CAIE P2 (Pure Mathematics 2) 2020 Specimen

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1. Tb p lm ial \(2 x ^ { 3 } + a x ^ { 2 } - a x - 2\) wh re \(a\) is a co tan, is d h ed \(\mathrm { y } \quad \mathrm { p } x\) ). It is g n th t \(( x + 1\) is a facto \(6 \quad ( x )\). Fid b le \(6 a\).
2. Wh n \(a \mathbf { b }\) s th s le , f in e remaid r wh \(\underline { \mathrm { p } } \quad x )\) is \(\dot { \mathbf { d } } \dot { \mathbf { v } } \mathbf { d }\) dt \(\quad x + \beta\).

2 Sb th equ \(\operatorname { tin } \sin 2 \theta \tan \theta = \Im\) o \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

3 It is g n th t \(a\) is a p itie co tan.

1. 1. Sketch sib ed ag am th g ad \(6 y = | 2 x - 3 a |\) ad \(y = | 2 x + 4 a |\).
   2. State th co dia tes

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1. Sb the eq tiந \({ } ^ { 2 x } + 5 ^ { x } = \frac { 13 } { 8 } \quad\) in wer co rect t \(\beta \quad\) sig fican fig es. [44
2. It is g vert \(\mathbf { h } \mathrm { t } \ln y + \overline { 5 } + \mathrm { n } y = 2 \ln x\). Eq ess \(y\) irt erms \(\mathbf { b } x\), irr fo m no id \(\mathbf { v }\) ng \(\mathbf { g }\) riths.

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\includegraphics[max width=\textwidth, alt={}, center]{d4bec1a9-2d24-4cf8-9991-9ab61ddbc865-08_430_990_260_539} Th id ag am sto cn \(\mathrm { y } = \frac { \sin 2 \mathrm { x } } { \mathrm { x } + 2 }\) fo \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Tb \(x\)-co \(\dot { \mathrm { d } } \mathbf { a }\) te 6 th max mm \(\dot { \mathrm { p } } n M\) is d t ed y \(\alpha\).

1. Fid \(\frac { \mathrm { dy } } { \mathrm { dx } }\) ad th t \(\alpha\) satisfies th eq tin \(\tan 2 x = 2 x + 4\) [4]
2. Stw alch atin \(\mathbf { b }\) t \(\alpha\) lies b tweerfd nd
3. Use th iterati fo mu a \(x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( 2 x _ { n } + 4 \right.\) to id b \& le \(6 \alpha\) co rect tod cimal p aces. Gie th resu to each teratio od cimal places.

6 Th \(\mathbf { p }\) rametric eq tion \(\mathbf { 6 }\) a cn \(\mathbf { e }\) are $$x = \mathrm { e } ^ { 2 t } , \quad y = 4 t \mathrm { e } ^ { t }$$

1. Stw th \(t \frac { d y } { d x } = \frac { 2 ( t + 1 ) } { e ^ { t } }\).
2. Fid b eq tin th \(\mathbf { n }\) mal to \(\mathbf { b }\) cn at te \(\dot { \mathbf { p } }\) n wh re \(t = 0\) [4]

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1. Shat that \(\tan ^ { 2 } x + \operatorname { co } { } ^ { 2 } x \equiv \sec ^ { 2 } x + \frac { 1 } { 2 } \mathrm { co } 2 x - \frac { 1 } { 2 }\) ach n e fid b ex ct le 6 $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \left( \tan ^ { 2 } x + \cos ^ { 2 } x \right) d x$$
2. 
   \includegraphics[max width=\textwidth, alt={}, center]{d4bec1a9-2d24-4cf8-9991-9ab61ddbc865-13_556_794_260_639} Th regn en lo edy th cn \(y = \tan x + \mathrm { co } x\) ad th lin \(\mathrm { s } x = 0 \quad x = \frac { 1 } { 4 } \pi\) ad \(y = 0\) is sw n in th d ag am. Fid th ex ct m e \(\mathbf { 6 }\) th sb idpd ed wh n this reg n is ro ated cm p etely abt th \(x\)-ax s. If B e th follw ig lin dpg to cm p ete th an wer(s) to ay q stin (s), th q stin \(\mathrm { m } \quad \mathbf { b } \quad \mathrm { r } ( \mathrm { s } )\) ms tb clearlys n n