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Edexcel P1 2023 October Q4
7 marks Moderate -0.3
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-08_687_775_248_646} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation \(y = \frac { 1 } { x + 2 }\)
  1. State the equation of the asymptote of \(C\) that is parallel to the \(y\)-axis.
  2. Factorise fully \(x ^ { 3 } + 4 x ^ { 2 } + 4 x\) A copy of Figure 1, labelled Diagram 1, is shown on the next page.
  3. On Diagram 1, add a sketch of the curve with equation $$y = x ^ { 3 } + 4 x ^ { 2 } + 4 x$$ On your sketch, state clearly the coordinates of each point where this curve cuts or meets the coordinate axes.
  4. Hence state the number of real solutions of the equation $$( x + 2 ) \left( x ^ { 3 } + 4 x ^ { 2 } + 4 x \right) = 1$$ giving a reason for your answer.
    \includegraphics[max width=\textwidth, alt={}]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-09_800_1700_1053_185}
    Only use the copy of Diagram 1 if you need to redraw your answer to part (c).
Edexcel P1 2023 October Q5
7 marks Standard +0.3
5. Figure 2 Diagram NOT accurately drawn Figure 2 shows the plan view of a frame for a flat roof.
The shape of the frame consists of triangle \(A B D\) joined to triangle \(B C D\).
Given that
  • \(B D = x \mathrm {~m}\)
  • \(C D = ( 1 + x ) \mathrm { m }\)
  • \(B C = 5 \mathrm {~m}\)
  • angle \(B C D = \theta ^ { \circ }\)
    1. show that \(\cos \theta ^ { \circ } = \frac { 13 + x } { 5 + 5 x }\)
Given also that
  • \(x = 2 \sqrt { 3 }\)
  • angle \(B A C = 30 ^ { \circ }\)
  • \(A D C\) is a straight line
  • find the area of triangle \(A B C\), giving your answer, in \(\mathrm { m } ^ { 2 }\), to one decimal place.
Edexcel P1 2023 October Q6
6 marks Standard +0.3
  1. In this question you must show all stages of your working.
\section*{Solutions relying on calculator technology are not acceptable.} The equation $$4 ( p - 2 x ) = \frac { 12 + 15 p } { x + p } \quad x \neq - p$$ where \(p\) is a constant, has two distinct real roots.
  1. Show that $$3 p ^ { 2 } - 10 p - 8 > 0$$
  2. Hence, using algebra, find the range of possible values of \(p\)
Edexcel P1 2023 October Q7
10 marks Moderate -0.3
  1. The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where \(x > 0\)
Given that
  • \(f ^ { \prime } ( x ) = \frac { 4 x ^ { 2 } + 10 - 7 x ^ { \frac { 1 } { 2 } } } { 4 x ^ { \frac { 1 } { 2 } } }\)
  • the point \(P ( 4 , - 1 )\) lies on \(C\)
    1. (i) find the value of the gradient of \(C\) at \(P\) (ii) Hence find the equation of the normal to \(C\) at \(P\), giving your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers to be found.
    2. Find \(\mathrm { f } ( x )\).
Edexcel P1 2023 October Q8
7 marks Standard +0.8
  1. In this question you must show all stages of your working.
\section*{Solutions relying on calculator technology are not acceptable.} The curve \(C _ { 1 }\) has equation $$x y = \frac { 15 } { 2 } - 5 x \quad x \neq 0$$ The curve \(C _ { 2 }\) has equation $$y = x ^ { 3 } - \frac { 7 } { 2 } x - 5$$
  1. Show that \(C _ { 1 }\) and \(C _ { 2 }\) meet when $$2 x ^ { 4 } - 7 x ^ { 2 } - 15 = 0$$ Given that \(C _ { 1 }\) and \(C _ { 2 }\) meet at points \(P\) and \(Q\)
  2. find, using algebra, the exact distance \(P Q\)
Edexcel P1 2023 October Q9
7 marks Standard +0.3
9. Diagram NOT accurately drawn \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-24_581_1491_340_296} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows the plan view of the area being used for a ball-throwing competition.
Competitors must stand within the circle \(C\) and throw a ball as far as possible into the target area, \(P Q R S\), shown shaded in Figure 3. Given that
  • circle \(C\) has centre \(O\)
  • \(P\) and \(S\) are points on \(C\)
  • \(O P Q R S O\) is a sector of a circle with centre \(O\)
  • the length of arc \(P S\) is 0.72 m
  • the size of angle \(P O S\) is 0.6 radians
    1. show that \(O P = 1.2 \mathrm {~m}\)
Given also that
  • the target area, \(P Q R S\), is \(90 \mathrm {~m} ^ { 2 }\)
  • length \(P Q = x\) metres
  • show that
$$5 x ^ { 2 } + 12 x - 1500 = 0$$
  • Hence calculate the total perimeter of the target area, \(P Q R S\), giving your answer to the nearest metre.
    •
    Edexcel P1 2023 October Q10
    6 marks
    10. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-28_538_652_255_708} \captionsetup{labelformat=empty} \caption{Figure 4}
    \end{figure} Figure 4 shows a sketch of part of the curve \(C _ { 1 }\) with equation $$y = 3 \cos \left( \frac { x } { n } \right) ^ { \circ } \quad x \geqslant 0$$ where \(n\) is a constant.
    The curve \(C _ { 1 }\) cuts the positive \(x\)-axis for the first time at point \(P ( 270,0 )\), as shown in Figure 4.
      1. State the value of \(n\)
      2. State the period of \(C _ { 1 }\) The point \(Q\), shown in Figure 4, is a minimum point of \(C _ { 1 }\)
    2. State the coordinates of \(Q\). The curve \(C _ { 2 }\) has equation \(y = 2 \sin x ^ { \circ } + k\), where \(k\) is a constant.
      The point \(R \left( a , \frac { 12 } { 5 } \right)\) and the point \(S \left( - a , - \frac { 3 } { 5 } \right)\), both lie on \(C _ { 2 }\) Given that \(a\) is a constant less than 90
    3. find the value of \(k\).
    Edexcel P1 2023 October Q11
    10 marks Easy -1.2
    11. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-30_595_869_255_568} \captionsetup{labelformat=empty} \caption{Figure 5}
    \end{figure} Figure 5 shows part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = 2 x ^ { 2 } - 12 x + 14$$
    1. Write \(2 x ^ { 2 } - 12 x + 14\) in the form $$a ( x + b ) ^ { 2 } + c$$ where \(a\), \(b\) and \(c\) are constants to be found. Given that \(C\) has a minimum at the point \(P\)
    2. state the coordinates of \(P\) The line \(l\) intersects \(C\) at \(( - 1,28 )\) and at \(P\) as shown in Figure 5.
    3. Find the equation of \(l\) giving your answer in the form \(y = m x + c\) where \(m\) and \(c\) are constants to be found. The finite region \(R\), shown shaded in Figure 5, is bounded by the \(x\)-axis, \(l\), the \(y\)-axis, and \(C\).
    4. Use inequalities to define the region \(R\).
    Edexcel P1 2018 Specimen Q1
    6 marks Easy -1.3
    1. Given that \(y = 4 x ^ { 3 } - \frac { 5 } { x ^ { 2 } } , x \neq 0\), find in their simplest form
      1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
      2. \(\int y \mathrm {~d} x\) a) \(y = 4 x ^ { 3 } - 5 x ^ { - 2 }\) \(\frac { d y } { d x } = 12 x ^ { 2 } + 10 x ^ { - 3 }\) b) \(\int 4 x ^ { 3 } - 5 x ^ { - 2 } d x\) \(= \frac { 4 x ^ { 4 } } { 4 } - \frac { 5 x ^ { - 1 } } { - 1 } + c = x ^ { 4 } + 5 x ^ { - 1 } + c\)
    Edexcel P1 2018 Specimen Q2
    5 marks Moderate -0.8
    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
    5 marks Moderate -0.8
    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
    8 marks Easy -1.2
    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
    7 marks Moderate -0.5
    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
    5 marks Moderate -0.8
    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
    10 marks Moderate -0.8
    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
    11 marks Moderate -0.8
    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
    12 marks Standard +0.3
    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
    Edexcel C12 2014 January Q1
    4 marks Easy -1.2
    1. Find the first 3 terms in ascending powers of \(x\) of
    $$\left( 2 - \frac { x } { 2 } \right) ^ { 6 }$$ giving each term in its simplest form.
    Edexcel C12 2014 January Q2
    7 marks Easy -1.2
    2. $$\mathrm { f } ( x ) = \frac { 8 } { x ^ { 2 } } - 4 \sqrt { x } + 3 x - 1 , \quad x > 0$$ Giving your answers in their simplest form, find
    1. \(\mathrm { f } ^ { \prime } ( x )\)
    2. \(\int \mathrm { f } ( x ) \mathrm { d } x\)
    Edexcel C12 2014 January Q3
    7 marks Moderate -0.8
    3. $$f ( x ) = 10 x ^ { 3 } + 27 x ^ { 2 } - 13 x - 12$$
    1. Find the remainder when \(\mathrm { f } ( x )\) is divided by
      1. \(x - 2\)
      2. \(x + 3\)
    2. Hence factorise \(\mathrm { f } ( x )\) completely.
    Edexcel C12 2014 January Q4
    7 marks Moderate -0.8
    4. Answer this question without the use of a calculator and show all your working.
    1. Show that $$\frac { 4 } { 2 \sqrt { 2 } - \sqrt { 6 } } = 2 \sqrt { 2 } ( 2 + \sqrt { 3 } )$$
    2. Show that $$\sqrt { 27 } + \sqrt { 21 } \times \sqrt { 7 } - \frac { 6 } { \sqrt { 3 } } = 8 \sqrt { 3 }$$
    Edexcel C12 2014 January Q5
    7 marks Standard +0.8
    5. A sequence is defined by $$\begin{aligned} u _ { 1 } & = 3 \\ u _ { n + 1 } & = 2 - \frac { 4 } { u _ { n } } , \quad n \geqslant 1 \end{aligned}$$ Find the exact values of
    1. \(u _ { 2 } , u _ { 3 }\) and \(u _ { 4 }\)
    2. \(u _ { 61 }\)
    3. \(\sum _ { i = 1 } ^ { 99 } u _ { i }\)
    Edexcel C12 2014 January Q6
    6 marks Standard +0.3
    6. Given that \(a\) and \(b\) are positive constants, solve the simultaneous equations $$\begin{gathered} a b = 25 \\ \log _ { 4 } a - \log _ { 4 } b = 3 \end{gathered}$$ Show each step of your working, giving exact values for \(a\) and \(b\).
    Edexcel C12 2014 January Q7
    5 marks Moderate -0.3
    7. (a) Show that $$12 \sin ^ { 2 } x - \cos x - 11 = 0$$ may be expressed in the form $$12 \cos ^ { 2 } x + \cos x - 1 = 0$$ (b) Hence, using trigonometry, find all the solutions in the interval \(0 \leqslant x \leqslant 360 ^ { \circ }\) of $$12 \sin ^ { 2 } x - \cos x - 11 = 0$$ Give each solution, in degrees, to 1 decimal place. \includegraphics[max width=\textwidth, alt={}, center]{e878227b-d625-4ef2-ac49-a9dc05c5321a-15_106_97_2615_1784}
    Edexcel C12 2014 January Q8
    7 marks Moderate -0.3
    8. Find the range of values of \(k\) for which the quadratic equation $$k x ^ { 2 } + 8 x + 2 ( k + 7 ) = 0$$ has no real roots.