Questions — Edexcel (10514 questions)

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Edexcel C34 2015 January Q7
9 marks Standard +0.8
7.
  1. Given that $$2 \cos ( x + 30 ) ^ { \circ } = \sin ( x - 30 ) ^ { \circ }$$ without using a calculator, show that $$\tan x ^ { \circ } = 3 \sqrt { 3 } - 4$$
  2. Hence or otherwise solve, for \(0 \leqslant \theta < 180\), $$2 \cos ( 2 \theta + 40 ) ^ { \circ } = \sin ( 2 \theta - 20 ) ^ { \circ }$$ Give your answers to one decimal place.
Edexcel C34 2015 January Q8
9 marks Standard +0.3
8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-13_743_1198_219_372} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} The value of Lin's car is modelled by the formula $$V = 18000 \mathrm { e } ^ { - 0.2 t } + 4000 \mathrm { e } ^ { - 0.1 t } + 1000 , \quad t \geqslant 0$$ where the value of the car is \(V\) pounds when the age of the car is \(t\) years.
A sketch of \(t\) against \(V\) is shown in Figure 1.
  1. State the range of \(V\). According to this model,
  2. find the rate at which the value of the car is decreasing when \(t = 10\) Give your answer in pounds per year.
  3. Calculate the exact value of \(t\) when \(V = 15000\)
Edexcel C34 2015 January Q9
12 marks Standard +0.3
9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-15_618_899_262_566} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} The curve \(C\) has parametric equations $$x = \ln ( t + 2 ) , \quad y = \frac { 4 } { t ^ { 2 } } \quad t > 0$$ The finite region \(R\), shown shaded in Figure 2, is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = \ln 3\) and \(x = \ln 5\)
  1. Show that the area of \(R\) is given by the integral $$\int _ { 1 } ^ { 3 } \frac { 4 } { t ^ { 2 } ( t + 2 ) } \mathrm { d } t$$
  2. Hence find an exact value for the area of \(R\). Write your answer in the form ( \(a + \ln b\) ), where \(a\) and \(b\) are rational numbers.
  3. Find a cartesian equation of the curve \(C\) in the form \(y = \mathrm { f } ( x )\).
Edexcel C34 2015 January Q10
10 marks Standard +0.3
10. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-17_598_736_223_603} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$ Point \(A\) is the minimum turning point on the curve.
  1. Show, by using calculus, that the \(x\) coordinate of point \(A\) is a solution of $$x = \frac { 6 } { 1 + \ln \left( x ^ { 2 } \right) }$$
  2. Starting with \(x _ { 0 } = 2.27\), use the iteration $$x _ { n + 1 } = \frac { 6 } { 1 + \ln \left( x _ { n } ^ { 2 } \right) }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
  3. Use your answer to part (b) to deduce the coordinates of point \(A\) to one decimal place.
Edexcel C34 2015 January Q11
12 marks Standard +0.3
11. With respect to a fixed origin \(O\) the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 14 \\ - 6 \\ - 13 \end{array} \right) + \lambda \left( \begin{array} { r } - 2 \\ 1 \\ 4 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } p \\ - 7 \\ 4 \end{array} \right) + \mu \left( \begin{array} { l } q \\ 2 \\ 1 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters and \(p\) and \(q\) are constants. Given that \(l _ { 1 }\) and \(l _ { 2 }\) are perpendicular,
  1. show that \(q = 3\) Given further that \(l _ { 1 }\) and \(l _ { 2 }\) intersect at point \(X\), find
  2. the value of \(p\),
  3. the coordinates of \(X\). The point \(A\) lies on \(l _ { 1 }\) and has position vector \(\left( \begin{array} { r } 6 \\ - 2 \\ 3 \end{array} \right)\) Given that point \(B\) also lies on \(l _ { 1 }\) and that \(A B = 2 A X\)
  4. find the two possible position vectors of \(B\). \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \section*{Question 11 continued} \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)
Edexcel C34 2015 January Q12
13 marks Standard +0.3
12. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-21_615_732_233_605} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$ The finite region \(S\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = 1\) and \(x = 3\)
  1. Complete the table below with the value of \(y\) corresponding to \(x = 2\). Give your answer to 4 decimal places.
    \(x\)11.522.53
    \(y\)21.30410.90891.2958
  2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(S\), giving your answer to 3 decimal places.
  3. Use calculus to find the exact area of \(S\). Give your answer in the form \(\frac { a } { b } + \ln c\), where \(a , b\) and \(c\) are integers.
  4. Hence calculate the percentage error in using your answer to part (b) to estimate the area of \(S\). Give your answer to one decimal place.
  5. Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of \(S\). \section*{Question 12 continued} \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) 13. (a) Express \(10 \cos \theta - 3 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\) Give the exact value of \(R\) and give the value of \(\alpha\) to 2 decimal places. Alana models the height above the ground of a passenger on a Ferris wheel by the equation $$H = 12 - 10 \cos ( 30 t ) ^ { \circ } + 3 \sin ( 30 t ) ^ { \circ }$$ where the height of the passenger above the ground is \(H\) metres at time \(t\) minutes after the wheel starts turning. \includegraphics[max width=\textwidth, alt={}, center]{03548211-79cb-4629-b6ca-aa9dfcc77a33-23_419_567_516_1160}
    (b) Calculate
    1. the maximum value of \(H\) predicted by this model,
    2. the value of \(t\) when this maximum first occurs. Give each answer to 2 decimal places.
      (c) Calculate the value of \(t\) when the passenger is 18 m above the ground for the first time. Give your answer to 2 decimal places.
      (d) Determine the time taken for the Ferris wheel to complete two revolutions. \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \section*{Question 13 continued}
Edexcel C34 2016 January Q1
4 marks Moderate -0.3
1. $$f ( x ) = ( 3 - 2 x ) ^ { - 4 } , \quad | x | < \frac { 3 } { 2 }$$ Find the binomial expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\), giving each coefficient as a simplified fraction.
Edexcel C34 2016 January Q2
6 marks Standard +0.3
  1. Show that $$\cot ^ { 2 } x - \operatorname { cosec } x - 11 = 0$$ may be expressed in the form \(\operatorname { cosec } ^ { 2 } x - \operatorname { cosec } x + k = 0\), where \(k\) is a constant.
  2. Hence solve for \(0 \leqslant x < 360 ^ { \circ }\) $$\cot ^ { 2 } x - \operatorname { cosec } x - 11 = 0$$ Give each solution in degrees to one decimal place.
    (Solutions based entirely on graphical or numerical methods are not acceptable.)
Edexcel C34 2016 January Q3
7 marks Standard +0.3
3. A curve \(C\) has equation $$3 ^ { x } + 6 y = \frac { 3 } { 2 } x y ^ { 2 }$$ Find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point on \(C\) with coordinates (2, 3). Give your answer in the form \(\frac { a + \ln b } { 8 }\), where \(a\) and \(b\) are integers.
Edexcel C34 2016 January Q4
7 marks Standard +0.3
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-05_504_844_255_543} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} The curve \(C\) with equation \(y = \frac { 2 } { ( 4 + 3 x ) } , x > - \frac { 4 } { 3 }\) is shown in Figure 1
The region bounded by the curve, the \(x\)-axis and the lines \(x = - 1\) and \(x = \frac { 2 } { 3 }\), is shown shaded in Figure 1 This region is rotated through 360 degrees about the \(x\)-axis.
  1. Use calculus to find the exact value of the volume of the solid generated. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-05_583_433_1398_753} \captionsetup{labelformat=empty} \caption{Figure 2}
    \end{figure} Figure 2 shows a candle with axis of symmetry \(A B\) where \(A B = 15 \mathrm {~cm}\). \(A\) is a point at the centre of the top surface of the candle and \(B\) is a point at the centre of the base of the candle. The candle is geometrically similar to the solid generated in part (a).
  2. Find the volume of this candle.
Edexcel C34 2016 January Q5
9 marks Moderate -0.3
5. $$f ( x ) = - x ^ { 3 } + 4 x ^ { 2 } - 6$$
  1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\)
  2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = \sqrt { \left( \frac { 6 } { 4 - x } \right) }$$
  3. Starting with \(x _ { 1 } = 1.5\) use the iteration \(x _ { n + 1 } = \sqrt { \left( \frac { 6 } { 4 - x _ { n } } \right) }\) to calculate the values of \(x _ { 2 }\), \(x _ { 3 }\) and \(x _ { 4 }\) giving all your answers to 4 decimal places.
  4. Using a suitable interval, show that 1.572 is a root of \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.
Edexcel C34 2016 January Q6
8 marks Moderate -0.8
6. A hot piece of metal is dropped into a cool liquid. As the metal cools, its temperature \(T\) degrees Celsius, \(t\) minutes after it enters the liquid, is modelled by $$T = 300 \mathrm { e } ^ { - 0.04 t } + 20 , \quad t \geqslant 0$$
  1. Find the temperature of the piece of metal as it enters the liquid.
  2. Find the value of \(t\) for which \(T = 180\), giving your answer to 3 significant figures. (Solutions based entirely on graphical or numerical methods are not acceptable.)
  3. Show, by differentiation, that the rate, in degrees Celsius per minute, at which the temperature of the metal is changing, is given by the expression $$\frac { 20 - T } { 25 }$$
    VIII SIHI NI I IVM I I ON OCVIIV SIHI NI JIIIM IONOOVI4V SIHI NI BIIIM ION OO
Edexcel C34 2016 January Q7
11 marks Moderate -0.3
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-10_543_817_278_584} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows part of the curve \(C\) with equation $$y = \frac { 3 \ln \left( x ^ { 2 } + 1 \right) } { \left( x ^ { 2 } + 1 \right) } , \quad x \in \mathbb { R }$$
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
  2. Using your answer to (a), find the exact coordinates of the stationary point on the curve \(C\) for which \(x > 0\). Write each coordinate in its simplest form.
    (5) The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\), the \(x\)-axis and the line \(x = 3\)
  3. Complete the table below with the value of \(y\) corresponding to \(x = 1\)
    \(x\)0123
    \(y\)0\(\frac { 3 } { 5 } \ln 5\)\(\frac { 3 } { 10 } \ln 10\)
  4. Use the trapezium rule with all the \(y\) values in the completed table to find an approximate value for the area of \(R\), giving your answer to 4 significant figures.
Edexcel C34 2016 January Q8
9 marks Standard +0.3
8. $$f ( \theta ) = 9 \cos ^ { 2 } \theta + \sin ^ { 2 } \theta$$
  1. Show that \(\mathrm { f } ( \theta ) = a + b \cos 2 \theta\), where \(a\) and \(b\) are integers which should be found.
  2. Using your answer to part (a) and integration by parts, find the exact value of $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \theta ^ { 2 } \mathrm { f } ( \theta ) \mathrm { d } \theta$$
Edexcel C34 2016 January Q9
10 marks Challenging +1.2
  1. Express \(\frac { 3 x ^ { 2 } - 4 } { x ^ { 2 } ( 3 x - 2 ) }\) in partial fractions.
  2. Given that \(x > \frac { 2 } { 3 }\), find the general solution of the differential equation $$x ^ { 2 } ( 3 x - 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = y \left( 3 x ^ { 2 } - 4 \right)$$ Give your answer in the form \(y = \mathrm { f } ( x )\).
Edexcel C34 2016 January Q10
12 marks Standard +0.3
10.
  1. Express \(3 \sin 2 x + 5 \cos 2 x\) in the form \(R \sin ( 2 x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\) Give the exact value of \(R\) and give the value of \(\alpha\) to 3 significant figures.
  2. Solve, for \(0 < x < \pi\), $$3 \sin 2 x + 5 \cos 2 x = 4$$ (Solutions based entirely on graphical or numerical methods are not acceptable.) $$g ( x ) = 4 ( 3 \sin 2 x + 5 \cos 2 x ) ^ { 2 } + 3$$
  3. Using your answer to part (a) and showing your working,
    1. find the greatest value of \(\mathrm { g } ( x )\),
    2. find the least value of \(\mathrm { g } ( x )\).
Edexcel C34 2016 January Q11
14 marks Standard +0.3
11. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{101ec3c2-699e-4c74-bfdc-d8c610646571-16_572_1338_278_239} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x ) , \quad x \in \mathbb { R }\) The curve meets the coordinate axes at the points \(A ( 0 , - 3 )\) and \(B \left( - \frac { 1 } { 3 } \ln 4,0 \right)\) and the curve has an asymptote with equation \(y = - 4\) In separate diagrams, sketch the graph with equation
  1. \(y = | f ( x ) |\)
  2. \(y = 2 \mathrm { f } ( x ) + 6\) On each sketch, give the exact coordinates of the points where the curve crosses or meets the coordinate axes and the equation of any asymptote. Given that $$\begin{array} { l l } \mathrm { f } ( x ) = \mathrm { e } ^ { - 3 x } - 4 , & x \in \mathbb { R } \\ \mathrm {~g} ( x ) = \ln \left( \frac { 1 } { x + 2 } \right) , & x > - 2 \end{array}$$
  3. state the range of f,
  4. find \(\mathrm { f } ^ { - 1 } ( x )\),
  5. express \(f g ( x )\) as a polynomial in \(x\).
Edexcel C34 2016 January Q12
14 marks Standard +0.3
  1. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations
$$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 12 \\ - 4 \\ 5 \end{array} \right) + \lambda \left( \begin{array} { r } 5 \\ - 4 \\ 2 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { l } 2 \\ 2 \\ 0 \end{array} \right) + \mu \left( \begin{array} { l } 0 \\ 6 \\ 3 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters.
  1. Show that \(l _ { 1 }\) and \(l _ { 2 }\) meet, and find the position vector of their point of intersection \(A\).
  2. Find, to the nearest \(0.1 ^ { \circ }\), the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\) The point \(B\) has position vector \(\left( \begin{array} { l } 7 \\ 0 \\ 3 \end{array} \right)\).
  3. Show that \(B\) lies on \(l _ { 1 }\)
  4. Find the shortest distance from \(B\) to the line \(l _ { 2 }\), giving your answer to 3 significant figures.
Edexcel C34 2016 January Q13
14 marks Standard +0.3
13. A curve \(C\) has parametric equations $$x = 6 \cos 2 t , \quad y = 2 \sin t , \quad - \frac { \pi } { 2 } < t < \frac { \pi } { 2 }$$
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \lambda \operatorname { cosec } t\), giving the exact value of the constant \(\lambda\).
  2. Find an equation of the normal to \(C\) at the point where \(t = \frac { \pi } { 3 }\) Give your answer in the form \(y = m x + c\), where \(m\) and \(c\) are simplified surds. The cartesian equation for the curve \(C\) can be written in the form $$x = f ( y ) , \quad - k < y < k$$ where \(\mathrm { f } ( y )\) is a polynomial in \(y\) and \(k\) is a constant.
  3. Find \(\mathrm { f } ( y )\).
  4. State the value of \(k\).
Edexcel C34 2017 January Q1
6 marks Standard +0.3
  1. Find an equation of the tangent to the curve
$$x ^ { 3 } + 3 x ^ { 2 } y + y ^ { 3 } = 37$$ at the point \(( 1,3 )\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
(6)
Edexcel C34 2017 January Q2
7 marks Moderate -0.3
2. $$f ( x ) = x ^ { 3 } - 5 x + 16$$
  1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be rewritten as $$x = ( a x + b ) ^ { \frac { 1 } { 3 } }$$ giving the values of the constants \(a\) and \(b\). The equation \(\mathrm { f } ( x ) = 0\) has exactly one real root \(\alpha\), where \(\alpha = - 3\) to one significant figure.
  2. Starting with \(x _ { 1 } = - 3\), use the iteration $$x _ { n + 1 } = \left( a x _ { n } + b \right) ^ { \frac { 1 } { 3 } }$$ with the values of \(a\) and \(b\) found in part (a), to calculate the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving all your answers to 3 decimal places.
  3. Using a suitable interval, show that \(\alpha = - 3.17\) correct to 2 decimal places.
Edexcel C34 2017 January Q3
9 marks Standard +0.3
3.
  1. Express \(\frac { 9 + 11 x } { ( 1 - x ) ( 3 + 2 x ) }\) in partial fractions.
  2. Hence, or otherwise, find the series expansion of $$\frac { 9 + 11 x } { ( 1 - x ) ( 3 + 2 x ) } , \quad | x | < 1$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).
    Give each coefficient as a simplified fraction.
Edexcel C34 2017 January Q4
10 marks Standard +0.3
  1. Given that
$$\begin{array} { l l } \mathrm { f } ( x ) = \frac { 4 } { 3 x + 5 } , & x > 0 \\ \mathrm {~g} ( x ) = \frac { 1 } { x } , & x > 0 \end{array}$$
  1. state the range of f,
  2. find \(\mathrm { f } ^ { - 1 } ( x )\),
  3. find \(\mathrm { fg } ( x )\).
  4. Show that the equation \(\mathrm { fg } ( x ) = \mathrm { gf } ( x )\) has no real solutions.
Edexcel C34 2017 January Q5
9 marks Moderate -0.3
5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e30f0c28-1695-40a1-8e9a-6ea7e29042bf-08_579_1038_258_452} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of part of the curve \(C\) with equation $$y = x \cos x , \quad x \in \mathbb { R }$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve \(C\) and the \(x\)-axis for \(\frac { 3 \pi } { 2 } \leqslant x \leqslant \frac { 5 \pi } { 2 }\)
  1. Complete the table below with the exact value of \(y\) corresponding to \(x = \frac { 7 \pi } { 4 }\) and with the exact value of \(y\) corresponding to \(x = \frac { 9 \pi } { 4 }\)
    \(x\)\(\frac { 3 \pi } { 2 }\)\(\frac { 7 \pi } { 4 }\)\(2 \pi\)\(\frac { 9 \pi } { 4 }\)\(\frac { 5 \pi } { 2 }\)
    \(y\)0\(2 \pi\)0
  2. Use the trapezium rule, with all five \(y\) values in the completed table, to find an approximate value for the area of \(R\), giving your answer to 4 significant figures.
  3. Find $$\int x \cos x d x$$
  4. Using your answer from part (c), find the exact area of the region \(R\).
Edexcel C34 2017 January Q6
6 marks Standard +0.3
  1. Differentiate \(y = 5 x ^ { 2 } \ln 3 x , \quad x > 0\)
  2. Given that $$y = \frac { x } { \sin x + \cos x } , \quad - \frac { \pi } { 4 } < x < \frac { 3 \pi } { 4 }$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 1 + x ) \sin x + ( 1 - x ) \cos x } { 1 + \sin 2 x } , \quad - \frac { \pi } { 4 } < x < \frac { 3 \pi } { 4 }$$ \includegraphics[max width=\textwidth, alt={}, center]{e30f0c28-1695-40a1-8e9a-6ea7e29042bf-11_99_104_2631_1781}