Question
Let \(f(x) = 3{(x + 1)^2} – 12\) .
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.
Answer/Explanation
Markscheme
\(f(x) = 3({x^2} + 2x + 1) – 12\) 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]
Question
Part of the graph of a function f is shown in the diagram below.
On the same diagram sketch the graph of \(y = – f(x)\) .
Let \(g(x) = f(x + 3)\) .
(i) Find \(g( – 3)\) .
(ii) Describe fully the transformation that maps the graph of f to the graph of g.
Answer/Explanation
Markscheme
M1A1 N2
Note: Award M1 for evidence of reflection in x-axis, A1 for correct vertex and all intercepts approximately correct.
(i) \(g( – 3) = f(0)\) (A1)
\(f(0) = – 1.5\) A1 N2
(ii) translation (accept shift, slide, etc.) of \(\left( {\begin{array}{*{20}{c}}
{ – 3}\\
0
\end{array}} \right)\) A1A1 N2
[4 marks]
Question
Let \(f(t) = a\cos b(t – c) + d\) , \(t \ge 0\) . Part of the graph of \(y = f(t)\) is given below.
When \(t = 3\) , there is a maximum value of 29, at M.
When \(t = 9\) , there is a minimum value of 15.
(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 .
Answer/Explanation
Markscheme
(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]
Question
Let \(f(x) = {x^2}\) and \(g(x) = 2{(x – 1)^2}\) .
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.
Answer/Explanation
Markscheme
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
Question
Let \(f(x) = \frac{{ax}}{{{x^2} + 1}}\) , \( – 8 \le x \le 8\) , \(a \in \mathbb{R}\) .The graph of f is shown below.
The region between \(x = 3\) and \(x = 7\) is shaded.
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 \(\int_4^8 {2f(x – 1){\rm{d}}x} \) .
Answer/Explanation
Markscheme
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]
Question
The diagram below shows the graph of a function \(f(x)\) , for \( – 2 \le x \le 4\) .
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.
Answer/Explanation
Markscheme
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]
Question
Let \(f(x) = \frac{1}{2}{x^3} – {x^2} – 3x\) . Part of the graph of f is shown below.
There is a maximum point at A and a minimum point at B(3, − 9) .
Find the coordinates of A.
Write down the coordinates of
(i) the image of B after reflection in the y-axis;
(ii) the image of B after translation by the vector \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
5
\end{array}} \right)\) ;
(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor \(\frac{1}{2}\) .
Answer/Explanation
Markscheme
\(f(x) = {x^2} – 2x – 3\) A1A1A1
evidence of solving \(f'(x) = 0\) (M1)
e.g. \({x^2} – 2x – 3 = 0\)
evidence of correct working A1
e.g. \((x + 1)(x – 3)\) , \(\frac{{2 \pm \sqrt {16} }}{2}\)
\(x = – 1\) (ignore \(x = 3\) ) (A1)
evidence of substituting their negative x-value into \(f(x)\) (M1)
e.g. \(\frac{1}{3}{( – 1)^3} – {( – 1)^2} – 3( – 1)\) , \( – \frac{1}{3} – 1 + 3\)
\(y = \frac{5}{3}\) A1
coordinates are \(\left( { – 1,\frac{5}{3}} \right)\) N3
[8 marks]
(i) \(( – 3{\text{, }} – 9)\) A1 N1
(ii) \((1{\text{, }} – 4)\) A1A1 N2
(iii) reflection gives \((3{\text{, }}9)\) (A1)
stretch gives \(\left( {\frac{3}{2}{\text{, }}9} \right)\) A1A1 N3
[6 marks]
Question
Let \(f(x) = {x^2} + 4\) and \(g(x) = x – 1\) .
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.
Answer/Explanation
Markscheme
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]
Question
Let \(f(x) = 3\ln x\) and \(g(x) = \ln 5{x^3}\) .
Express \(g(x)\) in the form \(f(x) + \ln a\) , where \(a \in {{\mathbb{Z}}^ + }\) .
The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.
Answer/Explanation
Markscheme
attempt to apply rules of logarithms (M1)
e.g. \(\ln {a^b} = b\ln a\) , \(\ln ab = \ln a + \ln b\)
correct application of \(\ln {a^b} = b\ln a\) (seen anywhere) A1
e.g. \(3\ln x = \ln {x^3}\)
correct application of \(\ln ab = \ln a + \ln b\) (seen anywhere) A1
e.g. \(\ln 5{x^3} = \ln 5 + \ln {x^3}\)
so \(\ln 5{x^3} = \ln 5 + 3\ln x\)
\(g(x) = f(x) + \ln 5\) (accept \(g(x) = 3\ln x + \ln 5\) ) A1 N1
[4 marks]
transformation with correct name, direction, and value A3
e.g. translation by \(\left( {\begin{array}{*{20}{c}}
0\\
{\ln 5}
\end{array}} \right)\) , shift up by \(\ln 5\) , vertical translation of \(\ln 5\)
[3 marks]
Examiners report
This question was very poorly done by the majority of candidates. While candidates seemed to have a vague idea of how to apply the rules of logarithms in part (a), very few did so successfully. The most common error in part (a) was to begin incorrectly with \(\ln 5{x^3} = 3\ln 5x\) . This error was often followed by other errors.
In part (b), very few candidates were able to describe the transformation as a vertical translation (or shift). Many candidates attempted to describe numerous incorrect transformations, and some left part (b) entirely blank.
Question
The diagram below shows the graph of a function \(f(x)\) , for \( – 2 \le x \le 3\) .
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 .
Answer/Explanation
Markscheme
A2 N2
[2 marks]
\(a = – 2,b = – 1\) A2A2 N4
Note: Award A1 for \(a = 2\) , A1 for \(b = 1\) .
[4 marks]
Question
Let \(f(x) = 3{x^2} – 6x + p\). The equation \(f(x) = 0\) has two equal roots.
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 \(g(x) = A{x^2} + Bx + C\).
Answer/Explanation
Markscheme
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]
Question
The following diagram shows the graph of a function \(f\).
Find \({f^{ – 1}}( – 1)\).
Find \((f \circ f)( – 1)\).
On the same diagram, sketch the graph of \(y = f( – x)\).
Answer/Explanation
Markscheme
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]
Question
The following diagram shows part of the graph of a quadratic function \(f\).
The vertex is at \((1,{\text{ }} – 9)\), and the graph crosses the y–axis at the point \((0,{\text{ }}c)\).
The function can be written in the form \(f(x) = {(x – h)^2} + k\).
Write down the value of \(h\) and of \(k\).
Find the value of \(c\).
Let \(g(x) = – {(x – 3)^2} + 1\). The graph of \(g\) is obtained by a reflection of the graph of \(f\) in the \(x\)-axis, followed by a translation of \(\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right)\).
Find the value of \(p\) and of \(q\).
Find the x-coordinates of the points of intersection of the graphs of \(f\) and \(g\).
Answer/Explanation
Markscheme
\(h = 1,{\text{ }}k = – 9\;\;\;\left( {{\text{accept }}{{(x – 1)}^2} – 9} \right)\) A1A1 N2
[2 marks]
METHOD 1
attempt to substitute \(x = 0\) into their quadratic function (M1)
eg\(\;\;\;f(0),{\text{ }}{(0 – 1)^2} – 9\)
\(c = – 8\) A1 N2
METHOD 2
attempt to expand their quadratic function (M1)
eg\(\;\;\;{x^2} – 2x + 1 – 9,{\text{ }}{x^2} – 2x – 8\)
\(c = – 8\) A1 N2
[2 marks]
evidence of correct reflection A1
eg\(\;\;\; – \left( {{{(x – 1)}^2} – 9} \right)\), vertex at \((1,{\text{ }}9)\), y-intercept at \((0,{\text{ }}8)\)
valid attempt to find horizontal shift (M1)
eg\(\;\;\;1 + p = 3,{\text{ }}1 \to 3\)
\(p = 2\) A1 N2
valid attempt to find vertical shift (M1)
eg\(\;\;\;9 + q = 1,{\text{ }}9 \to 1,{\text{ }} – 9 + q = 1\)
\(q = – 8\) A1 N2
Notes: An error in finding the reflection may still allow the correct values of \(p\) and \(q\) to be found, as the error may not affect subsequent working. In this case, award A0 for the reflection, M1A1 for \(p = 2\), and M1A1 for \(q = – 8\).
If no working shown, award N0 for \(q = 10\).
[5 marks]
valid approach (check FT from (a)) M1
eg\(\;\;\;f(x) = g(x),{\text{ }}{(x – 1)^2} – 9 = – {(x – 3)^2} + 1\)
correct expansion of both binomials (A1)
eg\(\;\;\;{x^2} – 2x + 1,{\text{ }}{x^2} – 6x + 9\)
correct working (A1)
eg\(\;\;\;{x^2} – 2x – 8 = – {x^2} + 6x – 8\)
correct equation (A1)
eg\(\;\;\;2{x^2} – 8x = 0,{\text{ }}2{x^2} = 8x\)
correct working (A1)
eg\(\;\;\;2x(x – 4) = 0\)
\(x = 0,{\text{ }}x = 4\) A1A1 N3
[7 marks]
Total [16 marks]
Question
Let \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), for \(0 < x < 6\).
The graph of \(f\) has a maximum point at P.
The \(y\)-coordinate of P is \(\ln 27\).
Find the \(x\)-coordinate of P.
Find \(f(x)\), expressing your answer as a single logarithm.
The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).
Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).
Answer/Explanation
Markscheme
recognizing \(f'(x) = 0\) (M1)
correct working (A1)
eg\(\,\,\,\,\,\)\(6 – 2x = 0\)
\(x = 3\) A1 N2
[3 marks]
evidence of integration (M1)
eg\(\,\,\,\,\,\)\(\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} } \)
using substitution (A1)
eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x – {x^2}\)
correct integral A1
eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x – {x^2})\)
substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\)) (M1)
eg\(\,\,\,\,\,\)\(\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27\)
correct working (A1)
eg\(\,\,\,\,\,\)\(c = \ln 27 – \ln 9\)
EITHER
\(c = \ln 3\) (A1)
attempt to substitute their value of \(c\) into \(f(x)\) (M1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 3\) A1 N4
OR
attempt to substitute their value of \(c\) into \(f(x)\) (M1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9\)
correct use of a log law (A1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9\)
\(f(x) = \ln \left( {3(6x – {x^2})} \right)\) A1 N4
[8 marks]
\(a = 3\) A1 N1
correct working A1
eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)
correct use of log law (A1)
eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)
\(b = 3\) A1 N2
[4 marks]
Question
The following diagram shows the graph of a function \(f\), for −4 ≤ x ≤ 2.
On the same axes, sketch the graph of \(f\left( { – x} \right)\).
Another function, \(g\), can be written in the form \(g\left( x \right) = a \times f\left( {x + b} \right)\). The following diagram shows the graph of \(g\).
Write down the value of a and of b.
Answer/Explanation
Markscheme
A2 N2
[2 marks]
recognizing horizontal shift/translation of 1 unit (M1)
eg b = 1, moved 1 right
recognizing vertical stretch/dilation with scale factor 2 (M1)
eg a = 2, y ×(−2)
a = −2, b = −1 A1A1 N2N2
[4 marks]
Question
Let \(f(x) = 3{(x + 1)^2} – 12\) .
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 .
Answer/Explanation
Markscheme
\(f(x) = 3({x^2} + 2x + 1) – 12\) 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]