Find the exact value of each of the zeros of \(f\).

Expand the expression for \(f(x)\).

Find \(f’(x)\).

Use your answer to part (b)(ii) to find the values of \(x\) for which \(f\) is increasing.

Draw the graph of \(f\) for \( - 3 \leqslant x \leqslant 3\) and \( - 40 \leqslant y \leqslant 20\). Use a scale of 2 cm to represent 1 unit on the \(x\)-axis and 1 cm to represent 5 units on the \(y\)-axis.

Write down the coordinates of the point of intersection.

\( - 1,{\text{ }}\sqrt 5 ,{\text{ }} - \sqrt 5 \)     (A1)(A1)(A1)

Note:     Award (A1) for –1 and each exact value seen. Award at most (A1)(A0)(A1) for use of 2.23606… instead of \(\sqrt 5 \).

[3 marks]

\(10x - 2{x^3} + 10 - 2{x^2}\)     (A1)

Notes:     The expansion may be seen in part (b)(ii).

[1 mark]

\(10 - 6{x^2} - 4x\)     (A1)(ft)(A1)(ft)(A1)(ft)

Notes:     Follow through from part (b)(i). Award (A1)(ft) for each correct term. Award at most (A1)(ft)(A1)(ft)(A0) if extra terms are seen.

[3 marks]

\(10 - 6{x^2} - 4x > 0\)     (M1)

Notes:     Award (M1) for their \(f’(x) > 0\). Accept equality or weak inequality.

\( - 1.67 < x < 1{\text{ }}\left( { - \frac{5}{3} < x < 1,{\text{ }} - 1.66666 \ldots  < x < 1} \right)\)     (A1)(ft)(A1)(ft)(G2)

Notes:     Award (A1)(ft) for correct endpoints, (A1)(ft) for correct weak or strict inequalities. Follow through from part (b)(ii). Do not award any marks if there is no answer in part (b)(ii).

[3 marks]

     (A1)(A1)(ft)(A1)(ft)(A1)

Notes:     Award (A1) for correct scale; axes labelled and drawn with a ruler.

Award (A1)(ft) for their correct \(x\)-intercepts in approximately correct location.

Award (A1) for correct minimum and maximum points in approximately correct location.

Award (A1) for a smooth continuous curve with approximate correct shape. The curve should be in the given domain.

Follow through from part (a) for the \(x\)-intercepts.

[4 marks]

\((1.49,{\text{ }}13.9){\text{ }}\left( {(1.48702 \ldots ,{\text{ }}13.8714 \ldots )} \right)\)     (G1)(ft)(G1)(ft)

Notes:     Award (G1) for 1.49 and (G1) for 13.9 written as a coordinate pair. Award at most (G0)(G1) if parentheses are missing. Accept \(x = 1.49\) and \(y = 13.9\). Follow through from part (b)(i).

[2 marks]

Write down the value of \(c\).

Find the value of \(a\).

Hence write down the equation of the quadratic function which models the edge of the water tank.

The water tank is shown below. It is partially filled with water.

Calculate the value of y when \(x = 2.4{\text{ m}}\).

The water tank is shown below. It is partially filled with water.

State what the value of \(x\) and the value of \(y\) represent for this water tank.

The water tank is shown below. It is partially filled with water.

Find the value of \(x\) when the height of water in the tank is \(0.9\) m.

The water tank is shown below. It is partially filled with water.

The water tank has a length of 5 m.

When the water tank is filled to a height of \(0.9\) m, the front cross-sectional area of the water is \({\text{2.55 }}{{\text{m}}^2}\).

(i)     Calculate the volume of water in the tank.

The total volume of the tank is \({\text{36 }}{{\text{m}}^3}\).

(ii)     Calculate the percentage of water in the tank.

\(0\)     (A1)(G1)

[1 mark]

\(1.8 = a{(3)^2} + 0\)     (M1)

OR

\(1.8 = a{( - 3)^2} + 0\)     (M1)

Note: Award (M1) for substitution of \(y = 1.8\) or \(x = 3\) and their value of \(c\) into equation. \(0\) may be implied.

\(a = 0.2\)   \(\left( {\frac{1}{5}} \right)\)     (A1)(ft)(G1)

Note: Follow through from their answer to part (a).

     Award (G1) for a correct answer only.

[2 marks]

\(y = 0.2{x^2}\)     (A1)(ft)

Note: Follow through from their answers to parts (a) and (b).

     Answer must be an equation.

[1 mark]

\(0.2 \times {(2.4)^2}\)     (M1)

\( = 1.15{\text{ (m)}}\)   \((1.152)\)     (A1)(ft)(G1)

Notes: Award (M1) for correctly substituted formula, (A1) for correct answer. Follow through from their answer to part (c).

     Award (G1) for a correct answer only.

[2 marks]

\(y\) is the height     (A1)

positive value of \(x\) is half the width (or equivalent)     (A1)

[2 marks]

\(0.9 = 0.2{x^2}\)     (M1)

Note: Award (M1) for setting their equation equal to \(0.9\).

\(x =  \pm 2.12{\text{ (m)}}\)   \(\left( { \pm \frac{3}{2}\sqrt 2 ,{\text{ }} \pm \sqrt {4.5} ,{\text{ }} \pm {\text{2.12132}} \ldots } \right)\)     (A1)(ft)(G1)

Note: Accept \(2.12\). Award (G1) for a correct answer only.

[2 marks]

(i)     \(2.55 \times 5\)     (M1)

Note: Award (M1) for correct substitution in formula.

\( = 12.8{\text{ (}}{{\text{m}}^3}{\text{)}}\)   \(\left( {{\text{12.75 (}}{{\text{m}}^3}{\text{)}}} \right)\)     (A1)(G2)

[2 marks]

(ii)     \(\frac{{12.75}}{{36}} \times 100\)     (M1)

Note: Award (M1) for correct quotient multiplied by \(100\).

\( = 35.4 (\%)\)  \((35.4166 \ldots )\)     (A1)(ft)(G2)

Note: Award (G2) for \(35.6 (\%) (35.5555… (\%))\).

     Follow through from their answer to part (g)(i).

[2 marks]

Write down an expression for \(f ′(x)\).

Consider the graph of f. The graph of f passes through the point P(1, 4).

Find the value of c.

There is a local minimum at the point Q.

Find the coordinates of Q.

There is a local minimum at the point Q.

Find the set of values of x for which the function is decreasing.

Let T be the tangent to the graph of f at P.

Show that the gradient of T is –7.

Let T be the tangent to the graph of f at P.

Find the equation of T.

T intersects the graph again at R. Use your graphic display calculator to find the coordinates of R.

\(f'(x) = \frac{{ - 10}}{{{x^3}}} + 3\)     (A1)(A1)(A1)(A1)

Note: Award (A1) for −10, (A1) for x³ (or x⁻³), (A1) for 3, (A1) for no other constant term.

[4 marks]

4 = 5 + 3 + c     (M1)

Note: Award (M1) for substitution in f (x).

c = −4     (A1)(G2)

[2 marks]

\(f '(x) = 0\)     (M1)

\(0 = \frac{{ - 10}}{{{x^3}}} + 3\)     (A1)(ft)

(1.49, 2.72)   (accept x = 1.49  y = 2.72)     (A1)(ft)(A1)(ft)(G3)

Notes: If answer is given as (1.5, 2.7) award (A0)(AP)(A1).

Award at most (M1)(A1)(A1)(A0) if parentheses not included. (ft) from their (a).

If no working shown award (G2)(G0) if parentheses are not included.

OR

Award (M2) for sketch, (A1)(ft)(A1)(ft) for correct coordinates. (ft) from their (b).     (M2)(A1)(ft)(A1)(ft)

Note: Award at most (M2)(A1)(ft)(A0) if parentheses not included.

[4 marks]

0 < x < 1.49  OR  0 < x ≤ 1.49     (A1)(A1)(ft)(A1)

Notes: Award (A1) for 0, (A1)(ft) for 1.49 and (A1) for correct inequality signs.

(ft) from their x value in (c) (i).

[3 marks]

For P(1, 4) \(f '(1) = - 10 + 3\)     (M1)(A1)

\(= -7\)     (AG)

Note: Award (M1) for substituting \(x = 1\) into their \(f '(x)\). (A1) for \(-10 + 3\).

\(-7\) must be seen for (A1) to be awarded.

[2 marks]

\(4 = -7 \times 1 + c\)     \(11 = c\)     (A1)

\(y = -7 x + 11\)     (A1)

[2 marks]

Point of intersection is R(−0.5, 14.5)     (A1)(ft)(A1)(ft)(G2)(ft)

Notes: Award (A1) for the x coordinate, (A1) for the y coordinate.

Allow (ft) from candidate’s (d)(ii) equation and their (b) even with no working seen.

Award (A1)(ft)(A0) if brackets not included and not previously penalised.

[2 marks]

A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.

Most candidates could score 4 marks.

A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.

A good number of candidates correctly substituted into the original function.

A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.

Very few managed to answer this part algebraically. Those candidates who were aware that they could read the values from their GDC gained some easy marks.

A large number of candidates were not able to answer this question well because of the firstterm of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.

Proved to be difficult for many candidates as increasing/decreasing intervals were not well understood by many.

A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.

As this part was linked to part (a), candidates who could not find the correct derivative could not show that the gradient of T was −7. Many candidates did not realize that they had to substitute into the first derivative. For those who did, finding the equation of T was a simple task.

A large number of candidates were not able to answer this question well because of the firstterm of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.

As this part was linked to part (a), candidates who could not find the correct derivative could not show that the gradient of T was −7. Many candidates did not realize that they had to substitute into the first derivative. For those who did, finding the equation of T was a simple task.

A large number of candidates were not able to answer this question well because of the first term of the function’s negative index. Many candidates who did make an attempt lacked understanding of the topic.

For those who had part (d) (ii) correct, this allowed them to score easy marks. For the others, it proved a difficult task because their equation in (d) would not be a tangent.

Sketch the graph of y = 2^x for \( - 2 \leqslant x \leqslant 3\). Indicate clearly where the curve intersects the y-axis.

Write down the equation of the asymptote of the graph of y = 2^x.

On the same axes sketch the graph of y = 3 + 2x − x². Indicate clearly where this curve intersects the x and y axes.

Using your graphic display calculator, solve the equation 3 + 2x − x² = 2^x.

Write down the maximum value of the function f (x) = 3 + 2x − x².

Use Differential Calculus to verify that your answer to (e) is correct.

The curve y = px² + qx − 4 passes through the point (2, –10).

Use the above information to write down an equation in p and q.

The gradient of the curve \(y = p{x^2} + qx - 4\) at the point (2, –10) is 1.

Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).

The gradient of the curve \(y = p{x^2} + qx - 4\) at the point (2, –10) is 1.

Hence, find a second equation in p and q.

The gradient of the curve \(y = p{x^2} + qx - 4\) at the point (2, –10) is 1.

Solve the equations to find the value of p and of q.

     (A1)(A1)(A1)

Note: Award (A1) for correct domain, (A1) for smooth curve, (A1) for y-intercept clearly indicated.

[3 marks]

y = 0     (A1)(A1)

Note: Award (A1) for y = constant, (A1) for 0.

[2 marks]

Note: Award (A1) for smooth parabola,

(A1) for vertex (maximum) in correct quadrant.

(A1) for all clearly indicated intercepts x = −1, x = 3 and y = 3.

The final mark is to be applied very strictly.     (A1)(A1)(A1)

[3 marks]

x = −0.857   x = 1.77     (G1)(G1)

Note: Award a maximum of (G1) if x and y coordinates are both given.

[2 marks]

4     (G1)

Note: Award (G0) for (1, 4).

[1 mark]

\(f'(x) = 2 - 2x\)     (A1)(A1)

Note: Award (A1) for each correct term.

Award at most (A1)(A0) if any extra terms seen.

\(2 - 2x = 0\)     (M1)

Note: Award (M1) for equating their gradient function to zero.

\(x = 1\)     (A1)(ft)

\(f (1) = 3 + 2(1) - (1)^2 = 4\)     (A1)

Note: The final (A1) is for substitution of x = 1 into \(f (x)\) and subsequent correct answer. Working must be seen for final (A1) to be awarded.

[5 marks]

2² × p + 2q − 4 = −10     (M1)

Note: Award (M1) for correct substitution in the equation.

4p + 2q = −6     or     2p + q = −3     (A1)

Note: Accept equivalent simplified forms.

[2 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2px + q\)     (A1)(A1)

Note: Award (A1) for each correct term.

Award at most (A1)(A0) if any extra terms seen.

[2 marks]

4p + q = 1     (A1)(ft)

[1 mark]

4p + 2q = −6

4p + q = 1     (M1)

Note: Award (M1) for sensible attempt to solve the equations.

p = 2, q = −7     (A1)(A1)(ft)(G3)

[3 marks]

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

The most common error was using the incorrect domain.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

Many had little idea of asymptotes. Others did not write their answer as an equation.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

The intercepts being inexact or unlabelled was the most frequent cause of loss of marks.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

Often, only one solution to the equation was given. Elsewhere, a lack of appreciation that the solutions were the x coordinates was a common mistake.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

The maximum is the y coordinate only; again a common misapprehension was the answer “(1, 4)”.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

This was a major discriminator in the paper. Many candidates were unable to follow the analytic approach to finding a maximum point.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

This part was challenging to the majority, with a large number not attempting the question at all. However, there were a pleasing number of correct attempts that showed a fine understanding of the calculus.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

This part was challenging to the majority, with a large number not attempting the question at all. However, there were a pleasing number of correct attempts that showed a fine understanding of the calculus.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

This part was challenging to the majority, with a large number not attempting the question at all. However, there were a pleasing number of correct attempts that showed a fine understanding of the calculus.

Undoubtedly, this question caused the most difficulty in terms of its content. Where there was no alternative to using the calculus, the majority of candidates struggled. However, for those with a sound grasp of the topic, there were many very successful attempts.

This part was challenging to the majority, with a large number not attempting the question at all. However, there were a pleasing number of correct attempts that showed a fine understanding of the calculus.

Calculate \(f(2)\) .

Sketch the graph of the function \(y = f(x)\) for \( - 5 \leqslant x \leqslant 5\) and \( - 200 \leqslant y \leqslant 200\) .

Find \(f'(x)\) .

Find \(f'(2)\) .

Write down the coordinates of the local maximum point on the graph of \(f\) .

Find the gradient of the tangent to the graph of \(f\) at \(x = 1\).

There is a second point on the graph of \(f\) at which the tangent is parallel to the tangent at \(x = 1\).

Find the \(x\)-coordinate of this point.

\(f(2) = {2^3} + \frac{{48}}{2}\)     (M1)

\(= 32\)     (A1)(G2)

[2 marks]

(A1) for labels and some indication of scale in an appropriate window
(A1) for correct shape of the two unconnected and smooth branches
(A1) for maximum and minimum in approximately correct positions
(A1) for asymptotic behaviour at \(y\)-axis     (A4)

Notes: Please be rigorous.
The axes need not be drawn with a ruler.
The branches must be smooth: a single continuous line that does not deviate from its proper direction.
The position of the maximum and minimum points must be symmetrical about the origin.
The \(y\)-axis must be an asymptote for both branches. Neither branch should touch the axis nor must the curve approach the
asymptote then deviate away later.

[4 marks]

\(f'(x) = 3{x^2} - \frac{{48}}{{{x^2}}}\)     (A1)(A1)(A1)

Notes: Award (A1) for \(3{x^2}\) , (A1) for \( - 48\) , (A1) for \({x^{ - 2}}\) . Award a maximum of (A1)(A1)(A0) if extra terms seen.

[3 marks]

\(f'(2) = 3{(2)^2} - \frac{{48}}{{{{(2)}^2}}}\)     (M1)

Note: Award (M1) for substitution of \(x = 2\) into their derivative.

\(= 0\)     (A1)(ft)(G1)

[2 marks]

\(( - 2{\text{, }} - 32)\) or \(x = - 2\), \(y = - 32\)     (G1)(G1)

Notes: Award (G0)(G0) for \(x = - 32\), \(y = - 2\) . Award at most (G0)(G1) if parentheses are omitted.

[2 marks]

\(\{ y \geqslant 32\}  \cup \{ y \leqslant - 32\} \)     (A1)(A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) \(y \geqslant 32\) or \(y > 32\) seen, (A1)(ft) for \(y \leqslant - 32\) or \(y < - 32\) , (A1) for weak (non-strict) inequalities used in both of the above.
Accept use of \(f\) in place of \(y\). Accept alternative interval notation.
Follow through from their (a) and (e).
If domain is given award (A0)(A0)(A0).
Award (A0)(A1)(ft)(A1)(ft) for \([ - 200{\text{, }} - 32]\) , \([32{\text{, }}200]\).
Award (A0)(A1)(ft)(A1)(ft) for \(] - 200{\text{, }} - 32]\) , \([32{\text{, }}200[\).

[3 marks]

\(f'(1) = - 45\)     (M1)(A1)(ft)(G2)

Notes: Award (M1) for \(f'(1)\) seen or substitution of \(x = 1\) into their derivative. Follow through from their derivative if working is seen.

[2 marks]

\(x = - 1\)     (M1)(A1)(ft)(G2)

Notes: Award (M1) for equating their derivative to their \( - 45\) or for seeing parallel lines on their graph in the approximately correct position.

[2 marks]

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

As usual and by intention, this question caused the most difficulty in terms of its content; however, for those with a sound grasp of the topic, there were many very successful attempts. Much of the question could have been answered successfully by using the GDC, however, it was also clear that a number of candidates did not connect the question they were attempting with the curve that they had either sketched or were viewing on their GDC. Where there was no alternative to using the calculus, many candidates struggled.

The majority of sketches were drawn sloppily and with little attention to detail. Teachers must impress on their students that a mathematical sketch is designed to illustrate the main points of a curve – the smooth nature by which it changes, any symmetries (reflectional or rotational), positions of turning points, intercepts with axes and the behaviour of a curve as it approaches an asymptote. There must also be some indication of the dimensions used for the “window”.

Differentiation of terms with negative indices remains a testing process for the majority; it will continue to be tested.

It was also evident that some centres do not teach the differential calculus.

Write down the values of x where the graph of f (x) intersects the x-axis.

Write down f ′(x).

Find the value of the local maximum of y = f (x).

Let P be the point where the graph of f (x) intersects the y axis.

Write down the coordinates of P.

Let P be the point where the graph of f (x) intersects the y axis.

Find the gradient of the curve at P.

The line, L, is the tangent to the graph of f (x) at P.

Find the equation of L in the form y = mx + c.

There is a second point, Q, on the curve at which the tangent to f (x) is parallel to L.

Write down the gradient of the tangent at Q.

There is a second point, Q, on the curve at which the tangent to f (x) is parallel to L.

Calculate the x-coordinate of Q.

–1.10, 0.218, 3.13     (A1)(A1)(A1)

[3 marks]

f ′(x) = 12x² – 18x – 12     (A1)(A1)(A1)

Note: Award (A1) for each correct term and award maximum of (A1)(A1) if other terms seen.

[3 marks]

f ′(x) = 0     (M1)
x = –0.5, 2
x = –0.5     (A1)

Note: If x = –0.5 not stated, can be inferred from working below.

y = 4(–0.5)³ – 9(–0.5)² – 12(–0.5) + 3     (M1)
y = 6.25     (A1)(G3)

Note: Award (M1) for their value of x substituted into f (x).

Award (M1)(G2) if sketch shown as method. If coordinate pair given then award (M1)(A1)(M1)(A0). If coordinate pair given with no working award (G2).

[4 marks]

(0, 3)     (A1)

Note: Accept x = 0, y = 3.

[1 mark]

f ′(0) = –12     (M1)(A1)(ft)(G2)

Note: Award (M1) for substituting x = 0 into their derivative.

[2 marks]

Tangent: y = –12x + 3     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for their gradient, (A1) for intercept = 3.

Award (A1)(A0) if y = not seen.

[2 marks]

–12     (A1)(ft)

Note: Follow through from their part (e).

[1 mark]

12x² – 18x – 12 = –12     (M1)

12x² – 18x = 0     (M1)

x = 1.5, 0

At Q x = 1.5     (A1)(ft)(G2)

Note: Award (M1)(G2) for 12x² – 18x – 12 = –12 followed by x = 1.5.

Follow through from their part (g).

[3 marks]

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

This question was either very well done – by the majority – or very poor and incomplete attempts were seen. This would perhaps indicate a lack of preparation in this area of the syllabus from some centres, though it is recognised that the differential calculus is one of the more problematic topics for the candidature.

It was however disappointing to note the number of candidates who do not use the GDC to good effect; in part (a) for example, the zeros were not found accurately due to “trace” being used; this is not a suitable approach – there is a built-in zero finder which should be used. Much of the question was accessible via a GDC approach, a sketch was given that could have been verified on the GDC; this was lost on many.

Copy the tree diagram below and fill in all the probabilities, where NL represents not late, to represent this information.

Find the probability that Geraldine travels by bus and is late.

Find the probability that Geraldine is late.

Find the probability that Geraldine travelled by train, given that she is late.

Sketch the curve of the function \(f (x) = x^3 − 2x^2 + x − 3\) for values of \(x\) from −2 to 4, giving the intercepts with both axes.

On the same diagram, sketch the line \(y = 7 − 2x\) and find the coordinates of the point of intersection of the line with the curve.

Find the value of the gradient of the curve where \(x = 1.7\) .

Award (A1) for 0.5 at B, (A1) for 0.3 at T, then (A1) for each correct pair. Accept fractions or percentages.     (A5)

[5 marks]

0.06 (accept \(0.5 \times 0.12\) or 6%)     (A1)(ft)

[1 mark]

for a relevant two-factor product, either \(C \times L\) or \(T \times L\)     (M1)

for summing three two-factor products     (M1)

\((0.2 \times 0.05 + 0.06 + 0.3 \times 0.08)\)

0.094     (A1)(ft)(G2)

[3 marks]

\(\frac{{0.3 \times 0.08}}{{0.094}}\)     (M1)(A1)(ft)

award (M1) for substituted conditional probability formula seen, (A1)(ft) for correct substitution

= 0.255     (A1)(ft)(G2)

[3 marks]

     (G3)

[3 marks]

line drawn with –ve gradient and +ve y-intercept     (G1)

(2.45, 2.11)     (G1)(G1)

[3 marks]

\(f ' (1.7) = 3(1.7)^2 - 4(1.7) + 1\)     (M1)

award (M1) for substituting in their \(f' (x)\)

2.87     (A1)(G2)

[2 marks]

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.

Find the initial temperature of the coffee.

Write down the equation of the horizontal asymptote.

Find the room temperature.

Find the temperature of the coffee after 10 minutes.

Find the temperature of Robert’s coffee after being heated in the microwave for 30 seconds after it has reached the temperature in part (d).

Calculate the length of time it would take a similar cup of coffee, initially at 20°C, to be heated in the microwave to reach 100°C.

Calculate correct to 2 decimal places the amount of euros he receives.

He buys 1 kilogram of Belgian chocolates at 1.35 euros per 100 g.

Calculate the cost of his chocolates in GBP correct to 2 decimal places.

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

(UP) 90°C     (A1)

[1 mark]

y = 16     (A1)

[1 mark]

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

(UP) 16°C (ft) from answer to part (b)     (A1)(ft)

[1 mark]

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

(UP) 25.4°C     (A1)

[1 mark]  

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

for seeing 2^0.75 or equivalent     (A1)

for multiplying their (d) by their 2^0.75     (M1)

(UP) 42.8°C     (A1)(ft)(G2)

[3 marks]

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

for seeing  \(20 \times 2^{1.5t} = 100\)     (A1)

for seeing a value of t between 1.54 and 1.56 inclusive     (M1)(A1)

(UP) 1.55 minutes or 92.9 seconds     (A1)(G3)

[4 marks]

Financial accuracy penalty (FP) is applicable in part (ii) only.

\(120 - 3 = 117\)

(FP) \(117 \times 1.37\)     (A1)

= 160.29 euros (correct answer only)     (M1)

first (A1) for 117 seen, (M1) for multiplying by 1.37     (A1)(G2)

[3 marks]

Financial accuracy penalty (FP) is applicable in part (ii) only.

(FP) \(\frac{{13.5}}{{1.37}}\)     (A1)(M1)

9.85 GBP (answer correct to 2dp only)

first (A1) is for 13.5 seen, (M1) for dividing by 1.37     (A1)(ft)(G3)

[3 marks]

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

The financial part was the best done question in the paper and a large majority of candidates gained full marks here.

The financial part was the best done question in the paper and a large majority of candidates gained full marks here.

Given that \(f(1) = 2\), show that \(k = 4\).

Write down the values of \(q\) and \(r\) for the following table.

As \(x\) increases from \( - 1\), the graph of \(y = f(x)\) reaches a maximum value and then decreases, behaving asymptotically.

Draw the graph of \(y = f(x)\) for \( - 1 \leqslant x \leqslant 8\). Use a scale of \({\text{1 cm}}\) to represent 1 unit on both axes. The position of the maximum, \({\text{M}}\), the \(y\)-intercept and the asymptotic behaviour should be clearly shown.

Using your graphic display calculator, find the coordinates of \({\text{M}}\), the maximum point on the graph of \(y = f(x)\).

Write down the equation of the horizontal asymptote to the graph of \(y = f(x)\).

(i) Draw and label the line \( y = 1\) on your graph.

(ii) The equation \(f(x) = 1\) has two solutions. One of the solutions is \(x = 4\). Use your graph to find the other solution.

Find \(C'(x)\).

Show that the cost per person is a minimum when \(10\) people are invited to the party.

Calculate the minimum cost per person.

\(f(1) = \frac{k}{{{2^1}}}\)     (M1)

Note: (M1) for substituting \(x = 1\) into the formula.

\(\frac{k}{2} = 2\)     (M1)

Note: (M1) for equating to 2.

\(k = 4\)     (AG)

[2 marks]

\(q = 2\), \(r = 0.125\)     (A1)(A1)

[2 marks]

     (A4)

Notes: (A1) for scales and labels.
(A1) for accurate smooth curve passing through \((0, 0)\) drawn at least in the given domain.
(A1) for asymptotic behaviour (curve must not go up or cross the \(x\)-axis).
(A1) for indicating the position of the maximum point.

[4 marks]

\({\text{M}}\) (\(1.44\), \(2.12\))     (G1)(G1)

Note: Brackets required, if missing award (G1)(G0). Accept \(x = 1.44\) and \(y = 2.12\).

[2 marks]

\(y = 0\)     (A1)(A1)

Note: (A1) for ‘\(y = \)’ provided the right hand side is a constant. (A1) for 0.

[2 marks]

(i) See graph     (A1)(A1)

Note: (A1) for correct line, (A1) for label.

(ii) \(x = 0.3\) (ft) from candidate’s graph.     (A2)(ft)

Notes: Accept \( \pm 0.1\) from their x. For \(0.310\) award (G1)(G0). For other answers taken from the GDC and not given correct to 3 significant figures award (G0)(AP)(G0) or (G1)(G0) if (AP) already applied.

[4 marks]

\(C'(x) = 1 - \frac{{100}}{{{x^2}}}\)     (A1)(A1)(A1)

Note: (A1) for 1, (A1) for \( - 100\) , (A1) for \({x^2}\) as denominator or \({{x^{ - 2}}}\) as numerator. Award a maximum of (A2) if an extra term is seen.

[3 marks]

For studying signs of the derivative at either side of \(x = 10\)     (M1)

For saying there is a change of sign of the derivative     (M1)(AG)

OR

For putting \(x = 10\) into \(C'\) and getting zero     (M1)

For clear sketch of the function or for mentioning that the function changes from decreasing to increasing at \(x = 10\)     (M1)(AG)

OR

For solving \(C'(x) = 0\) and getting \(10\)     (M1)

For clear sketch of the function or for mentioning that the function changes from decreasing to increasing at \(x = 10\)     (M1)(AG)

Note: For a sketch with a clear indication of the minimum or for a table with values of \(x\) at either side of \(x = 10\) award (M1)(M0).

[2 marks]

\(C(10) = 10 + \frac{{100}}{{10}}\)     (M1)

\(C(10) = 20\)     (A1)(G2)

[2 marks]

There were many well drawn graphs using correctly scaled and labelled axes with a good curve drawn. A number of students did not label the maximum point. Although many students showed in their graph the asymptotic behaviour of the curve, they did not know how to describe the asymptote. It was noticed that some students were tracing the curve to find the coordinates of the maximum instead of finding the maximum directly. The intersection between the line \(y = 1\) and the curve was not always read from their graph but from their GDC's graph.

There were many well drawn graphs using correctly scaled and labelled axes with a good curve drawn. A number of students did not label the maximum point. Although many students showed in their graph the asymptotic behaviour of the curve, they did not know how to describe the asymptote. It was noticed that some students were tracing the curve to find the coordinates of the maximum instead of finding the maximum directly. The intersection between the line \(y = 1\) and the curve was not always read from their graph but from their GDC's graph.

There were many well drawn graphs using correctly scaled and labelled axes with a good curve drawn. A number of students did not label the maximum point. Although many students showed in their graph the asymptotic behaviour of the curve, they did not know how to describe the asymptote. It was noticed that some students were tracing the curve to find the coordinates of the maximum instead of finding the maximum directly. The intersection between the line \(y = 1\) and the curve was not always read from their graph but from their GDC's graph.

There were many well drawn graphs using correctly scaled and labelled axes with a good curve drawn. A number of students did not label the maximum point. Although many students showed in their graph the asymptotic behaviour of the curve, they did not know how to describe the asymptote. It was noticed that some students were tracing the curve to find the coordinates of the maximum instead of finding the maximum directly. The intersection between the line \(y = 1\) and the curve was not always read from their graph but from their GDC's graph.

There were many well drawn graphs using correctly scaled and labelled axes with a good curve drawn. A number of students did not label the maximum point. Although many students showed in their graph the asymptotic behaviour of the curve, they did not know how to describe the asymptote. It was noticed that some students were tracing the curve to find the coordinates of the maximum instead of finding the maximum directly. The intersection between the line \(y = 1\) and the curve was not always read from their graph but from their GDC's graph.

There were many well drawn graphs using correctly scaled and labelled axes with a good curve drawn. A number of students did not label the maximum point. Although many students showed in their graph the asymptotic behaviour of the curve, they did not know how to describe the asymptote. It was noticed that some students were tracing the curve to find the coordinates of the maximum instead of finding the maximum directly. The intersection between the line \(y = 1\) and the curve was not always read from their graph but from their GDC's graph.

Finding the derivative was done at least partially correctly by most of the candidates. However, using it to find the minimum and to justify why it is a minimum was troublesome for the majority of the candidates. Even those who used a graph in their reasoning neglected to mention the change from decreasing to increasing or to supply a sign diagram. Many candidates recovered in the last part of the question when finding the minimum cost.

Finding the derivative was done at least partially correctly by most of the candidates. However, using it to find the minimum and to justify why it is a minimum was troublesome for the majority of the candidates. Even those who used a graph in their reasoning neglected to mention the change from decreasing to increasing or to supply a sign diagram. Many candidates recovered in the last part of the question when finding the minimum cost.

Finding the derivative was done at least partially correctly by most of the candidates. However, using it to find the minimum and to justify why it is a minimum was troublesome for the majority of the candidates. Even those who used a graph in their reasoning neglected to mention the change from decreasing to increasing or to supply a sign diagram. Many candidates recovered in the last part of the question when finding the minimum cost.

Sketch the curve for −1 < x < 3 and −2 < y < 12.

A teacher asks her students to make some observations about the curve.

Three students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.

State the name of the student who made an incorrect observation.

Find the value of y when x = 1 .

Find \(\frac{{{\text{dy}}}}{{{\text{dx}}}}\).

Show that the stationary points of the curve are at x = 1 and x = 2.

Given that y = 2x³ − 9x² + 12x + 2 = k has three solutions, find the possible values of k.

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1^st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.

[4 marks]

Rick     (A1)

Note: Award (A0) if extra names stated.

[1 mark]

2(1)³ − 9(1)² + 12(1) + 2     (M1)

Note: Award (M1) for correct substitution into equation.

= 7     (A1)(G2)

[2 marks]

6x² − 18x + 12     (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

6x² − 18x + 12 = 0     (M1)

Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).

6(x  − 1)(x − 2) = 0  (or equivalent)      (M1)

Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.

Award (M0)(M0) for substitution of 1 and of 2 in their derivative.

x = 1, x = 2 (AG)

[2 marks]

6 < k < 7     (A1)(A1)(ft)(A1)

Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).

[3 marks]

Find the time, in seconds, Paco takes to run his fifth lap.

Paco runs his last lap in 260 seconds.

Find how many laps he has run on Monday.

Find the total time, in minutes, run by Paco on Monday.

On Wednesday Paco takes Lola to train. They both run the first lap of the track in 120 seconds. Each lap Lola runs takes 1.06 times as long as her previous lap.

Find the time, in seconds, Lola takes to run her third lap.

Find the total time, in seconds, Lola takes to run her first four laps.

Each lap Paco runs again takes him 10 seconds longer than his previous lap. After a certain number of laps Paco takes less time per lap than Lola.

Find the number of the lap when this happens.

\(120 + 10 \times 4\)     (M1)(A1)

Notes: Award (M1) for substituted AP formula, (A1) for correct substitutions. Accept a list of 4 correct terms.

= 160     (A1)(G3)

\(120 + (n - 1) \times 10 = 260\)     (M1)(M1)

Notes: Award (M1) for correctly substituted AP formula, (M1) for equating to 260. Accept a list of correct terms showing at least the 14^th and 15^th terms.

= 15     (A1)(G2)

\(\frac{{15}}{2}(120 + 260)\) or \(\frac{{15}}{2}(2 \times 120 + (15 - 1) \times 10)\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted AP sum formula, (A1)(ft) for correct substitutions. Accept a sum of a list of 15 correct terms. Follow through from their answer to part (b).

2850 seconds     (A1)(ft)(G2)

Note: Award (G2) for 2850 seen with no working shown.

47.5 minutes     (A1)(ft)(G3)

Notes: A final (A1)(ft) can be awarded for correct conversion from seconds into minutes of their incorrect answer. Follow through from their answer to part (b).

\(120 \times {1.06^{3 - 1}}\)     (M1)(A1)

Notes: Award (M1) for substituted GP formula, (A1) for correct substitutions. Accept a list of 3 correct terms.

= 135 (134.832)     (A1)(G2)

\({S_4} = \frac{{120({{1.06}^4} - 1)}}{{(1.06 - 1)}}\)     (M1)(A1)

Notes: Award (M1) for substituted GP sum formula, (A1) for correct substitutions. Accept a sum of a list of 4 correct terms.

= 525 (524.953...)     (A1)(G2)

\(120 + (n - 1) \times 10 < 120 \times {1.06^{n - 1}}\)     (M1)(M1)

Notes: Award (M1) for correct left hand side, (M1) for correct right hand side. Accept an equation. Follow through from their expressions given in parts (a) and (d).

OR

List of at least 2 terms for both sequences (120, 130, … and 120, 127.2, …)     (M1)

List of correct 12^th and 13^th terms for both sequences (..., 230, 240 and …, 227.8, 241.5)     (M1)

OR

A sketch with a line and an exponential curve,     (M1)

An indication of the correct intersection point     (M1)

13^th lap     (A1)(ft)(G2)

Note: Do not award the final (A1)(ft) if final answer is not a positive integer.

Write down the \(y\)-intercept of the graph.

Find \(f'(x)\).

Show that \(a = 8\).

Find \(f(2)\).

Write down the \(x\)-coordinates of these two points;

Write down the intervals where the gradient of the graph of \(y = f(x)\) is positive.

Write down the range of \(f(x)\).

Write down the number of possible solutions to the equation \(f(x) = 5\).

The equation \(f(x) = m\), where \(m \in \mathbb{R}\), has four solutions. Find the possible values of \(m\).

5     (A1)

Note:     Accept an answer of \((0,{\text{ }}5)\).

[1 mark]

\(\left( {f'(x) = } \right) - 4{x^3} + 2ax\)     (A1)(A1)

Note:     Award (A1) for \( - 4{x^3}\) and (A1) for \( + 2ax\). Award at most (A1)(A0) if extra terms are seen.

[2 marks]

\( - 4 \times {2^3} + 2a \times 2 = 0\)     (M1)(M1)

Note:     Award (M1) for substitution of \(x = 2\) into their derivative, (M1) for equating their derivative, written in terms of \(a\), to 0 leading to a correct answer (note, the 8 does not need to be seen).

\(a = 8\)     (AG)

[2 marks]

\(\left( {f(2) = } \right) - {2^4} + 8 \times {2^2} + 5\)     (M1)

Note:     Award (M1) for correct substitution of \(x = 2\) and  \(a = 8\) into the formula of the function.

21     (A1)(G2)

[2 marks]

\((x = ){\text{ }} - 2,{\text{ }}(x = ){\text{ 0}}\)     (A1)(A1)

Note:     Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as \(( - 2{\text{ }},21)\) and \((0,{\text{ }}5)\) or \(( - 2,{\text{ }}0)\) and \((0,{\text{ }}0)\).

[2 marks]

\(x <  - 2,{\text{ }}0 < x < 2\)     (A1)(ft)(A1)(ft)

Note:     Award (A1)(ft) for \(x <  - 2\), follow through from part (d)(i) provided their value is negative.

Award (A1)(ft) for \(0 < x < 2\), follow through only from their 0 from part (d)(i); 2 must be the upper limit.

Accept interval notation.

[2 marks]

\(y \leqslant 21\)     (A1)(ft)(A1)

Notes:     Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “\(y \leqslant \)”.

Accept interval notation.

Accept \( - \infty  < y \leqslant 21\) or \(f(x) \leqslant 21\).

Follow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if \(x\) is seen instead of \(y\). Do not award the second (A1) if a (finite) lower limit is seen.

[2 marks]

3 (solutions)     (A1)

[1 mark]

\(5 < m < 21\) or equivalent     (A1)(ft)(A1)

Note:     Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).

Accept interval notation.

[2 marks]

Write down a formula for \(A\), the surface area to be coated.

Express this volume in \({\text{c}}{{\text{m}}^3}\).

Write down, in terms of \(r\) and \(h\), an equation for the volume of this water container.

Show that \(A = \pi {r^2}\frac{{1\,000\,000}}{r}\).

Show that \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\).

Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).

Using your answer to part (e), find the value of \(r\) which minimizes \(A\).

Find the value of this minimum area.

Find the least number of cans of water-resistant material that will coat the area in part (g).

\((A = ){\text{ }}\pi {r^2} + 2\pi rh\)    (A1)(A1)

Note:     Award (A1) for either \(\pi {r^2}\) OR \(2\pi rh\) seen. Award (A1) for two correct terms added together.

[2 marks]

\(500\,000\)    (A1)

Notes:     Units not required.

[1 mark]

\(500\,000 = \pi {r^2}h\)    (A1)(ft)

Notes:     Award (A1)(ft) for \(\pi {r^2}h\) equating to their part (b).

Do not accept unless \(V = \pi {r^2}h\) is explicitly defined as their part (b).

[1 mark]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

\(2\pi r - \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}\)    (A1)(A1)(A1)

Note:     Award (A1) for \(2\pi r\), (A1) for \(\frac{1}{{{r^2}}}\) or \({r^{ - 2}}\), (A1) for \( - {\text{1}}\,{\text{000}}\,{\text{000}}\).

[3 marks]

\(2\pi r - \frac{{1\,000\,000}}{{{r^2}}} = 0\)    (M1)

Note:     Award (M1) for equating their part (e) to zero.

\({r^3} = \frac{{1\,000\,000}}{{2\pi }}\) OR \(r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}\)     (M1)

Note:     Award (M1) for isolating \(r\).

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

\((r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )\)    (A1)(ft)(G2)

[3 marks]

\(\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}\)    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

\( = 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )\)    (A1)(ft)(G2)

[2 marks]

\(\frac{{27\,679.0 \ldots }}{{2000}}\)    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

\( = 13.8395 \ldots \)    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their \(13.8395 \ldots \) to the next integer.

[3 marks]

Find the value of \({u_{20}}\).

Find the sum of the first 10 terms of the sequence.

Now consider the sequence \({v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots \) where

\[{v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24\]

This sequence continues in the same manner.

Find the exact value of \({v_{10}}\).

Now consider the sequence \({v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots \) where

\[{v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24\]

This sequence continues in the same manner.

Find the sum of the first 8 terms of this sequence.

\(k\) is the smallest value of \(n\) for which \({v_n}\) is greater than \({u_n}\).

Calculate the value of \(k\).

\(600 + (20 - 1) \times 17\)     (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 20 correct terms seen.

\( = 923\)     (A1)(G3)

[3 marks]

\(\frac{{10}}{2}\left[ {2 \times 600 + (10 - 1) \times 17} \right]\)     (M1)(A1)

Note: Award (M1) for substituted arithmetic series formula, (A1) for their correct substitutions. Follow through from part (a). For consistent use of geometric series formula in part (b) with the geometric sequence formula in part (a) award a maximum of (M1)(A1)(A0) since their final answer cannot be an integer.

OR

\({u_{10}} = 600 + (10 - 1)17 = 753\)     (M1)

\({S_{10}} = \frac{{10}}{2}\left( {600 + {\text{their }}{u_{10}}} \right)\)     (M1)

Note: Award (M1) for their correctly substituted arithmetic sequence formula, (M1) for their correctly substituted arithmetic series formula. Follow through from part (a) and within part (b).

Note: If a list is used, award (M1) for at least 10 correct terms seen, award (A1) for these terms being added.

\( = 6765\)   (accept \(6770\))     (A1)(ft)(G2)

[3 marks]

\(3 \times {2^9}\)     (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 8 correct terms seen.

\( = 1536\)     (A1)(G3)

Note: Exact answer only. If both exact and rounded answer seen, award the final (A1).

[3 marks]

\(\frac{{3 \times \left( {{2^8} - 1} \right)}}{{2 - 1}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted geometric series formula, (A1) for their correct substitutions. Follow through from part (c). If a list is used, award (M1) for at least 8 correct terms seen, award (A1) for these 8 correct terms being added. For consistent use of arithmetic series formula in part (d) with the arithmetic sequence formula in part (c) award a maximum of (M1)(A1)(A1).

\( = 765\)     (A1)(ft)(G2)

[3 marks]

\(3 \times {2^{k - 1}} > 600 + (k - 1)(17)\)     (M1)

Note: Award (M1) for their correct inequality; allow equation. 

Follow through from parts (a) and (c). Accept sketches of the two functions as a valid method.

\(k > 8.93648 \ldots \)   (may be implied)     (A1)(ft)

Note: Award (A1) for \(8.93648…\) seen. The GDC gives answers of \(-34.3\) and \(8.936\) to the inequality; award (M1)(A1) if these are seen with working shown.

OR

\({v_8} = 384\)     \({u_8} = 719\)     (M1)

\({v_9} = 768\)     \({u_9} = 736\)     (M1)

Note: Award (M1) for \({v_8}\) and \({u_8}\) both seen, (M1) for \({v_9}\) and \({u_9}\) both seen.

\(k = 9\)     (A1)(ft)(G2)

Note: Award (G1) for \(8.93648…\) and \(-34.3\) seen as final answer without working. Accept use of \(n\).

[3 marks]

On graph paper, draw a scatter diagram for these data. Use a scale of 2 cm to represent 5 hours on the \(x\)-axis and 2 cm to represent 10 points on the \(y\)-axis.

(i)     \({\bar x}\), the mean number of hours spent on social media;

(ii)     \({\bar y}\), the mean number of IB Diploma points.

Plot the point \((\bar x,{\text{ }}\bar y)\) on your scatter diagram and label this point M.

Write down the value of \(r\), the Pearson’s product–moment correlation coefficient, for these data.

Write down the equation of the regression line \(y\) on \(x\) for these eight male students.

Draw the regression line, from part (e), on your scatter diagram.

Use the given equation of the regression line to estimate the number of IB Diploma points that this girl obtained.

Write down a reason why this estimate is not reliable.

     (A4)

Notes:     Award (A1) for correct scale and labelled axes.

Award (A3) for 7 or 8 points correctly plotted,

(A2) for 5 or 6 points correctly plotted,

(A1) for 3 or 4 points correctly plotted.

Award at most (A0)(A3) if axes reversed.

Accept \(x\) and \(y\) sufficient for labelling.

If graph paper is not used, award (A0).

If an inconsistent scale is used, award (A0). Candidates’ points should be read from this scale where possible and awarded accordingly.

A scale which is too small to be meaningful (ie mm instead of cm) earns (A0) for plotted points.

[4 marks]

(i)     \(\bar x = 21\)     (A1)

(ii)    \(\bar y = 31\)     (A1)

[2 marks]

\((\bar x,{\text{ }}\bar y)\) correctly plotted on graph     (A1)(ft)

this point labelled M     (A1)

Note:     Follow through from parts (b)(i) and (b)(ii).

Only accept M for labelling.

[2 marks]

\( - 0.973{\text{ }}( - 0.973388 \ldots )\)    (G2)

Note:     Award (G1) for 0.973, without minus sign.

[2 marks]

\(y =  - 0.761x + 47.0{\text{ }}(y =  - 0.760638 \ldots x + 46.9734 \ldots )\)    (A1)(A1)(G2)

Notes:     Award (A1) for \( - 0.761x\) and (A1) \( + 47.0\). Award a maximum of (A1)(A0) if answer is not an equation.

[2 marks]

line on graph     (A1)(ft)(A1)(ft)

Notes:     Award (A1)(ft) for straight line that passes through their M, (A1)(ft) for line (extrapolated if necessary) that passes through \((0,{\text{ }}47.0)\).

If M is not plotted or labelled, follow through from part (e).

[2 marks]

\(y =  - \frac{2}{3}(34) + \frac{{125}}{3}\)    (M1)

Note:     Award (M1) for correct substitution.

19 (points)     (A1)(G2)

[2 marks]

extrapolation     (R1)

OR

34 hours is outside the given range of data     (R1)

Note:     Do not accept ‘outlier’.

[1 mark]

Find \(f( - 2)\).

Find \(f'(x)\).

The graph of the function \(f(x)\) has a local minimum at the point where \(x =  - 2\).

Using your answer to part (b), show that there is a second local minimum at \(x = 3\).

The graph of the function \(f(x)\) has a local minimum at the point where \(x =  - 2\).

Sketch the graph of the function \(f(x)\) for \( - 5 \leqslant x \leqslant 5\) and \( - 40 \leqslant y \leqslant 50\). Indicate on your

sketch the coordinates of the \(y\)-intercept.

The graph of the function \(f(x)\) has a local minimum at the point where \(x =  - 2\).

Write down the coordinates of the local maximum.

Let \(T\) be the tangent to the graph of the function \(f(x)\) at the point \((2, –12)\).

Find the gradient of \(T\).

The line \(L\) passes through the point \((2, −12)\) and is perpendicular to \(T\).

\(L\) has equation \(x + by + c = 0\), where \(b\) and \(c \in \mathbb{Z}\).

Find

(i)     the gradient of \(L\);

(ii)     the value of \(b\) and the value of \(c\).

\(\frac{3}{4}{( - 2)^4} - {( - 2)^3} - 9{( - 2)^2} + 20\)     (M1)

Note: Award (M1) for substituting \(x =  - 2\) in the function.

\(= 4\)     (A1)(G2)

Note: If the coordinates \(( - 2,{\text{ }}4)\) are given as the answer award, at most, (M1)(A0). If no working shown award (G1).

     If \(x =  - 2,{\text{ }}y = 4\) seen then award full marks.

[2 marks]

\(3{x^3} - 3{x^2} - 18x\)     (A1)(A1)(A1)

Note: Award (A1) for each correct term, award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

\(f'(3) = 3 \times {(3)^3} - 3 \times {(3)^2} - 18 \times 3\)     (M1)

Note: Award (M1) for substitution in their \(f'(x)\) of \(x = 3\).

\( = 0\)     (A1)

OR

\(3{x^3} - 3{x^2} - 18x = 0\)     (M1)

Note: Award (M1) for equating their \(f'(x)\) to zero.

\(x = 3\)     (A1)

\(f'({x_1}) = 3 \times {({x_1})^3} - 3 \times {({x_1})^2} - 18 \times {x_1} < 0\) where \(0 < {x_1} < 3\)     (M1)

Note: Award (M1) for substituting a value of \({x_1}\) in the range \(0 < {x_1} < 3\) into their \(f'\) and showing it is negative (decreasing).

\(f'({x_2}) = 3 \times {({x_2})^3} - 3 \times {({x_2})^2} - 18 \times {x_2} > 0\) where \({x_2} > 3\)     (M1)

Note: Award (M1) for substituting a value of \({x_2}\) in the range \({x_2} > 3\) into their \(f'\) and showing it is positive (increasing).

OR

With or without a sketch:

Showing \(f({x_1}) > f(3)\) where \({x_1} < 3\) and \({x_1}\) is close to 3.     (M1)

Showing \(f({x_2}) > f(3)\) where \({x_2} > 3\) and \({x_2}\) is close to 3.     (M1)

Note: If a sketch of \(f(x)\) is drawn in this part of the question and \(x = 3\) is identified as a stationary point on the curve, then

     (i) award, at most, (M1)(A1)(M1)(M0) if the stationary point has been found;

     (ii) award, at most, (M0)(A0)(M1)(M0) if the stationary point has not been previously found.

Since the gradients go from negative (decreasing) through zero to positive (increasing) it is a local minimum     (R1)(AG)

Note: Only award (R1) if the first two marks have been awarded ie \(f'(3)\) has been shown to be equal to \(0\).

[5 marks]

     (A1)(A1)(A1)(A1)

Notes: Award (A1) for labelled axes and indication of scale on both axes.

     Award (A1) for smooth curve with correct shape.

     Award (A1) for local minima in \({2^{{\text{nd}}}}\) and \({4^{{\text{th}}}}\) quadrants.

     Award (A1) for y intercept \((0, 20)\) seen and labelled. Accept \(20\) on \(y\)-axis.

     Do not award the third (A1) mark if there is a turning point on the \(x\)-axis.

     If the derivative function is sketched then award, at most, (A1)(A0)(A0)(A0).

     For a smooth curve (with correct shape) there should be ONE continuous thin line, no part of which is straight and no (one to many) mappings of \(x\).

[4 marks]

\((0, 20)\)     (G1)(G1)

Note: If parentheses are omitted award (G0)(G1).

OR

\(x = 0,{\text{ }}y = 20\)     (G1)(G1)

Note: If the derivative function is sketched in part (d), award (G1)(ft)(G1)(ft) for \((–1.12, 12.2)\).

[2 marks]

\(f'(2) = 3{(2)^3} - 3{(2)^2} - 18(2)\)     (M1)

Notes: Award (M1) for substituting \(x = 2\) into their \(f'(x)\).

\( =  - 24\)     (A1)(ft)(G2)

[2 marks]

(i)     Gradient of perpendicular \( = \frac{1}{{24}}\)   \((0.0417, 0.041666…)\)     (A1)(ft)(G1)

Note: Follow through from part (f).

(ii)     \(y + 12 = \frac{1}{{24}}(x - 2)\)     (M1)(M1)

Note: Award (M1) for correct substitution of \((2, –12)\), (M1) for correct substitution of their perpendicular gradient into equation of line.

OR

\( - 12 = \frac{1}{{24}} \times 2 + d\)     (M1)

\(d =  - \frac{{145}}{{12}}\)

\(y = \frac{1}{{24}}x - \frac{{145}}{{12}}\)     (M1)

Note: Award (M1) for correct substitution of \((2, –12)\) and gradient into equation of a straight line, (M1) for correct substitution of the perpendicular gradient and correct substitution of \(d\)into equation of line.

\(b =  - 24,{\text{ }}c =  - 290\)     (A1)(ft)(A1)(ft)(G3)

Note: Follow through from parts (f) and g(i).

     To award (ft) marks, \(b\) and \(c\) must be integers.

     Where candidate has used \(0.042\) from g(i), award (A1)(ft) for \(–288\).

[5 marks]

Surprisingly, a correct method for substituting the value of –2 into the given function led many candidates to a variety of incorrect answers. This suggests a poor handling of negative signs and/or poor use of the graphic display calculator. Many correct answers were seen in part (b) as candidates seemed to be well-drilled in the process of differentiation. Part (c), however, proved to be quite a discriminator. There were 5 marks for this part of the question and simply showing that \(x - 3\) is a turning point was not sufficient for all of these marks. Many simply scored only two marks by substituting \(x - 3\) into their answer to part (b). Once they had shown that there was a turning point at \(x - 3\), candidates were not expected to use the second derivative but to show that the function decreases for \(x < 3\) and increases for \(x > 3\). Part (d) required a sketch which could have either been done on lined paper or on graph paper. The majority of candidates obtained at least two marks here with the most common errors seen being incomplete labelled axes and curves which were far from being smooth. In part (e), many candidates identified the correct coordinates for the two marks available. But for many candidates, this is where responses stopped as, in part (f), connecting the gradient function found in part (b) to the given coordinates proved problematic and only a significant minority of candidates were able to arrive at the required answer of –24. Indeed, there were many NR (no responses) to this part and the final part of the question. As many candidates found part (f) difficult, even more candidates found getting beyond the gradient of L very difficult indeed. A minority of candidates wrote down the gradient of their perpendicular but then did not seem to know where to proceed from there. Substituting their gradient for b and the coordinates (2, –12) into the equation \(x + by + c = 0\) was a popular, but erroneous, method. It was a rare event indeed to see a script with a correct answer for this part of the question.

Surprisingly, a correct method for substituting the value of –2 into the given function led many candidates to a variety of incorrect answers. This suggests a poor handling of negative signs and/or poor use of the graphic display calculator. Many correct answers were seen in part (b) as candidates seemed to be well-drilled in the process of differentiation. Part (c), however, proved to be quite a discriminator. There were 5 marks for this part of the question and simply showing that \(x - 3\) is a turning point was not sufficient for all of these marks. Many simply scored only two marks by substituting \(x - 3\) into their answer to part (b). Once they had shown that there was a turning point at \(x - 3\), candidates were not expected to use the second derivative but to show that the function decreases for \(x < 3\) and increases for \(x > 3\). Part (d) required a sketch which could have either been done on lined paper or on graph paper. The majority of candidates obtained at least two marks here with the most common errors seen being incomplete labelled axes and curves which were far from being smooth. In part (e), many candidates identified the correct coordinates for the two marks available. But for many candidates, this is where responses stopped as, in part (f), connecting the gradient function found in part (b) to the given coordinates proved problematic and only a significant minority of candidates were able to arrive at the required answer of –24. Indeed, there were many NR (no responses) to this part and the final part of the question. As many candidates found part (f) difficult, even more candidates found getting beyond the gradient of L very difficult indeed. A minority of candidates wrote down the gradient of their perpendicular but then did not seem to know where to proceed from there. Substituting their gradient for b and the coordinates (2, –12) into the equation \(x + by + c = 0\) was a popular, but erroneous, method. It was a rare event indeed to see a script with a correct answer for this part of the question.

Surprisingly, a correct method for substituting the value of –2 into the given function led many candidates to a variety of incorrect answers. This suggests a poor handling of negative signs and/or poor use of the graphic display calculator. Many correct answers were seen in part (b) as candidates seemed to be well-drilled in the process of differentiation. Part (c), however, proved to be quite a discriminator. There were 5 marks for this part of the question and simply showing that \(x - 3\) is a turning point was not sufficient for all of these marks. Many simply scored only two marks by substituting \(x - 3\) into their answer to part (b). Once they had shown that there was a turning point at \(x - 3\), candidates were not expected to use the second derivative but to show that the function decreases for \(x < 3\) and increases for \(x > 3\). Part (d) required a sketch which could have either been done on lined paper or on graph paper. The majority of candidates obtained at least two marks here with the most common errors seen being incomplete labelled axes and curves which were far from being smooth. In part (e), many candidates identified the correct coordinates for the two marks available. But for many candidates, this is where responses stopped as, in part (f), connecting the gradient function found in part (b) to the given coordinates proved problematic and only a significant minority of candidates were able to arrive at the required answer of –24. Indeed, there were many NR (no responses) to this part and the final part of the question. As many candidates found part (f) difficult, even more candidates found getting beyond the gradient of L very difficult indeed. A minority of candidates wrote down the gradient of their perpendicular but then did not seem to know where to proceed from there. Substituting their gradient for b and the coordinates (2, –12) into the equation \(x + by + c = 0\) was a popular, but erroneous, method. It was a rare event indeed to see a script with a correct answer for this part of the question.

Surprisingly, a correct method for substituting the value of –2 into the given function led many candidates to a variety of incorrect answers. This suggests a poor handling of negative signs and/or poor use of the graphic display calculator. Many correct answers were seen in part (b) as candidates seemed to be well-drilled in the process of differentiation. Part (c), however, proved to be quite a discriminator. There were 5 marks for this part of the question and simply showing that \(x - 3\) is a turning point was not sufficient for all of these marks. Many simply scored only two marks by substituting \(x - 3\) into their answer to part (b). Once they had shown that there was a turning point at \(x - 3\), candidates were not expected to use the second derivative but to show that the function decreases for \(x < 3\) and increases for \(x > 3\). Part (d) required a sketch which could have either been done on lined paper or on graph paper. The majority of candidates obtained at least two marks here with the most common errors seen being incomplete labelled axes and curves which were far from being smooth. In part (e), many candidates identified the correct coordinates for the two marks available. But for many candidates, this is where responses stopped as, in part (f), connecting the gradient function found in part (b) to the given coordinates proved problematic and only a significant minority of candidates were able to arrive at the required answer of –24. Indeed, there were many NR (no responses) to this part and the final part of the question. As many candidates found part (f) difficult, even more candidates found getting beyond the gradient of L very difficult indeed. A minority of candidates wrote down the gradient of their perpendicular but then did not seem to know where to proceed from there. Substituting their gradient for b and the coordinates (2, –12) into the equation \(x + by + c = 0\) was a popular, but erroneous, method. It was a rare event indeed to see a script with a correct answer for this part of the question.

Surprisingly, a correct method for substituting the value of –2 into the given function led many candidates to a variety of incorrect answers. This suggests a poor handling of negative signs and/or poor use of the graphic display calculator. Many correct answers were seen in part (b) as candidates seemed to be well-drilled in the process of differentiation. Part (c), however, proved to be quite a discriminator. There were 5 marks for this part of the question and simply showing that \(x - 3\) is a turning point was not sufficient for all of these marks. Many simply scored only two marks by substituting \(x - 3\) into their answer to part (b). Once they had shown that there was a turning point at \(x - 3\), candidates were not expected to use the second derivative but to show that the function decreases for \(x < 3\) and increases for \(x > 3\). Part (d) required a sketch which could have either been done on lined paper or on graph paper. The majority of candidates obtained at least two marks here with the most common errors seen being incomplete labelled axes and curves which were far from being smooth. In part (e), many candidates identified the correct coordinates for the two marks available. But for many candidates, this is where responses stopped as, in part (f), connecting the gradient function found in part (b) to the given coordinates proved problematic and only a significant minority of candidates were able to arrive at the required answer of –24. Indeed, there were many NR (no responses) to this part and the final part of the question. As many candidates found part (f) difficult, even more candidates found getting beyond the gradient of L very difficult indeed. A minority of candidates wrote down the gradient of their perpendicular but then did not seem to know where to proceed from there. Substituting their gradient for b and the coordinates (2, –12) into the equation \(x + by + c = 0\) was a popular, but erroneous, method. It was a rare event indeed to see a script with a correct answer for this part of the question.

Surprisingly, a correct method for substituting the value of –2 into the given function led many candidates to a variety of incorrect answers. This suggests a poor handling of negative signs and/or poor use of the graphic display calculator. Many correct answers were seen in part (b) as candidates seemed to be well-drilled in the process of differentiation. Part (c), however, proved to be quite a discriminator. There were 5 marks for this part of the question and simply showing that \(x - 3\) is a turning point was not sufficient for all of these marks. Many simply scored only two marks by substituting \(x - 3\) into their answer to part (b). Once they had shown that there was a turning point at \(x - 3\), candidates were not expected to use the second derivative but to show that the function decreases for \(x < 3\) and increases for \(x > 3\). Part (d) required a sketch which could have either been done on lined paper or on graph paper. The majority of candidates obtained at least two marks here with the most common errors seen being incomplete labelled axes and curves which were far from being smooth. In part (e), many candidates identified the correct coordinates for the two marks available. But for many candidates, this is where responses stopped as, in part (f), connecting the gradient function found in part (b) to the given coordinates proved problematic and only a significant minority of candidates were able to arrive at the required answer of –24. Indeed, there were many NR (no responses) to this part and the final part of the question. As many candidates found part (f) difficult, even more candidates found getting beyond the gradient of L very difficult indeed. A minority of candidates wrote down the gradient of their perpendicular but then did not seem to know where to proceed from there. Substituting their gradient for b and the coordinates (2, –12) into the equation \(x + by + c = 0\) was a popular, but erroneous, method. It was a rare event indeed to see a script with a correct answer for this part of the question.

Surprisingly, a correct method for substituting the value of –2 into the given function led many candidates to a variety of incorrect answers. This suggests a poor handling of negative signs and/or poor use of the graphic display calculator. Many correct answers were seen in part (b) as candidates seemed to be well-drilled in the process of differentiation. Part (c), however, proved to be quite a discriminator. There were 5 marks for this part of the question and simply showing that \(x - 3\) is a turning point was not sufficient for all of these marks. Many simply scored only two marks by substituting \(x - 3\) into their answer to part (b). Once they had shown that there was a turning point at \(x - 3\), candidates were not expected to use the second derivative but to show that the function decreases for \(x < 3\) and increases for \(x > 3\). Part (d) required a sketch which could have either been done on lined paper or on graph paper. The majority of candidates obtained at least two marks here with the most common errors seen being incomplete labelled axes and curves which were far from being smooth. In part (e), many candidates identified the correct coordinates for the two marks available. But for many candidates, this is where responses stopped as, in part (f), connecting the gradient function found in part (b) to the given coordinates proved problematic and only a significant minority of candidates were able to arrive at the required answer of –24. Indeed, there were many NR (no responses) to this part and the final part of the question. As many candidates found part (f) difficult, even more candidates found getting beyond the gradient of L very difficult indeed. A minority of candidates wrote down the gradient of their perpendicular but then did not seem to know where to proceed from there. Substituting their gradient for b and the coordinates (2, –12) into the equation \(x + by + c = 0\) was a popular, but erroneous, method. It was a rare event indeed to see a script with a correct answer for this part of the question.

Write down \(f'(x)\).

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Show that \(k = 3\).

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Find \(f(2)\).

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Find \(f'(2)\)

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Find the equation of the normal to the graph of \(y = f(x)\) at the point where \(x = 2\).

Give your answer in the form \(ax + by + d = 0\) where \(a,{\text{ }}b,{\text{ }}d \in \mathbb{Z}\).

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Sketch the graph of \(y = f(x)\), for \( - 5 \leqslant x \leqslant 10\) and \( - 10 \leqslant y \leqslant 100\).

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Write down the coordinates of the point where the graph of \(y = f(x)\) intersects the \(x\)-axis.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

State the values of \(x\) for which \(f(x)\) is decreasing.

\(\frac{{ - 192}}{{{x^3}}} + k\)     (A1)(A1)(A1)

Note: Award (A1) for \(-192\), (A1) for \({x^{ - 3}}\), (A1) for \(k\) (only).

at local minimum \(f'(x) = 0\)     (M1)

Note: Award (M1) for seeing \(f'(x) = 0\) (may be implicit in their working).

\(\frac{{ - 192}}{{{4^3}}} + k = 0\)     (A1)

\(k = 3\)     (AG)

Note: Award (A1) for substituting \(x = 4\) in their \(f'(x) = 0\), provided it leads to \(k = 3\). The conclusion \(k = 3\) must be seen for the (A1) to be awarded.

\(\frac{{96}}{{{2^2}}} + 3(2)\)     (M1)

Note: Award (M1) for substituting \(x = 2\) and \(k = 3\) in \(f(x)\).

\( = 30\)     (A1)(G2)

\(\frac{{ - 192}}{{{2^3}}} + 3\)     (M1)

Note: Award (M1) for substituting \(x = 2\) and \(k = 3\) in their \(f'(x)\).

\( =  - 21\)     (A1)(ft)(G2)

Note: Follow through from part (a).

\(y - 30 = \frac{1}{{21}}(x - 2)\)     (A1)(ft)(M1)

Notes: Award (A1)(ft) for their \(\frac{1}{{21}}\) seen, (M1) for the correct substitution of their point and their normal gradient in equation of a line.

Follow through from part (c) and part (d).

OR

gradient of normal \( = \frac{1}{{21}}\)     (A1)(ft)

\(30 = \frac{1}{{21}} \times 2 + c\)     (M1)

\(c = 29\frac{{19}}{{21}}\)

\(y = \frac{1}{{21}}x + 29\frac{{19}}{{21}}\;\;\;(y = 0.0476x + 29.904)\)

\(x - 21y + 628 = 0\)     (A1)(ft)(G2)

Notes:  Accept equivalent answers.

     (A1)(A1)(A1)(A1)

Notes: Award (A1) for correct window (at least one value, other than zero, labelled on each axis), the axes must also be labelled; (A1) for a smooth curve with the correct shape (graph should not touch \(y\)-axis and should not curve away from the \(y\)-axis), on the given domain; (A1) for axis intercept in approximately the correct position (nearer \(-5\) than zero); (A1) for local minimum in approximately the correct position (first quadrant, nearer the \(y\)-axis than \(x = 10\)).

If there is no scale, award a maximum of (A0)(A1)(A0)(A1) – the final (A1) being awarded for the zero and local minimum in approximately correct positions relative to each other.

\(( - 3.17,{\text{ }}0)\;\;\;\left( {( - 3.17480 \ldots ,{\text{ 0)}}} \right)\)     (G1)(G1)

Notes: If parentheses are omitted award (G0)(G1)(ft).

Accept \(x =  - 3.17,{\text{ }}y = 0\). Award (G1) for \(-3.17\) seen.

\(0 < x \leqslant 4{\text{ or }}0 < x < 4\)     (A1)(A1)

Notes: Award (A1) for correct end points of interval, (A1) for correct notation (note: lower inequality must be strict).

Award a maximum of (A1)(A0) if \(y\) or \(f(x)\) used in place of \(x\).

Differentiation of terms including negative indices remains a testing process; it will continue to be tested. There was, however, a noticeable improvement in responses compared to previous years. The parameter k was problematic for a number of candidates.

In part (b), the manipulation of the derivative to find the local minimum point caused difficulties for all but the most able; note that a GDC approach is not accepted in such questions and that candidates are expected to be able to apply the theory of the calculus as appropriate. Further, once a parameter is given, candidates are expected to use this value in subsequent parts.

Parts (c) and (d) were accessible and all but the weakest candidates scored well.

Parts (c) and (d) were accessible and all but the weakest candidates scored well.

Part (e) discriminated at the highest level; the gradient of the normal often was not used, the form of the answer not given correctly.

Curve sketching is a skill that most candidates find very difficult; axes must be labelled and some indication of the window must be present; care must be taken with the domain and the range; any asymptotic behaviour must be indicated. It was very rare to see sketches that attained full marks, yet this should be a skill that all can attain. There were many no attempts seen, yet some of these had correct answers to part (g).

Curve sketching is a skill that most candidates find very difficult; axes must be labelled and some indication of the window must be present; care must be taken with the domain and the range; any asymptotic behaviour must be indicated. It was very rare to see sketches that attained full marks, yet this should be a skill that all can attain. There were many no attempts seen, yet some of these had correct answers to part (g).

Part (h) was not well attempted in the main; decreasing (and increasing) functions is a testing concept for the majority.

Show that the width of the plot, in metres, is given by \(100 - x\).

Write down the area of the plot in terms of \(x\).

Find the value of \(x\) that maximizes the area of the plot.

Show that Violeta earns 5000 BGN from selling the flowers grown on the plot.

Find the amount of money that Violeta would have after 6 years. Give your answer correct to two decimal places.

Find how long it would take for the interest earned to be 2000 BGN.

Find the lowest possible value for \(r\).

\(\frac{{200 - 2x}}{2}\) (or equivalent)     (M1)

OR

\(2x + 2y = 200\) (or equivalent)     (M1)

Note:     Award (M1) for a correct expression leading to \(100 - x\) (the \(100 - x\) does not need to be seen). The 200 must be seen for the (M1) to be awarded. Do not accept \(100 - x\) substituted in the perimeter of the rectangle formula.

\(100 - x\)     (AG)

[1 mark]

\(({\text{area}} = ){\text{ }}x(100 - x)\)\(\,\,\,\)OR\(\,\,\,\)\( - {x^2} + 100x\) (or equivalent)     (A1)

[1 mark]

\(x = \frac{{ - 100}}{{ - 2}}\)\(\,\,\,\)OR\(\,\,\,\)\( - 2x + 100 = 0\)\(\,\,\,\)OR\(\,\,\,\)graphical method     (M1)

Note:     Award (M1) for use of axis of symmetry formula or first derivative equated to zero or a sketch graph.

\(x = 50\)     (A1)(ft)(G2)

Note:     Follow through from part (b), provided x is positive and less than 100.

[2 marks]

\(50(100 - 50) \times 2\)     (M1)(M1)

Note:     Award (M1) for substituting their \(x\) into their formula for area (accept “\(50 \times 50\)” for the substituted formula), and (M1) for multiplying by 2. Award at most (M0)(M1) if their calculation does not lead to 5000 (BGN), although the 5000 (BGN) does not need to be seen explicitly.

Substitution of 50 into area formula may be seen in part (c).

5000 (BGN)     (AG)

[2 marks]

\(5000{\left( {1 + \frac{4}{{2 \times 100}}} \right)^{2 \times 6}}\)     (M1)(A1)

Note:     Award (M1) for substitution into compound interest formula, (A1) for correct substitution.

OR

\({\text{N}} = 6\)

\({\text{I}}\%  = 4\)

\({\text{PV}} =  - 5000\)

\({\text{P/Y}} = 1\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for \({\text{C/Y}} = 2\)  seen, (M1) for other correct entries.

OR

\({\text{N}} = 12\)

\({\text{I}}\%  = 4\)

\({\text{PV}} =  - 5000\)

\({\text{P/Y}} = 2\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for \({\text{C/Y}} = 2\) seen, (M1) for other correct entries.

6341.21 (BGN)     (A1)(G3)

Note:     Award (A1) for correct answer, to two decimal places only.

Award (G2) for 6341.20 or a correct, unrounded final answer if no working is seen (6341.2089…).

[3 marks]

\(5000{\left( {1 + \frac{4}{{2 \times 100}}} \right)^{2 \times t}} = 7000\)     (M1)(A1)(ft)

Note:     Award (M1) for using the compound interest formula with a variable for time, (A1)(ft) for substituting the correct values and equating to 7000. Follow through for their “2” from part (e)(i).

OR

\({\text{I% }} = 4\)

\({\text{PV}} = ( \pm )5000\)

\({\text{FV}} =  \mp 7000\)

\({\text{P/Y}} = 1\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for 7000 seen, (M1) for the other correct entries.

Award (M1) for their C/Y from part (e)(i).

OR

\({\text{I% }} = 4\)

\({\text{PV}} = ( \pm )5000\)

\({\text{FV}} =  \mp 7000\)

\({\text{P/Y}} = 2\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for 7000 seen, (M1) for the other correct entries.

Award (M1) for their C/Y from part (e)(i).

OR

     (M1)(A1)(ft)

Note:     Award (M1) for a sketch with a straight line intercepted by appropriate curve, (A1)(ft) for numerical answer in the range of 8.4 and 8.5.

Follow through from their part (e)(i).

\(t = 8.50{\text{ (years) }}(8.49564 \ldots )\)     (A1)(ft)(G3)

Note:     Award only (A1) if 16.9912… is seen without working. If working is seen, award at most (M1)(A1)(A0).

[3 marks]

\(5000{\left( {1 + \frac{r}{{100}}} \right)^{12}} = 10000\)     (M1)

Note: Award (M1) for correct substitution into compound interest formula with 10 000 seen. 

OR

\(2 = {\left( {1 + \frac{r}{{100}}} \right)^{12}}\)     (M1)

Note:     Award (M1) for correct substitution and simplification of compound interest formula, equating to 2.

\(r = 5.95{\text{ (% ) }}(5.94630 \ldots )\)     (A1)(G2)

[2 marks]

Calculate \(f(1)\).

Sketch the graph of \(y = f(x)\) for \( - 7 \leqslant x \leqslant 4\) and \( - 30 \leqslant y \leqslant 30\).

Write down the equation of the vertical asymptote.

Write down the coordinates of the \(x\)-intercept.

Write down the possible values of \(x\) for which \(x < 0\) and \(f’(x) > 0\).

Find the solution of \(f(x) = g(x)\).

\(0.3{(1)^3} + \frac{{10}}{1} + {2^{ - 1}}\)     (M1)

Note:     Award (M1) for correct substitution into function.

\( = 10.8\)     (A1)(G2)

[2 marks]

     (A1)(A1)(A1)(A1)

Note:     Award (A1) for indication of correct window and labelled axes.

Award (A1) for correct shape and position for \(x < 0\) (with the local maximum, local minimum and \(x\)-intercept in relative approximate location in \({{\text{3}}^{{\text{rd}}}}\) quadrant).

Award (A1) for correct shape and position for \(x > 0\) (with the local minimum in relative approximate location in \({{\text{1}}^{{\text{st}}}}\) quadrant).

Award (A1) for smooth curve with indication of asymptote (graph should not touch \(y\)-axis and should not curve away from the \(y\)-axis). The asymptote is only assessed in this mark.

[4 marks]

\(x = 0\)     (A2)

Note:     Award (A1) for “\(x = {\text{(a constant)}}\)” and (A1) for “\({\text{(a constant)}} = 0\)”.

The answer must be an equation.

[2 marks]

\(( - 6.18,{\text{ }}0){\text{ }}( - 6.17516 \ldots ,{\text{ }}0)\)     (A1)(A1)

Note:     Award (A1) for each correct coordinate. Award (A0)(A1) if parentheses are missing.

[2 marks]

\( - 4.99 < x <  - 2.47{\text{ }}( - 4.98688 \ldots  < x <  - 2.46635 \ldots )\)     (A1)(A1)

Note:     Award (A1) for both correct end points, (A1) for strict inequalities used with 2 endpoints.

[2 marks]

\(0.3{x^3} + \frac{{10}}{x} + {2^{ - x}} = 2x - 3\)     (M1)

Note:     Award (M1) for equating the expressions for \(f\) and \(g\) or for the line \(y = 2x - 3\) sketched (positive gradient, negative \(y\)-intercept) on their graph from part (a).

\((x = ){\text{ }} - 1.34{\text{ }}( - 1.33650 \ldots )\)     (A1)(G2)

Note:     Award a maximum of (M1)(A0) or (G1) for coordinate pair seen as final answer.

[2 marks]

Sketch the graph of the function \(f(x)\), for \( - 10 \leqslant x \leqslant 10\) . Indicating clearly the axis intercepts and any asymptotes.

Write down the equation of the vertical asymptote.

On the same diagram as part (a) sketch the graph of \(g(x) = x + 0.5\) .

Using your graphical display calculator write down the coordinates of one of the points of intersection on the graphs of \(f\) and \(g\), giving your answer correct to five decimal places.

Write down the gradient of the line \(g(x) = x + 0.5\) .

The line \(L\) passes through the point with coordinates \(( - 2{\text{, }} - 3)\) and is perpendicular to the line \(g(x)\) . Find the equation of \(L\).

     (A6)

Notes: (A1) for labels and some idea of scale.
(A1) for \(x\)-intercept seen, (A1) for \(y\)-intercept seen in roughly the correct places (coordinates not required).
(A1) for vertical asymptote seen, (A1) for horizontal asymptote seen in roughly the correct places (equations of the lines not required).
(A1) for correct general shape.

[6 marks]

\(x = - 4\)     (A1)(A1)(ft)

Note: (A1) for \(x =\), (A1)(ft) for \( - 4\).

[2 marks]

     (A1)(A1)

Note: (A1) for correct axis intercepts, (A1) for straight line

[2 marks]

\(( - 2.85078{\text{, }} - 2.35078)\) OR \((0.35078{\text{, }}0.85078)\)     (G1)(G1)(A1)(ft)

Notes: (A1) for \(x\)-coordinate, (A1) for \(y\)-coordinate, (A1)(ft) for correct accuracy. Brackets required. If brackets not used award (G1)(G0)(A1)(ft).
Accept \(x = - 2.85078\), \(y = - 2.35078\) or \(x = 0.35078\), \(y = 0.85078\).

[3 marks]

\({\text{gradient}} = 1\)     (A1)

[1 mark]

\({\text{gradient of perpendicular}} = - 1\)     (A1)(ft)

(can be implied in the next step)

\(y = mx + c\)

\( - 3 = - 1 \times - 2 + c\)     (M1)

\(c = - 5\)

\(y = - x - 5\)     (A1)(ft)(G2)

OR

\(y + 3 = - (x + 2)\)     (M1)(A1)(ft)(G2)

Note: Award (G2) for correct answer with no working at all but (A1)(G1) if the gradient is mentioned as \( - 1\) then correct answer with no further working.

[3 marks]

This was not very well done. The graph was often correct but was so small that it was difficult to check if axes intercepts were correct or not. Often the vertical asymptote looked as if it were joined to the rest of the graph. Very few of the candidates put a scale and/or labels on their axes.

Reasonably well done. Some put \(y = - 4\) while others omitted the minus sign.

Fairly well done – but once again too small to check the axes intercepts properly. Also, many candidates did not appear to have a ruler to draw the straight line.

Well done.

Most could find the gradient of the line.

Many forgot to find the gradient of the perpendicular line. Others had problems with the equation of a line in general.

Find \(g'(x)\).

Show that \(k = 6\).

Find the equation of the tangent to the graph of \(y = g(x)\) at \(x = 2\). Give your answer in the form \(y = mx + c\).

Use your answer to part (a) and the value of \(k\), to find the \(x\)-coordinates of the stationary points of the graph of \(y = g(x)\).

Find \(g’( - 1)\).

Hence justify that \(g\) is decreasing at \(x =  - 1\).

Find the \(y\)-coordinate of the local minimum.

\(3{x^2} + 2kx - 15\)     (A1)(A1)(A1)

Note:     Award (A1) for \(3{x^2}\), (A1) for \(2kx\) and (A1) for \( - 15\). Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

\(21 = 3{(2)^2} + 2k(2) - 15\)     (M1)(M1)

Note:     Award (M1) for equating their derivative to 21. Award (M1) for substituting 2 into their derivative. The second (M1) should only be awarded if correct working leads to the final answer of \(k = 6\).

Substituting in the known value, \(k = 6\), invalidates the process; award (M0)(M0).

\(k = 6\)     (AG)

[2 marks]

\(g(2) = {(2)^3} + (6){(2)^2} - 15(2) + 5{\text{ }}( = 7)\)     (M1)

Note:     Award (M1) for substituting 2 into \(g\).

\(7 = 21(2) + c\)     (M1)

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient intercept form.

OR

\(y - 7 = 21(x - 2)\)     (M1)

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient point form.

\(y = 21x - 35\)     (A1)     (G2)

[3 marks]

\(3{x^2} + 12x - 15 = 0\) (or equivalent)     (M1)

Note:     Award (M1) for equating their part (a) (with \(k = 6\) substituted) to zero.

\(x =  - 5,{\text{ }}x = 1\)     (A1)(ft)(A1)(ft)

Note:     Follow through from part (a).

[3 marks]

\(3{( - 1)^2} + 12( - 1) - 15\)     (M1)

Note:     Award (M1) for substituting \( - 1\) into their derivative, with \(k = 6\) substituted. Follow through from part (a).

\( =  - 24\)     (A1)(ft)     (G2)

[2 marks]

\(g’( - 1) < 0\) (therefore \(g\) is decreasing when \(x =  - 1\))     (R1)

[1 marks]

\(g(1) = {(1)^3} + (6){(1)^2} - 15(1) + 5\)     (M1)

Note:     Award (M1) for correctly substituting 6 and their 1 into \(g\).

\( =  - 3\)     (A1)(ft)     (G2)

Note:     Award, at most, (M1)(A0) or (G1) if answer is given as a coordinate pair. Follow through from part (c).

[2 marks]

Show that A lies on \({L_1}\).

Find the coordinates of M.

Find the length of AC.

Show that the equation of \({L_2}\) is \(2y + x - 19 = 0\).

Find the coordinates of D.

Write down the length of MD correct to five significant figures.

Find the area of ABCD.

\(2 \times 4 - 1 - 7 = 0\) (or equivalent)     (R1)

Note:     For (R1) accept substitution of \(x = 1\) or \(y = 4\) into the equation followed by a confirmation that \(y = 4\) or \(x = 1\).

(since the point satisfies the equation of the line,) A lies on \({L_1}\)     (A1)

Note:     Do not award (A1)(R0).

[2 marks]

\(\frac{{1 + 5}}{2}\) OR \(\frac{{4 + 12}}{2}\) seen     (M1)

Note:     Award (M1) for at least one correct substitution into the midpoint formula.

\((3,{\text{ }}8)\)    (A1)(G2)

Notes:     Accept \(x = 3,{\text{ }}y = 8\).

Award (M1)(A0) for \(\left( {\frac{{1 + 5}}{2},{\text{ }}\frac{{4 + 12}}{2}} \right)\).

Award (G1) for each correct coordinate seen without working.

[2 marks]

\(\sqrt {{{(5 - 1)}^2} + {{(12 - 4)}^2}} \)    (M1)

Note:     Award (M1) for a correct substitution into distance between two points formula.

\( = 8.94{\text{ }}\left( {4\sqrt 5 ,{\text{ }}\sqrt {80} ,{\text{ }}8.94427 \ldots } \right)\)    (A1)(G2)

[2 marks]

gradient of \({\text{AC}} = \frac{{12 - 4}}{{5 - 1}}\)     (M1)

Note:     Award (M1) for correct substitution into gradient formula.

\( = 2\)    (A1)

Note:     Award (M1)(A1) for gradient of \({\text{AC}} = 2\) with or without working

gradient of the normal \( =  - \frac{1}{2}\)     (M1)

Note:     Award (M1) for the negative reciprocal of their gradient of AC.

\(y - 8 =  - \frac{1}{2}(x - 3)\) OR \(8 =  - \frac{1}{2}(3) + c\)     (M1)

Note:     Award (M1) for substitution of their point and gradient into straight line formula. This (M1) can only be awarded where \( - \frac{1}{2}\) (gradient) is correctly determined as the gradient of the normal to AC.

\(2y - 16 =  - (x - 3)\) OR \( - 2y + 16 = x - 3\) OR \(2y =  - x + 19\)     (A1)

Note:     Award (A1) for correctly removing fractions, but only if their equation is equivalent to the given equation.

\(2y + x - 19 = 0\)    (AG)

Note:     The conclusion \(2y + x - 19 = 0\) must be seen for the (A1) to be awarded.

Where the candidate has shown the gradient of the normal to \({\text{AC}} =  - 0.5\), award (M1) for \(2(8) + 3 - 19 = 0\) and (A1) for (therefore) \(2y + x - 19 = 0\).

Simply substituting \((3,{\text{ }}8)\) into the equation of \({L_2}\) with no other prior working, earns no marks.

[5 marks]

\((6,{\text{ }}6.5)\)    (A1)(A1)(G2)

Note:     Award (A1) for 6, (A1) for 6.5. Award a maximum of (A1)(A0) if answers are not given as a coordinate pair. Accept \(x = 6,{\text{ }}y = 6.5\).

Award (M1)(A0) for an attempt to solve the two simultaneous equations \(2y - x - 7 = 0\) and \(2y + x - 19 = 0\) algebraically, leading to at least one incorrect or missing coordinate.

[2 marks]

3.3541     (A1)

Note:     Answer must be to 5 significant figures.

[1 mark]

\(2 \times \frac{1}{2} \times \sqrt {80}  \times \frac{{\sqrt {45} }}{2}\)     (M1)(M1)

Notes:     Award (M1) for correct substitution into area of triangle formula.

If their triangle is a quarter of the rhombus then award (M1) for multiplying their triangle by 4.

If their triangle is a half of the rhombus then award (M1) for multiplying their triangle by 2.

OR

\(\frac{1}{2} \times \sqrt {80}  \times \sqrt {45} \)    (M1)(M1)

Notes:     Award (M1) for doubling MD to get the diagonal BD, (M1) for correct substitution into the area of a rhombus formula.

Award (M1)(M1) for \(\sqrt {80}  \times \) their (f).

\( = 30\)    (A1)(ft)(G3)

Notes:     Follow through from parts (c) and (f).

\(8.94 \times 3.3541 = 29.9856 \ldots \)

[3 marks]

Write down the equation of the vertical asymptote.

Find \(f'(x)\) .

Find the gradient of the graph of the function at \(x = - 1\).

Using your answer to part (c), decide whether the function \(f(x)\) is increasing or decreasing at \(x = - 1\). Justify your answer.

Sketch the graph of \(f(x)\) for \( - 10 \leqslant x \leqslant 10\) and \( - 20 \leqslant y \leqslant 20\) .

\({{\text{P}}_1}\) is the local maximum point and \({{\text{P}}_2}\) is the local minimum point on the graph of \(f(x)\) .

Using your graphic display calculator, write down the coordinates of

(i)     \({{\text{P}}_1}\) ;

(ii)    \({{\text{P}}_2}\) .

Using your sketch from (e), determine the range of the function \(f(x)\) for \( - 10 \leqslant x \leqslant 10\) .

\(x = 0\)     (A1)(A1)

Note: Award (A1) for \(x = {\text{constant}}\), (A1) for \(0\).

[2 marks]

\(f'(x) = 1.5 - \frac{6}{{{x^2}}}\)     (A1)(A1)(A1)

Notes: Award (A1) for \(1.5\), (A1) for \( - 6\), (A1) for \({x^{ - 2}}\) . Award (A1)(A1)(A0) at most if any other term present.

[3 marks]

\(1.5 - \frac{6}{{( - 1)}}\)     (M1)

\( =  - 4.5\)     (A1)(ft)(G2)

Note: Follow through from their derivative function.

[2 marks]

Decreasing, the derivative (gradient or slope) is negative (at \(x = - 1\))     (A1)(R1)(ft)

Notes: Do not award (A1)(R0). Follow through from their answer to part (c).

[2 marks]

     (A4)
Notes: Award (A1) for labels and some indication of scales and an appropriate window.

Award (A1) for correct shape of the two unconnected, and smooth branches.

Award (A1) for the maximum and minimum points in the approximately correct positions.

Award (A1) for correct asymptotic behaviour at \(x = 0\) .

Notes: Please be rigorous.

The axes need not be drawn with a ruler.

The branches must be smooth and single continuous lines that do not deviate from their proper direction.

The max and min points must be symmetrical about point \((0{\text{, }}4)\) .

The \(y\)-axis must be an asymptote for both branches.

[4 marks]

(i)     \(( - 2{\text{, }} - 2)\) or \(x = - 2\), \(y = - 2\)     (G1)(G1)

(ii)    \((2{\text{, }}10)\) or \(x = 2\), \(y = 10\)     (G1)(G1)

[4 marks]

\(\{  - 2 \geqslant y\} \) or \(\{ y \geqslant 10\} \)     (A1)(A1)(ft)(A1)

Notes: (A1)(ft) for \(y > 10\) or \(y \geqslant 10\) . (A1)(ft) for \(y < - 2\) or \(y \leqslant - 2\) . (A1) for weak (non-strict) inequalities used in both of the above. Follow through from their (e) and (f).

[3 marks]

Part a) was either answered well or poorly.

Most candidates found the first term of the derivative in part b) correctly, but the rest of the terms were incorrect.

The gradient in c) was for the most part correctly calculated, although some candidates substituted incorrectly in \(f(x)\) instead of in \(f'(x)\) .

Part d) had mixed responses.

Lack of labels of the axes, appropriate scale, window, incorrect maximum and minimum and incorrect asymptotic behaviour were the main problems with the sketches in e).

Part f) was also either answered correctly or entirely incorrectly. Some candidates used the trace function on the GDC instead of the min and max functions, and thus acquired coordinates with unacceptable accuracy. Some were unclear that a point of local maximum may be positioned on the coordinate system “below” the point of local minimum, and exchanged the pairs of coordinates of those points in f(i) and f(ii).

Very few candidates were able to identify the range of the function in (g) irrespective of whether or not they had the sketches drawn correctly.

Calculate the volume of this pan.

Find the radius of the sphere in cm, correct to one decimal place.

Find the value of \(a\).

Find the temperature that the pizza will be 5 minutes after it is taken out of the oven.

Calculate, to the nearest second, the time since the pizza was taken out of the oven until it can be eaten.

In the context of this model, state what the value of 19 represents.

\((V = ){\text{ }}\pi  \times {{\text{(17.5)}}^2} \times 0.5\)     (A1)(M1)

Notes:     Award (A1) for 17.5 (or equivalent) seen.

Award (M1) for correct substitutions into volume of a cylinder formula.

\( = 481{\text{ c}}{{\text{m}}^3}{\text{ }}(481.056 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}153.125\pi {\text{ c}}{{\text{m}}^3})\)     (A1)(G2)

[3 marks]

\(\frac{4}{3} \times \pi  \times {r^3} = 481.056 \ldots \)     (M1)

Note:     Award (M1) for equating their answer to part (a) to the volume of sphere.

\({r^3} = \frac{{3 \times 481.056 \ldots }}{{4\pi }}{\text{ }}( = 114.843 \ldots )\)     (M1)

Note:     Award (M1) for correctly rearranging so \({r^3}\) is the subject.

\(r = 4.86074 \ldots {\text{ (cm)}}\)     (A1)(ft)(G2)

Note:     Award (A1) for correct unrounded answer seen. Follow through from part (a).

\( = 4.9{\text{ (cm)}}\)     (A1)(ft)(G3)

Note:     The final (A1)(ft) is awarded for rounding their unrounded answer to one decimal place.

[4 marks]

\(230 = a{(2.06)^0} + 19\)     (M1)

Note:     Award (M1) for correct substitution.

\(a = 211\)     (A1)(G2)

[2 marks]

\((P = ){\text{ }}211 \times {(2.06)^{ - 5}} + 19\)      (M1)

Note:     Award (M1) for correct substitution into the function, \(P(t)\). Follow through from part (c). The negative sign in the exponent is required for correct substitution.

\( = 24.7\) (°C) \((24.6878 \ldots \) (°C))     (A1)(ft)(G2)

[2 marks]

\(45 = 211 \times {(2.06)^{ - t}} + 19\)     (M1)

Note:     Award (M1) for equating 45 to the exponential equation and for correct substitution (follow through for their \(a\) in part (c)).

\((t = ){\text{ }}2.89711 \ldots \)     (A1)(ft)(G1)

\(174{\text{ (seconds) }}\left( {173.826 \ldots {\text{ (seconds)}}} \right)\)     (A1)(ft)(G2)

Note:     Award final (A1)(ft) for converting their \({\text{2.89711}} \ldots \) minutes into seconds.

[3 marks]

the temperature of the (dining) room     (A1)

OR

the lowest final temperature to which the pizza will cool     (A1)

[1 mark]

(i)     Write down the Pearson’s product–moment correlation coefficient, \(r\), for these data.

(ii)     Hence comment on the result.

Write down the equation of the regression line \(y\) on \(x\) for these data, in the form \(y = ax + b\).

Estimate the total cost, to the nearest USD, of producing \(13\) bicycles on a particular day.

All the bicycles that are produced are sold. The bicycles are sold for 304 USD each.

Explain why the factory does not make a profit when producing \(13\) bicycles on a particular day.

All the bicycles that are produced are sold. The bicycles are sold for 304 USD each.

(i)     Write down an expression for the total selling price of \(x\) bicycles.

(ii)     Write down an expression for the profit the factory makes when producing \(x\) bicycles on a particular day.

(iii)     Find the least number of bicycles that the factory should produce, on a particular day, in order to make a profit.

(i)     \(r = 0.985\;\;\;(0.984905 \ldots )\)     (G2)

Notes: If unrounded answer is not seen, award (G1)(G0) for \(0.99\) or \(0.984\). Award (G2) for \(0.98\).

(ii)     strong, positive     (A1)(A1)

\(y = 259.909 \ldots x + 698.648 \ldots \;\;\;(y = 260x + 699)\)     (G1)(G1)

Notes: Award (G1) for \(260x\) and (G1) for \(699\). If the answer is not an equation award a maximum of (G1)(G0).

\(y = 259.909 \ldots  \times 13 + 698.648 \ldots \)     (M1)

Note: Award (M1) for substitution of \(13\) into their regression line equation from part (b).

\(y = 4077.47 \ldots \)     (A1)(ft)(G2)

\(y = 4077{\text{ (USD)}}\)     (A1)(ft)

Notes: Follow through from their answer to part (b). If rounded values from part (b) used, answer is \(4079\). Award the final (A1)(ft) for a correct rounding to the nearest USD of their answer. The unrounded answer may not be seen.

If answer is \(4077\) and no working is seen, award (G2).

\(13 \times 304 - (4077.47) =  - 125.477 \ldots \;\;\;( - 125)\;\;\;\)OR

\(4077.47 - (13 \times 304) = 125.477 \ldots \;\;\;(125)\)     (M1)

Notes: Award (M1) for calculating the difference between \(13 \times 304\) and their answer to part (c).

If rounded values are used in equation, answer is \( - 127\).

profit is negative\(\;\;\;\)OR\(\;\;\;{\text{cost}} > {\text{sales}}\)     (A1)

OR

\(13 \times 304 = 3952\)     (M1)

Note: Award (M1) for calculating the price of \(13\) bikes.

\(3952 < 4077.47\)     (A1)(ft)

Note: Award (A1) for showing \(3952\) is less than their part (c). This may be communicated in words. Follow through from part (c), but only if value is greater than \(3952\).

OR

\(\frac{{4077}}{{13}} = 313.62\)     (M1)

Note: Award (M1) for calculating the cost of \(1\) bicycle.

\(313.62 > 304\)     (A1)(ft)

Note: Award (A1) for showing \(313.62\) is greater than \(304\). This may be communicated in words. Follow through from part (c), but only if value is greater than \(304\).

OR

\(\frac{{4077}}{{304}} = 13.41\)     (M1)

Note: Award (M1) for calculating the number of bicycles that should have been be sold to cover total cost.

\(13.41 > 13\)     (A1)(ft)

Note: Award (A1) for showing \(13.41\) is greater than \(13\). This may be communicated in words. Follow through from part (c), but only if value is greater than \(13\).

(i)     \(304x\)     (A1)

(ii)     \(304x - (259.909 \ldots x + 698.648 \ldots )\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for difference between their answers to parts (b) and (e)(i), (A1)(ft) for correct expression.

(iii)     \(304x - (259.909 \ldots x + 698.648 \ldots ) > 0\)     (M1)

Notes: Award (M1) for comparing their expression in part (e)(ii) to \(0\). Accept an equation. Accept \(3040x - y > 0\) or equivalent.

\(x = 16{\text{ bicycles}}\)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (b). Answer must be a positive integer greater than \(13\) for the (A1)(ft) to be awarded.

Award (G1) for an answer of \(15.84\).

Find the gradient of the line L.

Differentiate \(f (x)\) .

Find the coordinates of the point where the tangent to P is parallel to the line L.

Find the coordinates of the point where the tangent to P is perpendicular to the line L.

Find

(i) the gradient of the tangent to P at the point with coordinates (2, − 6).

(ii) the equation of the tangent to P at this point.

State the equation of the axis of symmetry of P.

Find the coordinates of the vertex of P and state the gradient of the curve at this point.

for attempt at substituted \(\frac{{ydistance}}{{xdistance}}\)     (M1)

gradient = 2     (A1)(G2)

[2 marks]

\(2x - 3\)     (A1)(A1)

(A1) for \(2x\) , (A1) for \(-3\)

[2 marks]

for their \(2x - 3 =\) their gradient and attempt to solve     (M1)

\(x = 2.5\)     (A1)(ft)

\(y = -5.25\) ((ft) from their x value)     (A1)(ft)(G2)

[3 marks]

for seeing \(\frac{{ - 1}}{{their(a)}}\)     (M1)

solving \(2x – 3 = - \frac{1}{2}\) (or their value)     (M1)

x = 1.25     (A1)(ft)(G1)

y = – 6.1875     (A1)(ft)(G1)

[4 marks]

(i) \(2 \times 2 - 3 = 1\) ((ft) from (b))     (A1)(ft)(G1)

(ii) \(y = mx + c\) or equivalent method to find \(c \Rightarrow -6 = 2 + c\)     (M1)

\(y = x - 8\)     (A1)(ft)(G2)

[3 marks]

\(x = 1.5\)     (A1)

[1 mark]

for substituting their answer to part (f) into the equation of the parabola (1.5, −6.25) accept x = 1.5, y = −6.25     (M1)(A1)(ft)(G2)

gradient is zero (accept \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\))     (A1)

[3 marks]

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

Parts (a) and (b) were very well done. After that, only the stronger candidates were able to cope. The equation of the tangent at the point with coordinates (2, 6) was badly done but some candidates managed to find the equation of the tangent line from their GDC. The equation of the axis of symmetry was reasonably well done although many just wrote down 1.5 instead of x = 1.5.

Some forgot to write down that the gradient at the vertex was 0.

The coordinates of point A are (75, 450). Determine whether point A is on the bicycle track. Give a reason for your answer.

Find the derivative of \(y = \frac{{ - {x^2}}}{{10}} + \frac{{27}}{2}x\).

Use the answer in part (b) to determine if A (75, 450) is the point furthest north on the track between O and B. Give a reason for your answer.

(i) Write down the midpoint of the line segment OB.

(ii) Find the gradient of the line segment OB.

Scott starts from a point C(0,150) . He hikes along a straight road towards the bicycle track, parallel to the line segment OB.

Find the equation of Scott’s road. Express your answer in the form \(ax + by = c\), where \(a, b {\text{ and }} c \in \mathbb{R}\).

Use your graphic display calculator to find the coordinates of the point where Scott first crosses the bicycle track.

\(y =  - \frac{{{{75}^2}}}{{10}} + \frac{{27}}{2} \times 75\)     (M1)

Note: Award (M1) for substitution of 75 in the formula of the function.

= 450     (A1)

Yes, point A is on the bike track. (A1)

Note: Do not award the final (A1) if correct working is not seen.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{{2x}}{{10}} + \frac{{27}}{2}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 0.2x + 13.5} \right)\)     (A1)(A1)

Notes: Award (A1) for each correct term. If extra terms are seen award at most (A1)(A0). Accept equivalent forms.

\( - \frac{{2x}}{{10}} + \frac{{27}}{2} = 0\)     (M1)

Note: Award (M1) for equating their derivative from part (b) to zero.

\(x = 67.5\)     (A1)(ft)

Note: Follow through from their derivative from part (b).

\( {\text{(Their) }} 67.5 \ne 75\)      (R1)

Note: Award (R1) for a comparison of their 67.5 with 75. Comparison may be implied (eg 67.5 is the x-coordinate of the furthest north point).

OR

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{{2 \times (75)}}{{10}} + \frac{{27}}{2}\)     (M1)

Note: Award (M1) for substitution of 75 into their derivative from part (b).

\(= -1.5\)     (A1)(ft)

Note: Follow through from their derivative from part (b).

\({\text{(Their)}} -1.5 \ne 0\)     (R1)

Note: Award (R1) for a comparison of their –1.5 with 0. Comparison may be implied (eg The gradient of the parabola at the furthest north point (vertex) is 0).

Hence A is not the furthest north point.     (A1)(ft)   

Note: Do not award (R0)(A1)(ft). Follow through from their derivative from part (b).

(i) M(50,175)     (A1)

Note: If parentheses are omitted award (A0). Accept x = 50, y = 175.

(ii) \(\frac{{350 - 0}}{{100 - 0}}\)     (M1)

Note: Award (M1) for correct substitution in gradient formula.

\( = 3.5\left( {\frac{{350}}{{100}},\frac{7}{2}} \right)\)     (A1)(ft)(G2)

Note: Follow through from (d)(i) if midpoint is used to calculate gradient. Award (G1)(G0) for answer 3.5x without working.

\(y = 3.5x + 150\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for using their gradient from part (d), (A1)(ft) for correct equation of line.

\(3.5x - y = -150\) or \(7x - 2y = -300\) (or equivalent)     (A1)(ft)

Note: Award (A1)(ft) for expressing their equation in the form \(ax + by = c\).

(18.4, 214) (18.3772..., 214.320...)     (A1)(ft)(A1)(ft)(G2)(ft)

Notes: Follow through from their equation in (e). Coordinates must be positive for follow through marks to be awarded. If parentheses are omitted and not already penalized in (d)(i) award at most (A0)(A1)(ft). If coordinates of the two intersection points are given award (A0)(A1)(ft). Accept x = 18.4, y = 214.

Find the amount of money that Ruby will have in the bank after 3 years.

Show that Ying will have 7545 USD in the bank at the end of 10 years.

Find the number of complete years it will take for Ying’s investment to first exceed 6500 USD.

Find the number of complete years it will take for Ying’s investment to exceed Ruby’s investment.

Ruby moves from the USA to Italy. She transfers 6610 USD into an Italian bank which has an exchange rate of 1 USD = 0.735 Euros. The bank charges 1.8 % commission.

Calculate the amount of money Ruby will invest in the Italian bank after commission.

Ruby returns to the USA for a short holiday. She converts 800 Euros at a bank in Chicago and receives 1006.20 USD. The bank advertises an exchange rate of 1 Euro = 1.29 USD.

Calculate the percentage commission Ruby is charged by the bank.

5000 + 3 × 230 = 5690     (M1)(A1)(G2)

Note: Accept alternative method.

[2 marks]

\({\text{A}} = 5000{\left( {1 + \frac{{4.2}}{{100}}} \right)^{10}}\) or equivalent     (M1)(A1)

\( = 7544.79 \ldots \)     (A1)

\( = 7545{\text{ USD}}\)     (AG)

Note: Award (M1) for correct substituted compound interest formula, (A1) for correct substitutions, (A1) for unrounded answer seen.

If final line not seen award at most (M1)(A1)(A0).

[3 marks]

5000(1.042)ⁿ > 6500     (M1)(A1)

Notes: Award (M1) for setting up correct equation/inequality, (A1) for correct values.

Follow through from their formula in part (b).

OR

List of values seen with at least 2 terms     (M1)

Lists of values including at least the terms with n = 6 and n = 7     (A1)

Note: Follow through from their formula in part (b).

OR

Sketch showing 2 graphs, one exponential, the other a horizontal line     (M1)

Point of intersection identified or vertical line     (M1)

Note: Follow through from their formula in part (b).

n = 7     (A1)(ft)(G2)

[3 marks]

5000(1.042)ⁿ > 5000 + 230n     (M1)(A1)

Note: Award (M1) for setting up correct equation/inequality, (A1) for correct values.

OR

2 lists of values seen (at least 2 terms per list)     (M1)

Lists of values including at least the terms with n = 5 and n = 6     (A1)

Note: One of the lists may be written under (c).

OR

Sketch showing 2 graphs of correct shape     (M1)

Point of intersection identified or vertical line     (M1)

n = 6     (A1)(ft)(G2)

Note: Follow through from their formulae used in parts (a) and (b).

[3 marks]

6610 × 0.735     (M1)

= 4858.35     (A1)

4858.35 × 0.982(= 4770.8997...)     (M1)

= 4771 Euros     (A1)(ft)(G3)

Note: Accept alternative method.

[4 marks]

800 × 1.29 (= 1032 USD)     (M1)(A1)

Note: Award (M1) for multiplying by 1.29, (A1) for 1032. Award (G2) for 1032 if product not seen.

(1032 – 1006.20 = 25.8)

\(25.8 \times \frac{100}{1032} \%\)     (A1)(M1)

Note: Award (A1) for 25.8 seen, (M1) for multiplying by \(\frac{100}{1032}\).

OR

\(\frac{1006.20}{1032} = 0.975\)     (M1)(A1)

OR

\(\frac{1006.20}{1032} \times 100 = 97.5\)     (M1)(A1)

\( = 2.5{\text{ }}\% \)     (A1)(G3)

Notes: If working not shown award (G3) for 2.5.

Accept alternative method.

[5 marks]

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

It was nice to see many students recovering after part (d) and to gain full marks in the last two parts of the question.