Find \({f^{ - 1}}(x)\).

Let \(g\) be a function so that \((f \circ g)(x) = 8{x^6}\). Find \(g(x)\).

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg\(\;\;\;x = {(y - 5)^3}\)

evidence of correct manipulation     (A1)

eg\(\;\;\;y - 5 = \sqrt[3]{x}\)

\({f^{ - 1}}(x) = \sqrt[3]{x} + 5\;\;\;({\text{accept }}5 + {x^{\frac{1}{3}}},{\text{ }}y = 5 + \sqrt[3]{x})\)     A1     N2

Notes:     If working shown, and they do not interchange \(x\) and \(y\), award A1A1M0 for \(\sqrt[3]{y} + 5\).

If no working shown, award N1 for \(\sqrt[3]{y} + 5\).

METHOD 1

attempt to form composite (in any order)     (M1)

eg\(\;\;\;g\left( {{{(x - 5)}^3}} \right),{\text{ }}{\left( {g(x) - 5} \right)^3} = 8{x^6},{\text{ }}f(2{x^2} + 5)\)

correct working     (A1)

eg\(\;\;\;g - 5 = 2{x^2},{\text{ }}{\left( {(2{x^2} + 5) - 5} \right)^3}\)

\(g(x) = 2{x^2} + 5\)     A1     N2

METHOD 2

recognising inverse relationship     (M1)

eg\(\;\;\;{f^{ - 1}}(8{x^6}) = g(x),{\text{ }}{f^{ - 1}}(f \circ g)(x) = {f^{ - 1}}(8{x^6})\)

correct working

eg\(\;\;\;g(x) = \sqrt[3]{{(8{x^6})}} + 5\)     (A1)

\(g(x) = 2{x^2} + 5\)     A1     N2

Find p .

The graph of f is reflected in the line \(y = x\) to give the graph of g .

(i)     Write down the y-intercept of the graph of g .

(ii)    Sketch the graph of g , noting clearly any asymptotes and the image of A.

The graph of \(f\) is reflected in the line \(y = x\) to give the graph of \(g\) .

Find \(g(x)\) .

evidence of substituting the point A     (M1)

e.g. \(2 = {\log _p}(6 + 3)\)

manipulating logs     A1

e.g. \({p^2} = 9\)

\(p = 3\)     A2     N2

[4 marks]

(i) \(y = - 2\) (accept \((0{\text{, }} - 2))\)     A1     N1

(ii)

     A1A1A1A1     N4

Note: Award A1 for asymptote at \(y = - 3\) , A1 for an increasing function that is concave up, A1 for a positive x-intercept and a negative y-intercept, A1 for passing through the point \((2{\text{, }}6)\) .

[5 marks] 

METHOD 1

recognizing that \(g = {f^{ - 1}}\)     (R1)

evidence of valid approach     (M1)

e.g. switching x and y (seen anywhere), solving for x

correct manipulation     (A1)

e.g. \({3^x} = y + 3\)

\(g(x) = {3^x} - 3\)     A1     N3

METHOD 2

recognizing that \(g(x) = {a^x} + b\)     (R1) 

identifying vertical translation     (A1)

e.g. graph shifted down 3 units, \(f(x) - 3\)

evidence of valid approach     (M1)

e.g. substituting point to identify the base

\(g(x) = {3^x} - 3\)     A1     N3

[4 marks]

In part (a), many candidates successfully substituted the point A to find the base of the logarithm, although some candidates lost a mark for not showing their manipulation of the logarithm equation into the exponential equation.

A number of candidates who correctly stated the y-intercept was \( - 2\) had difficulty sketching the graph of the reflection in the line \(y = x\) . A number of candidates graphed directly on the question paper rather than sketching their own graph; candidates should be reminded to show all working for Section B on separate paper. Some correct sketches did not have the position of A indicated. Many candidates had difficulty reflecting the asymptote.

Part (c) was often well done, with candidates showing clear and correct working.

The most successful candidates clearly appreciated the linkage between the question parts.

Solve for \(0 \le x < 2\pi \)

(i)     \(6 + 6\sin x = 6\) ;

(ii)    \(6 + 6\sin x = 0\) .

Write down the exact value of the x-intercept of f , for \(0 \le x < 2\pi \) .

The area of the shaded region is k . Find the value of k , giving your answer in terms of \(\pi \) .

Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Given that \(\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x}  = k\) and \(0 \le p < 2\pi \) , write down the two values of p.

(i) \(\sin x = 0\)     A1

\(x = 0\) , \(x = \pi \)     A1A1     N2

(ii) \(\sin x = - 1\)     A1

\(x = \frac{{3\pi }}{2}\)    A1     N1

[5 marks]

\(\frac{{3\pi }}{2}\)     A1     N1

[1 mark]

evidence of using anti-differentiation     (M1)

e.g. \(\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x} \)

correct integral \(6x - 6\cos x\) (seen anywhere)     A1A1

correct substitution     (A1)

e.g. \(6\left( {\frac{{3\pi }}{2}} \right) - 6\cos \left( {\frac{{3\pi }}{2}} \right) - ( - 6\cos 0)\) , \(9\pi  - 0 + 6\)

\(k = 9\pi + 6\)     A1A1     N3

[6 marks]

translation of \(\left( {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right)\)     A1A1     N2

[2 marks]

recognizing that the area under g is the same as the shaded region in f     (M1)

\(p = \frac{\pi }{2}\) , \(p = 0\)     A1A1     N3

[3 marks]

Many candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not show sufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidates worked in degrees; the majority worked in radians.

While some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), most used their working from part (a)(ii) sometimes with follow-through on an incorrect answer.

Most candidates recognized the need for integration in part (c) but far fewer were able to see the solution through correctly to the end. Some did not show the full substitution of the limits, having incorrectly assumed that evaluating the integral at 0 would be 0; without this working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with the common trigonometric values.

While there was an issue in the wording of the question with the given domains, this did not appear to bother candidates in part (d). This part was often well completed with candidates using a variety of language to describe the horizontal translation to the right by \(\frac{\pi }{2}\) .

Most candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majority tried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in the wording of the question did not appear to have been noticed by candidates in this part either.

The graph of g can be obtained from the graph of f using two transformations.

Give a full geometric description of each of the two transformations.

The graph of g is translated by the vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 2}
\end{array}} \right)\) to give the graph of h.

The point \(( - 1{\text{, }}1)\) on the graph of f is translated to the point P on the graph of h.

Find the coordinates of P.

in any order

translated 1 unit to the right     A1     N1

stretched vertically by factor 2     A1     N1

[2 marks]

METHOD 1

finding coordinates of image on g     (A1)(A1)

e.g.  \( - 1 + 1 = 0\) , \(1 \times 2 = 2\) , \(( - 1{\text{, }}1) \to ( - 1 + 1{\text{, }}2 \times 1)\) , \((0{\text{, }}2)\)

P is (3, 0)     A1A1     N4

METHOD 2

\(h(x) = 2{(x - 4)^2} - 2\)     (A1)(A1)

P is \((3{\text{, }}0)\)     A1A1     N4

The translation was often described well as horizontal (or shift) one unit right. There was considerable difficulty describing the vertical stretch as it was often referred to as "stretch by 2" or "amplitude of 2". A full description should include the name (e.g. vertical stretch) and value for full marks. Candidates also had difficulty applying two consecutive transformations to a single point. Often the translations were applied directly to \(( - 1{\text{, }}1)\) instead of first mapping from f to g . A good number of candidates correctly found \(h(x)\), but most could not find P from this function.

The translation was often described well as horizontal (or shift) one unit right. There was considerable difficulty describing the vertical stretch as it was often referred to as "stretch by 2" or "amplitude of 2". A full description should include the name (e.g. vertical stretch) and value for full marks. Candidates also had difficulty applying two consecutive transformations to a single point. Often the translations were applied directly to \(( - 1{\text{, }}1)\) instead of first mapping from f to g . A good number of candidates correctly found \(h(x)\), but most could not find P from this function.

Show that the two zeros are 3 and \( - 6\).

Find the value of \(q\) and of \(r\).

recognition that the \(x\)-coordinate of the vertex is \( - 1.5\) (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)axis of symmetry is \( - 1.5\), sketch, \(f'( - 1.5) = 0\)

correct working to find the zeroes     A1

eg\(\,\,\,\,\,\)\( - 1.5 \pm 4.5\)

\(x =  - 6\) and \(x = 3\)     AG     N0

[2 marks]

METHOD 1 (using factors)

attempt to write factors     (M1)

eg\(\,\,\,\,\,\)\((x - 6)(x + 3)\)

correct factors     A1

eg\(\,\,\,\,\,\)\((x - 3)(x + 6)\)

\(q = 3,{\text{ }}r =  - 18\)    A1A1     N3

METHOD 2 (using derivative or vertex)

valid approach to find \(q\)     (M1)

eg\(\,\,\,\,\,\)\(f'( - 1.5) = 0,{\text{ }} - \frac{q}{{2a}} =  - 1.5\)

\(q = 3\)    A1

correct substitution     A1

eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( - 6)^2} + 3( - 6) + r = 0\)

\(r =  - 18\)    A1

\(q = 3,{\text{ }}r =  - 18\)    N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations     (M1)

eg\(\,\,\,\,\,\)\(9 + 3q + r = 0,{\text{ }}36 - 6q + r = 0\)

one correct value

eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r =  - 18\)     A1

correct substitution     A1

eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( - 6)^2} + 3( - 6) + r = 0,{\text{ }}{3^2} + 3q - 18 = 0,{\text{ }}36 - 6q - 18 = 0\)

second correct value     A1

eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r =  - 18\)

\(q = 3,{\text{ }}r =  - 18\)    N3

[4 marks]

As a ‘show that’ question, part a) required a candidate to independently find the answers. Again, too many candidates used the given answers (of 3 and \( - 6\)) to show that the two zeros were 3 and \( - 6\) (a circular argument). Those who were able to recognize that the \(x\)-coordinate of the vertex is \( - 1.5\) tended to then use the given answers and work backwards thus scoring no further marks in part a).

Answers to part b) were more successful with a good variety of methods used and correct solutions seen.

Write down the value of \(g(3)\) , of \(f'(3)\) , and of \(h''(2)\) .

Explain why P is a point of inflexion.

find the \(y\)-coordinate of P.

find the equation of the normal to the graph of \(h\) at P.

\(g(3) = - 18\) , \(f'(3) = 1\) , \(h''(2) = - 6\)     A1A1A1     N3

[3 marks]

\(h''(3) = 0\)     (A1)

valid reasoning     R1

eg   \({h''}\) changes sign at \(x = 3\) , change in concavity of \(h\) at \(x = 3\)

so P is a point of inflexion     AG     N0

[2 marks]

writing \(h(3)\) as a product of \(f(3)\) and \(g(3)\)     A1

eg   \(f(3) \times g(3)\) , \(3 \times ( - 18)\)

\(h(3) = - 54\)     A1 N1

[2 marks]

recognizing need to find derivative of \(h\)     (R1)

eg   \({h'}\) , \(h'(3)\)

attempt to use the product rule (do not accept \(h' = f' \times g'\) )     (M1)

eg   \(h' = fg' + gf'\) ,  \(h'(3) = f(3) \times g'(3) + g(3) \times f'(3)\)

correct substitution     (A1)

eg   \(h'(3) = 3( - 3) + ( - 18) \times 1\)

\(h'(3) = - 27\)    A1

attempt to find the gradient of the normal     (M1)

eg   \( - \frac{1}{m}\) , \( - \frac{1}{{27}}x\) 

attempt to substitute their coordinates and their normal gradient into the equation of a line     (M1)

eg   \( - 54 = \frac{1}{{27}}(3) + b\) , \(0 = \frac{1}{{27}}(3) + b\) , \(y + 54 = 27(x - 3)\) , \(y - 54 = \frac{1}{{27}}(x + 3)\)

correct equation in any form     A1     N4

eg   \(y + 54 = \frac{1}{{27}}(x - 3)\) , \(y = \frac{1}{{27}}x - 54\frac{1}{9}\)

[7 marks]

Nearly all candidates who attempted to answer parts (a) and (c) did so correctly, as these questions simply required them to understand the notation being used and to read the values from the given table.

In part (b), the majority of candidates earned one mark for stating that \(h''(x) = 0\) at point P. As this is not enough to determine a point of inflexion, very few candidates earned full marks on this question.

Nearly all candidates who attempted to answer parts (a) and (c) did so correctly, as these questions simply required them to understand the notation being used and to read the values from the given table.

Part (d) proved to be quite challenging for even the strongest candidates, as almost none of them used the product rule to find \(h'(3)\). The most common error was to say \(h'(3) = f'(3) \times g'(3)\). Despite this error, many candidates were able to earn further method marks for their work in finding the equation of the normal. There were also a small number of candidates who were able to find the equation for \(h'(x)\) , and from that \(h''(x)\). These candidates were often successful in earning full marks, although this method was quite time-consuming.

Find \({f^{ - 1}}(x)\) . 

Find \((f \circ g)(1)\) . 

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = 2y - 1\)

correct manipulation     (A1)

e.g. \(x + 1 = 2y\)

\({f^{ - 1}}(x) = \frac{{x + 1}}{2}\)      A1     N2

[3 marks]

METHOD 1

attempt to find or \(g(1)\) or \(f(1)\)     (M1)

\(g(1) = 5\)     (A1)

\(f(5) = 9\)     A1     N2 

[3 marks]

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(2(3{x^2} + 2) - 1\) , \(3{(2x - 1)^2} + 2\)

\((f \circ g)(1) = 2(3 \times {1^2} + 2) - 1\) \(( = 6 \times {1^2} + 3)\)     (A1)

\((f \circ g)(1) = 9\)     A1     N2

[3 marks]

This question was answered correctly by nearly all candidates.

This question was answered correctly by nearly all candidates. In part (b), there were a few who seemed unfamiliar with the notation for composition of functions, and attempted to multiply the functions rather than finding the composite, and there were a few who found the correct composite function but failed to substitute in \(x = 1\) to find the value.

Write down the value of \(f(2)\).

Write down the value of \({f^{ - 1}}( - 1)\) .

Sketch the graph of \({f^{ - 1}}\) on the grid below.

\(f(2) = 3\)     A1     N1

[1 mark]

\({f^{ - 1}}( - 1) = 0\)     A2     N2

[2 marks]

EITHER

attempt to draw \(y = x\) on grid     (M1)

OR

attempt to reverse x and y coordinates     (M1)

eg   writing or plotting at least two of the points

\(( - 2, - 1)\) , \(( - 1,0)\) , \((0,1)\) , \((3,2)\)

THEN

correct graph     A2     N3

[3 marks]

In part (a) of this question, most candidates were able to find the value of \(f(2)\) correctly, while some had trouble finding \({f^{ - 1}}( - 1)\). Many candidates tried to find an equation for the function, or to make tables of values to help them find their answers. The intent of this question was to read the answers from the given graph. Candidates should be reminded that when the command term is "write down", there is no need for them to do large amounts of working.

In part (b) of this question, candidates were generally successful in reversing the \(x\) and \(y\) coordinates of key points or reflecting in the \(y = x\) line to correctly sketch the graph of the inverse function. Common errors included not sketching the graph for the appropriate domain, or sketching the graph of \(f(-x)\) or the graph of \(-f(x)\).

Show that the equation of L is \(y = - 4x + 18\) .

Point A is the x-intercept of L . Find the x-coordinate of A.

Find an expression for the area of R .

The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .

finding derivative     (A1)

e.g. \(f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}\)

correct value of derivative or its negative reciprocal (seen anywhere)     A1

e.g. \(\frac{1}{{2\sqrt 4 }}\) , \(\frac{1}{4}\)

gradient of normal =  \(\frac{1}{{{\text{gradient of tangent}}}}\) (seen anywhere)     A1

e.g. \( - \frac{1}{{f'(4)}} = - 4\) , \( - 2\sqrt x \)

substituting into equation of line (for normal)     M1

e.g. \(y - 2 = - 4(x - 4)\)

\(y = - 4x + 18\)     AG     N0

[4 marks]

recognition that \(y = 0\) at A     (M1)

e.g. \( - 4x + 18 = 0\)

\(x = \frac{{18}}{4}\) \(\left( { = \frac{9}{2}} \right)\)     A1     N2

[2 marks]

splitting into two appropriate parts (areas and/or integrals)     (M1)

correct expression for area of R     A2     N3

e.g. area of R = \(\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( - 4x + 18){\rm{d}}x} \) , \(\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2\) (triangle)

Note: Award A1 if dx is missing.

[3 marks]

correct expression for the volume from \(x = 0\) to \(x = 4\)     (A1)

e.g. \(V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x\) , \({\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x\) , \(\int_0^4 {\pi x{\rm{d}}x} \)

\(V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4\)     A1

\(V = \pi \left( {\frac{1}{2} \times 16 - \frac{1}{2} \times 0} \right)\)     (A1)

\(V = 8\pi \)     A1

finding the volume from \(x = 4\) to \(x = 4.5\)

EITHER

recognizing a cone     (M1)

e.g. \(V = \frac{1}{3}\pi {r^2}h\)

\(V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}\)     (A1)

\( = \frac{{2\pi }}{3}\)     A1

total volume is \(8\pi  + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\)     A1     N4

OR

\(V = \pi \int_4^{4.5} {{{( - 4x + 18)}^2}{\rm{d}}x} \)     (M1)

\( = \int_4^{4.5} {\pi (16{x^2} - 144x + 324){\rm{d}}x} \)

\( = \pi \left[ {\frac{{16}}{3}{x^3} - 72{x^2} + 324x} \right]_4^{4.5}\)     A1

\( = \frac{{2\pi }}{3}\)     A1

total volume is \(8\pi  + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\)     A1     N4

[8 marks]

Parts (a) and (b) were well done by most candidates.

Parts (a) and (b) were well done by most candidates.

While quite a few candidates understood that both functions must be used to find the area in part (c), very few were actually able to write a correct expression for this area and this was due to candidates not knowing that they needed to integrate from \(0\) to \(4\) and then from \(4\) to \(4.5\).

On part (d), some candidates were able to earn follow through marks by setting up a volume expression, but most of these expressions were incorrect. If they did not get the expression for the area correct, there was little chance for them to get part (d) correct.

For those candidates who used their expression in part (c) for (d), there was a surprising amount of them who incorrectly applied distributive law of the exponent with respect to the addition or subtraction.

Show that \(f(x) = 3{x^2} + 6x - 9\) .

For the graph of f

(i)     write down the coordinates of the vertex;

(ii)    write down the y-intercept;

(iii)   find both x-intercepts.

Hence sketch the graph of f .

Let \(g(x) = {x^2}\) . The graph of f may be obtained from the graph of g by the following two transformations

a stretch of scale factor t in the y-direction,

followed by a translation of \(\left( \begin{array}{l}
p\\
q
\end{array} \right)\) .

Write down \(\left( \begin{array}{l}
p\\
q
\end{array} \right)\) and the value of t .

\(f(x) = 3({x^2} + 2x + 1) - 12\)     A1

\( = 3{x^2} + 6x + 3 - 12\)     A1

\( = 3{x^2} + 6x - 9\)     AG     N0

[2 marks]

(i) vertex is \(( - 1, - 12)\)     A1A1     N2

(ii) \(y = - 9\) , or \((0, - 9)\)     A1     N1

(iii) evidence of solving \(f(x) = 0\)     M1

e.g. factorizing, formula

correct working     A1

e.g. \(3(x + 3)(x - 1) = 0\) , \(x = \frac{{ - 6 \pm \sqrt {36 + 108} }}{6}\)

\(x = - 3\) , \(x = 1\) , or \(( - 3{\text{, }}0){\text{, }}(1{\text{, }}0)\)     A1A1     N2

[7 marks]

     A1A1A1     N3

Note: Award A1 for a parabola opening upward, A1 for vertex in approximately correct position, A1 for intercepts in approximately correct positions. Scale and labelling not required.

[3 marks]

\(\left( \begin{array}{l}
p\\
q
\end{array} \right) = \left( \begin{array}{l}
- 1\\
- 12
\end{array} \right)\) , \(t = 3\)     A1A1A1     N3

[3 marks]

Find \((g \circ f)(x)\) .

Write down \({g^{ - 1}}(x)\) .

Find \((f \circ {g^{ - 1}})(5)\) .

attempt to form composite     (M1)

e.g. \(g(7 - 2x)\) , \(7 - 2x + 3\)

\((g \circ f)(x) = 10 - 2x\)     A1     N2

[2 marks]

\({g^{ - 1}}(x) = x - 3\)     A1     N1

[1 mark]

METHOD 1

valid approach     (M1)

e.g. \({g^{ - 1}}(5)\) , \(2\) , \(f(5)\)

\(f(2) = 3\)     A1     N2

METHOD 2

attempt to form composite of f and \({g^{ - 1}}\)     (M1)

e.g. \((f \circ {g^{ - 1}})(x) = 7 - 2(x - 3)\) , \(13 - 2x\)

\((f \circ {g^{ - 1}})(5) = 3\)     A1     N2

[2 marks]

A majority of candidates found success in the opening question. Common errors in (a) were to give \(f \circ g\) or to multiply f by g. 

For (b) some gave the inverse as the reciprocal function \(\frac{1}{{x + 3}}\) , or wrote \(x = y + 3\) .

Most candidates chose to find a composite in (c), sometimes making simple errors when working with brackets and a negative sign. Only a handful used the more efficient \(f(2) = 3\) . Additionally, it was not uncommon for candidates to give a correct substitution but not complete the result. Simple expressions such as \((7 - 2x) + 3\) should be finished as \(10 - 2x\) .

Write down the value of \(f( - 3)\).

Write down the value of  \({f^{ - 1}}(1)\).

Find the domain of \({f^{ - 1}}\).

On the grid above, sketch the graph of \({f^{ - 1}}\).

\(f( - 3) =  - 1\)     A1     N1

[1 mark]

\({f^{ - 1}}(1) = 0\)   (accept \(y = 0\))     A1     N1

[1 mark]

domain of \({f^{ - 1}}\) is range of \(f\)     (R1)

eg     \({\text{R}}f = {\text{D}}{f^{ - 1}}\)

correct answer     A1     N2

eg     \( - 3 \leqslant x \leqslant 3,{\text{ }}x \in [ - 3,{\text{ }}3]{\text{   (accept }} - 3 < x < 3,{\text{ }} - 3 \leqslant y \leqslant 3)\)

[2 marks]

     A1A1     N2

Note: Graph must be approximately correct reflection in \(y = x\).

     Only if the shape is approximately correct, award the following:

     A1 for x-intercept at \(1\), and A1 for endpoints within circles.

[2 marks]

(a)     Write down the \(x\)-intercepts of the graph of \(f\) .

(b)     Find the coordinates of the vertex of the graph of \(f\) .

Write down the \(x\)-intercepts of the graph of \(f\) .

Find the coordinates of the vertex of the graph of \(f\) .

(a)     \(x = 1\) , \(x = - 3\) (accept (\(1\), \(0\)), (\( - 3\), \(0\)) )     A1A1     N2

[2 marks]

(b)     METHOD 1

attempt to find \(x\)-coordinate     (M1)

eg   \(\frac{{1 + - 3}}{2}\) , \(x = \frac{{ - b}}{{2a}}\) , \(f'(x) = 0\)

correct value, \(x = - 1\) (may be seen as a coordinate in the answer)     A1

attempt to find their \(y\)-coordinate     (M1)

eg   \(f( - 1)\) , \( - 2 \times 2\) , \(y = \frac{{ - D}}{{4a}}\)

\(y = - 4\)     A1

vertex (\( - 1\), \( - 4\))     N3  

METHOD 2

attempt to complete the square     (M1)

eg   \({x^2} + 2x + 1 - 1 - 3\) 

attempt to put into vertex form     (M1)

eg   \({(x + 1)^2} - 4\) , \({(x - 1)^2} + 4\)

vertex (\( - 1\), \( - 4\))     A1A1     N3

[4 marks]

\(x = 1\) , \(x = - 3\) (accept (\(1\), \(0\)), (\( - 3\), \(0\)) )     A1A1     N2

[2 marks]

METHOD 1

attempt to find \(x\)-coordinate     (M1)

eg   \(\frac{{1 + - 3}}{2}\) , \(x = \frac{{ - b}}{{2a}}\) , \(f'(x) = 0\)

correct value, \(x = - 1\) (may be seen as a coordinate in the answer)     A1

attempt to find their \(y\)-coordinate     (M1)

eg   \(f( - 1)\) , \( - 2 \times 2\) , \(y = \frac{{ - D}}{{4a}}\)

\(y = - 4\)     A1

vertex (\( - 1\), \( - 4\))     N3  

METHOD 2

attempt to complete the square     (M1)

eg   \({x^2} + 2x + 1 - 1 - 3\) 

attempt to put into vertex form     (M1)

eg   \({(x + 1)^2} - 4\) , \({(x - 1)^2} + 4\)

vertex (\( - 1\), \( - 4\))     A1A1     N3

[4 marks]

Most candidates recognized the values of the x-intercepts from the factorized form of the function. Candidates also showed little difficulty finding the vertex of the graph, and employed a variety of techniques: averaging \(x\)-intercepts, using \(x = \frac{{ - b}}{{2a}}\) , completing the square.

Most candidates recognized the values of the x-intercepts from the factorized form of the function. Candidates also showed little difficulty finding the vertex of the graph, and employed a variety of techniques: averaging \(x\)-intercepts, using \(x = \frac{{ - b}}{{2a}}\) , completing the square.

Most candidates recognized the values of the x-intercepts from the factorized form of the function. Candidates also showed little difficulty finding the vertex of the graph, and employed a variety of techniques: averaging \(x\)-intercepts, using \(x = \frac{{ - b}}{{2a}}\) , completing the square.

Write down f (14).

Find \(\left( {g \circ f} \right)\) (14).

Find g⁻¹(x).

f (14) = 4     A1 N1

[1 mark]

attempt to substitute     (M1)

eg   g (4), 3 × 4 − 7

5     A1 N2

[2 marks]

interchanging x and y (seen anywhere)     (M1)

eg   x = 3y − 7

evidence of correct manipulation     (A1)

eg   x + 7 = 3y

\({g^{ - 1}}\left( x \right) = \frac{{x + 7}}{3}\)     A1 N3

[3 marks]

Write down \(g(2)\).

Find \((f \circ g)(x)\).

Find \({f^{ - 1}}(x)\).

\(g(2) = 8\)    A1     N1

[1 mark]

attempt to form composite (in any order)     (M1)

eg\(\,\,\,\,\,\)\(f(4x),{\text{ }}4 \times (8x + 3)\)

\((f \circ g)(x) = 32x + 3\)     A1     N2

[2 marks]

interchanging \(x\) and \(y\) (may be seen at any time)     (M1)

eg\(\,\,\,\,\,\)\(x = 8y + 3\)

\({f^{ - 1}}(x) = \frac{{x - 3}}{8}\,\,\,\,\,\left( {{\text{accept }}\frac{{x - 3}}{8},{\text{ }}y = \frac{{x - 3}}{8}} \right)\)     A1     N2

[2 marks]

This question was successfully answered by most candidates. The inverse notation was sometimes mistakenly interpreted as derivative or reciprocal.

This question was successfully answered by most candidates. The inverse notation was sometimes mistakenly interpreted as derivative or reciprocal.

This question was successfully answered by most candidates. The inverse notation was sometimes mistakenly interpreted as derivative or reciprocal.

Find \({f^{ - 1}}(x)\) .

Find \((f \circ g)(1)\) .

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg   \(x = 4y - 2\)

evidence of correct manipulation     (A1)

eg   \(x + 2 = 4y\)

\({f^{ - 1}}(x) = \frac{{x + 2}}{4}\) (accept \(y = \frac{{x + 2}}{4}\) , \(\frac{{x + 2}}{4}\) , \({f^{ - 1}}(x) = \frac{1}{4}x + \frac{1}{2}\)     A1     N2

[3 marks]

METHOD 1

attempt to substitute \(1\) into \(g(x)\)     (M1)

eg   \(g(1) =  - 2 \times {1^2} + 8\)

\(g(1) = 6\)     (A1)

\(f(6) = 22\)     A1     N3

METHOD 2

attempt to form composite function (in any order)     (M1)

eg   \((f \circ g)(x) = 4( - 2{x^2} + 8) - 2\) \(( =  - 8{x^2} + 30)\)

correct substitution

eg   \((f \circ g)(1) = 4( - 2 \times {1^2} + 8) - 2\) , \( - 8 + 30\)

\(f(6) = 22\)     A1     N3

[3 marks]

Write down the value of the discriminant.

Hence, show that \(p = 3\).

The graph of \(f\)has its vertex on the \(x\)-axis.

Find the coordinates of the vertex of the graph of \(f\).

The graph of \(f\) has its vertex on the \(x\)-axis.

Write down the solution of \(f(x) = 0\).

The graph of \(f\) has its vertex on the \(x\)-axis.

The function can be written in the form \(f(x) = a{(x - h)^2} + k\). Write down the value of \(a\).

The graph of \(f\) has its vertex on the \(x\)-axis.

The function can be written in the form \(f(x) = a{(x - h)^2} + k\). Write down the value of \(h\).

The graph of \(f\) has its vertex on the \(x\)-axis.

The function can be written in the form \(f(x) = a{(x - h)^2} + k\). Write down the value of \(k\).

The graph of \(f\) has its vertex on the \(x\)-axis.

The graph of a function \(g\) is obtained from the graph of \(f\) by a reflection of \(f\) in the \(x\)-axis, followed by a translation by the vector \(\left( \begin{array}{c}0\\6\end{array} \right)\). Find \(g\), giving your answer in the form \(g(x) = A{x^2} + Bx + C\).

correct value \(0\), or \(36 - 12p\)     A2     N2

[2 marks]

correct equation which clearly leads to \(p = 3\)     A1

eg     \(36 - 12p = 0,{\text{ }}36 = 12p\)

\(p = 3\)     AG     N0

[1 mark]

METHOD 1

valid approach     (M1)

eg     \(x =  - \frac{b}{{2a}}\)

correct working     A1

eg     \( - \frac{{( - 6)}}{{2(3)}},{\text{ }}x = \frac{6}{6}\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

METHOD 2

valid approach     (M1)

eg     \(f(x) = 0\), factorisation, completing the square

correct working     A1

eg     \({x^2} - 2x + 1 = 0,{\text{ }}(3x - 3)(x - 1),{\text{ }}f(x) = 3{(x - 1)^2}\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

METHOD 3

valid approach using derivative     (M1)

eg     \(f'(x) = 0,{\text{ }}6x - 6\)

correct equation     A1

eg     \(6x - 6 = 0\)

correct answers     A1A1     N2

eg     \(x = 1,{\text{ }}y = 0;{\text{ }}(1,{\text{ }}0)\)

[4 marks]

\(x = 1\)     A1     N1

[1 mark]

\(a = 3\)     A1     N1

[1 mark]

\(h = 1\)     A1     N1

[1 mark]

\(k = 0\)     A1     N1

[1 mark]

attempt to apply vertical reflection     (M1)

eg     \( - f(x),{\text{ }} - 3{(x - 1)^2}\), sketch

attempt to apply vertical shift 6 units up     (M1)

eg     \( - f(x) + 6\), vertex \((1, 6)\)

transformations performed correctly (in correct order)     (A1)

eg     \( - 3{(x - 1)^2} + 6,{\text{ }} - 3{x^2} + 6x - 3 + 6\)

\(g(x) =  - 3{x^2} + 6x + 3\)     A1     N3

[4 marks]

Write down \(f'\left( 2 \right)\).

Find \(f\left( 2 \right)\).

Show that the graph of g has a gradient of 6 at P.

Let L₂ be the tangent to the graph of g at P. L₁ intersects L₂ at the point Q.

Find the y-coordinate of Q.

recognize that \(f'\left( x \right)\) is the gradient of the tangent at \(x\)     (M1)

eg   \(f'\left( x \right) = m\)

\(f'\left( 2 \right) = 3\)  (accept m = 3)     A1 N2

[2 marks]

recognize that \(f\left( 2 \right) = y\left( 2 \right)\)     (M1)

eg  \(f\left( 2 \right) = 3 \times 2 + 1\)

\(f\left( 2 \right) = 7\)     A1 N2

[2 marks]

recognize that the gradient of the graph of g is \(g'\left( x \right)\)      (M1)

choosing chain rule to find \(g'\left( x \right)\)      (M1)

eg  \(\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u' = 2x\)

\(g'\left( x \right) = f'\left( {{x^2} + 1} \right) \times 2x\)     A2

\(g'\left( 1 \right) = 3 \times 2\)     A1

\(g'\left( 1 \right) = 6\)     AG N0

[5 marks]

 at Q, L₁ = L₂ (seen anywhere)      (M1)

recognize that the gradient of L₂ is g'(1)  (seen anywhere)     (M1)
eg  m = 6

finding g (1)  (seen anywhere)      (A1)
eg  \(g\left( 1 \right) = f\left( 2 \right),\,\,g\left( 1 \right) = 7\)

attempt to substitute gradient and/or coordinates into equation of a straight line      M1
eg  \(y - g\left( 1 \right) = 6\left( {x - 1} \right),\,\,y - 1 = g'\left( 1 \right)\left( {x - 7} \right),\,\,7 = 6\left( 1 \right) + {\text{b}}\)

correct equation for L₂ 

eg  \(y - 7 = 6\left( {x - 1} \right),\,\,y = 6x + 1\)     A1

correct working to find Q       (A1)
eg   same y-intercept, \(3x = 0\)

\(y = 1\)     A1 N2

[7 marks]

(i)     Find the value of a.

(ii)    Show that \(b = \frac{\pi }{6}\) .

(iii)   Find the value of d.

(iv)   Write down a value for c.

The transformation P is given by a horizontal stretch of a scale factor of \(\frac{1}{2}\) , followed by a translation of \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 10}
\end{array}} \right)\) .

Let \({M'}\) be the image of M under P. Find the coordinates of \({M'}\) .

The graph of g is the image of the graph of f under P.

Find \(g(t)\) in the form \(g(t) = 7\cos B(t - c) + D\) .

The graph of g is the image of the graph of f under P.

Give a full geometric description of the transformation that maps the graph of g to the graph of f .

(i) attempt to substitute     (M1)

e.g. \(a = \frac{{29 - 15}}{2}\)

\(a = 7\) (accept \(a = - 7\) )     A1     N2

(ii) \({\text{period}} = 12\)     (A1)

\(b = \frac{{2\pi }}{{12}}\)    A1

\(b = \frac{\pi }{6}\)    AG     N0

(iii) attempt to substitute     (M1)

e.g. \(d = \frac{{29 + 15}}{2}\)

\(d = 22\)     A1     N2

(iv) \(c = 3\) (accept \(c = 9\) from \(a = - 7\) )     A1     N1

Note: Other correct values for c can be found, \(c = 3 \pm 12k\) , \(k \in \mathbb{Z}\) .

[7 marks]

stretch takes 3 to 1.5     (A1)

translation maps \((1.5{\text{, }}29)\) to \((4.5{\text{, }}19)\) (so \({M'}\) is \((4.5{\text{, }}19)\))     A1     N2

[2 marks]

\(g(t) = 7\cos \frac{\pi }{3}\left( {t - 4.5} \right) + 12\)    A1A2A1    N4

Note: Award A1 for \(\frac{\pi }{3}\) , A2 for 4.5, A1 for 12.

Other correct values for c can be found, \(c = 4.5 \pm 6k\) , \(k \in \mathbb{Z}\) .

[4 marks]

translation \(\left( {\begin{array}{*{20}{c}}
{ - 3}\\
{10}
\end{array}} \right)\)     (A1)

horizontal stretch of a scale factor of 2     (A1)

completely correct description, in correct order     A1     N3

e.g. translation \(\left( {\begin{array}{*{20}{c}}
{ - 3}\\
{10}
\end{array}} \right)\) then horizontal stretch of a scale factor of 2

[3 marks]

This question was the most difficult on the paper. Where candidates attempted this question, part (a) was answered satisfactorily.

Few answered part (b) correctly as most could not interpret the horizontal stretch.

Few answered part (b) correctly as most could not interpret the horizontal stretch. As a result, there were many who were unable to answer part (c) although follow through marks were often obtained from incorrect answers in both parts (a) and (b). The link between the answer in (b) and the value of C in part (c) was lost on all but the most attentive.

In part (d), some candidates could name the transformations required, although only a handful provided the correct order of the transformations to return the graph to its original state.

Write down the value of \(h\) and of \(k\).

Find the value of \(a\).

\(h = 2,{\text{ }}k = 3\)     A1A1     N2

[2 marks]

attempt to substitute \((1,7)\) in any order into their \(f(x)\)     (M1)

eg     \(7 = a{(1 - 2)^2} + 3{\text{, }}7 = a{(1 - 3)^2} + 2{\text{, }}1 = a{(7 - 2)^2} + 3\)

correct equation     (A1)

eg     \(7 = a + 3\)

a = 4     A1     N2

[3 marks]

Sketch the graph of \(f( - x)\) on the grid below.

The graph of f is transformed to obtain the graph of g . The graph of g is shown below.

The function g can be written in the form \(g(x) = af(x + b)\) . Write down the value of a and of b .

     A2     N2

[2 marks]

\(a = - 2,b = - 1\)     A2A2     N4

Note: Award A1 for \(a = 2\) , A1 for \(b = 1\) .

[4 marks]

In part (a) of this question, a large number of candidates correctly sketched the graph of \(f( - x)\) , as asked. A fairly common error, however, was to graph \( - f(x)\) .

In part (b), many candidates seemed to recognize that the value of a was related to a vertical stretch, though some omitted the negative required for the vertical reflection. Similarly, some candidates gave a positive value for b.

Find \((f \circ g)(x)\) .

The vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 1}
\end{array}} \right)\) translates the graph of \((f \circ g)\) to the graph of h .

Find the coordinates of the vertex of the graph of h .

The vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 1}
\end{array}} \right)\) translates the graph of \((f \circ g)\) to the graph of h .

Show that \(h(x) = {x^2} - 8x + 19\) .

The vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 1}
\end{array}} \right)\) translates the graph of \((f \circ g)\) to the graph of h .

The line \(y = 2x - 6\) is a tangent to the graph of h at the point P. Find the x-coordinate of P.

attempt to form composition (in any order)     (M1)

\((f \circ g)(x) = {(x - 1)^2} + 4\)    \(({x^2} - 2x + 5)\)     A1     N2

[2 marks]

METHOD 1

vertex of \(f \circ g\) at (1, 4)     (A1)

evidence of appropriate approach     (M1)

e.g. adding \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 1}
\end{array}} \right)\) to the coordinates of the vertex of \(f \circ g\)

vertex of h at (4, 3)     A1     N3

METHOD 2

attempt to find \(h(x)\)     (M1)

e.g. \({((x - 3) - 1)^2} + 4 - 1\) , \(h(x) = (f \circ g)(x - 3) - 1\)

\(h(x) = {(x - 4)^2} + 3\)     (A1)

vertex of h at (4, 3)     A1     N3

[3 marks]

evidence of appropriate approach     (M1)

e.g. \({(x - 4)^2} + 3\) ,\({(x - 3)^2} - 2(x - 3) + 5 - 1\)

simplifying     A1

e.g. \(h(x) = {x^2} - 8x + 16 + 3\) , \({x^2} - 6x + 9 - 2x + 6 + 4\)

\(h(x) = {x^2} - 8x + 19\)     AG     N0

[2 marks]

METHOD 1

equating functions to find intersection point     (M1)

e.g. \({x^2} - 8x + 19 = 2x - 6\) , \(y = h(x)\)

\({x^2} - 10x + 25 + 0\)     A1

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

appropriate working     A1

e.g. \({(x - 5)^2} = 0\)

\(x = 5\)  \((p = 5)\)     A1     N3

METHOD 2

attempt to find \(h'(x)\)     (M1)

\(h(x) = 2x - 8\)     A1

recognizing that the gradient of the tangent is the derivative     (M1)

e.g. gradient at \(p = 2\)

\(2x - 8 = 2\)  \((2x = 10)\)     A1

\(x = 5\)     A1     N3

[5 marks]

Candidates showed good understanding of finding the composite function in part (a).

There were some who did not seem to understand what the vector translation meant in part (b).

Candidates showed good understanding of manipulating the quadratic in part (c).

There was more than one method to solve for h in part (d), and a pleasing number of candidates were successful in this part of the question.

Let \(f(x) = m - \frac{1}{x}\), for \(x \ne 0\). The line \(y = x - m\) intersects the graph of \(f\) in two distinct points. Find the possible values of \(m\).

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f = y,{\text{ }}m - \frac{1}{x} = x - m\)

correct working to eliminate denominator     (A1)

eg\(\,\,\,\,\,\)\(mx - 1 = x(x - m),{\text{ }}mx - 1 = {x^2} - mx\)

correct quadratic equal to zero     A1

eg\(\,\,\,\,\,\)\({x^2} - 2mx + 1 = 0\)

correct reasoning     R1

eg\(\,\,\,\,\,\)for two solutions, \({b^2} - 4ac > 0\)

correct substitution into the discriminant formula     (A1)

eg\(\,\,\,\,\,\)\({( - 2m)^2} - 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(4{m^2} > 4,{\text{ }}{m^2} = 1\), sketch of positive parabola on the \(x\)-axis

correct interval     A1     N4

eg\(\,\,\,\,\,\)\(\left| m \right| > 1,{\text{ }}m <  - 1\) or \(m > 1\)

[7 marks]

(i)     Find \(g(0)\) .

(ii)    Find \((f \circ g)(0)\) .

Find \({f^{ - 1}}(x)\) .

(i) \(g(0) = {{\rm{e}}^0} - 2\)     (A1)

\( = - 1\)     A1     N2

(ii) METHOD 1

substituting answer from (i)     (M1)

e.g. \((f \circ g)(0) = f( - 1)\)

correct substitution \(f( - 1) = 2{( - 1)^3} + 3\)     (A1)

\(f( - 1) = 1\)     A1     N3

METHOD 2

attempt to find \((f \circ g)(x)\)     (M1)

e.g. \((f \circ g)(x) = f({{\rm{e}}^{3x}} - 2)\) \( = 2{({{\rm{e}}^{3x}} - 2)^3} + 3\)

correct expression for \((f \circ g)(x)\)     (A1)

e.g. \(2{({{\rm{e}}^{3x}} - 2)^3} + 3\)

\((f \circ g)(0) = 1\)     A1     N3

[5 marks]

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = 2{y^3} + 3\)

attempt to solve     (M1)

e.g. \({y^3} = \frac{{x - 3}}{2}\)

\({f^{ - 1}}(x) = \sqrt[3]{{\frac{{x - 3}}{2}}}\)     A1     N3

[3 marks]

This question was generally done well, although some students consider \({{\rm{e}}^0}\) to be 0, losing them a mark.

A few candidates composed in the wrong order. Most found the formula of the inverse correctly, even if in some cases there were errors when trying to isolate x (or y). A common incorrect solution found was to find \(y = \sqrt[3]{{\frac{{x - 3}}{2}}}\) .

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

Find the gradient of \(L\).

Show that the \(x\)-coordinate of B is \( - \frac{k}{2}\).

Find the area of triangle ABC, giving your answer in terms of \(k\).

Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).

\(f'(x) = 2x\)     A1     N1

[1 mark]

attempt to substitute \(x =  - k\) into their derivative     (M1)

gradient of \(L\) is \( - 2k\)     A1     N2

[2 marks]

METHOD 1 

attempt to substitute coordinates of A and their gradient into equation of a line     (M1)

eg\(\,\,\,\,\,\)\({k^2} =  - 2k( - k) + b\)

correct equation of \(L\) in any form     (A1)

eg\(\,\,\,\,\,\)\(y - {k^2} =  - 2k(x + k),{\text{ }}y =  - 2kx - {k^2}\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(y = 0\)

correct substitution into \(L\) equation     A1

eg\(\,\,\,\,\,\)\( - {k^2} =  - 2kx - 2{k^2},{\text{ }}0 =  - 2kx - {k^2}\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  - {k^2}\)

\(x =  - \frac{k}{2}\)     AG     N0

METHOD 2

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{gradient}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}},{\text{ }} - 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}\)

recognizing \(y = 0\) at B     (A1)

attempt to substitute coordinates of A and B into slope formula     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{k^2} - 0}}{{ - k - x}},{\text{ }}\frac{{ - {k^2}}}{{x + k}}\)

correct equation     A1

eg\(\,\,\,\,\,\)\(\frac{{{k^2} - 0}}{{ - k - x}} =  - 2k,{\text{ }}\frac{{ - {k^2}}}{{x + k}} =  - 2k,{\text{ }} - {k^2} =  - 2k(x + k)\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  - {k^2}\)

\(x =  - \frac{k}{2}\)     AG     N0

[5 marks]

valid approach to find area of triangle     (M1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)\)

area of \({\text{ABC}} = \frac{{{k^3}}}{4}\)     A1     N2

[2 marks]

METHOD 1 (\(\int {f - {\text{triangle}}} \))

valid approach to find area from \( - k\) to 0     (M1)

eg\(\,\,\,\,\,\)\(\int_{ - k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ - k} f } \)

correct integration (seen anywhere, even if M0 awarded)     A1

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ - k}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(0 - \frac{{{{( - k)}^3}}}{3}\), area from \( - k\) to 0 is \(\frac{{{k^3}}}{3}\)

Note:     Award M0 for substituting into original or differentiated function.

attempt to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ - k}^0 {f(x){\text{d}}x - {\text{ triangle}}} \)

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{3} - \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} - \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

METHOD 2 (\(\int {(f - L)} \))

valid approach to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ - k}^{ - \frac{k}{2}} {{x^2} - ( - 2kx - {k^2}){\text{d}}x + \int_{ - \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ - k}^{ - \frac{k}{2}} {(f - L) + \int_{ - \frac{k}{2}}^0 f } } } \)

correct integration (seen anywhere, even if M0 awarded)     A2

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ - k}^{ - \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ - \frac{k}{2}}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\left( {\frac{{{{\left( { - \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { - \frac{k}{2}} \right)}^2} + {k^2}\left( { - \frac{k}{2}} \right)} \right) - \left( {\frac{{{{( - k)}^3}}}{3} + k{{( - k)}^2} + {k^2}( - k)} \right) + (0) - \left( {\frac{{{{\left( { - \frac{k}{2}} \right)}^3}}}{3}} \right)\)

Note:     Award M0 for substituting into original or differentiated function.

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} - \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} - \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} - {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

[7 marks]

The function can be written in the form \(f(x) = a{(x - h)^2} + k\) .

(i)     Write down the value of h and of k .

(ii)    Show that \(a = 3\) .

Find \(f(x)\)  , giving your answer in the form \(A{x^2} + Bx + C\) .

Calculate the area enclosed by the graph of f , the x-axis, and the lines \(x = 2\) and \(x = 4\) .

(i) \(h = 2\) , \(k = 1\)     A1A1     N2

(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into f     M1

e.g. \(13 = a{(0 - 2)^2} + 1\)

working towards solution     A1

e.g. \(13 = 4a + 1\)

\(a = 3\)     AG     N0

[4 marks]

attempting to expand their binomial     (M1)

e.g. \(f(x) = 3({x^2} - 2 \times 2x + 4) + 1\) , \({(x - 2)^2} = {x^2} - 4x + 4\)

correct working     (A1)

e.g. \(f(x) = 3{x^2} - 12x + 12 + 1\)

\(f(x) = 3{x^2} - 12x + 13\) (accept \(A = 3\) , \(B = - 12\) , \(C = 13\) )     A1     N2

[3 marks]

METHOD 1

integral expression     (A1)

e.g. \(\int_2^4 {(3{x^2}}  - 12x + 13)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{x^3} - 6{x^2} + 13x]_2^4\)     A1A1A1

Note: Award A1 for \({x^3}\) , A1 for \( - 6{x^2}\) , A1 for \(13x\) .

correct substitution of correct limits into their expression     A1A1

e.g. \(({4^3} - 6 \times {4^2} + 13 \times 4) - ({2^3} - 6 \times {2^2} + 13 \times 2)\) , \(64 - 96 + 52 - (8 - 24 + 26)\)

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working     (A1)

e.g. \(64 - 96 + 52 - 8 + 24 - 26,20 - 10\)

\({\rm{Area}} = 10\)     A1     N3

[8 marks]

METHOD 2

integral expression     (A1)

e.g. \(\int_2^4 {(3{{(x - 2)}^2}}  + 1)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{(x - 2)^3} + x]_2^4\)     A2A1

Note: Award A2 for \({(x - 2)^3}\) , A1 for \(x\) .

correct substitution of correct limits into their expression     A1A1

e.g. \({(4 - 2)^3} + 4 - [{(2 - 2)^3} + 2]\) , \({2^3} + 4 - ({0^3} + 2)\) , \({2^3} + 4 - 2\)

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working     (A1)

e.g. \(8 + 4 - 2\)

\({\rm{Area}} = 10\)    A1     N3

[8 marks]

METHOD 3

recognizing area from 0 to 2 is same as area from 2 to 4     (R1)

e.g. sketch, \(\int_2^4 {f = \int_0^2 f } \)

integral expression     (A1)

e.g. \(\int_0^2 {(3{x^2}}  - 12x + 13)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{x^3} - 6{x^2} + 13x]_0^2\)     A1A1A1

Note: Award A1 for \({x^3}\) , A1 for \( - 6{x^2}\) , A1 for \(13x\) .

correct substitution of correct limits into their expression     A1(A1)

e.g. \(({2^3} - 6 \times {2^2} + 13 \times 2) - ({0^3} - 6 \times {0^2} + 13 \times 0)\) , \(8 - 24 + 26\)

Note: Award A1 for substituting 2, (A1) for substituting 0.

\({\rm{Area}} = 10\)     A1     N3

[8 marks]

In part (a), nearly all the candidates recognized that h and k were the coordinates of the vertex of the parabola, and most were able to successfully show that \(a = 3\) . Unfortunately, a few candidates did not understand the "show that" command, and simply verified that \(a = 3\) would work, rather than showing how to find \(a = 3\) .

In part (b), most candidates were able to find \(f(x)\) in the required form. For a few candidates, algebraic errors kept them from finding the correct function, even though they started with correct values for a, h and k.

In part (c), nearly all candidates knew that they needed to integrate to find the area, but errors in integration, and algebraic and arithmetic errors prevented many from finding the correct area.

Write down \(f(x)\) in the form \(f(x) = - 10(x - p)(x - q)\) .

Find another expression for \(f(x)\) in the form \(f(x) = - 10{(x - h)^2} + k\) .

Show that \(f(x)\) can also be written in the form \(f(x) = 240 + 20x - 10{x^2}\) .

A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ - 1}}\) , at time t seconds is given by \(v = 240 + 20t - 10{t^2}\) , for \(0 \le t \le 6\) .

(i)     Find the value of t when the speed of the particle is greatest.

(ii)    Find the acceleration of the particle when its speed is zero.

\(f(x) = - 10(x + 4)(x - 6)\)     A1A1     N2

[2 marks]

METHOD 1

attempting to find the x-coordinate of maximum point     (M1)

e.g. averaging the x-intercepts, sketch, \(y' = 0\) , axis of symmetry

attempting to find the y-coordinate of maximum point     (M1)

e.g. \(k = - 10(1 + 4)(1 - 6)\)

\(f(x) = - 10{(x - 1)^2} + 250\)     A1A1     N4

METHOD 2

attempt to expand \(f(x)\)     (M1)

e.g. \( - 10({x^2} - 2x - 24)\)

attempt to complete the square     (M1)

e.g. \( - 10({(x - 1)^2} - 1 - 24)\)

\(f(x) = - 10{(x - 1)^2} + 250\)     A1A1     N4

[4 marks]

attempt to simplify     (M1)

e.g. distributive property, \( - 10(x - 1)(x - 1) + 250\)

correct simplification     A1

e.g. \( - 10({x^2} - 6x + 4x - 24)\) , \( - 10({x^2} - 2x + 1) + 250\)

\(f(x) = 240 + 20x - 10{x^2}\)     AG     N0

[2 marks]

(i) valid approach     (M1)

e.g. vertex of parabola, \(v'(t) = 0\)

\(t = 1\)     A1     N2

(ii) recognizing \(a(t) = v'(t)\)     (M1)

\(a(t) = 20 - 20t\)     A1A1

speed is zero \( \Rightarrow t = 6\)     (A1)

\(a(6) = - 100\) (\({\text{m}}{{\text{s}}^{ - 2}}\))     A1     N3

[7 marks]

Parts (a) and (c) of this question were very well done by most candidates.

In part (b), many candidates attempted to use the method of completing the square, but were unsuccessful dealing with the coefficient of \( - 10\). Candidates who recognized that the x-coordinate of the vertex was 1, then substituted this value into the function from part (a), were generally able to earn full marks here.

Parts (a) and (c) of this question were very well done by most candidates.

In part (d), it was clear that many candidates were not familiar with the relationship between velocity and acceleration, and did not understand how those concepts were related to the graph which was given. A large number of candidates used time \(t = 1\) in part b(ii), rather than \(t = 6\) . To find the acceleration, some candidates tried to integrate the velocity function, rather than taking the derivative of velocity. Still others found the derivative in part b(i), but did not realize they needed to use it in part b(ii), as well.

Find the equation of the axis of symmetry of the graph of \(f\).

(i)     Write down the value of \(h\).

(ii)     Find the value of \(k\).

correct approach     (A1)

eg \(\frac{{ - ( - 4)}}{2},{\text{ }}f'(x) = 2x - 4 = 0,{\text{ (}}{x^2} - 4x + 4) + 5 - 4\)

\(x = 2\) (must be an equation)     A1     N2

[2 marks]

(i)     \(h = 2\)     A1     N1

(ii)     METHOD 1

valid attempt to find \(k\)     (M1)

eg\(\,\,\,\,\,\)\(f(2)\)

correct substitution into their function     (A1)

eg\(\,\,\,\,\,\)\({(2)^2} - 4(2) + 5\)

\(k = 1\)     A1     N2

METHOD 2

valid attempt to complete the square     (M1)

eg\(\,\,\,\,\,\)\({x^2} - 4x + 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(({x^2} - 4x + 4) - 4 + 5,{\text{ }}{(x - 2)^2} + 1\)

\(k = 1\)     A1     N2

[4 marks]

Find \(f\left( {\frac{\pi }{2}} \right)\) .

Find \((g \circ f)\left( {\frac{\pi }{2}} \right)\) .

Given that \((g \circ f)(x)\) can be written as \(\cos (kx)\) , find the value of k, \(k \in \mathbb{Z}\) .

\(f\left( {\frac{\pi }{2}} \right) = \cos \pi \)     (A1)

\( = - 1\)     A1     N2

[2 marks]

\((g \circ f)\left( {\frac{\pi }{2}} \right) = g( - 1)\) \(( = 2{( - 1)^2} - 1)\)    (A1)

\(= 1\)     A1     N2

[2 marks]

\((g \circ f)(x) = 2{(\cos (2x))^2} - 1\) \(( = 2{\cos ^2}(2x) - 1)\)     A1

evidence of \(2{\cos ^2}\theta - 1 = \cos 2\theta \) (seen anywhere)     (M1)

\((g \circ f)(x) = \cos 4x\)

\(k = 4\)     A1     N2

[3 marks]

In part (a), a number of candidates were not able to evaluate \(\cos \pi \) , either leaving it or evaluating it incorrectly.

Almost all candidates evaluated the composite function in part (b) in the given order, many earning follow-through marks for incorrect answers from part (a). On both parts (a) and (b), there were candidates who correctly used double-angle formulas to come up with correct answers; while this is a valid method, it required unnecessary additional work.

Candidates were not as successful in part (c). Many tried to use double-angle formulas, but either used the formula incorrectly or used it to write the expression in terms of \(\cos x\) and went no further. There were a number of cases in which the candidates "accidentally" came up with the correct answer based on errors or lucky guesses and did not earn credit for their final answer. Only a few candidates recognized the correct method of solution.

Find \({f^{ - 1}}(x)\) .

Let \(g(x) = {{\rm{e}}^x}\) .

Find \((g \circ f)(x)\) , giving your answer in the form \(ax + b\) , where \(a,b \in \mathbb{Z}\) .

METHOD 1

\(\ln (x + 5) + \ln 2 = \ln (2(x + 5))\) \(( = \ln (2x + 10))\)     (A1)

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \ln (2y + 10)\)

evidence of correct manipulation     (A1)

e.g. \({{\rm{e}}^x} = 2y + 10\)

\({f^{ - 1}}(x) = \frac{{{{\rm{e}}^x} - 10}}{2}\)    A1     N2

METHOD 2

\(y = \ln (x + 5) + \ln 2\)

\(y - \ln 2 = ln(x + 5)\)     (A1)

evidence of correct manipulation     (A1)

e.g. \({{\rm{e}}^{y - \ln 2}} = x + 5\)

interchanging x and y (seen anywhere)     (M1)

e.g. \({{\rm{e}}^{x - \ln 2}} = y + 5\)

\({f^{ - 1}}(x) = {{\rm{e}}^{x - \ln 2}} - 5\)     A1     N2

[4 marks]

METHOD 1

evidence of composition in correct order     (M1)

e.g. \((g \circ f)(x) = g(\ln (x + 5) + \ln 2)\)

\( = {{\rm{e}}^{\ln (2(x + 5))}} = 2(x + 5)\)

\((g \circ f)(x) = 2x + 10\)     A1A1     N2

METHOD 2

evidence of composition in correct order     (M1)

e.g. \((g \circ f)(x) = {{\rm{e}}^{\ln (x + 5) + \ln 2}}\)

\( = {{\rm{e}}^{\ln (x + 5)}} \times {{\rm{e}}^{\ln 2}} = (x + 5)2\)

\((g \circ f)(x) = 2x + 10\)     A1A1     N2

[3 marks]

This was one of the more difficult problems for the candidates. Knowledge of the laws of logarithms appeared weak as did the inverse nature of the exponential and logarithmic functions. There were a number of candidates who mistook the notation for the inverse to mean either the derivative or the reciprocal. The order of composition seemed well understood by most candidates but they were unable to simplify by the rules of indices to obtain the correct final answer.

This was one of the more difficult problems for the candidates. Knowledge of the laws of logarithms appeared weak as did the inverse nature of the exponential and logarithmic functions. There were a number of candidates who mistook the notation for the inverse to mean either the derivative or the reciprocal. The order of composition seemed well understood by most candidates but they were unable to simplify by the rules of indices to obtain the correct final answer.

Given that \({f^{ - 1}}(1) = 8\) , find the value of \(k\) .

Find \({f^{ - 1}}\left( {\frac{2}{3}} \right)\) .

METHOD 1

recognizing that \(f(8) = 1\)     (M1)

e.g. \(1 = k{\log _2}8\)

recognizing that \({\log _2}8 = 3\)     (A1)

e.g. \(1 = 3k\)

\(k = \frac{1}{3}\)     A1     N2

METHOD 2

attempt to find the inverse of \(f(x) = k{\log _2}x\)     (M1)

e.g. \(x = k{\log _2}y\) , \(y = {2^{\frac{x}{k}}}\)

substituting 1 and 8     (M1)

e.g. \(1 = k{\log _2}8\) , \({2^{\frac{1}{k}}} = 8\)

\(k = \frac{1}{{{{\log }_2}8}}\) \(\left( {k = \frac{1}{3}} \right)\)     A1     N2

[3 marks]

METHOD 1

recognizing that \(f(x) = \frac{2}{3}\)     (M1)

e.g. \(\frac{2}{3} = \frac{1}{3}{\log _2}x\)

\({\log _2}x = 2\)     (A1)

\({f^{ - 1}}\left( {\frac{2}{3}} \right) = 4\) (accept \(x = 4\))     A2     N3

METHOD 2

attempt to find inverse of \(f(x) = \frac{1}{3}{\log _2}x\)     (M1)

e.g. interchanging x and y , substituting \(k = \frac{1}{3}\) into \(y = {2^{\frac{x}{k}}}\)

correct inverse     (A1)

e.g. \({f^{ - 1}}(x) = {2^{3x}}\) , \({2^{3x}}\)

\({f^{ - 1}}\left( {\frac{2}{3}} \right) = 4\)     A2    N3

[4 marks]

A very poorly done question. Most candidates attempted to find the inverse function for \(f\) and used that to answer parts (a) and (b). Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able to actually get the correct inverse. Almost none recognized that if \({f^{ - 1}}(1) = 8\) , then \(f(8) = 1\) . Many thought that the letters "log" could be simply "cancelled out", leaving the \(2\) and the \(8\).

A very poorly done question. Most candidates attempted to find the inverse function for \(f\) and used that to answer parts (a) and (b). Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able to actually get the correct inverse. Almost none recognized that if \({f^{ - 1}}(1) = 8\) , then \(f(8) = 1\) . Many thought that the letters "log" could be simply "cancelled out", leaving the \(2\) and the \(8\).

Show that \(f'(x) = \frac{{\ln x}}{x}\).

There is a minimum on the graph of \(f\). Find the \(x\)-coordinate of this minimum.

Write down the value of \(p\).

The graph of \(g\) intersects the graph of \(f'\) when \(x = q\).

Find the value of \(q\).

The graph of \(g\) intersects the graph of \(f'\) when \(x = q\).

Let \(R\) be the region enclosed by the graph of \(f'\), the graph of \(g\) and the line \(x = p\).

Show that the area of \(R\) is \(\frac{1}{2}\).

METHOD 1

correct use of chain rule     A1A1

eg     \(\frac{{2\ln x}}{2} \times \frac{1}{x},{\text{ }}\frac{{2\ln x}}{{2x}}\)

Note: Award A1 for \(\frac{{2\ln x}}{{2x}}\), A1 for \( \times \frac{1}{x}\).

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

METHOD 2

correct substitution into quotient rule, with derivatives seen     A1

eg     \(\frac{{2 \times 2\ln x \times \frac{1}{x} - 0 \times {{(\ln x)}^2}}}{4}\)

correct working     A1

eg     \(\frac{{4\ln x \times \frac{1}{x}}}{4}\)

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

setting derivative \( = 0\)     (M1)

eg     \(f'(x) = 0,{\text{ }}\frac{{\ln x}}{x} = 0\)

correct working     (A1)

eg     \(\ln x = 0,{\text{ }}x = {{\text{e}}^0}\)

\(x = 1\)     A1     N2

[3 marks] 

intercept when \(f'(x) = 0\)     (M1)

\(p = 1\)     A1     N2

[2 marks]

equating functions     (M1)

eg     \(f' = g,{\text{ }}\frac{{\ln x}}{x} = \frac{1}{x}\)

correct working     (A1)

eg     \(\ln x = 1\)

\(q = {\text{e   (accept }}x = {\text{e)}}\)     A1     N2

[3 marks]

evidence of integrating and subtracting functions (in any order, seen anywhere)     (M1)

eg     \(\int_q^e {\left( {\frac{1}{x} - \frac{{\ln x}}{x}} \right){\text{d}}x{\text{, }}\int {f' - g} } \)

correct integration \(\ln x - \frac{{{{(\ln x)}^2}}}{2}\)     A2

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     \((\ln {\text{e}} - \ln 1) - \left( {\frac{{{{(\ln {\text{e}})}^2}}}{2} - \frac{{{{(\ln 1)}^2}}}{2}} \right)\)

Note: Do not award M1 if the integrated function has only one term.

correct working     A1

eg     \((1 - 0) - \left( {\frac{1}{2} - 0} \right),{\text{ }}1 - \frac{1}{2}\)

\({\text{area}} = \frac{1}{2}\)     AG     N0

Notes: Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the markscheme.

[5 marks]

Find \((g \circ f)(x)\).

Given that \(\mathop {\lim }\limits_{x \to  + \infty } (g \circ f)(x) =  - 3\), find the value of \(b\).

attempt to form composite     (M1)

eg\(\,\,\,\,\,\)\(g(1 + {{\text{e}}^{ - x}})\)

correct function     A1     N2

eg\(\,\,\,\,\,\)\((g \circ f)(x) = 2 + b + 2{{\text{e}}^{ - x}},{\text{ }}2(1 + {{\text{e}}^{ - x}}) + b\)

[2 marks]

evidence of \(\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ - x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ - x}})\)     (M1)

eg\(\,\,\,\,\,\)\(2 + b + 2{{\text{e}}^{ - \infty }}\), graph with horizontal asymptote when \(x \to \infty \)

Note:     Award M0 if candidate clearly has incorrect limit, such as \(x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}\).

evidence that \({{\text{e}}^{ - x}} \to 0\) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ - x}}) = 0,{\text{ }}1 + {{\text{e}}^{ - x}} \to 1,{\text{ }}2(1) + b =  - 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0\), graph of \(y = {{\text{e}}^{ - x}}\) or

\(y = 2{{\text{e}}^{ - x}}\) with asymptote \(y = 0\), graph of composite function with asymptote \(y =  - 3\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(2 + b =  - 3\)

\(b =  - 5\)     A1     N2

[4 marks]

Show that \({f^{ - 1}}(x) = {3^{2x}}\) .

Write down the range of \({f^{ - 1}}\) .

Let \(g(x) = {\log _3}x\) , for \(x > 0\) .

Find the value of \(({f^{ - 1}} \circ g)(2)\) , giving your answer as an integer.

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \log \sqrt y \) (accept any base)

evidence of correct manipulation     A1

e.g. \(3^x = \sqrt y \) , \({3^y} = {x^{\frac{1}{2}}}\) , \(x = \frac{1}{2}{\log _3}y\) , \(2y = {\log _3}x\)

\({f^{ - 1}}(x) = {3^{2x}}\)     AG     N0 

[2 marks]

\(y > 0\) , \({f^{ - 1}}(x) > 0\)     A1     N1

[1 mark]

METHOD 1

finding \(g(2) = lo{g_3}2\) (seen anywhere)     A1

attempt to substitute     (M1)

e.g. \(({f^{ - 1}} \circ g)(2) = {3^{2\log {_3}2}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ - 1}} \circ g)(2) = {3^{\log {_3}4}}\) , \({3^{{{\log }_3}2^2}}\)

\(({f^{ - 1}} \circ g)(2) = 4\)     A1     N1

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(({f^{ - 1}} \circ g)(x) = {3^{2{{\log }_3}x}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ - 1}} \circ g)(x) = {3^{{{\log }_3}{x^2}}}\) , \({3^{{{\log }_3}{x^{}}}}^2\)

\(({f^{ - 1}} \circ g)(x) = {x^2}\)     A1

\(({f^{ - 1}} \circ g)(2) = 4\)     A1     N1

[4 marks]

Candidates were generally skilled at finding the inverse of a logarithmic function.

Few correctly gave the range of this function, often stating “all real numbers” or “ \(y \ge 0\) ”, missing the idea that the range of an inverse is the domain of the original function.

Some candidates answered part (c) correctly, although many did not get beyond \({3^{2{{\log }_3}2}}\) . Some attempted to form the composite in the incorrect order. Others interpreted \(({f^{ - 1}} \circ g)(2)\) as multiplication by 2.

(i)     Write down the amplitude of \(f\).

(ii)     Find the period of \(f\).

On the following grid sketch the graph of \(f\).

(i)     3     A1     N1

(ii)     valid attempt to find the period     (M1)

eg\(\,\,\,\,\,\)\(\frac{{2\pi }}{b},{\text{ }}\frac{{2\pi }}{{\frac{\pi }{2}}}\)

period \( = 4\)     A1     N2

[3 marks]

     A1A1A1A1     N4

[4 marks]

Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.

Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.

On the same diagram, sketch the graph of \({f^{ - 1}}\) .

Write down the range of \({f^{ - 1}}\) .

Find \({f^{ - 1}}(x)\) .

     A1A1A1     N3

Note: Award A1 for approximately correct (reflected) shape, A1 for right end point in circle, A1 for through \((1{\text{, }}0)\) .

\(0 \le y \le 3.5\)     A1     N1

[1 mark]

interchanging x and y (seen anywhere)     M1

e.g. \(x = {e^{0.5y}}\)

evidence of changing to log form     A1

e.g. \(\ln x = 0.5y\) , \(\ln x = \ln {{\rm{e}}^{0.5y}}\) (any base), \(\ln x = 0.5y\ln {\rm{e}}\) (any base)

\({f^{ - 1}}(x) = 2\ln x\)     A1     N1

[3 marks]

There were a large number of candidates who were unaware of the geometric relationship between a function and its inverse. Those that had some idea of the shape of the graph often did not consider the specified domain. Many more students were able to use an analytical approach to finding the inverse of a function and had little problem using logarithms to solve for y. Candidates were clearly more comfortable with algebraic procedures than graphical interpretations.

There were a large number of candidates who were unaware of the geometric relationship between a function and its inverse. Those that had some idea of the shape of the graph often did not consider the specified domain. Many more students were able to use an analytical approach to finding the inverse of a function and had little problem using logarithms to solve for y. Candidates were clearly more comfortable with algebraic procedures than graphical interpretations.

There were a large number of candidates who were unaware of the geometric relationship between a function and its inverse. Those that had some idea of the shape of the graph often did not consider the specified domain. Many more students were able to use an analytical approach to finding the inverse of a function and had little problem using logarithms to solve for y. Candidates were clearly more comfortable with algebraic procedures than graphical interpretations.

Find \({f^{ - 1}}(x)\).

Find \((f \circ g)(7)\).

interchanging \(x\) and \(x\)     (M1)

eg\(\,\,\,\,\,\)\(x = 5y\)

\({f^{ - 1}}\left( x \right) = \frac{x}{5}\)     A1     N2

[2 marks]

METHOD 1

attempt to substitute 7 into \(g(x)\) or \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\({7^2} + 1,{\text{ }}5 \times 7\)

\(g(7) = 50\)     (A1)

\(f\left( {50} \right) = 250\)     A1     N2

METHOD 2

attempt to form composite function (in any order)     (M1)

eg\(\,\,\,\,\,\)\(5({x^2} + 1),{\text{ }}{(5x)^2} + 1\)

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(5 \times ({7^2} + 1)\)

\((f \circ g)(7) = 250\)     A1     N2

[3 marks]

Write down the range of \(f\).

Write down \(f(2)\);

Write down \({f^{ - 1}}(2)\).

On the grid, sketch the graph of \({f^{ - 1}}\).

correct range (do not accept \(0 \leqslant x \leqslant 7\))     A1     N1

eg\(\,\,\,\,\,\)\([0,{\text{ }}7],{\text{ }}0 \leqslant y \leqslant 7\)

[1 mark]

\(f(2) = 3\)     A1     N1

[1 mark]

\({f^{ - 1}}(2) = 0\)     A1     N1

[1 mark]

     A1A1A1     N3

Notes:     Award A1 for both end points within circles,

A1 for images of \((2,{\text{ }}3)\) and \((0,{\text{ }}2)\) within circles,

A1 for approximately correct reflection in \(y = x\), concave up then concave down shape (do not accept line segments).

[3 marks]

Find the values of k such that \(f(x) = 0\) has two equal roots.

Each value of k is equally likely for \( - 5 \le k \le 5\) . Find the probability that \(f(x) = 0\) has no roots.

METHOD 1

evidence of discriminant     (M1)

e.g. \({b^2} - 4ac\) , discriminant = 0

correct substitution into discriminant     A1

e.g. \({k^2} - 4 \times \frac{1}{2} \times 8\) , \({k^2} - 16 = 0\)

\(k = \pm 4\)     A1A1     N3

METHOD 2

recognizing that equal roots means perfect square     (R1)

e.g. attempt to complete the square, \(\frac{1}{2}({x^2} + 2kx + 16)\)

correct working

e.g. \(\frac{1}{2}{(x + k)^2}\) ,  \(\frac{1}{2}{k^2} = 8\)     A1

\(k = \pm 4\)     A1A1     N3

[4 marks]

evidence of appropriate approach     (M1)

e.g. \({b^2} - 4ac < 0\)

correct working for k     A1

e.g. \( - 4 < k < 4\) , \({k^2} < 16\) , list all correct values of k

\(p = \frac{7}{{11}}\)     A2     N3

[4 marks]

A good number of candidates were successful in using the discriminant to find the correct values of k in part (a), however, there were many who tried to use the quadratic formula without recognizing the significance of the discriminant.

Part (b) was very poorly done by nearly all candidates. Common errors included finding the wrong values for k, and not realizing that there were 11 possible values for k.

Find \({f^{ - 1}}(x)\).

Show that \(\left( {g \circ {f^{ - 1}}} \right)(x) = \frac{5}{{x + 2}}\).

Find the \(y\)-intercept of the graph of \(h\).

Hence, sketch the graph of \(h\).

For the graph of \({h^{ - 1}}\), write down the \(x\)-intercept;

For the graph of \({h^{ - 1}}\), write down the equation of the vertical asymptote.

Given that \({h^{ - 1}}(a) = 3\), find the value of \(a\).

interchanging \(x\) and \(y\)     (M1)

eg     \(x = 3y - 2\)

\({f^{ - 1}}(x) = \frac{{x + 2}}{3}{\text{   }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)\)     A1     N2

[2 marks]

attempt to form composite (in any order)     (M1)

eg     \(g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}\)

correct substitution     A1

eg     \(\frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}\)

\(\left( {g \circ {f^{ - 1}}} \right)(x) = \frac{5}{{x + 2}}\)     AG     N0

[2 marks]

valid approach     (M1)

eg     \(h(0),{\text{ }}\frac{5}{{0 + 2}}\)

\(y = \frac{5}{2}{\text{   }}\left( {{\text{accept (0, 2.5)}}} \right)\)     A1     N2

[2 marks]

     A1A2     N3

Notes:     Award A1 for approximately correct shape (reciprocal, decreasing, concave up).

     Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at \((0, 2.5)\), asymptotic to x-axis, correct domain \(x \geqslant 0\).

     If only two of these features are correct, award A1.

[3 marks]

\(x = \frac{5}{2}{\text{   }}\left( {{\text{accept (2.5, 0)}}} \right)\)     A1     N1

[1 mark]

\(x = 0\)   (must be an equation)     A1     N1

[1 mark]

METHOD 1

attempt to substitute \(3\) into \(h\) (seen anywhere)     (M1)

eg     \(h(3),{\text{ }}\frac{5}{{3 + 2}}\)

correct equation     (A1)

eg     \(a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a\)

\(a = 1\)     A1     N2

[3 marks]

METHOD 2

attempt to find inverse (may be seen in (d))     (M1)

eg     \(x = \frac{5}{{y + 2}},{\text{ }}{h^{ - 1}} = \frac{5}{x} - 2,{\text{ }}\frac{5}{x} + 2\)

correct equation, \(\frac{5}{x} - 2 = 3\)     (A1)

\(a = 1\)     A1     N2

[3 marks]

Write down the value of \(q\).

The graph of \(f\) has a \(y\)-intercept at \((0,{\text{ }}4)\).

Find the value of \(p\).

The graph of \(f\) has a \(y\)-intercept at \((0,{\text{ }}4)\).

Write down the equation of the horizontal asymptote of the graph of \(f\).

\(q = 3\)     A1     N1

[1 mark]

correct expression for \(f(0)\)     (A1)

eg\(\;\;\;p + \frac{9}{{0 - 3}},{\text{ }}4 = p + \frac{9}{{ - q}}\)

recognizing that \(f(0) = 4\;\;\;\)(may be seen in equation)     (M1)

correct working     (A1)

eg\(\;\;\;4 = p - 3\)

\(p = 7\)     A1     N3

[3 marks]

\(y = 7\;\;\;\)(must be an equation, do not accept \(p = 7\)     A1     N1

[1 mark]

Total [6 marks]

Parts (a) and (b) were generally well done. Some candidates incorrectly answered \(q =  - 3\), rather than \(q = 3\), in part (a), but then were able to earn follow-through marks in part (b).

Parts (a) and (b) were generally well done. Some candidates incorrectly answered \(q =  - 3\), rather than \(q = 3\), in part (a), but then were able to earn follow-through marks in part (b).

Many candidates did not recognize the connection between parts (b) and (c) of this question, and many did a good deal of unnecessary work in part (c) before giving the correct answer. In part (c), many candidates did not write the equation of the asymptote, but just wrote the number.

Find \(f'(x)\).

Given that \(f'(x) \geqslant 0\), show that \({p^2} \leqslant 3pq\).

\(f'(x) = 3p{x^2} + 2px + q\)     A2     N2

Note:     Award A1 if only 1 error.

[2 marks]

evidence of discriminant (must be seen explicitly, not in quadratic formula)     (M1)

eg     \({b^2} - 4ac\)

correct substitution into discriminant (may be seen in inequality)     A1

eg     \({(2p)^2} - 4 \times 3p \times q,{\text{ }}4{p^2} - 12pq\)

\(f'(x) \geqslant 0\) then \(f'\) has two equal roots or no roots     (R1)

recognizing discriminant less or equal than zero     R1

eg     \(\Delta  \leqslant 0,{\text{ }}4{p^2} - 12pq \leqslant 0\)

correct working that clearly leads to the required answer     A1

eg     \({p^2} - 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq\)

\({p^2} \leqslant 3pq\)     AG     N0

[5 marks]

Find the x-intercepts of the graph.

(i)     Write down the equation of the axis of symmetry.

(ii)    Find the y-coordinate of the vertex.

evidence of setting function to zero     (M1)

e.g. \(f(x) = 0\) ,  \(8x = 2{x^2}\)

evidence of correct working     A1

e.g. \(0 = 2x(4 - x)\) , \(\frac{{ - 8 \pm \sqrt {64} }}{{ - 4}}\)

x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0) , or \(x = 4\) , \(x = 0\) )     A1A1     N1N1

[4 marks]

(i) \(x = 2\) (must be equation)     A1     N1

(ii) substituting \(x = 2\) into \(f(x)\)     (M1)

\(y = 8\)     A1     N2

[3 marks]

This question was answered well by most candidates.

This question was answered well by most candidates. Some did not give an equation for their axis of symmetry.

Write down the velocity of the particle when \(t = 0\) .

When \(t = k\) , the acceleration is zero.

(i)     Show that \(k = \frac{\pi }{4}\) .

(ii)    Find the exact velocity when \(t = \frac{\pi }{4}\) .

When \(t < \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\) and when \(t > \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\)  .

Sketch a graph of v against t .

Let d be the distance travelled by the particle for \(0 \le t \le 1\) .

(i)     Write down an expression for d .

(ii)    Represent d on your sketch.

\(v = 1\)     A1     N1

[1 mark]

(i) \(\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2\)     A1

\(\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = - 2\sin 2t\)     A1A1

Note: Award A1 for coefficient 2 and A1 for \( - \sin 2t\) .

evidence of considering acceleration = 0     (M1)

e.g. \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0\) , \(2 - 2\sin 2t = 0\)

correct manipulation     A1

e.g. \(\sin 2k = 1\) , \(\sin 2t = 1\)

\(2k = \frac{\pi }{2}\) (accept \(2t = \frac{\pi }{2}\) )     A1

\(k = \frac{\pi }{4}\)     AG     N0

(ii) attempt to substitute \(t = \frac{\pi }{4}\) into v     (M1)

e.g. \(2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)\)

\(v = \frac{\pi }{2}\)     A1     N2

[8 marks]

     A1A1A2      N4

Notes: Award A1 for y-intercept at \((0{\text{, }}1)\) , A1 for curve having zero gradient at \(t = \frac{\pi }{4}\) , A2 for shape that is concave down to the left of \(\frac{\pi }{4}\) and concave up to the right of \(\frac{\pi }{4}\) . If a correct curve is drawn without indicating \(t = \frac{\pi }{4}\) , do not award the second A1 for the zero gradient, but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essential features need to be clear.

[4 marks]

(i) correct expression     A2

e.g. \(\int_0^1 {(2t + \cos 2t){\rm{d}}t} \) , \(\left[ {{t^2} + \frac{{\sin 2t}}{2}} \right]_0^1\) , \(1 + \frac{{\sin 2}}{2}\) , \(\int_0^1 {v{\rm{d}}t} \)

(ii)

     A1

Note: The line at \(t = 1\) needs to be clearly after \(t = \frac{\pi }{4}\) .

[3 marks]

Many candidates gave a correct initial velocity, although a substantial number of candidates answered that \(0 + \cos 0 = 0\) .

For (b), students commonly applied the chain rule correctly to achieve the derivative, and many recognized that the acceleration must be zero. Occasionally a student would use a double-angle identity on the velocity function before differentiating. This is not incorrect, but it usually caused problems when trying to show \(k = \frac{\pi }{4}\) . At times students would reach the equation \(\sin 2k = 1\) and then substitute the \(\frac{\pi }{4}\) , which does not satisfy the “show that” instruction.

The challenge in this question is sketching the graph using the information achieved and provided. This requires students to make graphical interpretations, and as typical in section B, to link the early parts of the question with later parts. Part (a) provides the y-intercept, and part (b) gives a point with a horizontal tangent. Plotting these points first was a helpful strategy. Few understood either the notation or the concept that the function had to be increasing on either side of the \(\frac{\pi }{4}\) , with most thinking that the point was either a max or min. It was the astute student who recognized that the derivatives being positive on either side of \(\frac{\pi }{4}\) creates a point of inflexion.

Additionally, important points should be labelled in a sketch. Indicating the \(\frac{\pi }{4}\) on the x-axis is a requirement of a clear graph. Although students were not penalized for not labelling the \(\frac{\pi }{2}\) on the y-axis, there should be a recognition that the point is higher than the y-intercept.

While some candidates recognized that the distance is the area under the velocity graph, surprisingly few included neither the limits of integration in their expression, nor the “dt”. Most unnecessarily attempted to integrate the function, often giving an answer with “+C”, and only earned marks if the limits were included with their result. Few recognized that a shaded area is an adequate representation of distance on the sketch, with most fruitlessly attempting to graph a new curve.

Show that the discriminant of \(f(x)\) is \(100 - 4{p^2}\).

Find the values of \(p\) so that \(f(x) = 0\) has two equal roots.

correct substitution into \({b^2} - 4ac\)     A1

eg\(\;\;\;{(10 - p)^2} - 4(p)\left( {\frac{5}{4}p - 5} \right)\)

correct expansion of each term     A1A1

eg\(\;\;\;100 - 20p + {p^2} - 5{p^2} + 20p,{\text{ }}100 - 20p + {p^2} - (5{p^2} - 20p)\)

\(100 - 4{p^2}\)     AG     N0

[3 marks]

recognizing discriminant is zero for equal roots     (R1)

eg\(\;\;\;D = 0,{\text{ }}4{p^2} = 100\)

correct working     (A1)

eg\(\;\;\;{p^2} = 25\), \(1\) correct value of \(p\)

both correct values \(p =  \pm 5\)     A1     N2

[3 marks]

Total [6 marks]

Many candidates were able to identify the discriminant correctly and continued with good algebraic manipulation. A commonly seen mistake was identifying the constant as \(\frac{5}{4}p\) instead of \(\frac{5}{4}p - 5\). Mostly a correct approach to part b) was seen \((\Delta  = 0)\), with the common error being only one answer given for \(p\), even though the question said values (plural).

Many candidates were able to identify the discriminant correctly and continued with good algebraic manipulation. A commonly seen mistake was identifying the constant as \(\frac{5}{4}p\) instead of \(\frac{5}{4}p - 5\). Mostly a correct approach to part b) was seen \((\Delta  = 0)\), with the common error being only one answer given for \(p\), even though the question said values (plural).

Write down the equation of the axis of symmetry.

The function f can be written in the form \(f(x) = a{(x - h)^2} + k\) .

Write down the value of h and of k .

The function f can be written in the form \(f(x) = a{(x - h)^2} + k\) .

Find a .

\(x = 4\) (must be an equation)     A1     N1

[1 mark]

\(h = 4\) , \(k = 2\)     A1A1     N2

[2 marks]

attempt to substitute coordinates of any point on the graph into f     (M1)

e.g. \(f(0) = 6\) , \(6 = a{(0 - 4)^2} + 2\) , \(f(4) = 2\)

correct equation (do not accept an equation that results from \(f(4) = 2\) )     (A1)

e.g. \(6 = a{( - 4)^2} + 2\) , \(6 = 16a + 2\)

\(a = \frac{4}{{16}}\left( { = \frac{1}{4}} \right)\)     A1     N2

[3 marks]

A surprising number of candidates missed part (a) of this question, which required them to write the equation of the axis of symmetry. Some candidates did not write their answer as an equation, while others simply wrote the formula \(x = - \frac{b}{{2a}}\) .

This was answered correctly by the large majority of candidates.

The rest of this question was answered correctly by the large majority of candidates. The mistakes seen in part (c) were generally due to either incorrect substitution of a point into the equation, or substitution of the vertex coordinates, which got the candidates nowhere.

Find both x-intercepts.

Find the x-coordinate of the vertex.

evidence of attempting to solve \(f(x) = 0\)     (M1)

evidence of correct working     A1

e.g. \((x + 1)(x - 2)\) , \(\frac{{1 \pm \sqrt 9 }}{2}\)

intercepts are \(( - 1{\text{, }}0)\) and \((2{\text{, }}0)\) (accept \(x = - 1\) , \(x = 2\) ) A1A1     N1N1

[4 marks]

evidence of appropriate method     (M1)

e.g. \({x_v} = \frac{{{x_1} + {x_2}}}{2}\) , \({x_v} = - \frac{b}{{2a}}\) , reference to symmetry

\({x_v} = 0.5\)    A1     N2

[2 marks]

This question was consistently the best handled one on the entire paper.

This question was consistently the best handled one on the entire paper.

Find the value of \(f(0)\) .

Find the set of values of \(x\) for which \(f\) is increasing.

(i)     Find \(f''(1)\) .

(ii)     Hence, show that there is no point of inflexion on the graph of \(f\) at \(x = 0\) .

There is a point of inflexion on the graph of \(f\) at \(x = \sqrt[4]{3}\) \((x = 1.316 \ldots )\) .

Sketch the graph of \(f\) , for \(x \ge 0\) .

substitute \(0\) into \(f\)     (M1)

eg   \(\ln (0 + 1)\) , \(\ln 1\)

\(f(0) = 0\)     A1 N2

[2 marks]

\(f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}\) (seen anywhere)     A1A1

Note: Award A1 for \(\frac{1}{{{x^4} + 1}}\) and A1 for \(4{x^3}\) .

recognizing \(f\) increasing where \(f'(x) > 0\) (seen anywhere)     R1

eg   \(f'(x) > 0\) , diagram of signs

attempt to solve \(f'(x) > 0\)     (M1)

eg   \(4{x^3} = 0\) , \({x^3} > 0\)

\(f\) increasing for \(x > 0\) (accept \(x \ge 0\) )     A1     N1

[5 marks]

(i)     substituting \(x = 1\) into \(f''\)     (A1)

eg   \(\frac{{4(3 - 1)}}{{{{(1 + 1)}^2}}}\) , \(\frac{{4 \times 2}}{4}\)

\(f''(1) = 2\)     A1     N2

(ii)     valid interpretation of point of inflexion (seen anywhere)     R1

eg   no change of sign in \(f''(x)\) , no change in concavity,

\(f'\) increasing both sides of zero

attempt to find \(f''(x)\) for \(x < 0\)     (M1)

eg   \(f''( - 1)\) , \(\frac{{4{{( - 1)}^2}(3 - {{( - 1)}^4})}}{{{{({{( - 1)}^4} + 1)}^2}}}\) , diagram of signs

correct working leading to positive value     A1

eg   \(f''( - 1) = 2\) , discussing signs of numerator and denominator

there is no point of inflexion at \(x = 0\)     AG     N0

[5 marks]

     A1A1A1     N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

    Only if this A1 is awarded, then award the following:

    A1 for curve through (\(0\), \(0\)) , A1 for increasing throughout.

    Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

Many candidates left their answer to part (a) as \(\ln 1\). While this shows an understanding for substituting a value into a function, it leaves an unfinished answer that should be expressed as an integer.

Candidates who attempted to consider where \(f\) is increasing generally understood the derivative is needed. However, a number of candidates did not apply the chain rule, which commonly led to answers such as “increasing for all \(x\)”. Many set their derivative equal to zero, while neglecting to indicate in their working that \(f'(x) > 0\) for an increasing function. Some created a diagram of signs, which provides appropriate evidence as long as it is clear that the signs represent \(f’\).

Finding \(f''(1)\) proved no challenge, however, using this value to show that no point of inflexion exists proved elusive for many. Some candidates recognized the signs must not change in the second derivative. Few candidates presented evidence in the form of a calculation, which follows from the “hence” command of the question. In this case, a sign diagram without numerical evidence was not sufficient.

Few candidates created a correct graph from the information given or found in the question. This included the point (\(0\), \(0\)), the fact that the function is always increasing for \(x > 0\) , the concavity at \(x = 1\) and the change in concavity at the given point of inflexion. Many incorrect attempts showed a graph concave down to the right of \(x = 0\) , changing to concave up.

Use the quotient rule to show that \(f'(x) = \frac{{2{x^2} - 2}}{{{{( - 2{x^2} + 5x - 2)}^2}}}\) .

Hence find the coordinates of B.

Given that the line \(y = k\) does not meet the graph of f , find the possible values of k .

correct derivatives applied in quotient rule     (A1)A1A1

\(1\), \( - 4x + 5\)

Note: Award (A1) for 1, A1 for \( - 4x\) and A1 for \(5\), only if it is clear candidates are using the quotient rule.

correct substitution into quotient rule     A1

e.g. \(\frac{{1 \times ( - 2{x^2} + 5x - 2) - x( - 4x + 5)}}{{{{( - 2{x^2} + 5x - 2)}^2}}}\) , \(\frac{{ - 2{x^2} + 5x - 2 - x( - 4x + 5)}}{{{{( - 2{x^2} + 5x - 2)}^2}}}\)

correct working     (A1)

e.g. \(\frac{{ - 2{x^2} + 5x - 2 - ( - 4{x^2} + 5x)}}{{{{( - 2{x^2} + 5x - 2)}^2}}}\)

expression clearly leading to the answer     A1

e.g. \(\frac{{ - 2{x^2} + 5x - 2 + 4{x^2} - 5x}}{{{{( - 2{x^2} + 5x - 2)}^2}}}\)

\(f'(x) = \frac{{2{x^2} - 2}}{{{{( - 2{x^2} + 5x - 2)}^2}}}\)    AG     N0

[6 marks]

evidence of attempting to solve \(f'(x) = 0\)     (M1)

e.g. \(2{x^2} - 2 = 0\)

evidence of correct working     A1

e.g. \({x^2} = 1,\frac{{ \pm \sqrt {16} }}{4}{\text{, }}2(x - 1)(x + 1)\)

correct solution to quadratic     (A1)

e.g. \(x = \pm 1\)

correct x-coordinate \(x = - 1\) (may be seen in coordinate form \(\left( { - 1,\frac{1}{9}} \right)\) )    A1     N2

attempt to substitute \( - 1\) into f (do not accept any other value)     (M1)

e.g. \(f( - 1) = \frac{{ - 1}}{{ - 2 \times {{( - 1)}^2} + 5 \times ( - 1) - 2}}\)

correct working

e.g. \(\frac{{ - 1}}{{ - 2 - 5 - 2}}\)     A1

correct y-coordinate \(y = \frac{1}{9}\) (may be seen in coordinate form \(\left( { - 1,\frac{1}{9}} \right)\) )    A1     N2

[7 marks]

recognizing values between max and min     (R1)

\(\frac{1}{9} < k < 1\)     A2     N3

[3 marks]

While most candidates answered part (a) correctly, there were some who did not show quite enough work for a "show that" question. A very small number of candidates did not follow the instruction to use the quotient rule.

In part (b), most candidates knew that they needed to solve the equation \(f'(x) = 0\) , and many were successful in answering this question correctly. However, some candidates failed to find both values of x, or made other algebraic errors in their solutions. One common error was to find only one solution for \({x^2} = 1\) ; another was to work with the denominator equal to zero, rather than the numerator.

In part (c), a significant number of candidates seemed to think that the line \(y = k\) was a vertical line, and attempted to find the vertical asymptotes. Others tried looking for a horizontal asymptote. Fortunately, there were still a good number of intuitive candidates who recognized the link with the graph and with part (b), and realized that the horizontal line must pass through the space between the given local minimum and the local maximum they had found in part (b).

Given that \({2^m} = 8\) and \({2^n} = 16\), write down the value of \(m\) and of \(n\).

Hence or otherwise solve \({8^{2x + 1}} = {16^{2x - 3}}\).

\(m = 3,{\text{ }}n = 4\)     A1A1     N2

[2 marks]

attempt to apply \({({2^a})^b} = {2^{ab}}\)     (M1)

eg\(\;\;\;6x + 3,{\text{ }}4(2x - 3)\)

equating their powers of \(2\) (seen anywhere)     M1

eg\(\;\;\;3(2x + 1) = 8x - 12\)

correct working     A1

eg\(\;\;\;8x - 12 = 6x + 3,{\text{ }}2x = 15\)

\(x = \frac{{15}}{2}\;\;\;(7.5)\)     A1     N2

[4 marks]

Total [6 marks]

Indices laws were well understood with many candidates solving the equation correctly. Some candidates used logs, which took longer, and errors crept in.

Indices laws were well understood with many candidates solving the equation correctly. Some candidates used logs, which took longer, and errors crept in.

Write down the value of q and of r.

Write down the equation of the axis of symmetry.

Find the value of p.

\(q = - 2\) , \(r = 4\) or \(q = 4\) , \(r = - 2\)     A1A1     N2

[2 marks]

\(x = 1\) (must be an equation)     A1     N1

[1 mark]

substituting \((0{\text{, }} -  4)\) into the equation     (M1)

e.g. \( - 4 = p(0 - ( - 2))(0 - 4)\) , \( - 4 = p( - 4)(2)\)

correct working towards solution     (A1)

e.g. \( - 4 = - 8p\)

\(p = \frac{4}{8}\) \(\left( { = \frac{1}{2}} \right)\)     A1     N2

[3 marks]

The majority of candidates were successful on some or all parts of this question, with some candidates using a mix of algebra and graphical reasoning and others ignoring the graph and working only algebraically. Some did not recognize that p and q are the roots of the quadratic function and hence gave the answers as 2 and \( - 4\).

A common error in part (b) was the absence of an equation. Some candidates wrote down the equation \(x = \frac{{ - b}}{{2a}}\) but were not able to substitute correctly. Those students did not realize that the axis of symmetry is always halfway between the x-intercepts.

More candidates had trouble with part (c) with erroneous substitutions and simplification mistakes commonplace.

Write \(h(x)\) in the form \(a\sin (bx)\), where \(a,{\text{ }}b \in \mathbb{Z}\).

Hence find the range of \(h\).

attempt to form composite in any order     (M1)

eg\(\,\,\,\,\,\)\(f\left( {g(x)} \right),{\text{ }}\cos \left( {6x\sqrt {1 - {x^2}} } \right)\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6\cos x\sqrt {1 - {{\cos }^2}x} \)

correct application of Pythagorean identity (do not accept \({\sin ^2}x + {\cos ^2}x = 1\))     (A1)

eg\(\,\,\,\,\,\)\({\sin ^2}x = 1 - {\cos ^2}x,{\text{ }}6\cos x\sin x,{\text{ }}6\cos x \left| \sin x\right|\)

valid approach (do not accept \(2\sin x\cos x = \sin 2x\))     (M1)

eg\(\,\,\,\,\,\)\(3(2\cos x\sin x)\)

\(h(x) = 3\sin 2x\)    A1     N3

[5 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)amplitude \( = 3\), sketch with max and min \(y\)-values labelled, \( - 3 < y < 3\)

correct range     A1     N2

eg\(\,\,\,\,\,\)\( - 3 \leqslant y \leqslant 3\), \([ - 3,{\text{ }}3]\) from \( - 3\) to 3

Note:     Do not award A1 for \( - 3 < y < 3\) or for “between \( - 3\) and 3”.

[2 marks]

In part (a), nearly all candidates found the correct composite function in terms of \(\cos x\), though many did not get any further than this first step in their solution to the question. While some candidates seemed to recognize the need to use trigonometric identities, most were unsuccessful in finding the correct expression in the required form.

In part (b), very few candidates were able to provide the correct range of the function.

Find the value of \(p\).

Find the value of \(a\).

The line \(y = kx - 5\) is a tangent to the curve of \(f\). Find the values of \(k\).

METHOD 1 (using x-intercept)

determining that 3 is an \(x\)-intercept     (M1)

eg\(\,\,\,\,\,\)\(x - 3 = 0\),

valid approach     (M1)

eg\(\,\,\,\,\,\)\(3 - 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5\)

\(p = 2\)     A1     N2

METHOD 2 (expanding f (x)) 

correct expansion (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a{x^2} - a(3 + p)x + 3ap,{\text{ }}{x^2} - (3 + p)x + 3p\)

valid approach involving equation of axis of symmetry     (M1)

eg\(\,\,\,\,\,\)\(\frac{{ - b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}\)

\(p = 2\)     A1     N2

METHOD 3 (using derivative)

correct derivative (accept absence of \(a\))     (A1)

eg\(\,\,\,\,\,\)\(a(2x - 3 - p),{\text{ }}2x - 3 - p\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f’(2.5) = 0\)

\(p = 2\)     A1     N2

[3 marks]

attempt to substitute \((0,{\text{ }} - 6)\)     (M1)

eg\(\,\,\,\,\,\)\( - 6 = a(0 - 2)(0 - 3),{\text{ }}0 = a( - 8)( - 9),{\text{ }}a{(0)^2} - 5a(0) + 6a =  - 6\)

correct working     (A1)

eg\(\,\,\,\,\,\)\( - 6 = 6a\)

\(a =  - 1\)     A1     N2

[3 marks]

METHOD 1 (using discriminant)

recognizing tangent intersects curve once     (M1)

recognizing one solution when discriminant = 0     M1

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx - 5 =  - {x^2} + 5x - 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} - 5x + kx + 1 = 0\)

correct discriminant (if seen explicitly, not just in quadratic formula)     A1

eg\(\,\,\,\,\,\)\({(k - 5)^2} - 4,{\text{ }}25 - 10k + {k^2} - 4\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(k - 5 =  \pm 2,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

METHOD 2 (using derivatives)

attempt to set up equation     (M1)

eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx - 5 =  - {x^2} + 5x - 6\)

recognizing derivative/slope are equal     (M1)

eg\(\,\,\,\,\,\)\(f’ = {m_T},{\text{ }}f' = k\)

correct derivative of \(f\)     (A1)

eg\(\,\,\,\,\,\)\( - 2x + 5\)

attempt to set up equation in terms of either \(x\) or \(k\)     M1

eg\(\,\,\,\,\,\)\(( - 2x + 5)x - 5 =  - {x^2} + 5x - 6,{\text{ }}k\left( {\frac{{5 - k}}{2}} \right) - 5 =  - {\left( {\frac{{5 - k}}{2}} \right)^2} + 5\left( {\frac{{5 - k}}{2}} \right) - 6\)

rearranging their equation to equal zero     (M1)

eg\(\,\,\,\,\,\)\({x^2} - 1 = 0,{\text{ }}{k^2} - 10k + 21 = 0\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(x =  \pm 1,{\text{ }}(k - 3)(k - 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 - 4 \times 21} }}{2}\)

\(k = 3,{\text{ }}7\)     A1A1     N0

[8 marks]

Write down the value of \(h\) and of \(k\).

Write down the value of \(a\) and of \(b\).

Find the \(y\)-intercept.

\(h = 3,{\text{ }}k =  - 1\)    A1A1     N2

[2 marks]

\(a = 2,{\text{ }}b = 4{\text{ }}({\text{or }}a = 4,{\text{ }}b = 2)\)    A1A1     N2

[2 marks]

attempt to substitute \(x = 0\) into their \(f\)     (M1)

eg\(\,\,\,\,\,\)\({(0 - 3)^2} - 1,{\text{ }}(0 - 2)(0 - 4)\)

\(y = 8\)    A1     N2

[2 marks]

Nearly all candidates performed well on this question, earning full marks on all three question parts.

Nearly all candidates performed well on this question, earning full marks on all three question parts. In part (b), there were some candidates who factored the quadratic expression correctly, but went on to give negative values for \(a\) and \(b\).

Nearly all candidates performed well on this question, earning full marks on all three question parts.

Show that \(f( - x) = - f(x)\) .

Given that \(f''(x) = \frac{{2ax({x^2} - 3)}}{{{{({x^2} + 1)}^3}}}\) , find the coordinates of all points of inflexion.

It is given that \(\int {f(x){\rm{d}}x = \frac{a}{2}} \ln ({x^2} + 1) + C\) .

(i)     Find the area of the shaded region, giving your answer in the form \(p\ln q\) .

(ii)    Find the value of \(\int_4^8 {2f(x - 1){\rm{d}}x} \) .

METHOD 1

evidence of substituting \( - x\) for \(x\)     (M1)

\(f( - x) = \frac{{a( - x)}}{{{{( - x)}^2} + 1}}\)     A1

\(f( - x) = \frac{{ - ax}}{{{x^2} + 1}}\) \(( = - f(x))\)     AG     N0

METHOD 2

\(y = - f(x)\) is reflection of \(y = f(x)\) in x axis

and \(y = f( - x)\) is reflection of \(y = f(x)\) in y axis     (M1)

sketch showing these are the same     A1

\(f( - x) = \frac{{ - ax}}{{{x^2} + 1}}\) \(( = - f(x))\)     AG     N0

[2 marks]

evidence of appropriate approach     (M1)

e.g. \(f''(x) = 0\)

to set the numerator equal to 0     (A1)

e.g. \(2ax({x^2} - 3) = 0\) ; \(({x^2} - 3) = 0\)

(0, 0) , \(\left( {\sqrt 3 ,\frac{{a\sqrt 3 }}{4}} \right)\) , \(\left( { - \sqrt 3 , - \frac{{a\sqrt 3 }}{4}} \right)\) (accept \(x = 0\) , \(y = 0\) etc)      A1A1A1A1A1     N5

[7 marks]

(i) correct expression     A2

e.g. \(\left[ {\frac{a}{2}\ln ({x^2} + 1)} \right]_3^7\) , \(\frac{a}{2}\ln 50 - \frac{a}{2}\ln 10\) , \(\frac{a}{2}(\ln 50 - \ln 10)\)

area = \(\frac{a}{2}\ln 5\)     A1A1     N2

(ii) METHOD 1

recognizing the shift that does not change the area     (M1)

e.g. \(\int_4^8 {f(x - 1){\rm{d}}x}  = \int_3^7 {f(x){\rm{d}}x} \) , \(\frac{a}{2}\ln 5\)

recognizing that the factor of 2 doubles the area     (M1)

e.g. \(\int_4^8 {2f(x - 1){\rm{d}}x = } 2\int_4^8 {f(x - 1){\rm{d}}x} \) \(\left( { = 2\int_3^7 {f(x){\rm{d}}x} } \right)\)

\(\int_4^8 {2f(x - 1){\rm{d}}x = a\ln 5} \) (i.e. \(2 \times \) their answer to (c)(i))     A1     N3

METHOD 2

changing variable

let \(w = x - 1\) , so \(\frac{{{\rm{d}}w}}{{{\rm{d}}x}} = 1\)

\(2\int {f(w){\rm{d}}w = } \frac{{2a}}{2}\ln ({w^2} + 1) + c\)     (M1)

substituting correct limits

e.g. \(\left[ {a\ln \left[ {{{(x - 1)}^2} + 1} \right]} \right]_4^8\) , \(\left[ {a\ln ({w^2} + 1)} \right]_3^7\) , \(a\ln 50 - a\ln 10\)     (M1)

\(\int_4^8 {2f(x - 1){\rm{d}}x = a\ln 5} \)     A1     N3

[7 marks]

Part (a) was achieved by some candidates, although brackets around the \( - x\) were commonly neglected. Some attempted to show the relationship by substituting a specific value for x . This earned no marks as a general argument is required.

Although many recognized the requirement to set the second derivative to zero in (b), a majority neglected to give their answers as ordered pairs, only writing the x-coordinates. Some did not consider the negative root.

For those who found a correct expression in (c)(i), many finished by calculating \(\ln 50 - \ln 10 = \ln 40\) . Few recognized that the translation did not change the area, although some factored the 2 from the integrand, appreciating that the area is double that in (c)(i).

(i)     Find the coordinates of A.

(ii)    Show that \(f'(x) = 0\) at A.

The second derivative \(f''(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} - 4)}^3}}}\) . Use this to

(i)     justify that the graph of f has a local maximum at A;

(ii)    explain why the graph of f does not have a point of inflexion.

Describe the behaviour of the graph of \(f\) for large \(|x|\) .

Write down the range of \(f\) .

(i) coordinates of A are \((0{\text{, }} - 2)\)     A1A1     N2

(ii) derivative of \({x^2} - 4 = 2x\) (seen anywhere)     (A1)

evidence of correct approach     (M1)

e.g. quotient rule, chain rule

finding \(f'(x)\)     A2

e.g. \(f'(x) = 20 \times ( - 1) \times {({x^2} - 4)^{ - 2}} \times (2x)\) , \(\frac{{({x^2} - 4)(0) - (20)(2x)}}{{{{({x^2} - 4)}^2}}}\)

substituting \(x = 0\) into \(f'(x)\) (do not accept solving \(f'(x) = 0\) )     M1

at A \(f'(x) = 0\)     AG     N0

[7 marks]

(i) reference to \(f'(x) = 0\) (seen anywhere)     (R1)

reference to \(f''(0)\) is negative (seen anywhere)     R1

evidence of substituting \(x = 0\) into \(f''(x)\)     M1

finding \(f''(0) = \frac{{40 \times 4}}{{{{( - 4)}^3}}}\) \(\left( { = - \frac{5}{2}} \right)\)     A1

then the graph must have a local maximum     AG

(ii) reference to \(f''(x) = 0\) at point of inflexion     (R1)

recognizing that the second derivative is never 0     A1     N2

e.g. \(40(3{x^2} + 4) \ne 0\) , \(3{x^2} + 4 \ne 0\) , \({x^2} \ne  - \frac{4}{3}\) , the numerator is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

correct (informal) statement, including reference to approaching \(y = 3\)     A1     N1

e.g. getting closer to the line \(y = 3\) , horizontal asymptote at \(y = 3\)

[1 mark]

correct inequalities, \(y \le - 2\) , \(y > 3\) , FT from (a)(i) and (c)     A1A1     N2

[2 marks]

Almost all candidates earned the first two marks in part (a) (i), although fewer were able to apply the quotient rule correctly.

Many candidates were able to state how the second derivative can be used to identify maximum and inflection points, but fewer were actually able to demonstrate this with the given function. For example, in (b)(ii) candidates often simply said "the second derivative cannot equal 0" but did not justify or explain why this was true.

Not too many candidates could do part (c) correctly.

In (d) even those who knew what the range was had difficulty expressing the inequalities correctly.

Solve \({\log _2}x + {\log _2}(x - 2) = 3\) , for \(x > 2\) .

recognizing \(\log a + \log b = \log ab\) (seen anywhere)     (A1)

e.g. \({\log _2}(x(x - 2))\) , \({x^2} - 2x\)

recognizing \({\log _a}b = x \Leftrightarrow {a^x} = b\)     (A1)

e.g. \({2^3} = 8\)

correct simplification     A1

e.g. \(x(x - 2) = {2^3}\) , \({x^2} - 2x - 8\)

evidence of correct approach to solve     (M1)

e.g. factorizing, quadratic formula

correct working     A1

e.g. \((x - 4)(x + 2)\) , \(\frac{{2 \pm \sqrt {36} }}{2}\)

\(x = 4\)     A2     N3

[7 marks]

Candidates secure in their understanding of logarithm properties usually had success with this problem, solving the resulting quadratic either by factoring or using the quadratic formula. The majority of successful candidates correctly rejected the solution that was not in the domain. A number of candidates, however, were unclear on logarithm properties. Some unsuccessful candidates were able to demonstrate understanding of one property but without both were not able to make much progress. A few candidates employed a “guess and check” strategy, but this did not earn full marks.

Let \(h(x) = f( - x)\) . Sketch the graph of \(h\) on the grid below.

Let \(g(x) = \frac{1}{2}f(x - 1)\) . The point \({\text{A}}(3{\text{, }}2)\) on the graph of \(f\) is transformed to the point P on the graph of \(g\) . Find the coordinates of P.

     A2     N2

[2 marks]

evidence of appropriate approach     (M1)

e.g. reference to any horizontal shift and/or stretch factor, \(x = 3 + 1\) , \(y = \frac{1}{2} \times 2\)

P is \((4{\text{, }}1)\) (accept \(x = 4\) , \(y = 1\))     A1A1     N3

[3 marks]

Part (a) was generally solved correctly. Students had no trouble in deciding what transformation had to be done to the graph, although some confused \(f( - x)\) with \( - f(x)\) . 

Part (b) was generally poorly done. They could not "read" that the transformation shifted the curve 1 unit to the right and stretched it in the \(y\)-direction with a scale factor of \(\frac{1}{2}\) . It was often seen that the shift was interpreted, but in the opposite direction. Also, the stretch was applied to both coordinates of the point. Those candidates who answered part (a) incorrectly often had trouble on (b) as well, indicating a difficulty with transformations in general. However, there were also candidates who solved part (a) correctly but could not interpret part (b). This would indicate that it is simpler for them to plot the transformation of an entire function than to find how a particular point is transformed.

Show that \(f(x) = 3{x^2} + 6x - 9\) .

For the graph of f

(i)     write down the coordinates of the vertex;

(ii)    write down the equation of the axis of symmetry;

(iii)   write down the y-intercept;

(iv)   find both x-intercepts.

Hence sketch the graph of f .

Let \(g(x) = {x^2}\) . The graph of f may be obtained from the graph of g by the two transformations:

a stretch of scale factor t in the y-direction

followed by a translation of \(\left( {\begin{array}{*{20}{c}}
p\\
q
\end{array}} \right)\) .

Find \(\left( {\begin{array}{*{20}{c}}
p\\
q
\end{array}} \right)\) and the value of t.

\(f(x) = 3({x^2} + 2x + 1) - 12\)     A1

\( = 3{x^2} + 6x + 3 - 12\)     A1

\( = 3{x^2} + 6x - 9\)     AG     N0

[2 marks]

(i) vertex is \(( - 1{\text{, }} - 12)\)     A1A1     N2

(ii) \(x = - 1\) (must be an equation)     A1     N1

(iii) \((0{\text{, }} - 9)\)     A1     N1

(iv) evidence of solving \(f(x) = 0\)     (M1)

e.g. factorizing, formula,

correct working     A1

e.g. \(3(x + 3)(x - 1) = 0\) , \(x = \frac{{ - 6 \pm \sqrt {36 + 108} }}{6}\)

\(( - 3{\text{, }}0)\), \((1{\text{, }}0)\)     A1A1     N1N1

[8 marks]

     A1A1     N2

Note: Award A1 for a parabola opening upward, A1 for vertex and intercepts in approximately correct positions.

[2 marks]

\(\left( {\begin{array}{*{20}{c}}
p\\
q
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 1}\\
{ - 12}
\end{array}} \right)\) , \(t = 3\) (accept \(p = - 1\) , \(q = - 12\) , \(t = 3\) )     A1A1A1     N3

[3 marks]

This problem was generally well done. The “show that” question in part (a) was done correctly by most candidates, with a few attempting to show it by working backwards, which earned no marks.

Most candidates were able to identify the vertex but were unable to write the equation for the axis of symmetry. There was a great deal of success with the x and y intercepts.

Some of the sketches of the graph left much to be desired even if they were technically correct; many were v-shaped.

The final part was poorly done, indicating that defining a graph in terms of stretch and translation was unfamiliar to many candidates.

The equation \({x^2} - 3x + {k^2} = 4\) has two distinct real roots. Find the possible values of k .

evidence of rearranged quadratic equation (may be seen in working)     A1

e.g. \({x^2} - 3x + {k^2} - 4 = 0\) , \({k^2} - 4\) 

evidence of discriminant (must be seen explicitly, not in quadratic formula)     (M1)

e.g. \({b^2} - 4ac\) , \(\Delta  = {( - 3)^2} - 4(1)({k^2} - 4)\)

recognizing that discriminant is greater than zero (seen anywhere, including answer)     R1

e.g. \({b^2} - 4ac > 0\) , \(9 + 16 - 4{k^2} > 0\)

correct working (accept equality)     A1

e.g. \(25 - 4{k^2} > 0\) , \(4{k^2} < 25\) , \({k^2} = \frac{{25}}{4}\)

both correct values (even if inequality never seen)     (A1)

e.g. \(\pm \sqrt{{\frac{{25}}{4}}}\) , \( \pm 2.5\)

correct interval     A1     N3

e.g. \( - \frac{5}{2} < k < \frac{5}{2}\) , \( - 2.5 < k < 2.5\)

Note: Do not award the final mark for unfinished values, or for incorrect or reversed inequalities, including \( \le \) , \(k > - 2.5\) , \(k < 2.5\) .

Special cases:

If working shown, and candidates attempt to rearrange the quadratic equation to equal zero, but find an incorrect value of c, award A1M1R1A0A0A0.

If working shown, and candidates do not rearrange the quadratic equation to equal zero, but find \(c = {k^2}\) or \(c = \pm 4\) , award A0M1R1A0A0A0.

[6 marks]

The majority of candidates who attempted to answer this question recognized the need to use the discriminant, however very few were able to answer the question successfully. The majority of candidates did not recognize that the quadratic equation must first be set equal to zero. In addition, many candidates simply set their discriminant equal to zero, instead of setting it greater than zero. Even many of the strongest candidates, who obtained the correct numerical values for k, were unable to write their final answers as a correct interval.

This question is a good example of candidates who reach for familiar methods, without really thinking about what the question is asking them to find. There were many candidates who attempted to solve for x using the quadratic formula or factoring, even though the question did not ask them to solve for x.

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

Solve \(f(x) = 0\).

On the following grid, sketch the graph of \(f\), for \( - 4 \le x \le 3\).

\(y\)-intercept is \( - 6,{\text{ }}(0,{\text{ }} - 6),{\text{ }}y =  - 6\)     A1

[1 mark]

valid attempt to solve     (M1)

eg\(\;\;\;(x - 2)(x + 3) = 0,{\text{ }}x = \frac{{ - 1 \pm \sqrt {1 + 24} }}{2}\), one correct answer

\(x = 2,{\text{ }}x =  - 3\)     A1A1     N3

[3 marks]

    A1A1A1

Note:     The shape must be an approximately correct concave up parabola. Only if the shape is correct, award the following:

A1 for the \(y\)-intercept in circle and the vertex approximately on \(x =  - \frac{1}{2}\), below \(y =  - 6\),

A1 for both the \(x\)-intercepts in circles,

A1 for both end points in ovals.

[3 marks]

Total [7 marks]

Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).

Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).

In part (c), although most candidates were familiar with the general parabolic shape of the graph, many placed the vertex at the \(y\)-intercept \((0,{\text{ }} - 6)\), and very few candidates considered the endpoints of the function with the given domain.

Show that \(f’(1) = 1\).

Find the equation of \(L\) in the form \(y = ax + b\).

Find the \(x\)-coordinate of Q.

Find the area of the region enclosed by the graph of \(f\) and the line \(L\).

\(f’(x) = 2x - 1\)     A1A1

correct substitution     A1

eg\(\,\,\,\,\,\)\(2(1) - 1,{\text{ }}2 - 1\)

\(f’(1) = 1\)     AG     N0

[3 marks]

correct approach to find the gradient of the normal     (A1)

eg\(\,\,\,\,\,\)\(\frac{{ - 1}}{{f'(1)}},{\text{ }}{m_1}{m_2} =  - 1,{\text{ slope}} =  - 1\)

attempt to substitute correct normal gradient and coordinates into equation of a line     (M1)

eg\(\,\,\,\,\,\)\(y - 0 =  - 1(x - 1),{\text{ }}0 =  - 1 + b,{\text{ }}b = 1,{\text{ }}L =  - x + 1\)

\(y =  - x + 1\)     A1     N2

[3 marks]

equating expressions     (M1)

eg\(\,\,\,\,\,\)\(f(x) = L,{\text{ }} - x + 1 = {x^2} - x\)

correct working (must involve combining terms)     (A1)

eg\(\,\,\,\,\,\)\({x^2} - 1 = 0,{\text{ }}{x^2} = 1,{\text{ }}x = 1\)

\(x =  - 1\,\,\,\,\,\left( {{\text{accept }}Q( - 1,{\text{ }}2)} \right)\)     A2     N3

[4 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\int {L - f,{\text{ }}\int_{ - 1}^1 {(1 - {x^2}){\text{d}}x} } \), splitting area into triangles and integrals

correct integration     (A1)(A1)

eg\(\,\,\,\,\,\)\(\left[ {x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1,{\text{ }} - \frac{{{x^3}}}{3} - \frac{{{x^2}}}{2} + \frac{{{x^2}}}{2} + x\)

substituting their limits into their integrated function and subtracting (in any order)     (M1)

eg\(\,\,\,\,\,\)\(1 - \frac{1}{3} - \left( { - 1 - \frac{{ - 1}}{3}} \right)\)

Note:     Award M0 for substituting into original or differentiated function.

area \( = \frac{4}{3}\)     A2     N3

[6 marks]

Find \({f^{ - 1}}( - 1)\).

Find \((f \circ f)( - 1)\).

On the same diagram, sketch the graph of \(y = f( - x)\).

valid approach     (M1)

eg\(\;\;\;\)horizontal line on graph at \( - 1,{\text{ }}f(a) =  - 1,{\text{ }}( - 1,5)\)

\({f^{ - 1}}( - 1) = 5\)     A1     N2

[2 marks]

attempt to find \(f( - 1)\)     (M1)

eg\(\;\;\;\)line on graph

\(f( - 1) = 2\)     (A1)

\((f \circ f)( - 1) = 1\)     A1     N3

[3 marks]

     A1A1     N2

Note:     The shape must be an approximately correct shape (concave down and increasing). Only if the shape is approximately correct, award the following for points in circles:

A1 for the \(y\)-intercept,

A1 for any two of these points \(( - 5,{\text{ }} - 1),{\text{ }}( - 2,{\text{ }}1),{\text{ }}(1,{\text{ }}2)\).

[2 marks]

Total [7 marks]

Typically candidates were more successful in finding the composite function than the inverse. Some students tried to find the function, rather than read values from the given graph. The sketch of \(f( - x)\) was often well done, with the most common error being a reflection in the x-axis.

Typically candidates were more successful in finding the composite function than the inverse. Some students tried to find the function, rather than read values from the given graph. The sketch of \(f( - x)\) was often well done, with the most common error being a reflection in the x-axis.

Typically candidates were more successful in finding the composite function than the inverse. Some students tried to find the function, rather than read values from the given graph. The sketch of \(f( - x)\) was often well done, with the most common error being a reflection in the x-axis.

Find the value of \({\log _2}40 - {\log _2}5\) .

Find the value of \({8^{{{\log }_2}5}}\) .

evidence of correct formula    (M1)

eg   \(\log a - \log b = \log \frac{a}{b}\) , \(\log \left( {\frac{{40}}{5}} \right)\) , \(\log 8 + \log 5 - \log 5\)

Note: Ignore missing or incorrect base.

correct working     (A1)

eg   \({\log _2}8\) , \({2^3} = 8\)

\({\log _2}40 - {\log _2}5 = 3\)     A1     N2

[3 marks]

attempt to write \(8\) as a power of \(2\) (seen anywhere)     (M1)

eg   \({({2^3})^{{{\log }_2}5}}\) , \({2^3} = 8\) , \({2^a}\)