Questions P1 (1374 questions)

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Edexcel P1 2018 Specimen Q2
2. (a) Given that \(3 ^ { - 1.5 } = a \sqrt { 3 }\) find the exact value of \(a\)
(b) Simplify fully \(\frac { \left( 2 x ^ { \frac { 1 } { 2 } } \right) ^ { 3 } } { 4 x ^ { 2 } }\)
a) \(a = \frac { 3 ^ { - 1.5 } } { \sqrt { 3 } } = \frac { 1 } { 9 }\)
b) \(\frac { \left( 2 x ^ { 1 / 2 } \right) ^ { 3 } } { 4 x ^ { 2 } } = \frac { 2 ^ { 3 } x ^ { 1 / 2 \times 3 } } { 4 x ^ { 2 } } = 2 x ^ { - 1 / 2 }\) Solve the simultaneous equations $$\begin{aligned} & y + 4 x + 1 = 0
& y ^ { 2 } + 5 x ^ { 2 } + 2 x = 0 \end{aligned}$$ $$\begin{aligned} & \text { (1) } y = - 4 x - 1
& \therefore ( - 4 x - 1 ) ^ { 2 } + 5 x ^ { 2 } + 2 x = 0
& \therefore 16 x ^ { 2 } + 8 x + 1 + 5 x ^ { 2 } + 2 x = 0
& \therefore 21 x ^ { 2 } + 10 x + 1 = 0
& x = \frac { - 10 \pm \sqrt { ( - 10 ) ^ { 2 } - 4 ( 21 ) ( 1 ) } } { 2 \times 21 }
&
& x = - 1 / 7 \quad x = - 1 / 3
& \begin{array} { r l } y = - 4 ( - 1 / 7 ) - 1 & y = - 4 ( - 1 / 3 ) - 1
= & - 1 / 7
( - 1 / 7 , - 3 / 7 ) & = 1 / 3 \end{array}
& ( - 1 / 3,1 / 3 ) \end{aligned}$$
Edexcel P1 2018 Specimen Q4
4. The straight line with equation \(y = 4 x + c\), where \(c\) is a constant, is a tangent to the curve with equation \(y = 2 x ^ { 2 } + 8 x + 3\) Calculate the value of \(c\) $$\begin{aligned} & y = m x + c \rightarrow y = 4 x + c \quad ( m = 4 )
& \therefore \frac { d y } { d x } = 4 x + 8 = 4 \quad \text { (Gradient equation) }
& 4 x + 8 = 4
& x = - 1 \rightarrow y = - 3
& \text { At } ( - 1 , - 3 ) - 3 = 4 ( - 1 ) + c
& \quad c = 1
& y = 4 x + 1 \end{aligned}$$ \section*{PMT PhysicsAndMathsTutor.com}
\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-09_2258_54_312_34}
Edexcel P1 2018 Specimen Q5
5. (a) On the same axes, sketch the graphs of \(y = x + 2\) and \(y = x ^ { 2 } - x - 6\) showing the coordinates of all points at which each graph crosses the coordinate axes.
(b) On your sketch, show, by shading, the region \(R\) defined by the inequalities $$y < x + 2 \text { and } y > x ^ { 2 } - x - 6$$ (c) Hence, or otherwise, find the set of values of \(x\) for which \(x ^ { 2 } - 2 x - 8 < 0\)
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-10_921_1287_699_260} \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-11_2260_48_313_37} Quadratic: \(y = x ^ { 2 } - x - 6 = ( x - 3 ) ( x + 2 )\) $$x = 3 , \quad x = - 2 @ y = 0$$ Linear: \(\quad y = x + 2\) $$\begin{array} { l l } x = 0 : & y = 2 \quad ( 0,2 )
y = 0 : & x = - 2 \quad ( - 2,0 ) \end{array}$$ c) \(\quad x ^ { 2 } - 2 x - 8 < 0\) $$\begin{aligned} \therefore ( x - 4 ) ( x + 2 ) & < 0
x = 4 \quad x = - 2 &
\therefore - 2 < x < 4 & < 4 \end{aligned}$$
Edexcel P1 2018 Specimen Q6
6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-12_440_679_269_630} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\) The curve \(C\) passes through the origin and through \(( 6,0 )\) The curve \(C\) has a minimum at the point \(( 3 , - 1 )\) On separate diagrams, sketch the curve with equation
  1. \(y = \mathrm { f } ( 2 x )\)
  2. \(y = \mathrm { f } ( x + p )\), where \(p\) is a constant and \(0 < p < 3\) On each diagram show the coordinates of any points where the curve intersects the \(x\)-axis and of any minimum or maximum points.
    a) \(( 6,0 ) \rightarrow ( 3,0 )\) $$( 3 , - 1 ) - 1 > ( 1.5 , - 1 )$$
    \includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-12_616_772_1624_781}
    $$( 1.5 , - 1 )$$ \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-13_2261_50_312_39}
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-13_2637_1835_118_116}
Edexcel P1 2018 Specimen Q7
7. A curve with equation \(y = \mathrm { f } ( x )\) passes through the point \(( 4,25 )\) Given that $$\mathrm { f } ^ { \prime } ( x ) = \frac { 3 } { 8 } x ^ { 2 } - 10 x ^ { - \frac { 1 } { 2 } } + 1 , \quad x > 0$$ find \(\mathrm { f } ( x )\), simplifying each term. $$\begin{aligned} & \therefore F ( x ) = \int F ^ { \prime } ( x ) = \int 3 / 8 x ^ { 2 } - 10 x ^ { - 1 / 2 } + 1 d x
& F ( x ) = \frac { 3 x ^ { 3 } } { 8 ( 3 ) } - \frac { 10 x ^ { 1 / 2 } } { 1 / 2 } + x + c
& F ( x ) = 1 / 8 x ^ { 3 } - 20 x ^ { 1 / 2 } + x + c
& 25 = 1 / 8 ( 4 ) ^ { 3 } - 20 ( 4 ) ^ { 1 / 2 } + 4 + c
& 25 = 8 - 40 + 4 + c
& C = 53
& F ( x ) = 1 / 8 x ^ { 3 } - 20 x ^ { 1 / 2 } + x + 53 \end{aligned}$$ \section*{PMT PhysicsAndMathsTutor.com}
Edexcel P1 2018 Specimen Q8
8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-16_769_979_269_479} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} The line \(l _ { 1 }\), shown in Figure 2 has equation \(2 x + 3 y = 26\)
The line \(l _ { 2 }\) passes through the origin \(O\) and is perpendicular to \(l _ { 1 }\)
  1. Find an equation for the line \(l _ { 2 }\) The line \(l _ { 2 }\) intersects the line \(l _ { 1 }\) at the point \(C\). Line \(l _ { 1 }\) crosses the \(y\)-axis at the point \(B\) as shown in Figure 2.
  2. Find the area of triangle \(O B C\). Give your answer in the form \(\frac { a } { b }\), where \(a\) and \(b\) are integers to be found.
    a) \(L _ { 1 } : 2 x + 3 y = 26\)
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-16_188_820_2039_159}
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-16_129_631_2268_468} $$\begin{gathered} y = 3 / 2 x + 0
    y = 3 / 2 x \end{gathered}$$ \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-17_2257_51_315_34}
    b) \(A = \frac { 6 x h } { 2 } \quad L _ { 1 } : 2 x + 3 y = 26\) $$L _ { 2 } : y = 3 / 2 x$$ At B: \(x = 0 : 0 + 3 y = 26 y = 26 / 3\) At C: \(2 x + 3 \left( \frac { 3 x } { 2 } \right) = 26\)
    \(\therefore x = 4\)
    \(A = \frac { 4 \times 26 } { 3 } = 52 / 3\)
    VIAV SIHI NI IIIIM IONOOVIIV SIHI NI JIIIM ION OCVEYV SIHI NI JIIYM ION OO
    \section*{PMT PhysicsAndMathsTutor.com}
    \includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-19_2255_54_312_34}
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-19_113_61_2604_1884}
Edexcel P1 2018 Specimen Q9
9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-20_693_1038_267_450} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} A sketch of part of the curve \(C\) with equation $$y = 20 - 4 x - \frac { 18 } { x } , \quad x > 0$$ is shown in Figure 3. Point \(A\) lies on \(C\) and has \(x\) coordinate equal to 2
  1. Show that the equation of the normal to \(C\) at \(A\) is \(y = - 2 x + 7\). The normal to \(C\) at \(A\) meets \(C\) again at the point \(B\), as shown in Figure 3 .
  2. Use algebra to find the coordinates of \(B\).
    a) \(y = 20 - 4 ( 2 ) - \frac { 18 } { 2 } \quad \therefore \quad y = 3\) $$\begin{aligned} & y = 20 - 4 x - 18 x ^ { - 1 }
    & \therefore \frac { d y } { d x } = - 4 + 18 x ^ { - 2 } \end{aligned}$$
  3. \(x = 2 \quad \frac { d y } { d x } = - 4 + \frac { 18 } { 2 ^ { 2 } } = 1 / 2\) $$m = - 2$$
    \includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-20_2258_50_313_1980}
    9 continued
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-21_2253_51_315_35} $$\begin{aligned} \therefore \quad y & = - 2 x + c
    3 & = - 2 ( 2 ) + c
    c & = 7
    y & = - 2 x + 7 \end{aligned}$$ b) \(20 - 4 x - \frac { 18 } { x } = - 2 x + 7\) $$\begin{aligned} \therefore & 13 - 2 x - \frac { 18 } { x } = 0
    & 13 x - 2 x ^ { 2 } - 18 = 0
    & 0 = 2 x ^ { 2 } - 13 x + 18
    \therefore & x = \frac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a }
    & a = 2 \quad b = - 13 \quad c = 18
    & x = 2 \quad x = a / 2
    \therefore & y = 3 \quad \therefore y = - 2
    B = & ( a / 2 , - 2 ) \end{aligned}$$ "
    VIIV SIHI NI IIIIM I I N O CVI4V SIHI NI IIIHM ION OOV34V SIHI NI JIIYM ION OC
Edexcel P1 2018 Specimen Q10
10.
\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-23_2255_50_315_37}
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-23_411_1065_252_277} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} The triangle \(X Y Z\) in Figure 4 has \(X Y = 6 \mathrm {~cm} , Y Z = 9 \mathrm {~cm} , Z X = 4 \mathrm {~cm}\) and angle \(Z X Y = \alpha\). The point \(W\) lies on the line \(X Y\). The circular arc \(Z W\), in Figure 4, is a major arc of the circle with centre \(X\) and radius 4 cm .
  1. Show that, to 3 significant figures, \(\alpha = 2.22\) radians.
  2. Find the area, in \(\mathrm { cm } ^ { 2 }\), of the major sector \(X Z W X\). The region, shown shaded in Figure 4, is to be used as a design for a logo. \section*{Calculate}
  3. the area of the logo
  4. the perimeter of the logo.
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-23_383_705_1813_182} Cosine Rule: $$\begin{aligned} & \cos A = \frac { b ^ { 2 } + c ^ { 2 } - a ^ { 2 } } { 2 b c }
    & \cos A = \frac { 4 ^ { 2 } + 6 ^ { 2 } - 9 ^ { 2 } } { 2 ( 4 ) ( 6 ) }
    & A = 2.22 \text { radians } \end{aligned}$$ b)
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-24_323_429_302_353} $$\begin{aligned} \text { Area } & = 1 / 2 r ^ { 2 } \theta
    & = 1 / 2 ( 4 ) ^ { 2 } ( 2 \pi - 2.22 )
    & = 32.5 \mathrm {~cm} ^ { 2 } \end{aligned}$$ c) Area (Logo) \(=\) Area ( \(\Delta\) ) + Area (Sector)
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-24_193_268_840_620}
    \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-24_186_390_934_1028} $$\begin{aligned} & = 1 / 2 ( 4 ) ( 6 ) \sin 2.22
    \text { Area } ( \log 0 ) & = 1 / 2 ( 4 ) ( 6 ) \sin 2.22 + 1 / 2 ( 4 ) ^ { 2 } ( 2 \pi - 2.22 )
    & = 42.1 \mathrm {~cm} ^ { 2 } \end{aligned}$$ d) \(P = 9 + 2 +\) Arc length $$\begin{aligned} P & = 9 + 2 + 4 ( 2 \pi - 2.22 )
    & = 27.3 \mathrm {~cm} \end{aligned}$$ \section*{PMT PhysicsAndMathsTutor.com}
    VIUV SIHI NI IIIUM ION OCVIUV SIHI NI JIIIM ION OCV34V SIHI NI EIIYM ION OC
CAIE P1 2021 March Q10
  1. For the case where angle \(B A C = \frac { 1 } { 6 } \pi\) radians, find \(k\) correct to 4 significant figures.
  2. For the general case in which angle \(B A C = \theta\) radians, where \(0 < \theta < \frac { 1 } { 2 } \pi\), it is given that \(\frac { \theta } { \sin \theta } > 1\). Find the set of possible values of \(k\).
CAIE P1 2022 March Q6
  1. Find, by calculation, the coordinates of \(A\) and \(B\).
  2. Find an equation of the circle which has its centre at \(C\) and for which the line with equation \(y = 3 x - 20\) is a tangent to the circle.
CAIE P1 2022 March Q10
  1. Find the perimeter of the shaded region.
  2. Find the area of the shaded region.
CAIE P1 2024 March Q10
  1. Find the equation of the tangent to the circle at the point \(( - 6,9 )\).
  2. Find the equation of the circle in the form \(x ^ { 2 } + y ^ { 2 } + a x + b y + c = 0\).
  3. Find the value of \(\theta\) correct to 4 significant figures.
  4. Find the perimeter and area of the segment shaded in the diagram.
CAIE P1 2020 November Q10
  1. Find \(C D\) in terms of \(r\) and \(\sin \theta\).
    It is now given that \(r = 4\) and \(\theta = \frac { 1 } { 6 } \pi\).
  2. Find the perimeter of sector \(C A B\) in terms of \(\pi\).
  3. Find the area of the shaded region in terms of \(\pi\) and \(\sqrt { 3 }\).
CAIE P1 2021 November Q6
  1. Find the perimeter of the plate, giving your answer in terms of \(\pi\).
  2. Find the area of the plate, giving your answer in terms of \(\pi\) and \(\sqrt { 3 }\).
CAIE P1 2022 November Q10
  1. Find the perimeter of the cross-section RASB, giving your answer correct to 2 decimal places.
  2. Find the difference in area of the two triangles \(A O B\) and \(A P B\), giving your answer correct to 2 decimal places.
  3. Find the area of the cross-section RASB, giving your answer correct to 1 decimal place.
CAIE P1 2022 November Q10
  1. Find the coordinates of \(A\).
  2. Find the volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Give your answer in the form \(\frac { \pi } { a } ( b \sqrt { c } - d )\), where \(a , b , c\) and \(d\) are integers.
  3. Find an exact expression for the perimeter of the shaded region.
CAIE P1 2017 June Q4
  1. Express the perimeter of the shaded region in terms of \(r\) and \(\theta\).
  2. In the case where \(r = 5\) and \(\theta = \frac { 1 } { 6 } \pi\), find the area of the shaded region.
CAIE P1 2018 June Q5
  1. Express each of the vectors \(\overrightarrow { D A }\) and \(\overrightarrow { C A }\) in terms of \(\mathbf { i } , \mathbf { j }\) and \(\mathbf { k }\).
  2. Use a scalar product to find angle \(C A D\).
CAIE P1 2017 March Q4
  1. Show that angle \(C B D = \frac { 9 } { 14 } \pi\) radians.
  2. Find the perimeter of the shaded region.
CAIE P1 2005 November Q5
  1. Express \(h\) in terms of \(r\) and hence show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of the cylinder is given by $$V = 12 \pi r ^ { 2 } - 2 \pi r ^ { 3 }$$
  2. Given that \(r\) varies, find the stationary value of \(V\).
CAIE P1 2014 November Q5
  1. Show that the equation \(1 + \sin x \tan x = 5 \cos x\) can be expressed as $$6 \cos ^ { 2 } x - \cos x - 1 = 0$$
  2. Hence solve the equation \(1 + \sin x \tan x = 5 \cos x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\). The equation of a curve is \(y = x ^ { 3 } + a x ^ { 2 } + b x\), where \(a\) and \(b\) are constants.
  3. In the case where the curve has no stationary point, show that \(a ^ { 2 } < 3 b\).
  4. In the case where \(a = - 6\) and \(b = 9\), find the set of values of \(x\) for which \(y\) is a decreasing function of \(x\).
    \includegraphics[max width=\textwidth, alt={}, center]{8952fc09-004a-4fb6-ad80-5312095a7057-3_634_711_952_717} The diagram shows a pyramid \(O A B C X\). The horizontal square base \(O A B C\) has side 8 units and the centre of the base is \(D\). The top of the pyramid, \(X\), is vertically above \(D\) and \(X D = 10\) units. The mid-point of \(O X\) is \(M\). The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are parallel to \(\overrightarrow { O A }\) and \(\overrightarrow { O C }\) respectively and the unit vector \(\mathbf { k }\) is vertically upwards.
  5. Express the vectors \(\overrightarrow { A M }\) and \(\overrightarrow { A C }\) in terms of \(\mathbf { i } , \mathbf { j }\) and \(\mathbf { k }\).
  6. Use a scalar product to find angle \(M A C\).
    (a) The sum, \(S _ { n }\), of the first \(n\) terms of an arithmetic progression is given by \(S _ { n } = 32 n - n ^ { 2 }\). Find the first term and the common difference.
    (b) A geometric progression in which all the terms are positive has sum to infinity 20 . The sum of the first two terms is 12.8 . Find the first term of the progression.
CAIE P1 2015 November Q10
  1. For the case where \(a = 2\), find the unit vector in the direction of \(\overrightarrow { P M }\).
  2. For the case where angle \(A T P = \cos ^ { - 1 } \left( \frac { 2 } { 7 } \right)\), find the value of \(a\).
CAIE P1 2016 November Q4
  1. Find the equation of the line \(C D\), giving your answer in the form \(y = m x + c\).
  2. Find the distance \(A D\).
CAIE P1 Specimen Q10
  1. For the case where \(a = 2\), find the unit vector in the direction of \(\overrightarrow { P M }\).
  2. For the case where angle \(A T P = \cos ^ { - 1 } \left( \frac { 2 } { 7 } \right)\), find the value of \(a\).