area }}\Delta {\text{ACD}} = \frac{1}{2}{\text{CD}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}$ (M1)

${\text{area }}\Delta {\text{ACB}} = \frac{1}{2}{\text{CB}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}$ (M1)

${\text{ratio }}\frac{{{\text{area }}\Delta {\text{ACD}}}}{{{\text{area }}\Delta {\text{ACB}}}} = \frac{{{\text{CD}}}}{{{\text{CB}}}} = \frac{1}{{13}}$ A1

$k = \frac{{{\text{area }}\Delta {\text{OAB}}}}{{{\text{area }}\Delta {\text{CAD}}}} = \frac{{13}}{{{\text{area }}\Delta {\text{CAB}}}} \times {\text{area }}\Delta {\text{OAB}}$ (M1)

$= 13 \times 2 = 26$ A1

METHOD 2

${\text{area }}\Delta {\text{OAB}} = \frac{1}{2}\left| {a \times b} \right|$ A1

${\text{area }}\Delta {\text{CAD}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CD}}} } \right|$ or $\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{AD}}} } \right|$ M1

$= \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)} \right|$

$= \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a} \right) + \frac{1}{2}a \times \frac{1}{{13}}b} \right|$ (M1)

$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{{13}}\left| {a \times b} \right|{\text{ }}\left( { = \frac{1}{{52}}\left| {a \times b} \right|} \right)$ A1

${\text{area }}\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})$

$\frac{1}{2}\left| {a \times b} \right| = k\frac{1}{{52}}\left| {a \times b} \right|$

$k = 26$ A1

[5 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Show that ${\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)^2} = 1 + {\text{sin}}\,2x$ .

[2]

Show that ${\text{sec}}\,2x + {\text{tan}}\,2x = \frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}$ .

[4]

Hence or otherwise find $\int_0^{\frac{\pi }{6}} {\left( {{\text{sec}}\,2x + {\text{tan}}\,2x} \right)} {\text{d}}x$ in the form ${\text{ln}}\left( {a + \sqrt b } \right)$ where $a$ , $b \in \mathbb{Z}$ .

[9]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)^2} = {\text{si}}{{\text{n}}^2}\,x + 2{\text{sin}}\,x\,{\text{cos}}\,x + {\text{co}}{{\text{s}}^2}\,x$ M1A1

Note: Do not award the M1 for just ${\text{si}}{{\text{n}}^2}\,x + {\text{co}}{{\text{s}}^2}\,x$ .

Note: Do not award A1 if correct expression is followed by incorrect working.

$= 1 + {\text{sin}}\,2x$ AG

[2 marks]

${\text{sec}}\,2x + {\text{tan}}\,2x = \frac{1}{{{\text{cos}}\,2x}} + \frac{{{\text{sin}}\,2x}}{{{\text{cos}}\,2x}}$ M1

Note: M1 is for an attempt to change both terms into sine and cosine forms (with the same argument) or both terms into functions of ${\text{tan}}\,x$ .

$= \frac{{{\text{1}} + {\text{sin}}\,2x}}{{{\text{cos}}\,2x}}$

$= \frac{{{{\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)}^2}}}{{{\text{co}}{{\text{s}}^2}\,x - {\text{si}}{{\text{n}}^2}\,x}}$ A1A1

Note: Award A1 for numerator, A1 for denominator.

$= \frac{{{{\left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)}^2}}}{{\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)\left( {{\text{cos}}\,x + {\text{sin}}\,x} \right)}}$ M1

$= \frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}$ AG

Note: Apply MS in reverse if candidates have worked from RHS to LHS.

Note: Alternative method using ${\text{tan}}\,2x$ and ${\text{sec}}\,2x$ in terms of ${\text{tan}}\,x$ .

[4 marks]

METHOD 1

$\int_0^{\frac{\pi }{6}} {\left( {\frac{{{\text{cos}}\,x + {\text{sin}}\,x}}{{{\text{cos}}\,x - {\text{sin}}\,x}}} \right)} {\text{d}}x$ A1

Note: Award A1 for correct expression with or without limits.

EITHER

$= \left[ { - {\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)} \right]_0^{\frac{\pi }{6}}$ or $\left[ {{\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)} \right]_{\frac{\pi }{6}}^0$ (M1)A1A1

Note: Award M1 for integration by inspection or substitution, A1 for ${{\text{ln}}\left( {{\text{cos}}\,x - {\text{sin}}\,x} \right)}$ , A1 for completely correct expression including limits.

$= - {\text{ln}}\left( {{\text{cos}}\,\frac{\pi }{6} - {\text{sin}}\,\frac{\pi }{6}} \right) + {\text{ln}}\left( {{\text{cos}}\,0 - {\text{sin}}\,0} \right)$ M1

Note: Award M1 for substitution of limits into their integral and subtraction.

$= - {\text{ln}}\left( {\frac{{\sqrt 3 }}{2} - \frac{1}{2}} \right)$ (A1)

let $u = {\text{cos}}\,x - {\text{sin}}\,x$ M1

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = - {\text{sin}}\,x - {\text{cos}}\,x = - \left( {{\text{sin}}\,x + {\text{cos}}\,x} \right)$

$- \int_1^{\frac{{\sqrt 3 }}{2} - \frac{1}{2}} {\left( {\frac{1}{u}} \right)} {\text{d}}u$ A1A1

Note: Award A1 for correct limits even if seen later, A1 for integral.

$= \left[ { - {\text{ln}}\,u} \right]_1^{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}$ or $\left[ {{\text{ln}}\,u} \right]_{\frac{{\sqrt 3 }}{2} - \frac{1}{2}}^1$ A1

$= - {\text{ln}}\left( {\frac{{\sqrt 3 }}{2} - \frac{1}{2}} \right)\left( {{\text{ + ln}}\,1} \right)$ M1

THEN

$= {\text{ln}}\left( {\frac{2}{{\sqrt 3 - 1}}} \right)$

Note: Award M1 for both putting the expression over a common denominator and for correct use of law of logarithms.

$= {\text{ln}}\left( {1 + \sqrt 3 } \right)$ (M1)A1

METHOD 2

$\left[ {\frac{1}{2}{\text{ln}}\left( {{\text{tan}}\,2x + {\text{sec}}\,2x} \right) - \frac{1}{2}{\text{ln}}\left( {{\text{cos}}\,2x} \right)} \right]_0^{\frac{\pi }{6}}$ A1A1

$= \frac{1}{2}{\text{ln}}\left( {\sqrt 3 + 2} \right) - \frac{1}{2}{\text{ln}}\left( {\frac{1}{2}} \right) - 0$ A1A1(A1)

$= \frac{1}{2}{\text{ln}}\left( {4 + 2\sqrt 3 } \right)$ M1

$= \frac{1}{2}{\text{ln}}\left( {{{\left( {1 + \sqrt 3 } \right)}^2}} \right)$ M1A1

$= {\text{ln}}\left( {1 + \sqrt 3 } \right)$ A1

[9 marks]

Examiners report

[N/A]

The following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).

The Cartesian equation of the plane ${\Pi _2}$ , passing through the points B , C and D , is $y + z = 1$ .

The plane ${\Pi _3}$ passes through O and is normal to the line BD.

${\Pi _3}$ cuts AD and BD at the points P and Q respectively.

Find the Cartesian equation of the plane ${\Pi _1}$ , passing through the points A , B and D.

[3]

Find the angle between the faces ABD and BCD.

[4]

Find the Cartesian equation of ${\Pi _3}$ .

[3]

Show that P is the midpoint of AD.

[4]

Find the area of the triangle OPQ.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognising normal to plane or attempting to find cross product of two vectors lying in the plane (M1)

for example, $\mathop {{\text{AB}}}\limits^ \to \,\, \times \mathop {{\text{AD}}}\limits^ \to = \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 0 \hfill \\ \end{gathered} \right) \times \left( \begin{gathered} - 1 \hfill \\ \,0 \hfill \\ \,1 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right)$ (A1)

${\Pi _1}\,{\text{:}}\,\,x + z = 1$ A1

[3 marks]

EITHER

$\left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) \bullet \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 1 \hfill \\ \end{gathered} \right) = 1 = \sqrt 2 \sqrt 2 \,{\text{cos}}\,\theta$ M1A1

$\left| {\left( \begin{gathered} 1 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) \times \left( \begin{gathered} 0 \hfill \\ 1 \hfill \\ 1 \hfill \\ \end{gathered} \right)} \right| = \sqrt 3 = \sqrt 2 \sqrt 2 \,{\text{sin}}\,\theta$ M1A1

Note: M1 is for an attempt to find the scalar or vector product of the two normal vectors.

$\Rightarrow \theta = 60^\circ \left( { = \frac{\pi }{3}} \right)$ A1

angle between faces is $20^\circ \left( { = \frac{{2\pi }}{3}} \right)$ A1

[4 marks]

$\mathop {{\text{DB}}}\limits^ \to = \left( \begin{gathered} \,1 \hfill \\ \,1 \hfill \\ - 1 \hfill \\ \end{gathered} \right)$ or $\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered} - 1 \hfill \\ - 1 \hfill \\ \,1 \hfill \\ \end{gathered} \right)$ (A1)

${\Pi _3}\,{\text{:}}\,\,x + y - z = k$ (M1)

${\Pi _3}\,{\text{:}}\,\,x + y - z = 0$ A1

[3 marks]

METHOD 1

line AD : (r =) $\left( \begin{gathered} 0 \hfill \\ 0 \hfill \\ 1 \hfill \\ \end{gathered} \right) + \lambda \left( \begin{gathered} \,1 \hfill \\ \,0 \hfill \\ - 1 \hfill \\ \end{gathered} \right)$ M1A1

intersects ${\Pi _3}$ when $\lambda - \left( {1 - \lambda } \right) = 0$ M1

so $\lambda = \frac{1}{2}$ A1

hence P is the midpoint of AD AG

METHOD 2

midpoint of AD is (0.5, 0, 0.5) (M1)A1

substitute into $x + y - z = 0$ M1

0.5 + 0.5 − 0.5 = 0 A1

hence P is the midpoint of AD AG

[4 marks]

METHOD 1

${\text{OP}} = \frac{1}{{\sqrt 2 }},\,\,{\text{O}}\mathop {\text{P}}\limits^ \wedge {\text{Q}} = 90^\circ ,\,\,{\text{O}}\mathop {\text{Q}}\limits^ \wedge {\text{P}} = 60^\circ$ A1A1A1

${\text{PQ}} = \frac{1}{{\sqrt 6 }}$ A1

area $= \frac{1}{{2\sqrt {12} }} = \frac{1}{{4\sqrt 3 }} = \frac{{\sqrt 3 }}{{12}}$ A1

METHOD 2

line BD : (r =) $\left( \begin{gathered} 1 \hfill \\ 1 \hfill \\ 0 \hfill \\ \end{gathered} \right) + \lambda \left( \begin{gathered} - 1 \hfill \\ - 1 \hfill \\ \,1 \hfill \\ \end{gathered} \right)$

$\Rightarrow \lambda = \frac{2}{3}$ (A1)

$\mathop {{\text{OQ}}}\limits^ \to = \left( \begin{gathered} \frac{1}{3} \hfill \\ \frac{1}{3} \hfill \\ \frac{2}{3} \hfill \\ \end{gathered} \right)$ A1

area = $\frac{1}{2}\left| {\mathop {{\text{OP}}}\limits^ \to \, \times \mathop {{\text{OQ}}}\limits^ \to } \right|$ M1

$\mathop {{\text{OP}}}\limits^ \to = \left( \begin{gathered} \frac{1}{2} \hfill \\ 0 \hfill \\ \frac{1}{2} \hfill \\ \end{gathered} \right)$ A1

Note: This A1 is dependent on M1.

area = $\frac{{\sqrt 3 }}{{12}}$ A1

[5 marks]

Examiners report

[N/A]

Find the value of $\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}$ .

[2]

Show that $\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .

[2]

Use the principle of mathematical induction to prove that

$\sin x + \sin 3x + \ldots + \sin (2n - 1)x = \frac{{1 - \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .

[9]

Hence or otherwise solve the equation $\sin x + \sin 3x = \cos x$ in the interval $0 < x < \pi$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}$ (M1)A1

Note: Award M1 for 5 equal terms with \) + \) or $-$ signs.

[2 marks]

$\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \frac{{1 - (1 - 2{{\sin }^2}x)}}{{2\sin x}}$ M1

$\equiv \frac{{2{{\sin }^2}x}}{{2\sin x}}$ A1

$\equiv \sin x$ AG

[2 marks]

let ${\text{P}}(n):\sin x + \sin 3x + \ldots + \sin (2n - 1)x \equiv \frac{{1 - \cos 2nx}}{{2\sin x}}$

if $n = 1$

${\text{P}}(1):\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x$ which is true (as proved in part (b)) R1

assume ${\text{P}}(k)$ true, $\sin x + \sin 3x + \ldots + \sin (2k - 1)x \equiv \frac{{1 - \cos 2kx}}{{2\sin x}}$ M1

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let $n = k$ ” only. Subsequent marks are independent of this M1.

consider ${\text{P}}(k + 1)$ :

${\text{P}}(k + 1):\sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x \equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}$

$LHS = \sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x$ M1

$\equiv \frac{{1 - \cos 2kx}}{{2\sin x}} + \sin (2k + 1)x$ A1

$\equiv \frac{{1 - \cos 2kx + 2\sin x\sin (2k + 1)x}}{{2\sin x}}$

$\equiv \frac{{1 - \cos 2kx + 2\sin x\cos x\sin 2kx + 2{{\sin }^2}x\cos 2kx}}{{2\sin x}}$ M1

$\equiv \frac{{1 - \left( {(1 - 2{{\sin }^2}x)\cos 2kx - \sin 2x\sin 2kx} \right)}}{{2\sin x}}$ M1

$\equiv \frac{{1 - (\cos 2x\cos 2kx - \sin 2x\sin 2kx)}}{{2\sin x}}$ A1

$\equiv \frac{{1 - \cos (2kx + 2x)}}{{2\sin x}}$ A1

$\equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}$

so if true for $n = k$ , then also true for $n = k + 1$

as true for $n = 1$ then true for all $n \in {\mathbb{Z}^ + }$ R1

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

[9 marks]

EITHER

$\sin x + \sin 3x = \cos x \Rightarrow \frac{{1 - \cos 4x}}{{2\sin x}} = \cos x$ M1

$\Rightarrow 1 - \cos 4x = 2\sin x\cos x,{\text{ }}(\sin x \ne 0)$ A1

$\Rightarrow 1 - (1 - 2{\sin ^2}2x) = \sin 2x$ M1

$\Rightarrow \sin 2x(2\sin 2x - 1) = 0$ M1

$\Rightarrow \sin 2x = 0$ or $\sin 2x = \frac{1}{2}$ A1

$2x = \pi ,{\text{ }}2x = \frac{\pi }{6}$ and $2x = \frac{{5\pi }}{6}$

$\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x$ M1A1

$\Rightarrow (2\sin 2x - 1)\cos x = 0,{\text{ }}(\sin x \ne 0)$ M1A1

$\Rightarrow \sin 2x = \frac{1}{2}$ of $\cos x = 0$ A1

$2x = \frac{\pi }{6},{\text{ }}2x = \frac{{5\pi }}{6}$ and $x = \frac{\pi }{2}$

THEN

$\therefore x = \frac{\pi }{2},{\text{ }}x = \frac{\pi }{{12}}$ and $x = \frac{{5\pi }}{{12}}$ A1

Note: Do not award the final A1 if extra solutions are seen.

[6 marks]

Examiners report

[N/A]

Use the binomial theorem to expand ${(\cos θ + i \sin θ)}^{4}$ . Give your answer in the form $a + b i$ where $a$ and $b$ are expressed in terms of $\sin θ$ and $\cos θ$ .

[3]

Use de Moivre’s theorem and the result from part (a) to show that $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ .

[5]

Use the identity from part (b) to show that the quadratic equation $x^{2} - 6 x + 1 = 0$ has roots ${cot}^{2} \frac{π}{8}$ and ${cot}^{2} \frac{3 π}{8}$ .

[5]

Hence find the exact value of ${cot}^{2} \frac{3 π}{8}$ .

[4]

Deduce a quadratic equation with integer coefficients, having roots ${cosec}^{2} \frac{π}{8}$ and ${cosec}^{2} \frac{3 π}{8}$ .

[3]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

uses the binomial theorem on ${(\cos θ + i \sin θ)}^{4}$ M1

$= {}_{4}C_{0} \cos^{4} θ + {}_{4}C_{1} \cos^{3} θ (i \sin θ) + {}_{4}C_{2} \cos^{2} θ (i^{2} \sin^{2} θ) + {}_{4}C_{3} \cos θ (i^{3} \sin^{3} θ) + {}_{4}C_{4} (i^{4} \sin^{4} θ)$ A1

$= (\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ A1

[3 marks]

(using de Moivre’s theorem with $n = 4$ gives) $\cos 4 θ + i \sin 4 θ$ (A1)

equates both the real and imaginary parts of $\cos 4 θ + i \sin 4 θ$ and $(\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ M1

$\cos 4 θ = \cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ$ and $\sin 4 θ = 4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ$

recognizes that $\cot 4 θ = \frac{\cos 4 θ}{\sin 4 θ}$ (A1)

substitutes for $\sin 4 θ$ and $\cos 4 θ$ into $\frac{\cos 4 θ}{\sin 4 θ}$ M1

$\cot 4 θ = \frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}$

divides the numerator and denominator by $\sin^{4} θ$ to obtain

$\cot 4 θ = \frac{\frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{\sin^{4} θ}}{\frac{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}{\sin^{4} θ}}$ A1

$\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ AG

[5 marks]

setting $\cot 4 θ = 0$ and putting $x = \cot^{2} θ$ in the numerator of $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ gives $x^{2} - 6 x + 1 = 0$ M1

attempts to solve $\cot 4 θ = 0$ for $θ$ M1

$4 θ = \frac{π}{2}, \frac{3 π}{2}, \dots (4 θ = \frac{1}{2} (2 n + 1) π, n = 0, 1, \dots)$ (A1)

$θ = \frac{π}{8}, \frac{3 π}{8}$ A1

Note: Do not award the final A1 if solutions other than $θ = \frac{π}{8}, \frac{3 π}{8}$ are listed.

finding the roots of $\cot 4 θ = 0 (θ = \frac{π}{8}, \frac{3 π}{8})$ corresponds to finding the roots of $x^{2} - 6 x + 1 = 0$ where $x = \cot^{2} θ$ R1

so the equation $x^{2} - 6 x + 1 = 0$ as roots ${cot}^{2} \frac{π}{8}$ and ${cot}^{2} \frac{3 π}{8}$ AG

[5 marks]

attempts to solve $x^{2} - 6 x + 1 = 0$ for $x$ M1

$x = 3 \pm 2 \sqrt{2}$ A1

since ${cot}^{2} \frac{π}{8} > {cot}^{2} \frac{3 π}{8}, {cot}^{2} \frac{3 π}{8}$ has the smaller value of the two roots R1

Note: Award R1 for an alternative convincing valid reason.

so ${cot}^{2} \frac{3 π}{8} = 3 -2 \sqrt{2}$ A1

[4 marks]

let $y = {cosec}^{2} θ$

uses $\cot^{2} θ = {cosec}^{2} θ - 1$ where $x = \cot^{2} θ$ (M1)

$x^{2} - 6 x + 1 = 0 \Rightarrow {(y - 1)}^{2} - 6 (y - 1) + 1 = 0$ M1

$y^{2} - 8 y + 8 = 0$ A1

[3 marks]

Examiners report

[N/A]

In the following diagram, the points ${\text{A}}$ , ${\text{B}}$ , ${\text{C}}$ and ${\text{D}}$ are on the circumference of a circle with centre ${\text{O}}$ and radius $r$ . $\left[ {{\text{AC}}} \right]$ is a diameter of the circle. ${\text{BC}} = r$ , ${\text{AD}} = {\text{CD}}$ and ${\text{A}}\mathop {\text{B}}\limits^ \wedge {\text{C}} = {\text{A}}\mathop {\text{D}}\limits^ \wedge {\text{C}} = 90^\circ$ .

Given that ${\text{cos}}\,75^\circ = q$ , show that ${\text{cos}}\,105^\circ = - q$ .

[1]

Show that ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{D}} = 75^\circ$ .

[3]

By considering triangle ${\text{ABD}}$ , show that ${\text{B}}{{\text{D}}^2} = 5{r^2} - 2{r^2}q\sqrt 6$ .

[4]

c.i.

By considering triangle ${\text{CBD}}$ , find another expression for ${\text{B}}{{\text{D}}^2}$ in terms of $r$ and $q$ .

[3]

c.ii.

Use your answers to part (c) to show that ${\text{cos}}\,75^\circ = \frac{1}{{\sqrt 6 + \sqrt 2 }}$ .

[3]

Markscheme

${\text{cos}}\,105^\circ = {\text{cos}}\left( {180^\circ - 75^\circ } \right) = - {\text{cos}}\,75^\circ$ R1

$= - q$ AG

Note: Accept arguments using the unit circle or graphical/diagrammatical considerations.

[1 mark]

${\text{AD}} = {\text{CD}} \Rightarrow {\text{C}}\mathop {\text{A}}\limits^ \wedge {\text{D}} = 45^\circ$ A1

valid method to find ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ (M1)

for example: ${\text{BC}} = r \Rightarrow {\text{B}}\mathop {\text{C}}\limits^ \wedge {\text{A}} = 60^\circ$

$\Rightarrow {\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}} = 30^\circ$ A1

hence ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{D}} = 45^\circ + 30^\circ = 75^\circ$ AG

[3 marks]

${\text{AB}} = r\sqrt 3$ , ${\text{AD}} = \left( {{\text{CD}}} \right) = r\sqrt 2$ A1A1

applying cosine rule (M1)

${\text{B}}{{\text{D}}^2} = {\left( {r\sqrt 3 } \right)^2} + {\left( {r\sqrt 2 } \right)^2} - 2\left( {r\sqrt 3 } \right)\left( {r\sqrt 2 } \right){\text{cos}}\,75^\circ$ A1

$= 3{r^2} + 2{r^2} - 2{r^2}\sqrt 6 \,{\text{cos}}\,75^\circ$

$5{r^2} - 2{r^2}q\sqrt 6$ AG

[4 marks]

c.i.

${\text{B}}\mathop {\text{C}}\limits^ \wedge {\text{D}} = 105^\circ$ (A1)

attempt to use cosine rule on $\Delta {\text{BCD}}$ (M1)

${\text{B}}{{\text{D}}^2} = {r^2} + {\left( {r\sqrt 2 } \right)^2} - 2r\left( {r\sqrt 2 } \right){\text{cos}}\,105^\circ$

$= 3{r^2} + 2{r^2}q\sqrt 2$ A1

[3 marks]

c.ii.

$5{r^2} - 2{r^2}q\sqrt 6 = 3{r^2} + 2{r^2}q\sqrt 2$ (M1)(A1)

$2{r^2} = 2{r^2}q\left( {\sqrt 6 + \sqrt 2 } \right)$ A1

Note: Award A1 for any correct intermediate step seen using only two terms.

$q = \frac{1}{{\sqrt 6 + \sqrt 2 }}$ AG

Note: Do not award the final A1 if follow through is being applied.

[3 marks]

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Given any two non-zero vectors, $a$ and $b$ , show that ${|a \times b|}^{2} = {|a|}^{2} {|b|}^{2} - {(a \cdot b)}^{2}$ .

Markscheme

METHOD 1

use of $|a \times b| = |a| |b| \sin θ$ on the LHS (M1)

${|a \times b|}^{2} = {|a|}^{2} {|b|}^{2} \sin^{2} θ$ A1

$= {|a|}^{2} {|b|}^{2} (1 - \cos^{2} θ)$ M1

$= {|a|}^{2} {|b|}^{2} - {|a|}^{2} {|b|}^{2} \cos^{2} θ$ OR $= {|a|}^{2} {|b|}^{2} - {(|a| |b| \cos θ)}^{2}$ A1

$= {|a|}^{2} {|b|}^{2} - {(a \cdot b)}^{2}$ AG

METHOD 2

use of $a \cdot b = |a| |b| \cos θ$ on the RHS (M1)

$= {|a|}^{2} {|b|}^{2} - {|a|}^{2} {|b|}^{2} \cos^{2} θ$ A1

$= {|a|}^{2} {|b|}^{2} (1 - \cos^{2} θ)$ M1

$= {|a|}^{2} {|b|}^{2} \sin^{2} θ$ OR $= {(|a| |b| \sin θ)}^{2}$ A1

$= {|a \times b|}^{2}$ AG

Note: If candidates attempt this question using cartesian vectors, e.g

$a = (\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix})$ and $b = (\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix})$ ,

award full marks if fully developed solutions are seen.
Otherwise award no marks.

[4 marks]

Examiners report

[N/A]

The following diagram shows the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ , with asymptotes at $y = - \frac{π}{4}$ and $y = \frac{3 π}{4}$ .

Describe a sequence of transformations that transforms the graph of $y = arctan x$ to the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ .

[3]

Show that $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ .

[4]

Verify that $arctan (2 x + 1) = arctan (\frac{x}{x + 1}) + \frac{π}{4}$ for $x \in ℝ, x > 0$ .

[3]

Using mathematical induction and the result from part (b), prove that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$ .

[9]

Markscheme

EITHER
horizontal stretch/scaling with scale factor $\frac{1}{2}$

Note: Do not allow ‘shrink’ or ‘compression’

followed by a horizontal translation/shift $\frac{1}{2}$ units to the left A2

Note: Do not allow ‘move’

horizontal translation/shift $1$ unit to the left

followed by horizontal stretch/scaling with scale factor $\frac{1}{2}$ A2

THEN

vertical translation/shift up by $\frac{π}{4}$ (or translation through $(\begin{matrix} 0 \\ \frac{π}{4} \end{matrix})$ A1
(may be seen anywhere)

[3 marks]

let $α = arctan p$ and $β = arctan q$ M1

$p = tan α$ and $q = tan β$ (A1)

$\tan (α + β) = \frac{p + q}{1 - p q}$ A1

$α + β = arctan (\frac{p + q}{1 - p q})$ A1

so $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ . AG

[4 marks]

METHOD 1

$\frac{π}{4} = arctan 1$ (or equivalent) A1

$arctan (\frac{x}{x + 1}) + arctan 1 = arctan (\frac{\frac{x}{x + 1} + 1}{1 - \frac{x}{x + 1} (1)})$ A1

$= arctan (\frac{\frac{x + x + 1}{x + 1}}{\frac{x + 1 - x}{x + 1}})$ A1

$= arctan (2 x + 1)$ AG

METHOD 2

$\tan \frac{π}{4} = 1$ (or equivalent) A1

Consider $arctan (2 x + 1) - arctan (\frac{x}{x + 1}) = \frac{π}{4}$

$\tan (arctan (2 x + 1) - arctan (\frac{x}{x + 1}))$

$= arctan (\frac{2 x +1 - \frac{x}{x + 1}}{1 + \frac{x (2 x + 1)}{x + 1}})$ A1

$= arctan (\frac{(2 x +1) (x + 1) - x}{x + 1 + x (2 x + 1)})$ A1

$= arctan 1$ AG

METHOD 3

$tan (arctan (2 x + 1)) = \tan (arctan (\frac{x}{x + 1}) + \frac{π}{4})$

$\tan \frac{π}{4} = 1$ (or equivalent) A1

$LHS = 2 x + 1$ A1

$RHS = \frac{\frac{x}{x + 1} + 1}{1 - \frac{x}{x + 1}} (= 2 x + 1)$ A1

[3 marks]

let $P (n)$ be the proposition that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$

consider $P (1)$

when $n = 1, Σ_{r = 1}^{1} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{1}{2}) = RHS$ and so $P (1)$ is true R1

assume $P (k)$ is true, ie. $Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{k}{k + 1}) (k \in ℤ^{+})$ M1

Note: Award M0 for statements such as “let $n = k$ ”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

consider $P (k + 1)$ :

$Σ_{r = 1}^{k + 1} arctan (\frac{1}{2 r^{2}}) = Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ (M1)

$= arctan (\frac{k}{k + 1}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ A1

$= arctan (\frac{\frac{k}{k + 1} + \frac{1}{2 {(k + 1)}^{2}}}{1 - (\frac{k}{k + 1}) (\frac{1}{2 {(k + 1)}^{2}})})$ M1

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{2 {(k + 1)}^{3} - k})$ A1

Note: Award A1 for correct numerator, with $(k + 1)$ factored. Denominator does not need to be simplified

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{2 k^{3} + 6 k^{2} + 5 k + 2})$ A1

Note: Award A1 for denominator correctly expanded. Numerator does not need to be simplified. These two A marks may be awarded in any order

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{(k + 2) (2 k^{2} + 2 k + 1)}) = arctan (\frac{k + 1}{k + 2})$ A1

Note: The word ‘arctan’ must be present to be able to award the last three A marks

$P (k + 1)$ is true whenever $P (k)$ is true and $P (1)$ is true, so

$P (n)$ is true for for $n \in ℤ^{+}$ R1

Note: Award the final R1 mark provided at least four of the previous marks have been awarded.
Note: To award the final R1, the truth of $P (k)$ must be mentioned. ‘ $P (k)$ implies $P (k + 1)$ ’ is insufficient to award the mark.

[9 marks]

Examiners report

[N/A]

Consider the lines ${l_1}$ and ${l_2}$ defined by

${l_1}:$ r $= \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)$ and ${l_2}:\frac{{6 - x}}{3} = \frac{{y - 2}}{4} = 1 - z$ where $a$ is a constant.

Given that the lines ${l_1}$ and ${l_2}$ intersect at a point P,

find the value of $a$ ;

[4]

determine the coordinates of the point of intersection P.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

${l_1}:$ r $= \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) = \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = - 3 + \beta } \\ {y = - 2 + 4\beta } \\ {z = a + 2\beta } \end{array}} \right.$ M1

$\frac{{6 - ( - 3 + \beta )}}{3} = \frac{{( - 2 + 4\beta ) - 2}}{4} \Rightarrow 4 = \frac{{4\beta }}{3} \Rightarrow \beta = 3$ M1A1

$\frac{{6 - ( - 3 + \beta )}}{3} = 1 - (a + 2\beta ) \Rightarrow 2 = - 5 - a \Rightarrow a = - 7$ A1

METHOD 2

$\left\{ {\begin{array}{*{20}{l}} { - 3 + \beta = 6 - 3\lambda } \\ { - 2 + 4\beta = 4\lambda + 2} \\ {a + 2\beta = 1 - \lambda } \end{array}} \right.$ M1

attempt to solve M1

$\lambda = 2,{\text{ }}\beta = 3$ A1

$a = 1 - \lambda - 2\beta = - 7$ A1

[4 marks]

$\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ { - 7} \end{array}} \right) + 3 \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)$ (M1)

$= \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { - 1} \end{array}} \right)$ A1

$\therefore {\text{P}}(0,{\text{ 10, }} - 1)$

[2 marks]

Examiners report

[N/A]

Consider a triangle OAB such that O has coordinates (0, 0, 0), A has coordinates (0, 1, 2) and B has coordinates (2 $b$ , 0, $b$ − 1) where $b$ < 0.

Let M be the midpoint of the line segment [OB].

Find, in terms of $b$ , a Cartesian equation of the plane Π containing this triangle.

[5]

Find, in terms of $b$ , the equation of the line L which passes through M and is perpendicular to the plane П.

[3]

Show that L does not intersect the $y$ -axis for any negative value of $b$ .

[7]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$n = \left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 2 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} {2b} \\ 0 \\ {b - 1} \end{array}} \right)$ (M1)

$= \left( {\begin{array}{*{20}{c}} {b - 1} \\ {4b} \\ { - 2b} \end{array}} \right)$ (M1)A1

(0, 0, 0) on Π so $\left( {b - 1} \right)x + 4by - 2bz = 0$ (M1)A1

METHOD 2

using equation of the form $px + qy + rz = 0$ (M1)

(0, 1, 2) on Π ⇒ $q + 2r = 0$

(2 $b$ , 0, $b$ − 1) on Π ⇒ $2bp + r\left( {b - 1} \right) = 0$ (M1)A1

Note: Award (M1)A1 for both equations seen.

solve for $p$ , $q$ and $r$ (M1)

$\left( {b - 1} \right)x + 4by - 2bz = 0$ A1

[5 marks]

M has coordinates $\left( {b,\,\,0,\,\,\frac{{b - 1}}{2}} \right)$ (A1)

r = $\left( {\begin{array}{*{20}{c}} b \\ 0 \\ {\frac{{b - 1}}{2}} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} {b - 1} \\ {4b} \\ { - 2b} \end{array}} \right)$ M1A1

Note: Award M1A0 if r = (or equivalent) is not seen.

Note: Allow equivalent forms such as $\frac{{x - b}}{{b - 1}} = \frac{y}{{4b}} = \frac{{2z - b + 1}}{{ - 4b}}$ .

[3 marks]

METHOD 1

$x = z = 0$ (M1)

Note: Award M1 for either $x = 0$ or $z = 0$ or both.

$b + \lambda \left( {b - 1} \right) = 0$ and $\frac{{b - 1}}{2} - 2\lambda b = 0$ A1

attempt to eliminate $\lambda$ M1

$\Rightarrow - \frac{b}{{b - 1}} = \frac{{b - 1}}{{4b}}$ (A1)

$- 4{b^2} = {\left( {b - 1} \right)^2}$ A1

EITHER

consideration of the signs of LHS and RHS (M1)

the LHS is negative and the RHS must be positive (or equivalent statement) R1

$- 4{b^2} = {b^2} - 2b + 1$

$\Rightarrow 5{b^2} - 2b + 1 = 0$

$\Delta = {\left( { - 2} \right)^2} - 4 \times 5 \times 1 = - 16\,\left( { < 0} \right)$ M1

$\therefore$ no real solutions R1

THEN

so no point of intersection AG

METHOD 2

$x = z = 0$ (M1)

Note: Award M1 for either $x = 0$ or $z = 0$ or both.

$b + \lambda \left( {b - 1} \right) = 0$ and $\frac{{b - 1}}{2} - 2\lambda b = 0$ A1

attempt to eliminate $b$ M1

$\Rightarrow \frac{\lambda }{{1 + \lambda }} = \frac{1}{{1 - 4\lambda }}$ (A1)

$- 4{\lambda ^2} = 1\left( { \Rightarrow {\lambda ^2} = - \frac{1}{4}} \right)$ A1

consideration of the signs of LHS and RHS (M1)

there are no real solutions (or equivalent statement) R1

so no point of intersection AG

[7 marks]

Examiners report

[N/A]

The lengths of two of the sides in a triangle are 4 cm and 5 cm. Let θ be the angle between the two given sides. The triangle has an area of $\frac{{5\sqrt {15} }}{2}$ cm².

Show that ${\text{sin}}\,\theta = \frac{{\sqrt {15} }}{4}$ .

[1]

Find the two possible values for the length of the third side.

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

EITHER

$\frac{{5\sqrt {15} }}{2} = \frac{1}{2} \times 4 \times 5\,{\text{sin}}\,\theta$ A1

height of triangle is $\frac{{5\sqrt {15} }}{4}$ if using 4 as the base or ${\sqrt {15} }$ if using 5 as the base A1

THEN

${\text{sin}}\,\theta = \frac{{\sqrt {15} }}{4}$ AG

[1 mark]

let the third side be $x$

${x^2} = {4^2} + {5^2} - 2 \times 4 \times 5 \times {\text{cos}}\,\theta$ M1

valid attempt to find ${\text{cos}}\,\theta$ (M1)

Note: Do not accept writing ${\text{cos}}\left( {{\text{arcsin}}\left( {\frac{{\sqrt {15} }}{4}} \right)} \right)$ as a valid method.

${\text{cos}}\,\theta = \pm \sqrt {1 - \frac{{15}}{{16}}}$

$= \frac{1}{4}{\text{,}}\,\, - \frac{1}{4}$ A1A1

${x^2} = 16 + 25 - 2 \times 4 \times 5 \times \pm \frac{1}{4}$

$x = \sqrt {31}$ or $\sqrt {51}$ A1A1

[6 marks]

Examiners report

[N/A]

The points A and B are given by ${\text{A}}(0,{\text{ }}3,{\text{ }} - 6)$ and ${\text{B}}(6,{\text{ }} - 5,{\text{ }}11)$ .

The plane Π is defined by the equation $4x - 3y + 2z = 20$ .

Find a vector equation of the line L passing through the points A and B.

[3]

Find the coordinates of the point of intersection of the line L with the plane Π.

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)$ (A1)

r = $\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)$ or r = $\left( {\begin{array}{*{20}{c}} 6 \\ { - 5} \\ {11} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)$ M1A1

Note: Award M1A0 if r = is not seen (or equivalent).

[3 marks]

substitute line L in $\Pi :4(6\lambda ) - 3(3 - 8\lambda ) + 2( - 6 + 17\lambda ) = 20$ M1

$82\lambda = 41$

$\lambda = \frac{1}{2}$ (A1)

r = $\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)$

so coordinate is $\left( {3,{\text{ }} - 1,{\text{ }}\frac{5}{2}} \right)$ A1

Note: Accept coordinate expressed as position vector $\left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)$ .

[3 marks]

Examiners report

[N/A]

The function $f$ is defined by $f\left( x \right) = {{\text{e}}^x}\,{\text{cos}}{\,^2}x$ , where 0 ≤ $x$ ≤ 5. The curve $y = f\left( x \right)$ is shown on the following graph which has local maximum points at A and C and touches the $x$ -axis at B and D.

Use integration by parts to show that $\int {{{\text{e}}^x}\,{\text{cos}}\,2x{\text{d}}x = } \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x + c$ , $c \in \mathbb{R}$ .

[5]

Hence, show that $\int {{{\text{e}}^x}\,{\text{cos}}{\,^2}x{\text{d}}x = } \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2} + c$ , $c \in \mathbb{R}$ .

[3]

Find the $x$ -coordinates of A and of C , giving your answers in the form $a + {\text{arctan}}\,b$ , where $a$ , $b \in \mathbb{R}$ .

[6]

Find the area enclosed by the curve and the $x$ -axis between B and D, as shaded on the diagram.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

attempt at integration by parts with $u = {{\text{e}}^x}$ , $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {\text{cos}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = } \frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x\,{\text{d}}x - \int {\frac{{{{\text{e}}^x}}}{2}} {\text{sin}}\,2x\,{\text{d}}x$ A1

= $\frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x - \frac{1}{2}\left( { - \frac{{{{\text{e}}^x}}}{2}{\text{cos}}\,2x + \int {\frac{{{{\text{e}}^x}}}{2}} {\text{cos}}\,2x} \right)$ M1A1

= $\frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{4}{\text{cos}}\,2x - \frac{1}{4}\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$

$\therefore \frac{5}{4}\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x\, + \frac{{{{\text{e}}^x}}}{4}{\text{cos}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

METHOD 2

attempt at integration by parts with $u = {\text{cos}}\,2x$ , $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {{\text{e}}^x}$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = } {{\text{e}}^x}\,{\text{cos}}\,2x + 2\int {{{\text{e}}^x}\,{\text{sin}}\,2x\,{\text{d}}x}$ A1

$= {{\text{e}}^x}\,{\text{cos}}\,2x + 2\left( {{{\text{e}}^x}\,{\text{sin}}\,2x - 2\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} } \right)$ M1A1

$= {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x - 4\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$

$\therefore 5\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

METHOD 3

attempt at use of table M1

A1A1

Note: A1 for first 2 lines correct, A1 for third line correct.

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = \,} {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x - 4\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$ M1

$\therefore 5\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

[5 marks]

$\int {{{\text{e}}^x}\,{\text{co}}{{\text{s}}^2}\,x{\text{d}}x = } \int {\frac{{{{\text{e}}^x}}}{2}} \left( {{\text{cos}}\,2x + 1} \right){\text{d}}x$ M1A1

$= \frac{1}{2}\left( {\frac{{{\text{2}}{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x} \right) + \frac{{{{\text{e}}^x}}}{2}$ A1

$= \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2}\left( { + c} \right)$ AG

Note: Do not accept solutions where the RHS is differentiated.

[3 marks]

$f'\left( x \right) = {{\text{e}}^x}\,{\text{co}}{{\text{s}}^{\text{2}}}\,x - 2{{\text{e}}^x}\,{\text{sin}}\,x\,{\text{cos}}\,x$ M1A1

Note: Award M1 for an attempt at both the product rule and the chain rule.

${{\text{e}}^x}\,{\text{cos}}\,x\left( {{\text{cos}}\,x - 2\,{\text{sin}}\,x} \right) = 0$ (M1)

Note: Award M1 for an attempt to factorise ${{\text{cos}}\,x}$ or divide by ${\text{cos}}\,x\left( {{\text{cos}}\,x \ne 0} \right)$ .

discount ${\text{cos}}\,x = 0$ (as this would also be a zero of the function)

$\Rightarrow {\text{cos}}\,x - 2\,{\text{sin}}\,x = 0$

$\Rightarrow {\text{tan}}\,x = \frac{1}{2}$ (M1)

$\Rightarrow x = {\text{arctan}}\left( {\frac{1}{2}} \right)$ (at A) and $x = \pi + {\text{arctan}}\left( {\frac{1}{2}} \right)$ (at C) A1A1

Note: Award A1 for each correct answer. If extra values are seen award A1A0.

[6 marks]

${\text{cos}}\,x = 0 \Rightarrow x = \frac{\pi }{2}$ or $\frac{{3\pi }}{2}$ A1

Note: The A1may be awarded for work seen in part (c).

$\int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{{\text{e}}^x}\,{\text{co}}{{\text{s}}^{\text{2}}}\,x} \right)} \,{\text{d}}x = \left[ {\frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2}} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}$ M1

$= \left( { - \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}}}{{10}} + \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}}}{2}} \right) - \left( { - \frac{{{{\text{e}}^{\frac{\pi }{2}}}}}{{10}} + \frac{{{{\text{e}}^{\frac{\pi }{2}}}}}{2}} \right)\left( { = \frac{{{\text{2}}{{\text{e}}^{\frac{{3\pi }}{2}}}}}{5} - \frac{{{\text{2}}{{\text{e}}^{\frac{\pi }{2}}}}}{5}} \right)$ M1(A1)A1

Note: Award M1 for substitution of the end points and subtracting, (A1) for ${\text{sin}}\,3\pi = {\text{sin}}\,\pi = 0$ and ${\text{cos}}\,3\pi = {\text{cos}}\,\pi = - 1$ and A1 for a completely correct answer.

[5 marks]

Examiners report

[N/A]

The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.

It is given that $\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to$ .

The position vectors $\mathop {{\text{OA}}}\limits^ \to$ , $\mathop {{\text{OB}}}\limits^ \to$ , $\mathop {{\text{OC}}}\limits^ \to$ and $\mathop {{\text{OD}}}\limits^ \to$ are given by

a = i + 2j − 3k

b = 3i − j + pk

c = qi + j + 2k

d = −i + rj − 2k

where p , q and r are constants.

The point where the diagonals of ABCD intersect is denoted by M.

The plane $\Pi$ cuts the x, y and z axes at X , Y and Z respectively.

Explain why ABCD is a parallelogram.

[1]

a.i.

Using vector algebra, show that $\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to$ .

[3]

a.ii.

Show that p = 1, q = 1 and r = 4.

[5]

Find the area of the parallelogram ABCD.

[4]

Find the vector equation of the straight line passing through M and normal to the plane $\Pi$ containing ABCD.

[4]

Find the Cartesian equation of $\Pi$ .

[3]

Find the coordinates of X, Y and Z.

[2]

f.i.

Find YZ.

[2]

f.ii.

Markscheme

a pair of opposite sides have equal length and are parallel R1

hence ABCD is a parallelogram AG

[1 mark]

a.i.

attempt to rewrite the given information in vector form M1

b − a = c − d A1

rearranging d − a = c − b M1

hence $\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to$ AG

Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.

[3 marks]

a.ii.

EITHER

use of $\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to$ (M1)

$\left( \begin{gathered} 2 \hfill \\ - 3 \hfill \\ p + 3 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} q + 1 \hfill \\ 1 - r \hfill \\ 4 \hfill \\ \end{gathered} \right)$ A1A1

use of $\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to$ (M1)

$\left( \begin{gathered} - 2 \hfill \\ r - 2 \hfill \\ 1 \hfill \\ \end{gathered} \right) = \left( \begin{gathered} q - 3 \hfill \\ 2 \hfill \\ 2 - p \hfill \\ \end{gathered} \right)$ A1A1

THEN

attempt to compare coefficients of i, j, and k in their equation or statement to that effect M1

clear demonstration that the given values satisfy their equation A1
p = 1, q = 1, r = 4 AG

[5 marks]

attempt at computing $\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to$ (or equivalent) M1

$\left( \begin{gathered} - 11 \hfill \\ - 10 \hfill \\ - 2 \hfill \\ \end{gathered} \right)$ A1

area $= \left| {\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to } \right|\left( { = \sqrt {225} } \right)$ (M1)

= 15 A1

[4 marks]

valid attempt to find $\mathop {{\text{OM}}}\limits^ \to = \left( {\frac{1}{2}\left( {a + c} \right)} \right)$ (M1)

$\left( \begin{gathered} 1 \hfill \\ \frac{3}{2} \hfill \\ - \frac{1}{2} \hfill \\ \end{gathered} \right)$ A1

the equation is

r = $\left( \begin{gathered} 1 \hfill \\ \frac{3}{2} \hfill \\ - \frac{1}{2} \hfill \\ \end{gathered} \right) + t\left( \begin{gathered} 11 \hfill \\ 10 \hfill \\ 2 \hfill \\ \end{gathered} \right)$ or equivalent M1A1

Note: Award maximum M1A0 if 'r = …' (or equivalent) is not seen.

[4 marks]

attempt to obtain the equation of the plane in the form ax + by + cz = d M1

11x + 10y + 2z = 25 A1A1

Note: A1 for right hand side, A1 for left hand side.

[3 marks]

putting two coordinates equal to zero (M1)

${\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)$ A1

[2 marks]

f.i.

${\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}}$ M1

$= \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)$ A1

[4 marks]

f.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

f.i.

[N/A]

f.ii.

A function $f$ is defined by $f (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x \neq - 1, x \neq 3$ .

A function $g$ is defined by $g (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x > 3$ .

The inverse of $g$ is $g^{- 1}$ .

A function $h$ is defined by $h (x) = arctan \frac{x}{2}$ , where $x \in ℝ$ .

Sketch the curve $y = f (x)$ , clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.

[6]

Show that $g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ .

[6]

b.i.

State the domain of $g^{- 1}$ .

[1]

b.ii.

Given that $(h \circ g) (a) = \frac{π}{4}$ , find the value of $a$ .

Give your answer in the form $p + \frac{q}{2} \sqrt{r}$ , where $p, q, r \in ℤ^{+}$ .

[7]

Markscheme

$y$ -intercept $(0, - \frac{1}{3})$ A1

Note: Accept an indication of $- \frac{1}{3}$ on the $y$ -axis.

vertical asymptotes $x = - 1$ and $x = 3$ A1

horizontal asymptote $y = 0$ A1

uses a valid method to find the $x$ -coordinate of the local maximum point (M1)

Note: For example, uses the axis of symmetry or attempts to solve $f' (x) = 0$ .

local maximum point $(1, - \frac{1}{4})$ A1

Note: Award (M1)A0 for a local maximum point at $x = 1$ and coordinates not given.

three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other A1

[6 marks]

$x = \frac{1}{y^{2} - 2 y - 3}$ M1

Note: Award M1 for interchanging $x$ and $y$ (this can be done at a later stage).

EITHER

attempts to complete the square M1

$y^{2} - 2 y - 3 = {(y - 1)}^{2} - 4$ A1

$x = \frac{1}{{(y - 1)}^{2} - 4}$

${(y - 1)}^{2} - 4 = \frac{1}{x} ({(y - 1)}^{2} = 4 + \frac{1}{x})$ A1

$y - 1 = \pm \sqrt{4 + \frac{1}{x}} (= \pm \sqrt{\frac{4 x + 1}{x}})$

attempts to solve $x y^{2} - 2 x y - 3 x - 1 = 0$ for $y$ M1

$y = \frac{- (- 2 x) \pm \sqrt{{(- 2 x)}^{2} + 4 x (3 x + 1)}}{2 x}$ A1

Note: Award A1 even if $-$ (in $\pm$ ) is missing

$= \frac{2 x \pm \sqrt{16 x^{2} + 4 x}}{2 x}$ A1

THEN

$= 1 \pm \frac{\sqrt{4 x^{2} + x}}{x}$ A1

$y > 3$ and hence $y = 1 - \frac{\sqrt{4 x^{2} + x}}{x}$ is rejected R1

Note: Award R1 for concluding that the expression for $y$ must have the ‘ $+$ ’ sign.
The R1 may be awarded earlier for using the condition $x > 3$ .

$y = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$

$g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ AG

[6 marks]

b.i.

domain of $g^{- 1}$ is $x > 0$ A1

[1 mark]

b.ii.

attempts to find $(h \circ g) (a)$ (M1)

$(h \circ g) (a) = arctan (\frac{g (a)}{2}) ((h \circ g) (a) = arctan (\frac{1}{2 (a^{2} - 2 a - 3)}))$ (A1)

$arctan (\frac{g (a)}{2}) = \frac{π}{4} (arctan (\frac{1}{2 (a^{2} - 2 a - 3)}) = \frac{π}{4})$

attempts to solve for $g (a)$ M1

$\Rightarrow g (a) = 2 (\frac{1}{(a^{2} - 2 a - 3)} = 2)$

EITHER

$\Rightarrow a = g^{- 1} (2)$ A1

attempts to find their $g^{- 1} (2)$ M1

$a = 1 + \frac{\sqrt{4 {(2)}^{2} + 2}}{2}$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

$\Rightarrow 2 a^{2} - 4 a - 7 = 0$ A1

attempts to solve their quadratic equation M1

$a = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} + 4 (2) (7)}}{4} (= \frac{4 \pm \sqrt{72}}{4})$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

THEN

$a = 1 + \frac{3}{2} \sqrt{2}$ (as $a > 3$ ) A1

$(p = 1, q = 3, r = 2)$

Note: Award A1 for $a = 1 + \frac{1}{2} \sqrt{18}$ $(p = 1, q = 1, r = 18)$

[7 marks]

Examiners report

Part (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.

Part (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.

Part (c) was well done in general, with some algebraic errors seen in occasions.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

By using the substitution $u = sec x$ or otherwise, find an expression for $\int_{0}^{\frac{π}{3}} {sec}^{n} x \tan x d x$ in terms of $n$ , where $n$ is a non-zero real number.

Markscheme

METHOD 1

$u = sec x \Rightarrow d u = sec x \tan x d x$ (A1)

attempts to express the integral in terms of $u$ M1

$\int_{1}^{2} u^{n - 1} d u$ A1

$= \frac{1}{n} {[u^{n}]}_{1}^{2} (= \frac{1}{n} {[{sec}^{n} x]}_{0}^{\frac{π}{3}})$ A1

Note: Condone the absence of or incorrect limits up to this point.

$= \frac{2^{n} - 1^{n}}{n}$ M1

$= \frac{2^{n} - 1}{n}$ A1

Note: Award M1 for correct substitution of their limits for $u$ into their antiderivative for $u$ (or given limits for $x$ into their antiderivative for $x$ ).

METHOD 2

$\int {sec}^{n} x \tan x d x = \int {sec}^{n - 1} x sec x \tan x d x$ (A1)

applies integration by inspection (M1)

$= \frac{1}{n} {[{sec}^{n} x]}_{0}^{\frac{π}{3}}$ A2

Note: Award A2 if the limits are not stated.

$= \frac{1}{n} ({sec}^{n} \frac{π}{3} - {sec}^{n} 0)$ M1

Note: Award M1 for correct substitution into their antiderivative.

$= \frac{2^{n} - 1}{n}$ A1

[6 marks]

Examiners report

[N/A]

A straight line, ${L_\theta }$ , has vector equation r $= \left( {\begin{array}{*{20}{c}} 5 \\ 0 \\ 0 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 5 \\ {{\text{sin}}\,\theta } \\ {{\text{cos}}\,\theta } \end{array}} \right){\text{, }}\lambda {\text{, }}\theta \in \mathbb{R}$ .

The plane ${\Pi _p}$ , has equation $x = p{\text{, }}p \in \mathbb{R}$ .

Show that the angle between ${L_\theta }$ and ${\Pi _p}$ is independent of both $\theta$ and $p$ .

Markscheme

a vector normal to ${\Pi _p}$ is $\left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \end{array}} \right)$ (A1)

Note: Allow any scalar multiple of $\left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \end{array}} \right)$ , including $\left( {\begin{array}{*{20}{c}} p \\ 0 \\ 0 \end{array}} \right)$

attempt to find scalar product (or vector product) of direction vector of line with any scalar multiple of $\left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \end{array}} \right)$ M1

$\left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 5 \\ {{\text{sin}}\,\theta } \\ {{\text{cos}}\,\theta } \end{array}} \right) = 5$ (or $\left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 5 \\ {{\text{sin}}\,\theta } \\ {{\text{cos}}\,\theta } \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ { - {\text{cos}}\,\theta } \\ {{\text{sin}}\,\theta } \end{array}} \right)$ ) A1

(if $\alpha$ is the angle between the line and the normal to the plane)

${\text{cos}}\,\alpha = \frac{5}{{1 \times \sqrt {25 + {\text{si}}{{\text{n}}^2}\,\theta + {\text{co}}{{\text{s}}^2}\,\theta } }}$ (or ${\text{sin}}\,\alpha = \frac{1}{{1 \times \sqrt {25 + {\text{si}}{{\text{n}}^2}\,\theta + {\text{co}}{{\text{s}}^2}\,\theta } }}$ ) A1

$\Rightarrow {\text{cos}}\,\alpha = \frac{5}{{\sqrt {26} }}$ or ${\text{sin}}\,\alpha = \frac{1}{{\sqrt {26} }}$ A1

this is independent of $p$ and $\theta$ , hence the angle between the line and the plane, $\left( {90 - \alpha } \right)$ , is also independent of $p$ and $\theta$ R1

Note: The final R mark is independent, but is conditional on the candidate obtaining a value independent of $p$ and $\theta$ .

[6 marks]

Examiners report

[N/A]

The lines $l_{1}$ and $l_{2}$ have the following vector equations where $λ, μ \in ℝ$ and $m \in ℝ$ .

$l_{1} : r_{1} = (\begin{matrix} 3 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 2 \\ 1 \\ m \end{matrix}) l_{2} : r_{2} = (\begin{matrix} - 1 \\ - 4 \\ - 2 m \end{matrix}) + μ (\begin{matrix} 2 \\ - 5 \\ - m \end{matrix})$

The plane $Π$ has Cartesian equation $x + 4 y - z = p$ where $p \in ℝ$ .

Given that $l_{1}$ and $Π$ have no points in common, find

Show that $l_{1}$ and $l_{2}$ are never perpendicular to each other.

[3]

the value of $m$ .

[2]

b.i.

the condition on the value of $p$ .

[2]

b.ii.

Markscheme

attempts to calculate $(\begin{matrix} 2 \\ 1 \\ m \end{matrix}) \cdot (\begin{matrix} 2 \\ - 5 \\ - m \end{matrix})$ (M1)

$= - 1 - m^{2}$ A1

since $m^{2} \geq 0, - 1 - m^{2} < 0$ for $m \in ℝ$ R1

so $l_{1}$ and $l_{2}$ are never perpendicular to each other AG

[3 marks]

(since $l_{1}$ is parallel to $Π$ , $l_{1}$ is perpendicular to the normal of $Π$ and so)

$(\begin{matrix} 2 \\ 1 \\ m \end{matrix}) \cdot (\begin{matrix} 1 \\ 4 \\ - 1 \end{matrix}) = 0$ R1

$2 + 4 - m = 0$

$m = 6$ A1

[2 marks]

b.i.

since there are no points in common, $(3, - 2, 0)$ does not lie in $Π$

EITHER

substitutes $(3, - 2, 0)$ into $x + 4 y - z (\neq p)$ (M1)

$(\begin{matrix} 3 \\ - 2 \\ 0 \end{matrix}) \cdot (\begin{matrix} 1 \\ 4 \\ - 1 \end{matrix}) (\neq p)$ (M1)

THEN

$p \neq - 5$ A1

[2 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

Consider the line $L_{1}$ defined by the Cartesian equation $\frac{x + 1}{2} = y = 3 - z$ .

Consider a second line $L_{2}$ defined by the vector equation $r = (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}) + t (\begin{matrix} a \\ 1 \\ - 1 \end{matrix})$ , where $t \in ℝ$ and $a \in ℝ$ .

Show that the point $(- 1, 0, 3)$ lies on $L_{1}$ .

[1]

a.i.

Find a vector equation of $L_{1}$ .

[3]

a.ii.

Find the possible values of $a$ when the acute angle between $L_{1}$ and $L_{2}$ is $45 °$ .

[8]

It is given that the lines $L_{1}$ and $L_{2}$ have a unique point of intersection, $A$ , when $a \neq k$ .

Find the value of $k$ , and find the coordinates of the point $A$ in terms of $a$ .

[7]

Markscheme

$\frac{- 1 + 1}{2} = 0 = 3 - 3$ A1

the point $(- 1, 0, 3)$ lies on $L_{1}$ . AG

[1 mark]

a.i.

attempt to set equal to a parameter or rearrange cartesian form (M1)

$\frac{x + 1}{2} = y = 3 - z = λ \Rightarrow x = 2 λ - 1, y = λ, z = 3 - λ$ OR $\frac{x + 1}{2} = \frac{y - 0}{1} = \frac{z - 3}{- 1}$

correct direction vector $(\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix})$ or equivalent seen in vector form (A1)

$r = (\begin{matrix} - 1 \\ 0 \\ 3 \end{matrix}) + λ (\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix})$ (or equivalent) A1

Note: Award A0 if $= r$ is omitted.

[3 marks]

a.ii.

attempt to use the scalar product formula (M1)

$(\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix}) ∙ (\begin{matrix} a \\ 1 \\ - 1 \end{matrix}) = (\pm) \sqrt{6} \sqrt{a^{2} + 2} \cos 45 °$ (A1)(A1)

Note: Award A1 for LHS and A1 for RHS

$2 a + 2 = \frac{(\pm) \sqrt{6} \sqrt{a^{2} + 2} \sqrt{2}}{2} (\Rightarrow 2 a + 2 = (\pm) \sqrt{3} \sqrt{a^{2} + 2})$ A1A1

Note: Award A1 for LHS and A1 for RHS

$4 a^{2} + 8 a + 4 = 3 a^{2} + 6$ A1

$a^{2} + 8 a - 2 = 0$ M1

attempt to solve their quadratic

$a = \frac{- 8 \pm \sqrt{64 + 8}}{2} = \frac{- 8 \pm \sqrt{72}}{2} (= - 4 \pm 3 \sqrt{2})$ A1

[8 marks]

METHOD 1

attempt to equate the parametric forms of $L_{1}$ and $L_{2}$ (M1)

$\{\begin{matrix} 2 λ - 1 = t a \\ λ = 1 + t \\ 3 - λ = 2 - t \end{matrix}$ A1

attempt to solve equations by eliminating $λ$ or $t$ (M1)

$2 + 2 t - 1 = t a \Rightarrow 1 = t (a - 2)$ or $2 λ - 1 = (λ - 1) a \Rightarrow a - 1 = λ (a - 2)$

Solutions exist unless $a - 2 = 0$

$k = 2$ A1

Note: This A1 is independent of the following marks.

$t = \frac{1}{a - 2}$ or $λ = \frac{a - 1}{a - 2}$ A1

$A$ has coordinates $(\frac{a}{a - 2}, 1 + \frac{1}{a - 2}, 2 - \frac{1}{a - 2}) (= (\frac{a}{a - 2}, \frac{a - 1}{a - 2}, \frac{2 a - 5}{a - 2}))$ A2

Note: Award A1 for any two correct coordinates seen or final answer in vector form.

METHOD 2

no unique point of intersection implies direction vectors of $L_{1}$ and $L_{2}$ parallel

$k = 2$ A1

Note: This A1 is independent of the following marks.

attempt to equate the parametric forms of $L_{1}$ and $L_{2}$ (M1)

$\{\begin{matrix} 2 λ - 1 = t a \\ λ = 1 + t \\ 3 - λ = 2 - t \end{matrix}$ A1

attempt to solve equations by eliminating $λ$ or $t$ (M1)

$2 + 2 t - 1 = t a \Rightarrow 1 = t (a - 2)$ or $2 λ - 1 = (λ - 1) a \Rightarrow a - 1 = λ (a - 2)$

$t = \frac{1}{a - 2}$ or $λ = \frac{a - 1}{a - 2}$ A1

$A$ has coordinates $(\frac{a}{a - 2}, 1 + \frac{1}{a - 2}, 2 - \frac{1}{a - 2}) (= (\frac{a}{a - 2}, \frac{a - 1}{a - 2}, \frac{2 a - 5}{a - 2}))$ A2

Note: Award A1 for any two correct coordinates seen or final answer in vector form.

[7 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Points ${\text{A}}$ (0 , 0 , 10) , ${\text{B}}$ (0 , 10 , 0) , ${\text{C}}$ (10 , 0 , 0) , ${\text{V}}$ ( $p$ , $p$ , $p$ ) form the vertices of a tetrahedron.

Consider the case where the faces ${\text{ABV}}$ and ${\text{ACV}}$ are perpendicular.

The following diagram shows the graph of $\theta$ against $p$ . The maximum point is shown by ${\text{X}}$ .

Show that $\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AV}}} = - 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right)$ and find a similar expression for $\overrightarrow {{\text{AC}}} \times \overrightarrow {{\text{AV}}}$ .

[3]

a.i.

Hence, show that, if the angle between the faces ${\text{ABV}}$ and ${\text{ACV}}$ is $\theta$ , then ${\text{cos}}\,\theta = \frac{{p\left( {3p - 20} \right)}}{{6{p^2} - 40p + 100}}$ .

[5]

a.ii.

Find the two possible coordinates of ${\text{V}}$ .

[3]

b.i.

Comment on the positions of ${\text{V}}$ in relation to the plane ${\text{ABC}}$ .

[1]

b.ii.

At ${\text{X}}$ , find the value of $p$ and the value of $\theta$ .

[3]

c.i.

Find the equation of the horizontal asymptote of the graph.

[2]

c.ii.

Markscheme

$\overrightarrow {{\text{AV}}} = \left( {\begin{array}{*{20}{c}} p \\ p \\ {p - 10} \end{array}} \right)$ A1

$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AV}}} = \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { - 10} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} p \\ p \\ {p - 10} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {10\left( {p - 10} \right) + 10p} \\ { - 10p} \\ { - 10p} \end{array}} \right)$ A1

$= \left( {\begin{array}{*{20}{c}} {20p - 100} \\ { - 10p} \\ { - 10p} \end{array}} \right) = - 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right)$ AG

$\overrightarrow {{\text{AC}}} \times \overrightarrow {{\text{AV}}} = \left( {\begin{array}{*{20}{c}} {10} \\ 0 \\ { - 10} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} p \\ p \\ {p - 10} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {10p} \\ {100 - 20p} \\ {10p} \end{array}} \right)\left( { = 10\left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right)} \right)$ A1

[3 marks]

a.i.

attempt to find a scalar product M1

$- 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right) \bullet 10\left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right) = 100\left( {3{p^2} - 20p} \right)$

OR $- \left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right) = 3{p^2} - 20p$ A1

attempt to find magnitude of either $\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AV}}}$ or $\overrightarrow {{\text{AC}}} \times \overrightarrow {{\text{AV}}}$ M1

$\left| { - 10\left( {\begin{array}{*{20}{c}} {10 - 2p} \\ p \\ p \end{array}} \right)} \right| = \left| {10\left( {\begin{array}{*{20}{c}} p \\ {10 - 2p} \\ p \end{array}} \right)} \right| = 10\sqrt {{{\left( {10 - 2p} \right)}^2} + 2{p^2}}$ A1

$100\left( {3{p^2} - 20p} \right) = 100{\left( {\sqrt {{{\left( {10 - 2p} \right)}^2} + 2{p^2}} } \right)^2}{\text{cos}}\,\theta$

${\text{cos}}\,\theta = \frac{{3{p^2} - 20p}}{{{{\left( {10 - 2p} \right)}^2} + 2{p^2}}}$ A1

Note: Award A1 for any intermediate step leading to the correct answer.

$= \frac{{p\left( {3p - 20} \right)}}{{6{p^2} - 40p + 100}}$ AG

Note: Do not allow FT marks from part (a)(i).

[8 marks]

a.ii.

$p\left( {3p - 20} \right) = 0 \Rightarrow p = 0$ or $p = \frac{{20}}{3}$ M1A1

coordinates are (0, 0, 0) and $\left( {\frac{{20}}{3}{\text{, }}\frac{{20}}{3}{\text{, }}\frac{{20}}{3}} \right)$ A1

Note: Do not allow column vectors for the final A mark.

[3 marks]

b.i.

two points are mirror images in the plane
or opposite sides of the plane
or equidistant from the plane
or the line connecting the two Vs is perpendicular to the plane R1

[1 mark]

b.ii.

geometrical consideration or attempt to solve $- 1 = \frac{{p\left( {3p - 20} \right)}}{{6{p^2} - 40p + 100}}$ (M1)

$p = \frac{{10}}{3}{\text{, }}\theta = \pi$ or $\theta = 180^\circ$ A1A1

[3 marks]

c.i.

$p \to \infty \Rightarrow {\text{cos}}\,\theta \to \frac{1}{2}$ M1

hence the asymptote has equation $\theta = \frac{\pi }{3}$ A1

[2 marks]

c.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

Find the coordinates of the point of intersection of the planes defined by the equations $x + y + z = 3,{\text{ }}x - y + z = 5$ and $x + y + 2z = 6$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

for eliminating one variable from two equations (M1)

eg, $\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {2x + 2z = 8} \\ {2x + 3z = 11} \end{array}} \right.$ A1A1

for finding correctly one coordinate

eg, $\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {(2x + 2z = 8)} \\ {z = 3} \end{array}} \right.$ A1

for finding correctly the other two coordinates A1

$\Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {z = 3} \end{array}} \right.$

the intersection point has coordinates $(1,{\text{ }} - 1,{\text{ }}3)$

METHOD 2

for eliminating two variables from two equations or using row reduction (M1)

eg, $\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ { - 2 = 2} \\ {z = 3} \end{array}} \right.$ or $\left( {\begin{array}{*{20}{c}} 1&1&1 \\ 0&{ - 2}&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 3 \\ 2 \\ 3 \end{array}} \right.} \right)$ A1A1

for finding correctly the other coordinates A1A1

$\Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {(z = 3)} \end{array}} \right.$ or $\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 3 \end{array}} \right.} \right)$

the intersection point has coordinates $(1,{\text{ }} - 1,{\text{ }}3)$

METHOD 3

$\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{ - 1}&1 \\ 1&1&2 \end{array}} \right| = - 2$ (A1)

attempt to use Cramer’s rule M1

$x = \frac{{\left| {\begin{array}{*{20}{c}} 3&1&1 \\ 5&{ - 1}&1 \\ 6&1&2 \end{array}} \right|}}{{ - 2}} = \frac{{ - 2}}{{ - 2}} = 1$ A1

$y = \frac{{\left| {\begin{array}{*{20}{c}} 1&3&1 \\ 1&5&1 \\ 1&6&2 \end{array}} \right|}}{{ - 2}} = \frac{2}{{ - 2}} = - 1$ A1

$z = \frac{{\left| {\begin{array}{*{20}{c}} 1&1&3 \\ 1&{ - 1}&5 \\ 1&1&6 \end{array}} \right|}}{{ - 2}} = \frac{{ - 6}}{{ - 2}} = 3$ A1

Note: Award M1 only if candidate attempts to determine at least one of the variables using this method.

[5 marks]

Examiners report

[N/A]

Consider the function $g\left( x \right) = 4\,{\text{cos}}\,x + 1$ , $a \leqslant x \leqslant \frac{\pi }{2}$ where $a < \frac{\pi }{2}$ .

For $a = - \frac{\pi }{2}$ , sketch the graph of $y = g\left( x \right)$ . Indicate clearly the maximum and minimum values of the function.

[3]

Write down the least value of $a$ such that $g$ has an inverse.

[1]

For the value of $a$ found in part (b), write down the domain of ${g^{ - 1}}$ .

[1]

c.i.

For the value of $a$ found in part (b), find an expression for ${g^{ - 1}}\left( x \right)$ .

[2]

c.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

concave down and symmetrical over correct domain A1

indication of maximum and minimum values of the function (correct range) A1A1

[3 marks]

$a$ = 0 A1

Note: Award A1 for $a$ = 0 only if consistent with their graph.

[1 mark]

$1 \leqslant x \leqslant 5$ A1

Note: Allow FT from their graph.

[1 mark]

c.i.

$y = 4\,{\text{cos}}\,x + 1$

$x = 4\,{\text{cos}}\,y + 1$

$\frac{{x - 1}}{4} = {\text{cos}}\,y$ (M1)

$\Rightarrow y = {\text{arccos}}\left( {\frac{{x - 1}}{4}} \right)$

$\Rightarrow {g^{ - 1}}\left( x \right) = {\text{arccos}}\left( {\frac{{x - 1}}{4}} \right)$ A1

[2 marks]

c.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

Consider the vectors a $=$ i $- {\text{ }}3$ j $- {\text{ }}2$ k, b $= - {\text{ }}3$ j $+ {\text{ }}2$ k.

Find a $\times$ b.

[2]

Hence find the Cartesian equation of the plane containing the vectors a and b, and passing through the point $(1,{\text{ }}0,{\text{ }} - 1)$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

a $\times$ b $= - 12$ i $- {\text{ }}2$ j $- {\text{ }}3$ k (M1)A1

[2 marks]

METHOD 1

$- 12x - 2y - 3z = d$ M1

$- 12 \times 1 - 2 \times 0 - 3( - 1) = d$ (M1)

$\Rightarrow d = - 9$ A1

$- 12x - 2y - 3z = - 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)$

METHOD 2

$\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 12} \\ { - 2} \\ { - 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ { - 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 12} \\ { - 2} \\ { - 3} \end{array}} \right)$ M1A1

$- 12x - 2y - 3z = - 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)$ A1

[3 marks]

Examiners report

[N/A]

$A$ and $B$ are acute angles such that ${\text{cos}}\,A = \frac{2}{3}$ and ${\text{sin}}\,B = \frac{1}{3}$ .

Show that ${\text{cos}}\left( {2A + B} \right) = - \frac{{2\sqrt 2 }}{{27}} - \frac{{4\sqrt 5 }}{{27}}$ .

Markscheme

attempt to use ${\text{cos}}\left( {2A + B} \right) = {\text{cos}}\,2A\,{\text{cos}}\,B - {\text{sin}}\,2A\,{\text{sin}}\,B$ (may be seen later) M1

attempt to use any double angle formulae (seen anywhere) M1

attempt to find either ${\text{sin}}\,A$ or ${\text{cos}}\,B$ (seen anywhere) M1

${\text{cos}}\,A = \frac{2}{3} \Rightarrow {\text{sin}}\,A\left( { = \sqrt {1 - \frac{4}{9}} } \right) = \frac{{\sqrt 5 }}{3}$ (A1)

${\text{sin}}\,B = \frac{1}{3} \Rightarrow {\text{cos}}\,B\left( { = \sqrt {1 - \frac{1}{9}} = \frac{{\sqrt 8 }}{3}} \right) = \frac{{2\sqrt 2 }}{3}$ A1

${\text{cos}}\,2A\left( { = 2\,{\text{co}}{{\text{s}}^2}\,A - 1} \right) = - \frac{1}{9}$ A1

${\text{sin}}\,2A\left( { = 2\,{\text{sin}}\,A\,{\text{cos}}\,A} \right) = \frac{{4\sqrt 5 }}{9}$ A1

So ${\text{cos}}\left( {2A + B} \right) = \left( { - \frac{1}{9}} \right)\left( {\frac{{2\sqrt 2 }}{3}} \right) - \left( {\frac{{4\sqrt 5 }}{9}} \right)\left( {\frac{1}{3}} \right)$

$= - \frac{{2\sqrt 2 }}{{27}} - \frac{{4\sqrt 5 }}{{27}}$ AG

[7 marks]

Examiners report

[N/A]

The acute angle between the vectors 3i − 4j − 5k and 5i − 4j + 3k is denoted by θ.

Find cos θ.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

cos θ = $\frac{{\left( {3i - 4j - 5k} \right) \bullet \left( {5i - 4j + 3k} \right)}}{{\left| {3i - 4j - 5k} \right|\left| {5i - 4j + 3k} \right|}}$ (M1)

$= \frac{{16}}{{\sqrt {50} \sqrt {50} }}$ A1A1

Note: A1 for correct numerator and A1 for correct denominator.

$= \frac{8}{{25}}\left( { = \frac{{16}}{{50}} = 0.32} \right)$ A1

[4 marks]

Examiners report

[N/A]

ABCD is a parallelogram, where $\overrightarrow {{\text{AB}}}$ = –i + 2j + 3k and $\overrightarrow {{\text{AD}}}$ = 4i – j – 2k.

Find the area of the parallelogram ABCD.

[3]

By using a suitable scalar product of two vectors, determine whether ${\rm{A\hat BC}}$ is acute or obtuse.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} = -$ i $+ 10$ j – 7k M1A1

${\text{area}} = \left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} } \right|{\text{ = }}\sqrt {{1^2} + {{10}^2} + {7^2}}$

$= 5\sqrt 6 \left( {\sqrt {150} } \right)$ A1

[3 marks]

METHOD 1

$\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} = - 4 - 2 - 6$ M1A1

$= - 12$

considering the sign of the answer

$\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} < 0$ , therefore angle ${\rm{D\hat AB}}$ is obtuse M1

(as it is a parallelogram), ${\rm{A\hat BC}}$ is acute A1

[4 marks]

METHOD 2

$\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} = + 4 + 2 + 6$ M1A1

$= 12$ considering the sign of the answer M1

$\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} > 0 \Rightarrow {\rm{A\hat BC}}$ is acute A1

[4 marks]

Examiners report

[N/A]

It is given that $cosec θ = \frac{3}{2}$ , where $\frac{π}{2} < θ < \frac{3 π}{2}$ . Find the exact value of $cot θ$ .

Markscheme

METHOD 1

attempt to use a right angled triangle M1

correct placement of all three values and $θ$ seen in the triangle (A1)

$cot θ < 0$ (since $cosec θ > 0$ puts $θ$ in the second quadrant) R1

$cot θ = - \frac{\sqrt{5}}{2}$ A1

Note: Award M1A1R0A0 for $cot θ = \frac{\sqrt{5}}{2}$ seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.

METHOD 2

Attempt to use $1 + {cot}^{2} θ = {cosec}^{2} θ$ M1

$1 + {cot}^{2} θ = \frac{9}{4}$

$cot^{2} θ = \frac{5}{4}$ (A1)

$cot θ = \pm \frac{\sqrt{5}}{2}$

$cot θ < 0$ (since $cosec θ > 0$ puts $θ$ in the second quadrant) R1

$cot θ = - \frac{\sqrt{5}}{2}$ A1

Note: Award M1A1R0A0 for $cot θ = \frac{\sqrt{5}}{2}$ seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.

METHOD 3

$sin θ = \frac{2}{3}$

attempt to use ${sin}^{2} θ + {cos}^{2} θ = 1$ M1

$\frac{4}{9} + {cos}^{2} θ = 1$

${cos}^{2} θ = \frac{5}{9}$ (A1)

$cos θ = \pm \frac{\sqrt{5}}{3}$

$cos θ < 0$ (since $cosec θ > 0$ puts $θ$ in the second quadrant) R1

$cos θ = - \frac{\sqrt{5}}{3}$

$cot θ = - \frac{\sqrt{5}}{2}$ A1

Note: Award M1A1R0A0 for $cot θ = \frac{\sqrt{5}}{2}$ seen as the final answer
The R1 should be awarded independently for a negative value only given as a final answer.

[4 marks]

Examiners report

[N/A]

Let a = $\left( {\begin{array}{*{20}{c}} 2 \\ k \\ { - 1} \end{array}} \right)$ and b = $\left( {\begin{array}{*{20}{c}} { - 3} \\ {k + 2} \\ k \end{array}} \right)$ , $k \in \mathbb{R}$ .

Given that a and b are perpendicular, find the possible values of $k$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

a • b = $\left( {\begin{array}{*{20}{c}} 2 \\ k \\ { - 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 3} \\ {k + 2} \\ k \end{array}} \right)$

$= - 6 + k\left( {k + 2} \right) - k$ A1

a • b = 0 (M1)

${k^2} + k - 6 = 0$

attempt at solving their quadratic equation (M1)

$\left( {k + 3} \right)\left( {k - 2} \right) = 0$

$k = - 3{\text{,}}\,\,2$ A1

Note: Attempt at solving using |a||b| = |a × b| will be M1A0A0A0 if neither answer found M1(A1)A1A0
for one correct answer and M1(A1)A1A1 for two correct answers.

[4 marks]

Examiners report

[N/A]

Three points in three-dimensional space have coordinates A(0, 0, 2), B(0, 2, 0) and C(3, 1, 0).

Find the vector $\overrightarrow {{\text{AB}}}$ .

[1]

a.i.

Find the vector $\overrightarrow {{\text{AC}}}$ .

[1]

a.ii.

Hence or otherwise, find the area of the triangle ABC.

[4]

Markscheme

$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right)$ A1

Note: Accept row vectors or equivalent.

[1 mark]

a.i.

$\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ { - 2} \end{array}} \right)$ A1

Note: Accept row vectors or equivalent.

[1 mark]

a.ii.

METHOD 1

attempt at vector product using $\overrightarrow {{\text{AB}}}$ and $\overrightarrow {{\text{AC}}}$ . (M1)

±(2i + 6j +6k) A1

attempt to use area $= \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|$ M1

$= \frac{{\sqrt {76} }}{2}\,\,\,\left( { = \sqrt {19} } \right)$ A1

METHOD 2

attempt to use $\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} = \left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|{\text{cos}}\,\theta$ M1

$\left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ { - 2} \end{array}} \right) = \sqrt {{0^2} + {2^2} + {{\left( { - 2} \right)}^2}} \sqrt {{3^2} + {1^2} + {{\left( { - 2} \right)}^2}} {\text{cos}}\,\theta$

$6 = \sqrt 8 \sqrt {14} \,{\text{cos}}\,\theta$ A1

${\text{cos}}\,\theta = \frac{6}{{\sqrt 8 \sqrt {14} }} = \frac{6}{{\sqrt {112} }}$

attempt to use area $= \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|{\text{sin}}\,\theta$ M1

$= \frac{1}{2}\sqrt 8 \sqrt {14} \sqrt {1 - \frac{{36}}{{112}}} \,\left( { = \frac{1}{2}\sqrt 8 \sqrt {14} \sqrt {\frac{{76}}{{112}}} } \right)$

$= \frac{{\sqrt {76} }}{2}\,\,\,\left( { = \sqrt {19} } \right)$ A1

[4 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Solve the equation ${\sec ^2}x + 2\tan x = 0,{\text{ }}0 \leqslant x \leqslant 2\pi$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

use of ${\sec ^2}x = {\tan ^2}x + 1$ M1

${\tan ^2}x + 2\tan x + 1 = 0$

${(\tan x + 1)^2} = 0$ (M1)

$\tan x = - 1$ A1

$x = \frac{{3\pi }}{4},{\text{ }}\frac{{7\pi }}{4}$ A1A1

METHOD 2

$\frac{1}{{{{\cos }^2}x}} + \frac{{2\sin x}}{{\cos x}} = 0$ M1

$1 + 2\sin x\cos x = 0$

$\sin 2x = - 1$ M1A1

$2x = \frac{{3\pi }}{2},{\text{ }}\frac{{7\pi }}{2}$

$x = \frac{{3\pi }}{4},{\text{ }}\frac{{7\pi }}{4}$ A1A1

Note: Award A1A0 if extra solutions given or if solutions given in degrees (or both).

[5 marks]

Examiners report

[N/A]

The lines $l_{1}$ and $l_{2}$ have the following vector equations where $λ, μ \in ℝ$ .

$l_{1} : r_{1} = (\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}) + λ (\begin{matrix} 2 \\ - 2 \\ 2 \end{matrix})$

$l_{2} : r_{2} = (\begin{matrix} 2 \\ 0 \\ 4 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$

Show that $l_{1}$ and $l_{2}$ do not intersect.

[3]

Find the minimum distance between $l_{1}$ and $l_{2}$ .

[5]

Markscheme

METHOD 1

setting at least two components of $l_{1}$ and $l_{2}$ equal M1

$3 + 2 λ = 2 + μ (1)$

$2 - 2 λ = - μ (2)$

$- 1 + 2 λ = 4 + μ (3)$

attempt to solve two of the equations eg. adding $(1)$ and $(2)$ M1

gives a contradiction (no solution), eg $5 = 2$ R1

so $l_{1}$ and $l_{2}$ do not intersect AG

Note: For an error within the equations award M0M1R0.
Note: The contradiction must be correct to award the R1.

METHOD 2

$l_{1}$ and $l_{2}$ are parallel, so $l_{1}$ and $l_{2}$ are either identical or distinct. R1

Attempt to subtract two position vectors from each line,

e.g. $(\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}) - (\begin{matrix} 2 \\ 0 \\ 4 \end{matrix}) (= (\begin{matrix} 1 \\ 2 \\ - 5 \end{matrix}))$ M1

$(\begin{matrix} 3 \\ 2 \\ - 1 \end{matrix}) \neq k (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$ A1

[3 marks]

METHOD 1

$l_{1}$ and $l_{2}$ are parallel (as $(\begin{matrix} 2 \\ - 2 \\ 2 \end{matrix})$ is a multiple of $(\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$ )

let $A$ be $(3, 2, - 1)$ on $l_{1}$ and let $B$ be $(2, 0, 4)$ on $l_{2}$

Attempt to find vector $\vec{AB} (= (\begin{matrix} - 1 \\ - 2 \\ 5 \end{matrix}))$ (M1)

Distance required is $\frac{|v \times \vec{AB}|}{|v|}$ M1

$= \frac{1}{\sqrt{3}} |(\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}) \times (\begin{matrix} - 1 \\ - 2 \\ 5 \end{matrix})|$ (A1)

$= \frac{1}{\sqrt{3}} |(\begin{matrix} 3 \\ 6 \\ 3 \end{matrix})|$ A1

minimum distance is $\sqrt{18} (= 3 \sqrt{2})$ A1

METHOD 2

$l_{1}$ and $l_{2}$ are parallel (as $(\begin{matrix} 2 \\ - 2 \\ 2 \end{matrix})$ is a multiple of $(\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix})$ )

let $A$ be a fixed point on $l_{1}$ eg $(3, 2, - 1)$ and let $B$ be a general point on $l_{2}$ $(2 + μ, - μ, 4 + μ)$

attempt to find vector $\vec{AB}$ (M1)

$\vec{AB} = (\begin{matrix} - 1 \\ - 2 \\ 5 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}) (μ \in ℝ)$ A1

$|\vec{AB}| = \sqrt{{(- 1 + μ)}^{2} + {(- 2 - μ)}^{2} + {(5 + μ)}^{2}} (= \sqrt{3 μ^{2} + 12 μ + 30})$ M1

EITHER

null A1

$|\vec{AB}| = \sqrt{3 {(μ + 2)}^{2} + 18}$ to obtain $μ = - 2$ A1

THEN

minimum distance is $\sqrt{18} (= 3 \sqrt{2})$ A1

METHOD 3

let $A$ be $(3, 2, - 1)$ on $l_{1}$ and let $B$ be $(2 + μ, - μ, 4 + μ)$ on $l_{2}$ (M1)

(or let $A$ be $(2, 0, 4)$ on $l_{2}$ and let $B$ be $(3 + 2 λ, 2 - 2 λ, - 1 + 2 λ)$ on $l_{1}$ )

$\vec{AB} = (\begin{matrix} - 1 \\ - 2 \\ 5 \end{matrix}) + μ (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}) (μ \in ℝ)$ (or $\vec{AB} = (\begin{matrix} 2 λ + 1 \\ - 2 λ + 2 \\ 2 λ - 5 \end{matrix})$ ) A1

$(\begin{matrix} μ - 1 \\ - μ - 2 \\ μ + 5 \end{matrix}) \cdot (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}) = 0$ (or $(\begin{matrix} 2 λ + 1 \\ - 2 λ + 2 \\ 2 λ - 5 \end{matrix}) \cdot (\begin{matrix} 1 \\ - 1 \\ 1 \end{matrix}) = 0$ ) M1

$μ = - 2$ or $λ = 1$ A1

minimum distance is $\sqrt{18} (= 3 \sqrt{2})$ A1

[5 marks]

Examiners report

[N/A]

Let $a = {\text{sin}}\,b,\,\,0 < b < \frac{\pi }{2}$ .

Find, in terms of b, the solutions of ${\text{sin}}\,2x = - a,\,\,0 \leqslant x \leqslant \pi$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\text{sin}}\,2x = - {\text{sin}}\,b$

EITHER

${\text{sin}}\,2x = {\text{sin}}\left( { - b} \right)$ or ${\text{sin}}\,2x = {\text{sin}}\left( {\pi + b} \right)$ or ${\text{sin}}\,2x = {\text{sin}}\left( {2\pi - b} \right)$ … (M1)(A1)

Note: Award M1 for any one of the above, A1 for having final two.

(M1)(A1)

Note: Award M1 for one of the angles shown with b clearly labelled, A1 for both angles shown. Do not award A1 if an angle is shown in the second quadrant and subsequent A1 marks not awarded.

THEN

$2x = \pi + b$ or $2x = 2\pi - b$ (A1)(A1)

$x = \frac{\pi }{2} + \frac{b}{2},\,\,x = \pi - \frac{b}{2}$ A1

[5 marks]

Examiners report

[N/A]

Consider the functions $f$ and $g$ defined on the domain $0 < x < 2\pi$ by $f\left( x \right) = 3\,{\text{cos}}\,2x$ and $g\left( x \right) = 4 - 11\,{\text{cos}}\,x$ .

The following diagram shows the graphs of $y = f\left( x \right)$ and $y = g\left( x \right)$

Find the $x$ -coordinates of the points of intersection of the two graphs.

[6]

Find the exact area of the shaded region, giving your answer in the form $p\pi + q\sqrt 3$ , where $p$ , $q \in \mathbb{Q}$ .

[5]

At the points A and B on the diagram, the gradients of the two graphs are equal.

Determine the $y$ -coordinate of A on the graph of $g$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$3\,{\text{cos}}\,2x = 4 - 11\,{\text{cos}}\,x$

attempt to form a quadratic in ${\text{cos}}\,x$ M1

$3\left( {2\,{\text{co}}{{\text{s}}^2}\,x - 1} \right) = 4 - 11\,{\text{cos}}\,x$ A1

$\left( {6\,{\text{co}}{{\text{s}}^2}\,x + 11\,{\text{cos}}\,x - 7 = 0} \right)$

valid attempt to solve their quadratic M1

$\left( {3\,{\text{cos}}\,x + 7} \right)\left( {2\,{\text{cos}}\,x - 1} \right) = 0$

${\text{cos}}\,x = \frac{1}{2}$ A1

$x = \frac{\pi }{3}{\text{,}}\,\,\frac{{5\pi }}{3}$ A1A1

Note: Ignore any “extra” solutions.

[6 marks]

consider (±) $\int\limits_{\frac{\pi }{3}}^{\frac{{5\pi }}{3}} {\left( {4 - 11\,{\text{cos}}\,x - 3\,{\text{cos}}\,2x} \right)} \,{\text{d}}x$ M1

$= \left( \pm \right)\left[ {4x - 11\,{\text{sin}}\,x - \frac{3}{2}{\text{sin}}\,2x} \right]_{\frac{\pi }{3}}^{\frac{{5\pi }}{3}}$ A1

Note: Ignore lack of or incorrect limits at this stage.

attempt to substitute their limits into their integral M1

$= \frac{{20\pi }}{3} - 11\,{\text{sin}}\frac{{5\pi }}{3} - \frac{3}{2}{\text{sin}}\frac{{10\pi }}{3} - \left( {\frac{{4\pi }}{3} - 11\,{\text{sin}}\frac{\pi }{3} - \frac{3}{2}{\text{sin}}\frac{{2\pi }}{3}} \right)$

$= \frac{{16\pi }}{3} + \frac{{11\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{4} + \frac{{11\sqrt 3 }}{2} + \frac{{3\sqrt 3 }}{4}$

$= \frac{{16\pi }}{3} + \frac{{25\sqrt 3 }}{2}$ A1A1

[5 marks]

attempt to differentiate both functions and equate M1

$- 6\,{\text{sin}}\,2x = 11\,{\text{sin}}\,x$ A1

attempt to solve for $x$ M1

$11\,{\text{sin}}\,x + 12\,{\text{sin}}\,x\,{\text{cos}}\,x = 0$

${\text{sin}}\,x\left( {11 + 12\,{\text{cos}}\,x} \right) = 0$

${\text{cos}}\,x = - \frac{{11}}{{12}}\,\,\left( {{\text{or}}\,\,{\text{sin}}\,x = 0\,} \right)$ A1

$\Rightarrow y = 4 - 11\left( { - \frac{{11}}{{12}}} \right)$ M1

$y = \frac{{169}}{{12}}\,\left( { = 14\frac{1}{{12}}} \right)$ A1

[6 marks]

Examiners report

[N/A]

Consider quadrilateral $PQRS$ where $[PQ]$ is parallel to $[SR]$ .

In $PQRS$ , $PQ = x$ , $SR = y$ , $R \hat{S} P = α$ and $Q \hat{R} S = β$ .

Find an expression for $PS$ in terms of $x, y, \sin β$ and $\sin (α + β)$ .

Markscheme

METHOD 1

from vertex $P$ , draws a line parallel to $[QR]$ that meets $[SR]$ at a point $X$ (M1)

uses the sine rule in $ΔPSX$ M1

$\frac{PS}{\sin β} = \frac{y - x}{\sin (180 ° - α - β)}$ A1

$\sin (180 ° - α - β) = \sin (α + β)$ (A1)

$PS = \frac{(y - x) \sin β}{\sin (α + β)}$ A1

METHOD 2

let the height of quadrilateral $PQRS$ be $h$

$h = PS \sin α$ A1

attempts to find a second expression for $h$ M1

$h = (y - x - PS \cos α) \tan β$

$PS \sin α = (y - x - PS \cos α) \tan β$

writes $\tan β$ as $\frac{\sin β}{\cos β}$ , multiplies through by $\cos β$ and expands the RHS M1

$PS \sin α \cos β = (y - x) \sin β - PS \cos α \sin β$

$PS = \frac{(y - x) \sin β}{\sin α \cos β + \cos α \sin β}$ A1

$PS = \frac{(y - x) \sin β}{\sin (α + β)}$ A1

[5 marks]

Examiners report

[N/A]

Consider the complex numbers $z_{1} = 1 + b i$ and $z_{2} = (1 - b^{2}) - 2 b i$ , where $b \in ℝ, b \neq 0$ .

Find an expression for $z_{1} z_{2}$ in terms of $b$ .

[3]

Hence, given that $arg (z_{1} z_{2}) = \frac{π}{4}$ , find the value of $b$ .

[3]

Markscheme

$z_{1} z_{2} = (1 + b i) ((1 - b^{2}) - (2 b) i)$

$= (1 - b^{2} - 2 i^{2} b^{2}) + i (- 2 b + b - b^{3})$ M1

$= (1 + b^{2}) + i (- b - b^{3})$ A1A1

Note: Award A1 for $1 + b^{2}$ and A1 for $- b i - b^{3} i$ .

[3 marks]

$arg (z_{1} z_{2}) = arctan (\frac{- b - b^{3}}{1 + b^{2}}) = \frac{π}{4}$ (M1)

EITHER
$arctan (- b) = \frac{π}{4}$ (since $1 + b^{2} \neq 0$ , for $b \in ℝ$ ) A1

$- b - b^{3} = 1 + b^{2}$ (or equivalent) A1

THEN

$b = - 1$ A1

[3 marks]

Examiners report

Part (a) was generally well done with many completely correct answers seen. Part (b) proved to be challenging with many candidates incorrectly equating the ratio of their imaginary and real parts to $\frac{π}{4}$ instead of $\tan \frac{π}{4}$ . Stronger candidates realized that when $θ = \frac{π}{4}$ , it forms an isosceles right-angled triangle and equated the real and imaginary parts to obtain the value of b .

[N/A]

The vectors a and b are defined by a = $\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ t \end{array}} \right)$ , b = $\left( {\begin{array}{*{20}{c}} 0 \\ { - t} \\ {4t} \end{array}} \right)$ , where $t \in \mathbb{R}$ .

Find and simplify an expression for a • b in terms of $t$ .

[2]

Hence or otherwise, find the values of $t$ for which the angle between a and b is obtuse .

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

a • b = $\left( {1 \times 0} \right) + \left( {1 \times - t} \right) + \left( {t \times 4t} \right)$ (M1)

= $- t + 4{t^2}$ A1

[2 marks]

recognition that a • b = |a||b|cos θ (M1)

a • b < 0 or $- t + 4{t^2}$ < 0 or cos θ < 0 R1

Note: Allow ≤ for R1.

attempt to solve using sketch or sign diagram (M1)

$0 < t < \frac{1}{4}$ A1

[4 marks]

Examiners report

[N/A]

Let $f (x) = \sqrt{1 + x}$ for $x > - 1$ .

Show that $f'' (x) = - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ .

[3]

Use mathematical induction to prove that $f^{(n)} (x) = {(- \frac{1}{4})}^{n - 1} \frac{(2 n - 3)!}{(n - 2)!} {(1 + x)}^{\frac{1}{2} - n}$ for $n \in ℤ, n \geq 2$ .

[9]

Let $g (x) = e^{m x}, m \in ℚ$ .

Consider the function $h$ defined by $h (x) = f (x) \times g (x)$ for $x > - 1$ .

It is given that the $x^{2}$ term in the Maclaurin series for $h (x)$ has a coefficient of $\frac{7}{4}$ .

Find the possible values of $m$ .

[8]

Markscheme

attempt to use the chain rule M1

$f' (x) = \frac{1}{2} {(1 + x)}^{- \frac{1}{2}}$ A1

$f'' (x) = - \frac{1}{4} {(1 + x)}^{- \frac{3}{2}}$ A1

$= - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ AG

Note: Award M1A0A0 for $f' (x) = \frac{1}{\sqrt{1 + x}}$ or equivalent seen

[3 marks]

let $n = 2$

$f^{''} (x) = (- \frac{1}{4 \sqrt{{(1 + x)}^{3}}} =) {(- \frac{1}{4})}^{1} \frac{1!}{0!} {(1 + x)}^{\frac{1}{2} - 2}$ R1

Note: Award R0 for not starting at $n = 2$ . Award subsequent marks as appropriate.

assume true for $n = k$ , (so $f^{(k)} (x) = {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k}$ ) M1

Note: Do not award M1 for statements such as “let $n = k$ ” or “ $n = k$ is true”. Subsequent marks can still be awarded.

consider $n = k + 1$

$LHS = f^{(k + 1)} (x) = \frac{d (f^{(k)} (x))}{d x}$ M1

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1}{2} - k) {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

EITHER

$RHS = f^{(k + 1)} (x) = {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for $\frac{(2 k - 1)!}{(k - 1)!} = \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} (= \frac{2 (2 k - 1) (2 k - 3)!}{(k - 2)!})$

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$(= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1})$

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= (\frac{1}{2} - k) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: The following A marks can be awarded in any order.

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1 - 2 k}{2}) {(1 + x)}^{\frac{1}{2} - k - 1}$

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for isolating $(2 k - 1)$ correctly.

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(2 k - 2) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for multiplying top and bottom by $(k - 1)$ or $2 (k - 1)$ .

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$= {(- \frac{1}{4})}^{(k + 1) - 1} \frac{(2 (k + 1) - 3)!}{((k + 1) - 2)!} {(1 + x)}^{\frac{1}{2} - (k + 1)} = RHS$

THEN

since true for $n = 2$ , and true for $n = k + 1$ if true for $n = k$ , the statement is true for all, $n \in ℤ, n \geq 2$ by mathematical induction R1

Note: To obtain the final R1, at least four of the previous marks must have been awarded.

[9 marks]

METHOD 1

$h (x) = \sqrt{1 + x} e^{m x}$

using product rule to find $h' (x)$ (M1)

$h' (x) = \sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}$ A1

$h'' (x) = m (\sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}) + \frac{1}{2 \sqrt{1 + x}} m e^{m x} - \frac{1}{4 \sqrt{{(1 + x)}^{3}}} e^{m x}$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = m^{2} + \frac{1}{2} m + \frac{1}{2} m - \frac{1}{4} (= m^{2} + m - \frac{1}{4})$ A1

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 2

EITHER

attempt to find $f (0), f' (0), f'' (0)$ (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} f (0) = 1$

$f' (x) = \frac{1}{2} {(1 + x)}^{- \frac{1}{2}} f' (0) = \frac{1}{2}$

$f'' (x) = - \frac{1}{4} {(1 + x)}^{- \frac{3}{2}} f'' (0) = - \frac{1}{4}$

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

attempt to apply binomial theorem for rational exponents (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} = 1 + \frac{1}{2} x + \frac{(\frac{1}{2}) (- \frac{1}{2})}{2!} x^{2} \dots$

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

THEN

$g (x) = 1 + m x + \frac{m^{2}}{2} x^{2} + \dots$ (A1)

$h (x) = (1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots) (1 + m x + \frac{m^{2}}{2} x^{2} + \dots)$ (M1)

coefficient of $x^{2}$ is $\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8}$ A1

attempt to set equal to $\frac{7}{4}$ and solve M1

$\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8} = \frac{7}{4}$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 3

$g' (x) = m e^{m x}$ and $g'' (x) = m^{2} e^{m x}$ (A1)

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

using product rule to find $h' (x)$ and $h'' (x)$ (M1)

$h' (x) = f (x) g' (x) + f' (x) g (x)$

$h'' (x) = f (x) g'' (x) + 2 f' (x) g' (x) + f'' (x) g (x)$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = f (0) g'' (0) + 2 g' (0) f' (0) + g (0) f'' (0)$

$= 1 \times m^{2} + 2 m \times \frac{1}{2} + 1 \times (- \frac{1}{4}) (= m^{2} + m - \frac{1}{4})$ A1

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

[8 marks]

Examiners report

[N/A]

Solve the equation $2 \cos^{2} x + 5 \sin x = 4, 0 \leq x \leq 2 π$ .

Markscheme

attempt to use $\cos^{2} x = 1 - \sin^{2} x$ M1

$2 \sin^{2} x - 5 \sin x + 2 = 0$ A1

EITHER

attempting to factorise M1

$(2 \sin x - 1) (\sin x - 2)$ A1

attempting to use the quadratic formula M1

$\sin x = \frac{5 \pm \sqrt{5^{2} - 4 \times 2 \times 2}}{4} (= \frac{5 \pm 3}{4})$ A1

THEN

$\sin x = \frac{1}{2}$ (A1)

$x = \frac{π}{6}, \frac{5 π}{6}$ A1A1

[7 marks]

Examiners report

[N/A]

A sector of a circle with radius $r$ cm , where $r$ > 0, is shown on the following diagram.
The sector has an angle of 1 radian at the centre.

Let the area of the sector be $A$ cm² and the perimeter be $P$ cm. Given that $A = P$ , find the value of $r$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$A = P$

use of the correct formula for area and arc length (M1)

perimeter is $r\theta + 2r$ (A1)

Note: A1 independent of previous M1.

$\frac{1}{2}{r^2}\left( 1 \right) = r\left( 1 \right) + 2r$ A1

${r^2} - 6r = 0$

$r = 6$ (as $r$ > 0) A1

Note: Do not award final A1 if $r = 0$ is included.

[4 marks]

Examiners report

[N/A]

The plane П has the Cartesian equation $2x + y + 2z = 3$

The line L has the vector equation r $= \left( {\begin{array}{*{20}{c}} 3 \\ { - 5} \\ 1 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \\ p \end{array}} \right){\text{,}}\,\,\mu {\text{,}}\,p \in \mathbb{R}$ . The acute angle between the line L and the plane П is 30°.

Find the possible values of $p$ .

Markscheme

recognition that the angle between the normal and the line is 60° (seen anywhere) R1

attempt to use the formula for the scalar product M1

cos 60° = $\frac{{\left| {\left( {\begin{array}{*{20}{c}} 2 \\ 1 \\ 2 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \\ p \end{array}} \right)} \right|}}{{\sqrt 9 \times \sqrt {1 + 4 + {p^2}} }}$ A1

$\frac{1}{2} = \frac{{\left| {2p} \right|}}{{3\sqrt {5 + {p^2}} }}$ A1

$3\sqrt {5 + {p^2}} = 4\left| p \right|$

attempt to square both sides M1

$9\left( {5 + {p^2}} \right) = 16{p^2} \Rightarrow 7{p^2} = 45$

$p = \pm 3\sqrt {\frac{5}{7}}$ (or equivalent) A1A1

[7 marks]

Examiners report

[N/A]