Edexcel P1 (Pure Mathematics 1) 2018 Specimen

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) \(\int 4 x ^ { 3 } - 5 x ^ { - 2 } d x\)
      \(= \frac { 4 x ^ { 4 } } { 4 } - \frac { 5 x ^ { - 1 } } { - 1 } + c = x ^ { 4 } + 5 x ^ { - 1 } + c\)

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 $$\begin{aligned} & y + 4 x + 1 = 0
& y ^ { 2 } + 5 x ^ { 2 } + 2 x = 0 \end{aligned}$$ $$\begin{aligned} & \text { (1) } y = - 4 x - 1
& \therefore ( - 4 x - 1 ) ^ { 2 } + 5 x ^ { 2 } + 2 x = 0
& \therefore 16 x ^ { 2 } + 8 x + 1 + 5 x ^ { 2 } + 2 x = 0
& \therefore 21 x ^ { 2 } + 10 x + 1 = 0
& x = \frac { - 10 \pm \sqrt { ( - 10 ) ^ { 2 } - 4 ( 21 ) ( 1 ) } } { 2 \times 21 }
&
& x = - 1 / 7 \quad x = - 1 / 3
& \begin{array} { r l } y = - 4 ( - 1 / 7 ) - 1 & y = - 4 ( - 1 / 3 ) - 1
= & - 1 / 7
( - 1 / 7 , - 3 / 7 ) & = 1 / 3 \end{array}
& ( - 1 / 3,1 / 3 ) \end{aligned}$$

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}

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) \(\quad x ^ { 2 } - 2 x - 8 < 0\) $$\begin{aligned} \therefore ( x - 4 ) ( x + 2 ) & < 0
x = 4 \quad x = - 2 &
\therefore - 2 < x < 4 & < 4 \end{aligned}$$

6. \begin{figure}[h] \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}

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}

8. \begin{figure}[h] \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: \(2 x + 3 \left( \frac { 3 x } { 2 } \right) = 26\)
   \(\therefore x = 4\)
   \(A = \frac { 4 \times 26 } { 3 } = 52 / 3\)

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   \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}

9. \begin{figure}[h] \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) \(20 - 4 x - \frac { 18 } { x } = - 2 x + 7\) $$\begin{aligned} \therefore & 13 - 2 x - \frac { 18 } { x } = 0
   & 13 x - 2 x ^ { 2 } - 18 = 0
   & 0 = 2 x ^ { 2 } - 13 x + 18
   \therefore & x = \frac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a }
   & a = 2 \quad b = - 13 \quad c = 18
   & x = 2 \quad x = a / 2
   \therefore & y = 3 \quad \therefore y = - 2
   B = & ( a / 2 , - 2 ) \end{aligned}$$ "
   

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10. \begin{figure}[h] \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}

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