Questions — CAIE P2 (699 questions)

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CAIE P2 2020 Specimen Q2
2 Sb th equ \(\operatorname { tin } \sin 2 \theta \tan \theta = \Im\) o \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
CAIE P2 2020 Specimen Q3
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
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
4 marks
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
4 marks
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
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
1 Solve the inequality \(| x + 2 | < | 5 - 2 x |\).
CAIE P2 2002 June Q2
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
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
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
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
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
1 Solve the inequality \(| x - 4 | > | x + 1 |\).
CAIE P2 2003 June Q2
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.
CAIE P2 2003 June Q3
3
\includegraphics[max width=\textwidth, alt={}, center]{a31a4b4e-83a6-47d9-9679-3471b3da1b6e-2_488_664_863_737} The diagram shows the curve \(y = \mathrm { e } ^ { 2 x }\). The shaded region \(R\) is bounded by the curve and by the lines \(x = 0 , y = 0\) and \(x = p\).
  1. Find, in terms of \(p\), the area of \(R\).
  2. Hence calculate the value of \(p\) for which the area of \(R\) is equal to 5 . Give your answer correct to 2 significant figures.
CAIE P2 2003 June Q4
4
  1. Show that the equation $$\tan \left( 45 ^ { \circ } + x \right) = 4 \tan \left( 45 ^ { \circ } - x \right)$$ can be written in the form $$3 \tan ^ { 2 } x - 10 \tan x + 3 = 0$$
  2. Hence solve the equation $$\tan \left( 45 ^ { \circ } + x \right) = 4 \tan \left( 45 ^ { \circ } - x \right)$$ for \(0 ^ { \circ } < x < 90 ^ { \circ }\).
CAIE P2 2003 June Q5
5
  1. By sketching a suitable pair of graphs, show that the equation $$\ln x = 2 - x ^ { 2 }$$ has exactly one root.
  2. Verify by calculation that the root lies between 1.0 and 1.4 .
  3. Use the iterative formula $$x _ { n + 1 } = \sqrt { } \left( 2 - \ln x _ { n } \right)$$ to determine the root correct to 2 decimal places, showing the result of each iteration.
CAIE P2 2003 June Q6
6 The equation of a curve is \(y = \frac { 1 } { 1 + \tan x }\).
  1. Show, by differentiation, that the gradient of the curve is always negative.
  2. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 } { 1 + \tan x } \mathrm {~d} x$$ giving your answer correct to 2 significant figures.

  3. \includegraphics[max width=\textwidth, alt={}, center]{a31a4b4e-83a6-47d9-9679-3471b3da1b6e-3_556_802_1384_708} The diagram shows a sketch of the curve for \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\). State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).
CAIE P2 2003 June Q7
7 The parametric equations of a curve are $$x = 2 \theta - \sin 2 \theta , \quad y = 2 - \cos 2 \theta$$
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot \theta\).
  2. Find the equation of the tangent to the curve at the point where \(\theta = \frac { 1 } { 4 } \pi\).
  3. For the part of the curve where \(0 < \theta < 2 \pi\), find the coordinates of the points where the tangent is parallel to the \(x\)-axis.
CAIE P2 2004 June Q1
1 Given that \(2 ^ { x } = 5 ^ { y }\), use logarithms to find the value of \(\frac { x } { y }\) correct to 3 significant figures.
CAIE P2 2004 June Q2
2 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 5 } \left( 4 x _ { n } + \frac { 306 } { x _ { n } ^ { 4 } } \right)$$ with initial value \(x _ { 1 } = 3\), converges to \(\alpha\).
  1. Use this iterative formula to find \(\alpha\) correct to 3 decimal places, showing the result of each iteration.
  2. State an equation satisfied by \(\alpha\), and hence show that the exact value of \(\alpha\) is \(\sqrt [ 5 ] { 306 }\).
CAIE P2 2004 June Q3
3 The cubic polynomial \(2 x ^ { 3 } + a x ^ { 2 } - 13 x - 6\) is denoted by \(\mathrm { f } ( x )\). It is given that ( \(x - 3\) ) is a factor of \(\mathrm { f } ( x )\).
  1. Find the value of \(a\).
  2. When \(a\) has this value, solve the equation \(\mathrm { f } ( x ) = 0\).
CAIE P2 2004 June Q4
4
  1. Express \(3 \sin \theta + 4 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) correct to 2 decimal places.
  2. Hence solve the equation $$3 \sin \theta + 4 \cos \theta = 4.5$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\), correct to 1 decimal place.
  3. Write down the least value of \(3 \sin \theta + 4 \cos \theta + 7\) as \(\theta\) varies.
CAIE P2 2004 June Q5
5
\includegraphics[max width=\textwidth, alt={}, center]{34177829-f05d-449e-8881-5ab4d852c4ce-3_458_643_285_751} The diagram shows the part of the curve \(y = x \mathrm { e } ^ { - x }\) for \(0 \leqslant x \leqslant 2\), and its maximum point \(M\).
  1. Find the \(x\)-coordinate of \(M\).
  2. Use the trapezium rule with two intervals to estimate the value of $$\int _ { 0 } ^ { 2 } x \mathrm { e } ^ { - x } \mathrm {~d} x$$ giving your answer correct to 2 decimal places.
  3. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii).
CAIE P2 2004 June Q6
6 The parametric equations of a curve are $$x = 2 t + \ln t , \quad y = t + \frac { 4 } { t }$$ where \(t\) takes all positive values.
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } - 4 } { t ( 2 t + 1 ) }\).
  2. Find the equation of the tangent to the curve at the point where \(t = 1\).
  3. The curve has one stationary point. Find the \(y\)-coordinate of this point, and determine whether this point is a maximum or a minimum.