Question
If \(f(x) = x – 3{x^{\frac{2}{3}}},{\text{ }}x > 0\) ,
(a) find the x-coordinate of the point P where \(f'(x) = 0\) ;
(b) determine whether P is a maximum or minimum point.
▶️Answer/Explanation
Markscheme
(a) \(f'(x) = 1 – \frac{2}{{{x^{\frac{1}{3}}}}}\) A1
\( \Rightarrow 1 – \frac{2}{{{x^{\frac{1}{3}}}}} = 0 \Rightarrow {x^{\frac{1}{3}}} = 2 \Rightarrow x = 8\) A1
(b) \(f”(x) = \frac{2}{{3{x^{\frac{4}{3}}}}}\) A1
\(f”(8) > 0 \Rightarrow {\text{ at }}x = 8,{\text{ }}f(x){\text{ has a minimum.}}\) M1A1
[5 marks]
Examiners report
Most candidates were able to correctly differentiate the function and find the point where \(f'(x) = 0\) . They were less successful in determining the nature of the point.
Question
If \(f(x) = x – 3{x^{\frac{2}{3}}},{\text{ }}x > 0\) ,
(a) find the x-coordinate of the point P where \(f'(x) = 0\) ;
(b) determine whether P is a maximum or minimum point.
▶️Answer/Explanation
Markscheme
(a) \(f'(x) = 1 – \frac{2}{{{x^{\frac{1}{3}}}}}\) A1
\( \Rightarrow 1 – \frac{2}{{{x^{\frac{1}{3}}}}} = 0 \Rightarrow {x^{\frac{1}{3}}} = 2 \Rightarrow x = 8\) A1
(b) \(f”(x) = \frac{2}{{3{x^{\frac{4}{3}}}}}\) A1
\(f”(8) > 0 \Rightarrow {\text{ at }}x = 8,{\text{ }}f(x){\text{ has a minimum.}}\) M1A1
[5 marks]
Examiners report
Most candidates were able to correctly differentiate the function and find the point where \(f'(x) = 0\) . They were less successful in determining the nature of the point.
Question
The function f is defined by \(f(x) = x{{\text{e}}^{2x}}\) .
It can be shown that \({f^{(n)}}(x) = ({2^n}x + n{2^{n – 1}}){{\text{e}}^{2x}}\) for all \(n \in {\mathbb{Z}^ + }\), where \({f^{(n)}}(x)\) represents the \({n^{{\text{th}}}}\) derivative of \(f(x)\) .
(a) By considering \({f^{(n)}}(x){\text{ for }}n = 1{\text{ and }}n = 2\) , show that there is one minimum point P on the graph of f , and find the coordinates of P.
(b) Show that f has a point of inflexion Q at x = −1.
(c) Determine the intervals on the domain of f where f is
(i) concave up;
(ii) concave down.
(d) Sketch f , clearly showing any intercepts, asymptotes and the points P and Q.
(e) Use mathematical induction to prove that \({f^{(n)}}(x) = ({2^n}x + n{2^{n – 1}}){{\text{e}}^{2x}}{\text{ for all }}n \in {\mathbb{Z}^ + },{\text{ where }}{f^{(n)}}{\text{ represents the }}{n^{{\text{th}}}}{\text{ derivative of }}f(x)\) .
▶️Answer/Explanation
Markscheme
(a) \(f'(x) = (1 + 2x){{\text{e}}^{2x}}\) A1
\(f'(x) = 0\) M1
\( \Rightarrow (1 + 2x){{\text{e}}^{2x}} = 0 \Rightarrow x = – \frac{1}{2}\) A1
\(f”(x) = ({2^2}x + 2 \times {2^{2 – 1}}){{\text{e}}^{2x}} = (4x + 4){{\text{e}}^{2x}}\) A1
\(f”\left( { – \frac{1}{2}} \right) = \frac{2}{{\text{e}}}\) A1
\(\frac{2}{{\text{e}}} > 0 \Rightarrow {\text{at }}x = – \frac{1}{2},{\text{ }}f(x){\text{ has a minimum.}}\) R1
\({\text{P}}\left( { – \frac{1}{2}, – \frac{1}{{2{\text{e}}}}} \right)\) A1
[7 marks]
(b) \(f”(x) = 0 \Rightarrow 4x + 4 = 0 \Rightarrow x = – 1\) M1A1
\({\text{Using the }}{{\text{2}}^{{\text{nd}}}}{\text{ derivative }}f”\left( { – \frac{1}{2}} \right) = \frac{2}{{\text{e}}}{\text{ and }}f”( – 2) = – \frac{4}{{{{\text{e}}^4}}},\) M1A1
the sign change indicates a point of inflexion. R1
[5 marks]
(c) (i) f(x) is concave up for \(x > – 1\) . A1
(ii) f(x) is concave down for \(x < – 1\) . A1
[2 marks]
(d)
A1A1A1A1
Note: Award A1 for P and Q, with Q above P,
A1 for asymptote at y = 0 ,
A1 for (0, 0) ,
A1 for shape.
[4 marks]
(e) Show true for n = 1 (M1)
\(f'(x) = {{\text{e}}^{2x}} + 2x{{\text{e}}^{2x}}\) A1
\( = {{\text{e}}^{2x}}(1 + 2x) = (2x + {2^0}){{\text{e}}^{2x}}\)
Assume true for \(n = k{\text{ , }}i.e.{\text{ }}{f^{(k)}}x = ({2^k}x + k \times {2^{k – 1}}){{\text{e}}^{2x}}{\text{, }}k \geqslant 1\) M1A1
Consider \(n = k + 1{\text{ , }}i.e.{\text{ an attempt to find }}\frac{{\text{d}}}{{{\text{d}}x}}\left( {{f^k}(x)} \right)\) . M1
\({F^{(k + 1)}}(x) = {2^k}{{\text{e}}^{2x}} + 2{{\text{e}}^{2x}}({2^k}x + k \times {2^{k – 1}})\) A1
\( = \left( {{2^k} + 2({2^k}x + k \times {2^{k – 1}})} \right){{\text{e}}^{2x}}\)
\( = (2 \times {2^k}x + {2^k} + k \times 2 \times {2^{k – 1}}){{\text{e}}^{2x}}\)
\( = ({2^{k + 1}}x + {2^k} + k \times {2^k}){{\text{e}}^{2x}}\) A1
\( = \left( {{2^{k + 1}}x + \left( {k + 1} \right){2^k}} \right){{\text{e}}^{2x}}\) A1
P(n) is true for \(k \Rightarrow P(n)\) is true for k + 1, and since true for n = 1, result proved by mathematical induction \(\forall n \in {\mathbb{Z}^ + }\)
Note: Only award R1 if a reasonable attempt is made to prove the \({(k + 1)^{{\text{th}}}}\) step.
[9 marks]
Total [27 marks]
Examiners report
This was the most accessible question in section B for these candidates. A majority of candidates produced partially correct answers to part (a), but a significant number struggled with demonstrating that the point is a minimum, despite the hint being given in the question. Part (b) started quite successfully but many students were unable to prove it is a point of inflexion or, more commonly, did not attempt to justify it. Correct answers were often seen for part (c). Part (d) was dependent on the successful completion of the first three parts. If candidates made errors in earlier parts, this often became obvious when they came to sketch the curve. However, few candidates realised that this part was a good way of checking that the above answers were at least consistent. The quality of curve sketching was rather weak overall, with candidates not marking points appropriately and not making features such as asymptotes clear. It is not possible to tell to what extent this was an effect of candidates not having a calculator, but it should be noted that asking students to sketch curves without a calculator will continue to appear on non-calculator papers. In part (e) the basic idea of proof by induction had clearly been taught with a significant majority of students understanding this. However, many candidates did not understand that they had to differentiate again to find the result for (k + 1) .
Question
Consider \(f(x) = \frac{{{x^2} – 5x + 4}}{{{x^2} + 5x + 4}}\).
(a) Find the equations of all asymptotes of the graph of f.
(b) Find the coordinates of the points where the graph of f meets the x and y axes.
(c) Find the coordinates of
(i) the maximum point and justify your answer;
(ii) the minimum point and justify your answer.
(d) Sketch the graph of f, clearly showing all the features found above.
(e) Hence, write down the number of points of inflexion of the graph of f.
▶️Answer/Explanation
Markscheme
(a) \({x^2} + 5x + 4 = 0 \Rightarrow x = – 1{\text{ or }}x = – 4\) (M1)
so vertical asymptotes are x = – 1 and x = – 4 A1
as \(x \to \infty \) then \(y \to 1\) so horizontal asymptote is y = 1 (M1)A1
[4 marks]
(b) \({x^2} – 5x + 4 = 0 \Rightarrow x = 1{\text{ or }}x = 4\) A1
\(x = 0 \Rightarrow y = 1\) A1
so intercepts are (1, 0), (4, 0) and (0,1)
[2 marks]
(c) (i) \(f'(x) = \frac{{({x^2} + 5x + 4)(2x – 5) – ({x^2} – 5x + 4)(2x + 5)}}{{{{({x^2} + 5x + 4)}^2}}}\) M1A1A1
\( = \frac{{10{x^2} – 40}}{{{{({x^2} + 5x + 4)}^2}}}\,\,\,\,\,\left( { = \frac{{10(x – 2)(x + 2)}}{{{{({x^2} + 5x + 4)}^2}}}} \right)\) A1
\(f'(x) = 0 \Rightarrow x = \pm 2\) M1
so the points under consideration are (–2, –9) and \(\left( {2, – \frac{1}{9}} \right)\) A1A1
looking at the sign either side of the points (or attempt to find \(f”(x)\)) M1
e.g. if \(x = – {2^ – }\) then \((x – 2)(x + 2) > 0\) and if \(x = – {2^ + }\) then \((x – 2)(x + 2) < 0\),
therefore (–2, –9) is a maximum A1
(ii) e.g. if \(x = {2^ – }\) then \((x – 2)(x + 2) < 0\) and if \(x = {2^ + }\) then \((x – 2)(x + 2) > 0\),
therefore \(\left( {2, – \frac{1}{9}} \right)\) is a minimum A1
Note: Candidates may find the minimum first.
[10 marks]
(d)
A3
Note: Award A1 for each branch consistent with and including the features found in previous parts.
[3 marks]
(e) one A1
[1 mark]
Total [20 marks]
Examiners report
This was the most successfully answered question in part B, in particular parts (a), (b) and (c). In part (a) the horizontal asymptote was often missing (or x = 4, x = 1 given). Part (b) was well done. Use of the quotient rule was well done in part (c) and many simplified correctly. There was knowledge of max/min and how to justify their answer, usually with a sign diagram but also with the second derivative. A common misconception was that, as \( – 9 < – \frac{1}{9}\), the minimum is at (–2, –9). In part (d) many candidates were unable to sketch the graph consistent with the main features that they had determined before. Very few candidates answered part (e) correctly.
Question
The function f is defined by \(f(x) = {{\text{e}}^{{x^2} – 2x – 1.5}}\).
(a) Find \(f'(x)\).
(b) You are given that \(y = \frac{{f(x)}}{{x – 1}}\) has a local minimum at x = a, a > 1. Find the
value of a.
▶️Answer/Explanation
Markscheme
(a) \(\left( {u = {x^2} – 2x – 1.5;\frac{{{\text{d}}u}}{{{\text{d}}x}} = 2x – 2} \right)\)
\(\frac{{{\text{d}}f}}{{{\text{d}}x}} = \frac{{{\text{d}}f}}{{{\text{d}}u}}\frac{{{\text{d}}u}}{{{\text{d}}x}} = {{\text{e}}^u}(2x – 2)\) (M1)
\( = 2(x – 1){{\text{e}}^{{x^2} – 2x – 1.5}}\) A1
(b) \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(x – 1) \times 2(x – 1){{\text{e}}^{{x^2} – 2x – 1.5}} – 1 \times {{\text{e}}^{{x^2} – 2x – 1.5}}}}{{{{(x – 1)}^2}}}\) M1A1
\( = \frac{{2{x^2} – 4x + 1}}{{{{(x – 1)}^2}}}{{\text{e}}^{{x^2} – 2x – 1.5}}\) (A1)
minimum occurs when \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) (M1)
\(x = 1 \pm \sqrt {\frac{1}{2}} \,\,\,\,\,\left( {{\text{accept }}x = \frac{{4 \pm \sqrt 8 }}{4}} \right)\) A1
\(a = 1 + \sqrt {\frac{1}{2}} \,\,\,\,\,\left( {{\text{accept }}a = \frac{{4 + \sqrt 8 }}{4}} \right)\) R1
[8 marks]
Examiners report
Part (a) was successfully answered by most candidates. Most candidates were able to make significant progress with part (b) but were then let down by being unable to simplify the expression or by not understanding the significance of being told that a > 1.
Question
Consider the function \(f(x) = \frac{{\ln x}}{x}\) , \(0 < x < {{\text{e}}^2}\) .
(i) Solve the equation \(f'(x) = 0\) .
(ii) Hence show the graph of \(f\) has a local maximum.
(iii) Write down the range of the function \(f\) .
Show that there is a point of inflexion on the graph and determine its coordinates.
Sketch the graph of \(y = f(x)\) , indicating clearly the asymptote, x-intercept and the local maximum.
Now consider the functions \(g(x) = \frac{{\ln \left| x \right|}}{x}\) and \(h(x) = \frac{{\ln \left| x \right|}}{{\left| x \right|}}\) , where \(0 < x < {{\text{e}}^2}\) .
(i) Sketch the graph of \(y = g(x)\) .
(ii) Write down the range of \(g\) .
(iii) Find the values of \(x\) such that \(h(x) > g(x)\) .
▶️Answer/Explanation
Markscheme
(i) \(f'(x) = \frac{{x\frac{1}{x} – \ln x}}{{{x^2}}}\) M1A1
\( = \frac{{1 – \ln x}}{{{x^2}}}\)
so \(f'(x) = 0\) when \(\ln x = 1\), i.e. \(x = {\text{e}}\) A1
(ii) \(f'(x) > 0\) when \(x < {\text{e}}\) and \(f'(x) < 0\) when \(x > {\text{e}}\) R1
hence local maximum AG
Note: Accept argument using correct second derivative.
(iii) \(y \leqslant \frac{1}{{\text{e}}}\) A1
[5 marks]
\(f”(x) = \frac{{{x^2}\frac{{ – 1}}{x} – \left( {1 – \ln x} \right)2x}}{{{x^4}}}\) M1
\( = \frac{{ – x – 2x + 2x\ln x}}{{{x^4}}}\)
\( = \frac{{ – 3 + 2\ln x}}{{{x^3}}}\) A1
Note: May be seen in part (a).
\(f”(x) = 0\) (M1)
\({ – 3 + 2\ln x = 0}\)
\(x = {{\text{e}}^{\frac{3}{2}}}\)
since \(f”(x) < 0\) when \(x < {{\text{e}}^{\frac{3}{2}}}\) and \(f”(x) > 0\) when \(x > {{\text{e}}^{\frac{3}{2}}}\) R1
then point of inflexion \(\left( {{{\text{e}}^{\frac{3}{2}}},\frac{3}{{2{{\text{e}}^{\frac{3}{2}}}}}} \right)\) A1
[5 marks]
A1A1A1
Note: Award A1 for the maximum and intercept, A1 for a vertical asymptote and A1 for shape (including turning concave up).
[3 marks]
(i)
A1A1
Note: Award A1 for each correct branch.
(ii) all real values A1
(iii)
(M1)(A1)
Note: Award (M1)(A1) for sketching the graph of h, ignoring any graph of g.
\( – {{\text{e}}^2} < x < – 1\) (accept \(x < – 1\) ) A1
[6 marks]
Examiners report
Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly.
Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly.
Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly.
Most candidates attempted parts (a), (b) and (c) and scored well, although many did not gain the reasoning marks for the justification of the existence of local maximum and inflexion point. The graph sketching was poorly done. A wide selection of range shapes were seen, in some cases showing little understanding of the relation between the derivatives of the function and its graph and difficulties with transformation of graphs. In some cases candidates sketched graphs consistent with their previous calculations but failed to label them properly.
Question
The quadratic function \(f(x) = p + qx – {x^2}\) has a maximum value of 5 when x = 3.
Find the value of p and the value of q .
The graph of f(x) is translated 3 units in the positive direction parallel to the x-axis. Determine the equation of the new graph.
▶️Answer/Explanation
Markscheme
METHOD 1
\(f'(x) = q – 2x = 0\) M1
\(f'(3) = q – 6 = 0\)
q = 6 A1
f(3) = p + 18 − 9 = 5 M1
p = −4 A1
METHOD 2
\(f(x) = – {(x – 3)^2} + 5\) M1A1
\( = – {x^2} + 6x – 4\)
q = 6, p = −4 A1A1
[4 marks]
\(g(x) = – 4 + 6(x – 3) – {(x – 3)^2}{\text{ }}( = – 31 + 12x – {x^2})\) M1A1
Note: Accept any alternative form which is correct.
Award M1A0 for a substitution of (x + 3) .
[2 marks]
Examiners report
In general candidates handled this question well although a number equated the derivative to the function value rather than zero. Most recognised the shift in the second part although a number shifted only the squared value and not both x values.
In general candidates handled this question well although a number equated the derivative to the function value rather than zero. Most recognised the shift in the second part although a number shifted only the squared value and not both x values.
Question
The curve C has equation \(y = \frac{1}{8}(9 + 8{x^2} – {x^4})\) .
Find the coordinates of the points on C at which \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) .
The tangent to C at the point P(1, 2) cuts the x-axis at the point T. Determine the coordinates of T.
The normal to C at the point P cuts the y-axis at the point N. Find the area of triangle PTN.
▶️Answer/Explanation
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x – \frac{1}{2}{x^3}\) A1
\(x\left( {2 – \frac{1}{2}{x^2}} \right) = 0\)
\(x = 0,{\text{ }} \pm 2\)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) at \(\left( {0,\frac{9}{8}} \right),{\text{ }}\left( { – 2,\frac{{25}}{8}} \right),{\text{ }}\left( {2,\frac{{25}}{8}} \right)\) A1A1A1
Note: Award A2 for all three x-values correct with errors/omissions in y-values.
[4 marks]
at x =1, gradient of tangent \( = \frac{3}{2}\) (A1)
Note: In the following, allow FT on incorrect gradient.
equation of tangent is \(y – 2 = \frac{3}{2}(x – 1)\,\,\,\,\,\left( {y = \frac{3}{2}x + \frac{1}{2}} \right)\) (A1)
meets x-axis when y = 0 , \( – 2 = \frac{3}{2}(x – 1)\) (M1)
\(x = – \frac{1}{3}\)
coordinates of T are \(\left( { – \frac{1}{3},0} \right)\) A1
[4 marks]
gradient of normal \( = – \frac{2}{3}\) (A1)
equation of normal is \(y – 2 = – \frac{2}{3}(x – 1)\,\,\,\,\,\left( {y = – \frac{2}{3}x + \frac{8}{3}} \right)\) (M1)
at x = 0 , \(y = \frac{8}{3}\) A1
Note: In the following, allow FT on incorrect coordinates of T and N.
lengths of \({\text{PN}} = \sqrt {\frac{{13}}{9}} \) , \({\text{PT}} = \sqrt {\frac{{52}}{9}} \) A1A1
area of triangle \({\text{PTN}} = \frac{1}{2} \times \sqrt {\frac{{13}}{9}} \times \sqrt {\frac{{52}}{9}} \) M1
\( = \frac{{13}}{9}\) (or equivalent e.g. \(\frac{{\sqrt {676} }}{{18}}\)) A1
[7 marks]
Examiners report
The whole of this question seemed to prove accessible to a high proportion of candidates.
(a) was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.
(b) was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).
The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded.
The whole of this question seemed to prove accessible to a high proportion of candidates.
(a) was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.
(b) was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).
The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded.
The whole of this question seemed to prove accessible to a high proportion of candidates.
(a) was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.
(b) was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).
The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded.
Question
The diagram shows the graph of the function defined by \(y = x{(\ln x)^2}{\text{ for }}x > 0\) .
The function has a local maximum at the point A and a local minimum at the point B.
Find the coordinates of the points A and B.
Given that the graph of the function has exactly one point of inflexion, find its coordinates.
▶️Answer/Explanation
Markscheme
\(f'(x) = {(\ln x)^2} + \frac{{2x\ln x}}{x}\left( { = {{(\ln x)}^2} + 2\ln x = \ln x(\ln x + 2)} \right)\) M1A1
\(f'(x) = 0{\text{ }}( \Rightarrow x = 1,{\text{ }}x = {e^{ – 2}})\) M1
Note: Award M1 for an attempt to solve \(f'(x) = 0\).
\(A({e^{ – 2}},\,4{e^{ – 2}})\) and B(1, 0) A1A1
Note: The final A1 is independent of prior working.
[5 marks]
\(f”(x) = \frac{2}{x}(\ln x + 1)\) A1
\(f”(x) = 0{\text{ }}\left( { \Rightarrow x = {e^{ – 1}}} \right)\) (M1)
inflexion point \(({e^{ – 1}},{\text{ }}{e^{ – 1}})\) A1
Note: M1 for attempt to solve \(f”(x) = 0\).
[3 marks]
Examiners report
This was answered very well. Candidates are very familiar with this type of question. Some lost a couple of marks by failing to find their final y coordinates, though only the weakest struggled with differentiation and so made little progress.
This was answered very well. Candidates are very familiar with this type of question. Some lost a couple of marks by failing to find their final y coordinates, though only the weakest struggled with differentiation and so made little progress.
Question
The function \(f\) is given by \(f(x) = x{{\text{e}}^{ – x}}{\text{ }}(x \geqslant 0)\).
(i) Find an expression for \(f'(x)\).
(ii) Hence determine the coordinates of the point A, where \(f'(x) = 0\).
Find an expression for \(f”(x)\) and hence show the point A is a maximum.
Find the coordinates of B, the point of inflexion.
The graph of the function \(g\) is obtained from the graph of \(f\) by stretching it in the x-direction by a scale factor 2.
(i) Write down an expression for \(g(x)\).
(ii) State the coordinates of the maximum C of \(g\).
(iii) Determine the x-coordinates of D and E, the two points where \(f(x) = g(x)\).
Sketch the graphs of \(y = f(x)\) and \(y = g(x)\) on the same axes, showing clearly the points A, B, C, D and E.
Find an exact value for the area of the region bounded by the curve \(y = g(x)\), the x-axis and the line \(x = 1\).
▶️Answer/Explanation
Markscheme
(i) \(f'(x) = {{\text{e}}^{ – x}} – x{{\text{e}}^{ – x}}\) M1A1
(ii) \(f'(x) = 0 \Rightarrow x = 1\)
coordinates \(\left( {1,{\text{ }}{{\text{e}}^{ – 1}}} \right)\) A1
[3 marks]
\(f”(x) = – {{\text{e}}^{ – x}} – {{\text{e}}^{ – x}} + x{{\text{e}}^{ – x}}{\text{ }}\left( { = – {{\text{e}}^{ – x}}(2 – x)} \right)\) A1
substituting \(x = 1\) into \(f”(x)\) M1
\(f”(1){\text{ }}\left( { = – {{\text{e}}^{ – 1}}} \right) < 0\) hence maximum R1AG
[3 marks]
\(f”(x) = 0{\text{ (}} \Rightarrow x = 2)\) M1
coordinates \(\left( {2,{\text{ 2}}{{\text{e}}^{ – 2}}} \right)\) A1
[2 marks]
(i) \(g(x) = \frac{x}{2}{{\text{e}}^{ – \frac{x}{2}}}\) A1
(ii) coordinates of maximum \(\left( {2,{\text{ }}{{\text{e}}^{ – 1}}} \right)\) A1
(iii) equating \(f(x) = g(x)\) and attempting to solve \(x{{\text{e}}^{ – x}} = \frac{x}{2}{{\text{e}}^{ – \frac{x}{2}}}\)
\( \Rightarrow x\left( {2{{\text{e}}^{\frac{x}{2}}} – {{\text{e}}^x}} \right) = 0\) (A1)
\( \Rightarrow x = 0\) A1
or \(2{{\text{e}}^{\frac{x}{2}}} = {{\text{e}}^x}\)
\( \Rightarrow {{\text{e}}^{\frac{x}{2}}} = 2\)
\( \Rightarrow x = 2\ln 2\) \((\ln 4)\) A1
Note: Award first (A1) only if factorisation seen or if two correct
solutions are seen.
Note: Award A1 for shape of \(f\), including domain extending beyond \(x = 2\).
Ignore any graph shown for \(x < 0\).
Award A1 for A and B correctly identified.
Award A1 for shape of \(g\), including domain extending beyond \(x = 2\).
Ignore any graph shown for \(x < 0\). Allow follow through from \(f\).
Award A1 for C, D and E correctly identified (D and E are interchangeable).
[4 marks]
\(A = \int_0^1 {\frac{x}{2}{{\text{e}}^{ – \frac{x}{2}}}{\text{d}}x} \) M1
\( = \left[ { – x{{\text{e}}^{ – \frac{x}{2}}}} \right]_0^1 – \int_0^1 { – {{\text{e}}^{ – \frac{x}{2}}}{\text{d}}x} \) A1
Note: Condone absence of limits or incorrect limits.
\( = – {{\text{e}}^{ – \frac{1}{2}}} – \left[ {2{{\text{e}}^{ – \frac{x}{2}}}} \right]_0^1\)
\( = – {{\text{e}}^{ – \frac{1}{2}}} – \left( {2{{\text{e}}^{ – \frac{1}{2}}} – 2} \right) = 2 – 3{{\text{e}}^{ – \frac{1}{2}}}\) A1
[3 marks]
Examiners report
Part a) proved to be an easy start for the vast majority of candidates.
Full marks for part b) were again likewise seen, though a small number shied away from considering the sign of their second derivative, despite the question asking them to do so.
Part c) again proved to be an easily earned 2 marks.
Full marks for part b) were again likewise seen, though a small number shied away from considering the sign of their second derivative, despite the question asking them to do so.
Part c) again proved to be an easily earned 2 marks.
Many candidates lost their way in part d). A variety of possibilities for \(g(x)\) were suggested, commonly \(2x{{\text{e}}^{ – 2x}}\), \(\frac{{x{{\text{e}}^{ – 1}}}}{2}\) or similar variations. Despite section ii) being worth only one mark, (and ‘state’ being present in the question), many laborious attempts at further differentiation were seen. Part diii was usually answered well by those who gave the correct function for \(g(x)\).
Part e) was also answered well by those who had earned full marks up to that point.
While the integration by parts technique was clearly understood, it was somewhat surprising how many careless slips were seen in this part of the question. Only a minority gained full marks for part f).
Question
Consider the function \(f(x) = \frac{{\ln x}}{x},{\text{ }}x > 0\).
The sketch below shows the graph of \(y = {\text{ }}f(x)\) and its tangent at a point A.
Show that \(f'(x) = \frac{{1 – \ln x}}{{{x^2}}}\).
Find the coordinates of B, at which the curve reaches its maximum value.
Find the coordinates of C, the point of inflexion on the curve.
The graph of \(y = {\text{ }}f(x)\) crosses the \(x\)-axis at the point A.
Find the equation of the tangent to the graph of \(f\) at the point A.
The graph of \(y = {\text{ }}f(x)\) crosses the \(x\)-axis at the point A.
Find the area enclosed by the curve \(y = f(x)\), the tangent at A, and the line \(x = {\text{e}}\).
▶️Answer/Explanation
Markscheme
\(f'(x) = \frac{{x \times \frac{1}{x} – \ln x}}{{{x^2}}}\) M1A1
\( = \frac{{1 – \ln x}}{{{x^2}}}\) AG
[2 marks]
\(\frac{{1 – \ln x}}{{{x^2}}} = 0\) has solution \(x = {\text{e}}\) M1A1
\(y = \frac{1}{{\text{e}}}\) A1
hence maximum at the point \(\left( {{\text{e, }}\frac{1}{{\text{e}}}} \right)\)
[3 marks]
\(f”(x) = \frac{{{x^2}\left( { – \frac{1}{x}} \right) – 2x(1 – \ln x)}}{{{x^4}}}\) M1A1
\( = \frac{{2\ln x – 3}}{{{x^3}}}\)
Note: The M1A1 should be awarded if the correct working appears in part (b).
point of inflexion where \(f”(x) = 0\) M1
so \(x = {{\text{e}}^{\frac{3}{2}}},{\text{ }}y = \frac{3}{2}{{\text{e}}^{\frac{{ – 3}}{2}}}\) A1A1
C has coordinates \(\left( {{{\text{e}}^{\frac{3}{2}}},{\text{ }}\frac{3}{2}{{\text{e}}^{\frac{{ – 3}}{2}}}} \right)\)
[5 marks]
\(f(1) = 0\) A1
\(f'(1) = 1\) (A1)
\(y = x + c\) (M1)
through (1, 0)
equation is \(y = x – 1\) A1
[4 marks]
METHOD 1
area \( = \int_1^{\text{e}} {x – 1 – \frac{{\ln x}}{x}{\text{d}}x} \) M1A1A1
Note: Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.
\(\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}} ( + c)\) (M1)A1
\(\int {(x – 1){\text{d}}x = \frac{{{x^2}}}{2} – x( + c)} \) A1
\( = \left[ {\frac{1}{2}{x^2} – x – \frac{1}{2}{{(\ln x)}^2}} \right]_1^{\text{e}}\)
\( = \left( {\frac{1}{2}{{\text{e}}^2} – {\text{e}} – \frac{1}{2}} \right) – \left( {\frac{1}{2} – 1} \right)\)
\( = \frac{1}{2}{{\text{e}}^2} – {\text{e}}\) A1
METHOD 2
area = area of triangle \( – \int_1^e {\frac{{\ln x}}{x}{\text{d}}x} \) M1A1
Note: A1 is for correct integral with limits and is dependent on the M1.
\(\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}( + c)} \) (M1)A1
area of triangle \( = \frac{1}{2}(e – 1)(e – 1)\) M1A1
\(\frac{1}{2}(e – 1)(e – 1) – \left( {\frac{1}{2}} \right) = \frac{1}{2}{{\text{e}}^2} – {\text{e}}\) A1
[7 marks]
Examiners report
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Question
Consider the function \(f(x) = \frac{{\ln x}}{x},{\text{ }}x > 0\).
The sketch below shows the graph of \(y = {\text{ }}f(x)\) and its tangent at a point A.
Show that \(f'(x) = \frac{{1 – \ln x}}{{{x^2}}}\).
Find the coordinates of B, at which the curve reaches its maximum value.
Find the coordinates of C, the point of inflexion on the curve.
The graph of \(y = {\text{ }}f(x)\) crosses the \(x\)-axis at the point A.
Find the equation of the tangent to the graph of \(f\) at the point A.
The graph of \(y = {\text{ }}f(x)\) crosses the \(x\)-axis at the point A.
Find the area enclosed by the curve \(y = f(x)\), the tangent at A, and the line \(x = {\text{e}}\).
▶️Answer/Explanation
Markscheme
\(f'(x) = \frac{{x \times \frac{1}{x} – \ln x}}{{{x^2}}}\) M1A1
\( = \frac{{1 – \ln x}}{{{x^2}}}\) AG
[2 marks]
\(\frac{{1 – \ln x}}{{{x^2}}} = 0\) has solution \(x = {\text{e}}\) M1A1
\(y = \frac{1}{{\text{e}}}\) A1
hence maximum at the point \(\left( {{\text{e, }}\frac{1}{{\text{e}}}} \right)\)
[3 marks]
\(f”(x) = \frac{{{x^2}\left( { – \frac{1}{x}} \right) – 2x(1 – \ln x)}}{{{x^4}}}\) M1A1
\( = \frac{{2\ln x – 3}}{{{x^3}}}\)
Note: The M1A1 should be awarded if the correct working appears in part (b).
point of inflexion where \(f”(x) = 0\) M1
so \(x = {{\text{e}}^{\frac{3}{2}}},{\text{ }}y = \frac{3}{2}{{\text{e}}^{\frac{{ – 3}}{2}}}\) A1A1
C has coordinates \(\left( {{{\text{e}}^{\frac{3}{2}}},{\text{ }}\frac{3}{2}{{\text{e}}^{\frac{{ – 3}}{2}}}} \right)\)
[5 marks]
\(f(1) = 0\) A1
\(f'(1) = 1\) (A1)
\(y = x + c\) (M1)
through (1, 0)
equation is \(y = x – 1\) A1
[4 marks]
METHOD 1
area \( = \int_1^{\text{e}} {x – 1 – \frac{{\ln x}}{x}{\text{d}}x} \) M1A1A1
Note: Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.
\(\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}} ( + c)\) (M1)A1
\(\int {(x – 1){\text{d}}x = \frac{{{x^2}}}{2} – x( + c)} \) A1
\( = \left[ {\frac{1}{2}{x^2} – x – \frac{1}{2}{{(\ln x)}^2}} \right]_1^{\text{e}}\)
\( = \left( {\frac{1}{2}{{\text{e}}^2} – {\text{e}} – \frac{1}{2}} \right) – \left( {\frac{1}{2} – 1} \right)\)
\( = \frac{1}{2}{{\text{e}}^2} – {\text{e}}\) A1
METHOD 2
area = area of triangle \( – \int_1^e {\frac{{\ln x}}{x}{\text{d}}x} \) M1A1
Note: A1 is for correct integral with limits and is dependent on the M1.
\(\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}( + c)} \) (M1)A1
area of triangle \( = \frac{1}{2}(e – 1)(e – 1)\) M1A1
\(\frac{1}{2}(e – 1)(e – 1) – \left( {\frac{1}{2}} \right) = \frac{1}{2}{{\text{e}}^2} – {\text{e}}\) A1
[7 marks]
Examiners report
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Question
The graph of the function \(f(x) = \frac{{x + 1}}{{{x^2} + 1}}\) is shown below.
The point (1, 1) is a point of inflexion. There are two other points of inflexion.
Find \(f'(x)\).
Hence find the \(x\)-coordinates of the points where the gradient of the graph of \(f\) is zero.
Find \(f”(x)\) expressing your answer in the form \(\frac{{p(x)}}{{{{({x^2} + 1)}^3}}}\), where \(p(x)\) is a polynomial of degree 3.
Find the \(x\)-coordinates of the other two points of inflexion.
Find the area of the shaded region. Express your answer in the form \(\frac{\pi }{a} – \ln \sqrt b \), where \(a\) and \(b\) are integers.
▶️Answer/Explanation
Markscheme
(a) \(f'(x) = \frac{{\left( {{x^2} + 1} \right) – 2x(x + 1)}}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{ }}\left( { = \frac{{ – {x^2} – 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)\) M1A1
[2 marks]
\(\frac{{ – {x^2} – 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0\)
\(x = – 1 \pm \sqrt 2 \) A1
[1 mark]
\(f”(x) = \frac{{( – 2x – 2){{\left( {{x^2} + 1} \right)}^2} – 2(2x)\left( {{x^2} + 1} \right)\left( { – {x^2} – 2x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}}\) A1A1
Note: Award A1 for \(( – 2x – 2){\left( {{x^2} + 1} \right)^2}\) or equivalent.
Note: Award A1 for \( – 2(2x)\left( {{x^2} + 1} \right)\left( { – {x^2} – 2x + 1} \right)\) or equivalent.
\( = \frac{{( – 2x – 2)\left( {{x^2} + 1} \right) – 4x\left( { – {x^2} – 2x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}\)
\( = \frac{{2{x^3} + 6{x^2} – 6x – 2}}{{{{\left( {{x^2} + 1} \right)}^3}}}\) A1
\(\left( { = \frac{{2\left( {{x^3} + 3{x^2} – 3x – 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}} \right)\)
[3 marks]
recognition that \((x – 1)\) is a factor (R1)
\((x – 1)\left( {{x^2} + bx + c} \right) = \left( {{x^3} + 3{x^2} – 3x – 1} \right)\) M1
\( \Rightarrow {x^2} + 4x + 1 = 0\) A1
\(x = – 2 \pm \sqrt 3 \) A1
Note: Allow long division / synthetic division.
[4 marks]
\(\int_{ – 1}^0 {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x} \) M1
\(\int {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x = \int {\frac{x}{{{x^2} + 1}}{\text{d}}x + \int {\frac{1}{{{x^2} + 1}}{\text{d}}x} } } \) M1
\( = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)\) A1A1
\( = \left[ {\frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)} \right]_{ – 1}^0 = \frac{1}{2}\ln 1 + \arctan 0 – \frac{1}{2}\ln 2 – \arctan ( – 1)\) M1
\( = \frac{\pi }{4} – \ln \sqrt 2 \) A1
[6 marks]
Examiners report
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Question
A tranquilizer is injected into a muscle from which it enters the bloodstream.
The concentration \(C\) in \({\text{mg}}{{\text{l}}^{ – 1}}\), of tranquilizer in the bloodstream can be modelled by the function \(C(t) = \frac{{2t}}{{3 + {t^2}}},{\text{ }}t \ge 0\) where \(t\) is the number of minutes after the injection.
Find the maximum concentration of tranquilizer in the bloodstream.
▶️Answer/Explanation
Markscheme
use of the quotient rule or the product rule M1
\(C'(t) = \frac{{(3 + {t^2}) \times 2 – 2t \times 2t}}{{{{\left( {3 + {t^2}} \right)}^2}}}\;\;\;\left( { = \frac{{6 – 2{t^2}}}{{{{\left( {3 + {t^2}} \right)}^2}}}} \right)\;\;\;{\text{or}}\;\;\;\frac{2}{{3 + {t^2}}} – \frac{{4{t^2}}}{{{{\left( {3 + {t^2}} \right)}^2}}}\) A1A1
Note: Award A1 for a correct numerator and A1 for a correct denominator in the quotient rule, and A1 for each correct term in the product rule.
attempting to solve \(C'(t) = 0\;\;\;{\text{for }}t\) (M1)
\(t = \pm \sqrt 3 \;\;\;{\text{(minutes)}}\) A1
\(C\left( {\sqrt 3 } \right) = \frac{{\sqrt 3 }}{3}\;\;\;\left( {{\text{mg}}{{\text{l}}^{ – 1}}} \right)\;\;\;{\text{or equivalent.}}\) A1
[6 marks]
Examiners report
This question was generally well done. A significant number of candidates did not calculate the maximum value of \(C\).
Question
Let \(y(x) = x{e^{3x}},{\text{ }}x \in \mathbb{R}\).
Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).
Prove by induction that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = n{3^{n – 1}}{{\text{e}}^{3x}} + x{3^n}{{\text{e}}^{3x}}\) for \(n \in {\mathbb{Z}^ + }\).
Find the coordinates of any local maximum and minimum points on the graph of \(y(x)\).
Justify whether any such point is a maximum or a minimum.
Find the coordinates of any points of inflexion on the graph of \(y(x)\). Justify whether any such point is a point of inflexion.
Hence sketch the graph of \(y(x)\), indicating clearly the points found in parts (c) and (d) and any intercepts with the axes.
▶️Answer/Explanation
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = ({{\text{e}}^{3x}} + 3x{{\text{e}}^{3x}})\) M1A1
[2 marks]
let \(P(n)\) be the statement \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = n{3^{n – 1}}{{\text{e}}^{3x}} + x{3^n}{{\text{e}}^{3x}}\)
prove for \(n = 1\) M1
\(LHS\) of \(P(1)\) is \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) which is \(1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}}\) and \(RHS\) is \({3^0}{{\text{e}}^{3x}} + x{3^1}{{\text{e}}^{3x}}\) R1
as \({\text{LHS}} = {\text{RHS, }}P(1)\) is true
assume \(P(k)\) is true and attempt to prove \(P(k + 1)\) is true M1
assuming \(\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = k{3^{k – 1}}{{\text{e}}^{3x}} + x{3^k}{{\text{e}}^{3x}}\)
\(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}}} \right)\) (M1)
\( = k{3^{k – 1}} \times 3{{\text{e}}^{3x}} + 1 \times {3^k}{{\text{e}}^{3x}} + x{3^k} \times 3{{\text{e}}^{3x}}\) A1
\( = (k + 1){3^k}{{\text{e}}^{3x}} + x{3^{k + 1}}{{\text{e}}^{3x}}\;\;\;\)(as required) A1
Note: Can award the A marks independent of the M marks
since \(P(1)\) is true and \(P(k)\) is true \( \Rightarrow P(k + 1)\) is true
then (by \(PMI\)), \(P(n)\) is true \((\forall n \in {\mathbb{Z}^ + })\) R1
Note: To gain last R1 at least four of the above marks must have been gained.
[7 marks]
\({{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = 0 \Rightarrow 1 + 3x = 0 \Rightarrow x = – \frac{1}{3}\) M1A1
point is \(\left( { – \frac{1}{3},{\text{ }} – \frac{1}{{3{\text{e}}}}} \right)\) A1
EITHER
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\)
when \(x = – \frac{1}{3},{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} > 0\) therefore the point is a minimum M1A1
OR
nature table shows point is a minimum M1A1
[5 marks]
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\) A1
\(2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}} = 0 \Rightarrow 2 + 3x = 0 \Rightarrow x = – \frac{2}{3}\) M1A1
point is \(\left( { – \frac{2}{3},{\text{ }} – \frac{2}{{3{{\text{e}}^2}}}} \right)\) A1
since the curvature does change (concave down to concave up) it is a point of inflection R1
Note: Allow \({3^{{\text{rd}}}}\) derivative is not zero at \( – \frac{2}{3}\)
[5 marks]
(general shape including asymptote and through origin) A1
showing minimum and point of inflection A1
Note: Only indication of position of answers to (c) and (d) required, not coordinates.
[2 marks]
Total [21 marks]
Examiners report
Well done.
The logic of an induction proof was not known well enough. Many candidates used what they had to prove rather than differentiating what they had assumed. They did not have enough experience in doing Induction proofs.
Good, some forgot to test for min/max, some forgot to give the \(y\) value.
Again quite good, some forgot to check for change in curvature and some forgot the \(y\) value.
Some accurate sketches, some had all the information from earlier parts but could not apply it. The asymptote was often missed.
Question
In triangle \({\text{ABC, BC}} = \sqrt 3 {\text{ cm}}\), \({\rm{A\hat BC}} = \theta \) and \({\rm{B\hat CA}} = \frac{\pi }{3}\).
Show that length \({\text{AB}} = \frac{3}{{\sqrt 3 \cos \theta + \sin \theta }}\).
Given that \(AB\) has a minimum value, determine the value of \(\theta \) for which this occurs.
▶️Answer/Explanation
Markscheme
any attempt to use sine rule M1
\(\frac{{{\text{AB}}}}{{\sin \frac{\pi }{3}}} = \frac{{\sqrt 3 }}{{\sin \left( {\frac{{2\pi }}{3} – \theta } \right)}}\) A1
\( = \frac{{\sqrt 3 }}{{\sin \frac{{2\pi }}{3}\cos \theta – \cos \frac{{2\pi }}{3}\sin \theta }}\) A1
Note: Condone use of degrees.
\( = \frac{{\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}\cos \theta + \frac{1}{2}\sin \theta }}\) A1
\(\frac{{{\text{AB}}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{\sqrt 3 }}{{\frac{{\sqrt 3 }}{2}\cos \theta + \frac{1}{2}\sin \theta }}\)
\(\therefore {\text{AB}} = \frac{3}{{\sqrt 3 \cos \theta + \sin \theta }}\) AG
[4 marks]
METHOD 1
\(({\text{AB}})’ = \frac{{ – 3\left( { – \sqrt 3 \sin \theta + \cos \theta } \right)}}{{{{\left( {\sqrt 3 \cos \theta + \sin \theta } \right)}^2}}}\) M1A1
setting \(({\text{AB}})’ = 0\) M1
\(\tan \theta = \frac{1}{{\sqrt 3 }}\)
\(\theta = \frac{\pi }{6}\) A1
METHOD 2
\({\text{AB}} = \frac{{\sqrt 3 \sin \frac{\pi }{3}}}{{\sin \left( {\frac{{2\pi }}{3} – \theta } \right)}}\)
\(AB\) minimum when \(\sin \left( {\frac{{2\pi }}{3} – \theta } \right)\) is maximum M1
\(\sin \left( {\frac{{2\pi }}{3} – \theta } \right) = 1\) (A1)
\(\frac{{2\pi }}{3} – \theta = \frac{\pi }{2}\) M1
\(\theta = \frac{\pi }{6}\) A1
METHOD 3
shortest distance from \(B\) to \(AC\) is perpendicular to \(AC\) R1
\(\theta = \frac{\pi }{2} – \frac{\pi }{3} = \frac{\pi }{6}\) M1A2
[4 marks]
Total [8 marks]
Examiners report
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Question
Consider the function defined by \(f(x) = x\sqrt {1 – {x^2}} \) on the domain \( – 1 \le x \le 1\).
Show that \(f\) is an odd function.
Find \(f'(x)\).
Hence find the \(x\)-coordinates of any local maximum or minimum points.
Find the range of \(f\).
Sketch the graph of \(y = f(x)\) indicating clearly the coordinates of the \(x\)-intercepts and any local maximum or minimum points.
Find the area of the region enclosed by the graph of \(y = f(x)\) and the \(x\)-axis for \(x \ge 0\).
Show that \(\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > \left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right|} \).
▶️Answer/Explanation
Markscheme
\(f( – x) = ( – x)\sqrt {1 – {{( – x)}^2}} \) M1
\( = – x\sqrt {1 – {x^2}} \)
\( = – f(x)\) R1
hence \(f\) is odd AG
[2 marks]
\(f'(x) = x \bullet \frac{1}{2}{(1 – {x^2})^{ – \frac{1}{2}}} \bullet – 2x + {(1 – {x^2})^{\frac{1}{2}}}\) M1A1A1
[3 marks]
\(f'(x) = \sqrt {1 – {x^2}} – \frac{{{x^2}}}{{\sqrt {1 – {x^2}} }}\;\;\;\left( { = \frac{{1 – 2{x^2}}}{{\sqrt {1 – {x^2}} }}} \right)\) A1
Note: This may be seen in part (b).
Note: Do not allow FT from part (b).
\(f'(x) = 0 \Rightarrow 1 – 2{x^2} = 0\) M1
\(x = \pm \frac{1}{{\sqrt 2 }}\) A1
[3 marks]
\(y\)-coordinates of the Max Min Points are \(y = \pm \frac{1}{2}\) M1A1
so range of \(f(x)\) is \(\left[ { – \frac{1}{2},{\text{ }}\frac{1}{2}} \right]\) A1
Note: Allow FT from (c) if values of \(x\), within the domain, are used.
[3 marks]
Shape: The graph of an odd function, on the given domain, s-shaped,
where the max(min) is the right(left) of \(0.5{\text{ }}( – 0.5)\) A1
\(x\)-intercepts A1
turning points A1
[3 marks]
\({\text{area}} = \int_0^1 {x\sqrt {1 – {x^2}} {\text{d}}x} \) (M1)
attempt at “backwards chain rule” or substitution M1
\( = – \frac{1}{2}\int_0^1 {( – 2x)\sqrt {1 – {x^2}} {\text{d}}x} \)
Note: Condone absence of limits for first two marks.
\( = \left[ {\frac{2}{3}{{(1 – {x^2})}^{\frac{3}{2}}} \bullet – \frac{1}{2}} \right]_0^1\) A1
\( = \left[ { – \frac{1}{3}{{(1 – {x^2})}^{\frac{3}{2}}}} \right]_0^1\)
\( = 0 – \left( { – \frac{1}{3}} \right) = \frac{1}{3}\) A1
[4 marks]
\(\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > 0} \) R1
\(\left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right| = 0\) R1
so \(\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > \left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right|} \) AG
[2 marks]
Total [20 marks]
Examiners report
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Question
Let \(y = {{\text{e}}^x}\sin x\).
Consider the function \(f\) defined by \(f(x) = {{\text{e}}^x}\sin x,{\text{ }}0 \leqslant x \leqslant \pi \).
The curvature at any point \((x,{\text{ }}y)\) on a graph is defined as \(\kappa = \frac{{\left| {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right|}}{{{{\left( {1 + {{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}\).
Find an expression for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).
Show that \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^x}\cos x\).
Show that the function \(f\) has a local maximum value when \(x = \frac{{3\pi }}{4}\).
Find the \(x\)-coordinate of the point of inflexion of the graph of \(f\).
Sketch the graph of \(f\), clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.
Find the area of the region enclosed by the graph of \(f\) and the \(x\)-axis.
The curvature at any point \((x,{\text{ }}y)\) on a graph is defined as \(\kappa = \frac{{\left| {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right|}}{{{{\left( {1 + {{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}\).
Find the value of the curvature of the graph of \(f\) at the local maximum point.
Find the value \(\kappa \) for \(x = \frac{\pi }{2}\) and comment on its meaning with respect to the shape of the graph.
▶️Answer/Explanation
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x{\text{ }}\left( { = {{\text{e}}^x}(\sin x + \cos x)} \right)\) M1A1
[2 marks]
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = {{\text{e}}^x}(\sin x + \cos x) + {{\text{e}}^x}(\cos x – \sin x)\) M1A1
\( = 2{{\text{e}}^x}\cos x\) AG
[2 marks]
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^{\frac{{3\pi }}{4}}}\left( {\sin \frac{{3\pi }}{4} + \cos \frac{{3\pi }}{4}} \right) = 0\) R1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} < 0\) R1
hence maximum at \(x = \frac{{3\pi }}{4}\) AG
[2 marks]
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0 \Rightarrow 2{{\text{e}}^x}\cos x = 0\) M1
\( \Rightarrow x = \frac{\pi }{2}\) A1
Note: Award M1A0 if extra zeros are seen.
[2 marks]
correct shape and correct domain A1
max at \(x = \frac{{3\pi }}{4}\), point of inflexion at \(x = \frac{\pi }{2}\) A1
zeros at \(x = 0\) and \(x = \pi \) A1
Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.
[3 marks]
EITHER
\(\int_0^x {{{\text{e}}^x}\sin x{\text{d}}x = [{{\text{e}}^x}\sin x]_0^\pi – \int_0^\pi {{{\text{e}}^x}\cos x{\text{d}}x} } \) M1A1
\(\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = [{{\text{e}}^x}\sin x]_0^\pi – \left( {[{{\text{e}}^x}\cos x]_0^x + \int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x} } \right)} \) A1
OR
\(\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = [ – {{\text{e}}^x}\cos x]_0^\pi + \int_0^\pi {{{\text{e}}^x}\cos x{\text{d}}x} } \) M1A1
\(\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = [ – {{\text{e}}^x}\cos x]} _0^\pi + \left( {[{{\text{e}}^x}\sin x]_0^\pi – \int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x} } \right)\) A1
THEN
\(\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}\left( {[{{\text{e}}^x}\sin x]_0^x – [{{\text{e}}^x}\cos x]_0^x} \right)} \) M1A1
\(\int_0^\pi {{{\text{e}}^x}\sin x{\text{d}}x = \frac{1}{2}({{\text{e}}^x} + 1)} \) A1
[6 marks]
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) (A1)
\(\frac{{{d^2}y}}{{d{x^2}}} = 2{e^{\frac{{3\pi }}{4}}}\cos \frac{{3\pi }}{4} = – \sqrt 2 {e^{\frac{{3\pi }}{4}}}\) (A1)
\(\kappa = \frac{{\left| { – \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}} \right|}}{1} = \sqrt 2 {{\text{e}}^{\frac{{3\pi }}{4}}}\) A1
[3 marks]
\(\kappa = 0\) A1
the graph is approximated by a straight line R1
[2 marks]
Examiners report
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Question
Let \(f\left( x \right) = \frac{{2 – 3{x^5}}}{{2{x^3}}},\,\,x \in \mathbb{R},\,\,x \ne 0\).
The graph of \(y = f\left( x \right)\) has a local maximum at A. Find the coordinates of A.
Show that there is exactly one point of inflexion, B, on the graph of \(y = f\left( x \right)\).
The coordinates of B can be expressed in the form B\(\left( {{2^a},\,b \times {2^{ – 3a}}} \right)\) where a, b\( \in \mathbb{Q}\). Find the value of a and the value of b.
Sketch the graph of \(y = f\left( x \right)\) showing clearly the position of the points A and B.
▶️Answer/Explanation
Markscheme
attempt to differentiate (M1)
\(f’\left( x \right) = – 3{x^{ – 4}} – 3x\) A1
Note: Award M1 for using quotient or product rule award A1 if correct derivative seen even in unsimplified form, for example \(f’\left( x \right) = \frac{{ – 15{x^4} \times 2{x^3} – 6{x^2}\left( {2 – 3{x^5}} \right)}}{{{{\left( {2{x^3}} \right)}^2}}}\).
\( – \frac{3}{{{x^4}}} – 3x = 0\) M1
\( \Rightarrow {x^5} = – 1 \Rightarrow x = – 1\) A1
\({\text{A}}\left( { – 1,\, – \frac{5}{2}} \right)\) A1
[5 marks]
\(f”\left( x \right) = 0\) M1
\(f”\left( x \right) = 12{x^{ – 5}} – 3\left( { = 0} \right)\) A1
Note: Award A1 for correct derivative seen even if not simplified.
\( \Rightarrow x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)\) A1
hence (at most) one point of inflexion R1
Note: This mark is independent of the two A1 marks above. If they have shown or stated their equation has only one solution this mark can be awarded.
\(f”\left( x \right)\) changes sign at \(x = \sqrt[5]{4}\left( { = {2^{\frac{2}{5}}}} \right)\) R1
so exactly one point of inflexion
[5 marks]
\(x = \sqrt[5]{4} = {2^{\frac{2}{5}}}\left( { \Rightarrow a = \frac{2}{5}} \right)\) A1
\(f\left( {{2^{\frac{2}{5}}}} \right) = \frac{{2 – 3 \times {2^2}}}{{2 \times {2^{\frac{6}{5}}}}} = – 5 \times {2^{ – \frac{6}{5}}}\left( { \Rightarrow b = – 5} \right)\) (M1)A1
Note: Award M1 for the substitution of their value for \(x\) into \(f\left( x \right)\).
[3 marks]
A1A1A1A1
A1 for shape for x < 0
A1 for shape for x > 0
A1 for maximum at A
A1 for POI at B.
Note: Only award last two A1s if A and B are placed in the correct quadrants, allowing for follow through.
[4 marks]
Examiners report
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Question
Consider the curve \(y = \frac{1}{{1 – x}} + \frac{4}{{x – 4}}\).
Find the x-coordinates of the points on the curve where the gradient is zero.
▶️Answer/Explanation
Markscheme
valid attempt to find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{{{\left( {1 – x} \right)}^2}}} – \frac{4}{{{{\left( {x – 4} \right)}^2}}}\) A1A1
attempt to solve \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1
\(x = 2,\,\,x = – 2\) A1A1
[6 marks]
Examiners report
Question
Consider the function
\(f(x)=\frac{e^{x}-1}{e^{x}+1}\)
(a) Find \({f}'(x)\) and deduce the monotony of the function.
(b) Explain why \(f^{-1}(x)\) exists and find its expression.
▶️Answer/Explanation
Ans
(a) \({f}'(x)=\frac{2e^{x}}{(e^{2}+1)^{2}}> 0\) , it is increasing.
(b) since the function is increasing it is 1-1 (horizontal line test). \(f^{-1}(x)=\textup{In}\left ( \frac{1+y}{1-y} \right )\)
Question
For each of the functions
\(f(x)=e^{x}+e^{-x}\) and \(g(x)=e^{x}-e^{-x}\)
(a) Find the stationary points (if any) and determine their nature.
(b) Find the points of inflection (if any); justify your answer.
▶️Answer/Explanation
Ans
For \(f(x)=e^{x}+e^{-x}\) , min at (0,2), no points of inflection
For \(f(x)=e^{x}-e^{-x}\) , no stationary points, point of inflection at (0,2)
Question
Consider \(f(x)=(x-1)^{2}(x-4)^{3}\)
(a) Find the stationary points and determine their nature.
(b) Find the points of inflection; justify your answer.
▶️Answer/Explanation
Ans
(a) Max at \(x=1\), min at \(x=11/5\), (\(x=4\) stationary)
(b) Points of inflection at \(x=1.47\), \(x=2.93\), \(x=4\)
Question
A cubic function has a maximum at A(0,5) and a point of inflexion at B(1,1). Find
(a) an expression of the cubic function.
(b) the coordinates of the minimum point; justify that it is a minimum.
▶️Answer/Explanation
Ans
(a) \(f(x)=2x^{3}-6x^{2}+5\) (b) min at \((2,-3)\)
Question
If \(f(x)=x-3x^{\frac{2}{3}},x> 0,\)
(a) find the x-coordinate of the point P where \({f}'(x)=0;\)
(b) determine whether P is a maximum or minimum point. (Total 5 marks)
▶️Answer/Explanation
Ans
(a) \({f}'(x)=1-\frac{2}{x^{\frac{1}{3}}}\) A1
\(\Rightarrow 1-\frac{2}{x^{\frac{1}{3}}}=0\Rightarrow x^{\frac{1}{3}}=2\Rightarrow x=8\) A1
(b) \({f}”(x)=\frac{2}{3x^{\frac{4}{3}}}\) A1
\({f}”(8)> 0\Rightarrow\) at \(x=8,f(x)\) has a minimum. M1A1 [5]
Question
Find the x-coordinate of the point of inflexion on the graph of \(y=xe^{x},-3\leq x\leq 1\). (Total 6 marks)
▶️Answer/Explanation
Ans
METHOD 1
\(y=xe^{x}\; \; \; \frac{dy}{dx}=xe^{x}+e^{x}\) (M1)(A1)
\(\frac{d^{2}y}{dx^{2}}=xe^{x}+2e^{x}=e^{x}(x+2)\) (M1)(A1)(A1)
Therefore the x-coordinate of the point of inflexion is \(x=-2\) (A1) (C6)
METHOD 2
Sketching \(y={f}'(x)\)
(G4)
\({f}'(x)\) has a minimum when \(x=-2\) (G1)
Thus, \(f(x)\) has point of inflexion when \(x=-2\) (G1) [6]
Question
Let \(f:x \mapsto e^{\textup{sin}\, x}\). There is a point of inflexion on the graph of \(f\), for \(0< x< 1\).
(a) Find \({f}'(x)\).
(b) Write down, but do not solve, an equation in terms of x, that would allow you to find the
value of x at this point of inflexion. (Total 3 marks)
▶️Answer/Explanation
Ans
(a) Given \(f(x)=e^{\textup{sin}\, x}\) Then \(f{}'(x)=\textup{cos}\, x\times e^{\textup{sin}x}\) (A1) (C1)
(b) \({f}”(x)=\textup{cos}^{2}x\times e^{\textup{sin}x}-\textup{sin}\, x\times e^{\textup{sin}x}\)
\(=e^{\textup{sin}x}(cos^{2}\, x-sin\, x)\) (M1)
For the point of inflexion, put \({f}”(x)=0\)
\(\Rightarrow e^{\textup{sin}\, x}(\textup{cos}^{2}\, x-\textup{sin}\, x)=0\) (or equivalent) (A1) (C2)
Note: Award (C1) if the candidate only writes \({f}”(x)=0\) [3]
Question
Find the x-coordinate, between –2 and 0, of the point of inflexion on the graph of the function \(f:x \mapsto x^2 e^{x}\). Give your answer to 3 decimal places. (Total 3 marks)
▶️Answer/Explanation
Ans
If \(f:(x) \mapsto x^2 e^{x}\) then \({f}'(x)=x^2 e^{x}+2xe^{x}\)
\({f}”(x)=x^{2}e^{x}+4xe^{x}+2e^{x}=e^{x}(x^{2}+4x+2)\) (A1)
For a point of inflexion solve \({f}”(x)=0\)
\({f}”(x)=0\) at \(x=-0.586\) (using a GDC or the quadratic formula) (A1)
(Since \({f}'(x)\neq 0\) at this value, then it is a point of inflexion.)
Note: Some candidates may find the value of x from f′(x) by finding the minimum turning point using a graphic display calculator [3]
Question
Consider the curve with equation \(f(x)=e^{-2x^{2}}\) for \(x< 0\).
Find the cordinates of the point of inflexion and justify that it is a point of inflexion. (Total 7 marks)
▶️Answer/Explanation
Ans
EITHER OR
Using the graph of \(y={f}'(x)\) Using the graph of \(y={f}”(x)\) (M1)
A1
The maximum of \({f}'(x)\) occurs at \(x=-0.5\). The zero of \({f}”(x)\) occurs at \(x=-0.5\). A1
THEN
Note: Do not award this A1 for stating \(x=\pm 0.5\) as the final answer for \(x\). |
\(f(-0.5)=0.607(=e^{-0.5})\) A2
EITHER
Correctly labelled graph of \({f}'(x)\) for \(x< 0\) denoting the maximum \({f}'(x)\) R1
(eg. \({f}'(-0.6)=1.17\) and \({f}'(-0.4)=1.16\) stated) A1 N2
OR
Correctly labelled graph of \({f}”(x)\) for \(x< 0\) denoting the maximum \({f}'(x)\) R1
(e.g. \({f}”(-0.6)=0.857\) and \({f}”(-0.4)=-1.05\) stated) A1 N2
OR
\({f}'(0.5)\approx 1.21. {f}'(x)< 1.21\) just to the left of \(x=-\frac{1}{2}\)
and \({f}'(x)< 1.21\) just to the right of \(x=-\frac{1}{2}\) R1
(e.g. \({f}'(-0.6)=1.17\) and \({f}'(-0.4)=1.16\) stated) A1 N2
OR
\({f}”(x)> 0\) just to the left of \(x=-\frac{1}{2}\) and \({f}”(x)< 0\) just to the right of \(x=-\frac{1}{2}\) R1
(e.g. \({f}”(-0.6)=0.857\) and \({f}”(-0.4)=-1.05\) stated) A1 N2 [7]
Question
The function \(f\) is given by \(f(x)=\frac{x^{5}+2}{x},x\neq 0\). There is a point of inflexion on the graph of \(f\) at the point P. Find the coordinates of P. (Total 6 marks)
▶️Answer/Explanation
Ans
\({f}'(x)=4x^{3}-\frac{2}{x^{2}}\) \({f}”(x)=12x^{2}+\frac{4}{x^{3}}\) (A1)(M1)(A1)
\({f}”(x)=0\Rightarrow x=-\frac{1}{\sqrt[5]{3}}=-0.803\) and \(y=-2.08\) (accept \(-2.07\)) (A1)(A1) (M1)
The point of inflexion is \((-0.803,-2.08)\left ( \textup{or}\left ( -\frac{1}{\sqrt[5]{3}}-\frac{5}{3}\sqrt[5]{3} \right ) \right )\) (C5)(C1) [6]
Question
The function \(f\) is given by \(f:x \mapsto e^{(1+\textup{sin}\, \pi x)} ,x\geq 0.\)
(a) Find \({f}'(x)\).
Let \(x_{n}\) be the value of x where the \((n+1)^{th}\) maximum or minimum point occurs, \(n\: \epsilon \) \(\mathbb{N}\). (i.e. \(x_{0}\) is the value of x where the first maximum or minimum occurs, x1 is the value of x where the second maximum or minimum occurs, etc).
(b) Find xn in terms of n. (Total 3 marks)
▶️Answer/Explanation
Ans
(a) \({f}'(x)=\pi \textup{cos}(\pi x )e^{(1+\textup{sin}\pi x)}\) (A1) (C1)
(b) For maximum or minimum points, \({f}'(x)=0,\textup{cos}\, \pi x=0\) (M1)
\(\pi x=\frac{2k+1}{2}\pi\) then \(x_{n}=\frac{2n+1}{2}\) (A1) (C2) [3]
Question
The function \(f\) is defined by \(f(x)=\frac{2x}{x^{2}+6}\), for \(x\geq b\), where \(b\: \epsilon\, R\).
(a) Show that \({f}'(x)=\frac{12-2x^{2}}{(x^{2}+6)^{2}}\)
(b) Hence find the smallest exact value of b for which the inverse function \(f^{-1}\) exists. Justify your answer. (Total 6 marks)
▶️Answer/Explanation
Ans
(a) Use of quotient (or product) rule (M1)
\({f}'(x)=\frac{2(x^{2}+6)-(2x\times 2x)}{(x^{2}+6)^{2}}\) \(2x(-1)(x^{2}+6)^{-2}(2x)+2(x^{2}+6)^{-1}\) A1
\(=\frac{12-2x^{2}}{(x^{2}+6)^{2}}\) (AG) N0
(b) Solving \({f}'(x)=0\) for \(x\) (M1)
\(x=\pm \sqrt{6}\) A1
\(f\) has to be 1-1 for \(f^{-1}\) to exist and so the least value of b is the larger of the two x-coordinates (accept a labelled sketch) R1
Hence \(b=\sqrt{6}\) A1 N2 [6]
Question
The curve \(y=\frac{x^{3}}{3}-x^{2}-3x+4\) has a local maximum point at P and a local minimum point at Q.
Determine the equation of the straight line passing through P and Q, in the form \(ax+by+c=0\), where \(a,b,c\, \epsilon \, R\). (Total 6 marks)
▶️Answer/Explanation
Ans
\(\frac{dy}{dx}=x^{2}-2x-3\) (M1)
\( \textup{at}\, \frac{dy}{dx}=0,(x-3)(x+1)=0\) (M1)
\(x=3,-1;y=5,\frac{17}{3}\)
\(\textup{So P}(3,-5)\, \textup{and Q}\left ( -1,\frac{17}{3} \right )\) (A1)(A1)
\(\textup{Equation of (PQ) is} \frac{y+5}{\left ( \frac{17}{3}+5 \right )}=\frac{x-3}{-1-3}\) (M1)
\(\frac{3y+15}{32}=\frac{x-3}{-4}\)
\(\frac{3y+15}{8}=\frac{x-3}{-1}\)
\(-3y-15=8x-24\)
\(8x+3y-9=0\) (A1) (C6) [6]
Question
A function \(f\) is defined by \(f(x)=ax^{3}+bx^{2}+30x+c\) where a, b and c are constants. The graph has a maximum at (1,7) and a point of inflexion when \(x=3\). Find the value of a, of b and of c. (Total 6 marks)
▶️Answer/Explanation
Ans
\(f(x)=ax^{3}+bx^{2}+30x+c\)
\({f}'(x)=3ax^{2}+2bx+30,{f}'(1)=0\Rightarrow 3a+2b+30=0\) (M1)
\({f}”(x)=6ax+2b,{f}”(3)=0\Rightarrow 18a+2b=0\) (M1)
\(a=2\) (A1)
\(b=-18\) (A1)
\(f(1)=7\Rightarrow 2-18+30+c=7\) (M1)
\(c=-7\) (A1) [6]