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CAIE P2 2024 November Q2
6 marks Moderate -0.3
2 Let \(\mathrm { f } ( x ) = 4 \sin ^ { 2 } 3 x\).
  1. Find the value of \(\mathrm { f } ^ { \prime } \left( \frac { 1 } { 4 } \pi \right)\).
  2. Find \(\int \mathrm { f } ( x ) \mathrm { d } x\). \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-05_2723_35_101_20}
CAIE P2 2024 November Q3
6 marks Standard +0.3
3 A curve has equation \(6 \mathrm { e } ^ { - x } y ^ { 2 } + \mathrm { e } ^ { 2 x } - 12 y + 7 = 0\).
Find the gradient of the curve at the point \(( \ln 3,2 )\).
CAIE P2 2024 November Q5
10 marks Standard +0.3
5 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = a x ^ { 3 } + b x ^ { 2 } - a x + 8$$ where \(a\) and \(b\) are constants.It is given that \(( x + 2 )\) is a factor of \(\mathrm { p } ( x )\) ,and that the remainder is 24 when \(\mathrm { p } ( x )\) is divided by \(( x - 2 )\) .
  1. Find the values of \(a\) and \(b\) . \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-09_2723_35_101_20}
  2. Factorise \(\mathrm { p } ( x )\) and hence show that the equation \(\mathrm { p } ( x ) = 0\) has exactly one real root.
  3. Solve the equation \(\mathrm { p } \left( \frac { 1 } { 2 } \operatorname { cosec } \theta \right) = 0\) for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\). \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-10_499_696_264_680} The diagram shows the curves with equations \(y = \sqrt [ 3 ] { 5 x ^ { 2 } + 7 }\) and \(y = \frac { 27 } { 2 x + 5 }\) for \(x \geqslant 0\).
    The curves meet at the point \(( 2,3 )\).
    Region \(A\) is bounded by the curve \(y = \sqrt [ 3 ] { 5 x ^ { 2 } + 7 }\) and the straight lines \(x = 0 , x = 2\) and \(y = 0\).
    Region \(B\) is bounded by the two curves and the straight line \(x = 0\).
CAIE P2 2024 November Q7
9 marks Standard +0.3
7
  1. Express \(4 \sin \theta \sin \left( \theta + 60 ^ { \circ } \right)\) in the form $$a + R \sin ( 2 \theta - \alpha ) ,$$ where \(a\) and \(R\) are positive integers and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-13_2723_33_99_21}
  2. Hence find the smallest positive value of \(\theta\) satisfying the equation $$\frac { 1 } { 5 } + 4 \sin \theta \sin \left( \theta + 60 ^ { \circ } \right) = 0 .$$ If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-14_2714_38_109_2010}
CAIE P2 2024 November Q1
5 marks Moderate -0.3
1 The variables \(x\) and \(y\) satisfy the equation \(a ^ { 2 y } = \mathrm { e } ^ { 3 x + k }\), where \(a\) and \(k\) are constants.
The graph of \(y\) against \(x\) is a straight line.
  1. Use logarithms to show that the gradient of the straight line is \(\frac { 3 } { 2 \ln a }\).
  2. Given that the straight line passes through the points \(( 0.4,0.95 )\) and \(( 3.3,3.80 )\), find the values of \(a\) and \(k\). \includegraphics[max width=\textwidth, alt={}, center]{468efb3f-be7b-4f9e-b8c3-c6fd40d7714a-03_2723_33_99_21}
CAIE P2 2024 November Q2
4 marks Standard +0.3
2 Solve the inequality \(| x - 7 | > 4 x + 3\).
CAIE P2 2024 November Q3
7 marks Moderate -0.3
3 The function f is defined by \(\mathrm { f } ( x ) = \tan ^ { 2 } \left( \frac { 1 } { 2 } x \right)\) for \(0 \leqslant x < \pi\).
  1. Find the exact value of \(\mathrm { f } ^ { \prime } \left( \frac { 2 } { 3 } \pi \right)\). \includegraphics[max width=\textwidth, alt={}, center]{468efb3f-be7b-4f9e-b8c3-c6fd40d7714a-05_2726_33_97_22}
  2. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } ( \mathrm { f } ( x ) + \sin x ) \mathrm { d } x\).
CAIE P2 2024 November Q5
8 marks Standard +0.3
5 It is given that \(\int _ { a } ^ { a ^ { 3 } } \frac { 10 } { 2 x + 1 } \mathrm {~d} x = 7\), where \(a\) is a constant greater than 1 .
  1. Show that \(a = \sqrt [ 3 ] { 0.5 \mathrm { e } ^ { 1.4 } ( 2 a + 1 ) - 0.5 }\). \includegraphics[max width=\textwidth, alt={}, center]{468efb3f-be7b-4f9e-b8c3-c6fd40d7714a-09_2725_35_99_20}
  2. Use an iterative formula, based on the equation in part (a), to find the value of \(a\) correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures.
CAIE P2 2024 November Q6
7 marks Standard +0.3
6 A curve has parametric equations $$x = \frac { \mathrm { e } ^ { 2 t } - 2 } { \mathrm { e } ^ { 2 t } + 1 } , \quad y = \mathrm { e } ^ { 3 t } + 1$$
  1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\).
  2. Find the exact gradient of the curve at the point where the curve crosses the \(y\)-axis.
CAIE P2 2024 November Q7
11 marks Standard +0.3
7
  1. Prove that \(\cos \left( \theta + 30 ^ { \circ } \right) \cos \left( \theta + 60 ^ { \circ } \right) \equiv \frac { 1 } { 4 } \sqrt { 3 } - \frac { 1 } { 2 } \sin 2 \theta\).
  2. Solve the equation \(5 \cos \left( 2 \alpha + 30 ^ { \circ } \right) \cos \left( 2 \alpha + 60 ^ { \circ } \right) = 1\) for \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
  3. Show that the exact value of \(\cos 20 ^ { \circ } \cos 50 ^ { \circ } + \cos 40 ^ { \circ } \cos 70 ^ { \circ }\) is \(\frac { 1 } { 2 } \sqrt { 3 }\).
    If you use the following page to complete the answer to any question, the question number must be clearly shown.
CAIE P2 2020 Specimen Q1
4 marks Moderate -0.8
1
  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 \(\quad x + \beta\).
CAIE P2 2020 Specimen Q2
4 marks Standard +0.3
2 Sb th equ \(\operatorname { tin } \sin 2 \theta \tan \theta = \Im\) o \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P2 2020 Specimen Q3
6 marks Moderate -0.8
3 It is g n th t \(a\) is a p itie co tan.
    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
CAIE P2 2020 Specimen Q4
8 marks Moderate -0.3
4
  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.
CAIE P2 2020 Specimen Q5
9 marks Standard +0.3
5 \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.
CAIE P2 2020 Specimen Q6
8 marks Standard +0.3
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]
CAIE P2 2020 Specimen Q7
11 marks Standard +0.3
7
  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
CAIE P2 2002 June Q1
4 marks Standard +0.3
1 Solve the inequality \(| x + 2 | < | 5 - 2 x |\).
CAIE P2 2002 June Q2
5 marks Moderate -0.8
2 The cubic polynomial \(3 x ^ { 3 } + a x ^ { 2 } - 2 x - 8\) is denoted by \(\mathrm { f } ( x )\).
  1. Given that ( \(x + 2\) ) is a factor of \(\mathrm { f } ( x )\), find the value of \(a\).
  2. When \(a\) has this value, factorise \(\mathrm { f } ( x )\) completely.
CAIE P2 2002 June Q3
5 marks Moderate -0.8
3 Two variable quantities \(x\) and \(y\) are related by the equation $$y = A x ^ { n }$$ where \(A\) and \(n\) are constants. \includegraphics[max width=\textwidth, alt={}, center]{9b103197-7ba0-427a-b983-34edb51b6cca-2_422_697_977_740} When a graph is plotted showing values of \(\ln y\) on the vertical axis and values of \(\ln x\) on the horizontal axis, the points lie on a straight line. This line crosses the vertical axis at the point ( \(0,2.3\) ) and also passes through the point (4.0,1.7), as shown in the diagram. Find the values of \(A\) and \(n\).
CAIE P2 2002 June Q4
8 marks Moderate -0.3
4
  1. Express \(3 \cos \theta + 2 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), stating the exact value of \(R\) and giving the value of \(\alpha\) correct to 1 decimal place.
  2. Solve the equation $$3 \cos \theta + 2 \sin \theta = 3.5$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
  3. The graph of \(y = 3 \cos \theta + 2 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\), has one stationary point. State the coordinates of this point. \includegraphics[max width=\textwidth, alt={}, center]{9b103197-7ba0-427a-b983-34edb51b6cca-3_421_823_299_662} The diagram shows the curve \(y = 2 x \mathrm { e } ^ { - x }\) and its maximum point \(P\). Each of the two points \(Q\) and \(R\) on the curve has \(y\)-coordinate equal to \(\frac { 1 } { 2 }\).
CAIE P2 2002 June Q6
10 marks Moderate -0.3
6
    1. Show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos 2 x \mathrm {~d} x = \frac { 1 } { 2 }\).
    2. By using an appropriate trigonometrical identity, find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sin ^ { 2 } x \mathrm {~d} x\).
    1. Use the trapezium rule with 2 intervals to estimate the value of \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec x d x\), giving your answer correct to 2 significant figures.
    2. Determine, by sketching the appropriate part of the graph of \(y = \sec x\), whether the trapezium rule gives an under-estimate or an over-estimate of the true value.
CAIE P2 2002 June Q7
10 marks Standard +0.3
7 The parametric equations of a curve are $$x = t + 2 \ln t , \quad y = 2 t - \ln t$$ where \(t\) takes all positive values.
  1. Express \(\frac { d y } { d x }\) in terms of \(t\).
  2. Find the equation of the tangent to the curve at the point where \(t = 1\).
  3. The curve has one stationary point. Show that the \(y\)-coordinate of this point is \(1 + \ln 2\) and determine whether this point is a maximum or a minimum.
CAIE P2 2003 June Q1
4 marks Moderate -0.3
1 Solve the inequality \(| x - 4 | > | x + 1 |\).
CAIE P2 2003 June Q2
6 marks Moderate -0.3
2 The polynomial \(x ^ { 4 } - 9 x ^ { 2 } - 6 x - 1\) is denoted by \(\mathrm { f } ( x )\).
  1. Find the value of the constant \(a\) for which $$f ( x ) \equiv \left( x ^ { 2 } + a x + 1 \right) \left( x ^ { 2 } - a x - 1 \right)$$
  2. Hence solve the equation \(\mathrm { f } ( x ) = 0\), giving your answers in an exact form.