Question
Consider the function \(f(x) = \frac{{\ln x}}{x}\) , \(0 < x < {{\text{e}}^2}\) .
a.(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 \(f\) .[5]
b.Show that there is a point of inflexion on the graph and determine its coordinates.[5]
c.Sketch the graph of \(y = f(x)\) , indicating clearly the asymptote, x-intercept and the local maximum.[3]
d.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 \(h(x) > g(x)\) .[6]
▶️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 diagram below shows a circular lake with centre O, diameter AB and radius 2 km.
Jorg needs to get from A to B as quickly as possible. He considers rowing to point P and then walking to point B. He can row at \(3{\text{ km}}\,{{\text{h}}^{ – 1}}\) and walk at \(6{\text{ km}}\,{{\text{h}}^{ – 1}}\). Let \({\rm{P\hat AB}} = \theta \) radians, and t be the time in hours taken by Jorg to travel from A to B.
a.Show that \(t = \frac{2}{3}(2\cos \theta + \theta )\).[3]
b.Find the value of \(\theta \) for which \(\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = 0\).[2]
c.What route should Jorg take to travel from A to B in the least amount of time?
Give reasons for your answer.[3]
▶️Answer/Explanation
Markscheme
angle APB is a right angle
\( \Rightarrow \cos \theta = \frac{{{\text{AP}}}}{4} \Rightarrow {\text{AP}} = 4\cos \theta \) A1
Note: Allow correct use of cosine rule.
\({\text{arc PB}} = 2 \times 2\theta = 4\theta \) A1
\(t = \frac{{{\text{AP}}}}{3} + \frac{{{\text{PB}}}}{6}\) M1
Note: Allow use of their AP and their PB for the M1.
\( \Rightarrow t = \frac{{4\cos \theta }}{3} + \frac{{4\theta }}{6} = \frac{{4\cos \theta }}{3} + \frac{{2\theta }}{3} = \frac{2}{3}(2\cos \theta + \theta )\) AG
[3 marks]
\(\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = \frac{2}{3}( – 2\sin \theta + 1)\) A1
\(\frac{2}{3}( – 2\sin \theta + 1) = 0 \Rightarrow \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi }{6}\) (or 30 degrees) A1
[2 marks]
\(\frac{{{{\text{d}}^2}t}}{{{\text{d}}{\theta ^2}}} = – \frac{4}{3}\cos \theta < 0\,\,\,\,\left( {{\text{at }}\theta = \frac{\pi }{6}} \right)\) M1
\( \Rightarrow t\) is maximized at \(\theta = \frac{\pi }{6}\) R1
time needed to walk along arc AB is \(\frac{{2\pi }}{6}{\text{ (}} \approx {\text{1 hour)}}\)
time needed to row from A to B is \(\frac{4}{3}{\text{ (}} \approx {\text{1.33 hour)}}\)
hence, time is minimized in walking from A to B R1
[3 marks]
Examiners report
The fairly easy trigonometry challenged a large number of candidates.
Part (b) was very well done.
Satisfactory answers were very rarely seen for (c). Very few candidates realised that a minimum can occur at the beginning or end of an interval.
Question
Let \(f(x) = \sqrt {\frac{x}{{1 – x}}} ,{\text{ }}0 < x < 1\).
a.Show that \(f'(x) = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}\) and deduce that f is an increasing function.[5]
b.Show that the curve \(y = f(x)\) has one point of inflexion, and find its coordinates.[6]
c. Use the substitution \(x = {\sin ^2}\theta \) to show that \(\int {f(x){\text{d}}x} = \arcsin \sqrt x – \sqrt {x – {x^2}} + c\) .[11]
▶️Answer/Explanation
Markscheme
EITHER
derivative of \(\frac{x}{{1 – x}}\) is \(\frac{{(1 – x) – x( – 1)}}{{{{(1 – x)}^2}}}\) M1A1
\(f'(x) = \frac{1}{2}{\left( {\frac{x}{{1 – x}}} \right)^{ – \frac{1}{2}}}\frac{1}{{{{(1 – x)}^2}}}\) M1A1
\( = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}\) AG
\(f'(x) > 0\) (for all \(0 < x < 1\)) so the function is increasing R1
OR
\(f(x) = \frac{{{x^{\frac{1}{2}}}}}{{{{(1 – x)}^{\frac{1}{2}}}}}\)
\(f'(x) = \frac{{{{(1 – x)}^{\frac{1}{2}}}\left( {\frac{1}{2}{x^{ – \frac{1}{2}}}} \right) – \frac{1}{2}{x^{\frac{1}{2}}}{{(1 – x)}^{ – \frac{1}{2}}}( – 1)}}{{1 – x}}\) M1A1
\( = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{1}{2}}} + \frac{1}{2}{x^{\frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}\) A1
\( = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}[1 – x + x]\) M1
\( = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}\) AG
\(f'(x) > 0\) (for all \(0 < x < 1\)) so the function is increasing R1
[5 marks]
\(f'(x) = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}\)
\( \Rightarrow f”(x) = -\frac{1}{4}{x^{ – \frac{3}{2}}}{(1 – x)^{ – \frac{3}{2}}} + \frac{3}{4}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{5}{2}}}\) M1A1
\( = -\frac{1}{4}{x^{ – \frac{3}{2}}}{(1 – x)^{ – \frac{5}{2}}}[1 – 4x]\)
\(f”(x) = 0 \Rightarrow x = \frac{1}{4}\) M1A1
\(f”(x)\) changes sign at \(x = \frac{1}{4}\) hence there is a point of inflexion R1
\(x = \frac{1}{4} \Rightarrow y = \frac{1}{{\sqrt 3 }}\) A1
the coordinates are \(\left( {\frac{1}{4},\frac{1}{{\sqrt 3 }}} \right)\)
[6 marks]
\(x = {\sin ^2}\theta \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sin \theta \cos \theta \) M1A1
\(\int {\sqrt {\frac{x}{{1 – x}}} {\text{d}}x = \int {\sqrt {\frac{{{{\sin }^2}\theta }}{{1 – {{\sin }^2}\theta }}} 2\sin \theta \cos \theta {\text{d}}\theta } } \) M1A1
\( = \int {2{{\sin }^2}\theta {\text{d}}\theta } \) A1
\( = \int {1 – \cos 2\theta } {\text{d}}\theta \) M1A1
\( = \theta – \frac{1}{2}\sin 2\theta + c\) A1
\(\theta = \arcsin \sqrt x \) A1
\(\frac{1}{2}\sin 2\theta = \sin \theta \cos \theta = \sqrt x \sqrt {1 – x} = \sqrt {x – {x^2}} \) M1A1
hence \(\int {\sqrt {\frac{x}{{1 – x}}} {\text{d}}x = \arcsin \sqrt x } – \sqrt {x – {x^2}} + c\) AG
[11 marks]
Examiners report
Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.
Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.
Part (a) was generally well done, although few candidates made the final deduction asked for. Those that lost other marks in this part were generally due to mistakes in algebraic manipulation. In part (b) whilst many students found the second derivative and set it equal to zero, few then confirmed that it was a point of inflexion. There were several good attempts for part (c), even though there were various points throughout the question that provided stopping points for other candidates.
Question
Consider the curve defined by the equation \({x^2} + \sin y – xy = 0\) .
a.Find the gradient of the tangent to the curve at the point \((\pi ,{\text{ }}\pi )\) .[6]
b.Hence, show that \(\tan \theta = \frac{1}{{1 + 2\pi }}\), where \(\theta \) is the acute angle between the tangent to the curve at \((\pi ,{\text{ }}\pi )\) and the line y = x .[3]
▶️Answer/Explanation
Markscheme
attempt to differentiate implicitly M1
\(2x + \cos y\frac{{{\text{d}}y}}{{{\text{d}}x}} – y – x\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) A1A1
Note: A1 for differentiating \({x^2}\) and sin y ; A1 for differentiating xy.
substitute x and y by \(\pi \) M1
\(2\pi – \frac{{{\text{d}}y}}{{{\text{d}}x}} – \pi – \pi \frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{\pi }{{1 + \pi }}\) M1A1
Note: M1 for attempt to make dy/dx the subject. This could be seen earlier.
[6 marks]
\(\theta = \frac{\pi }{4} – \arctan \frac{\pi }{{1 + \pi }}\) (or seen the other way) M1
\(\tan \theta = \tan \left( {\frac{\pi }{4} – \arctan \frac{\pi }{{1 + \pi }}} \right) = \frac{{1 – \frac{\pi }{{1 + \pi }}}}{{1 + \frac{\pi }{{1 + \pi }}}}\) M1A1
\(\tan \theta = \frac{1}{{1 + 2\pi }}\) AG
[3 marks]
Examiners report
Part a) proved an easy 6 marks for most candidates, while the majority failed to make any headway with part b), with some attempting to find the equation of their line in the form y = mx + c . Only the best candidates were able to see their way through to the given answer.
Part a) proved an easy 6 marks for most candidates, while the majority failed to make any headway with part b), with some attempting to find the equation of their line in the form y = mx + c . Only the best candidates were able to see their way through to the given answer.
Question
The function \(f\) is given by \(f(x) = x{{\text{e}}^{ – x}}{\text{ }}(x \geqslant 0)\).
a(i)(ii).(i) Find an expression for \(f'(x)\).
(ii) Hence determine the coordinates of the point A, where \(f'(x) = 0\).[3]
b.Find an expression for \(f”(x)\) and hence show the point A is a maximum.[3]
(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 \(f(x) = g(x)\).[5]
e.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.[4]
f.Find an exact value for the area of the region bounded by the curve \(y = g(x)\), the x-axis and the line \(x = 1\).[3]
▶️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 following functions:
a. \(h(x) = \arctan (x),{\text{ }}x \in \mathbb{R}\)
\(g(x) = \frac{1}{x}\), \(x\in \mathbb{R}\), \({\text{ }}x \ne 0\)
Sketch the graph of \(y = h(x)\).[2]
b.Find an expression for the composite function \(h \circ g(x)\) and state its domain.[2]
c.Given that \(f(x) = h(x) + h \circ g(x)\),
(i) find \(f'(x)\) in simplified form;
(ii) show that \(f(x) = \frac{\pi }{2}\) for \(x > 0\).[7]
d.Nigel states that \(f\) is an odd function and Tom argues that \(f\) is an even function.
(i) State who is correct and justify your answer.
(ii) Hence find the value of \(f(x)\) for \(x < 0\).[3]
▶️Answer/Explanation
Markscheme
A1A1
Note: A1 for correct shape, A1 for asymptotic behaviour at \(y = \pm \frac{\pi }{2}\).
[2 marks]
\(h \circ g(x) = \arctan \left( {\frac{1}{x}} \right)\) A1
domain of \(h \circ g\) is equal to the domain of \(g:x \in \circ ,{\text{ }}x \ne 0\) A1
[2 marks]
(i) \(f(x) = \arctan (x) + \arctan \left( {\frac{1}{x}} \right)\)
\(f'(x) = \frac{1}{{1 + {x^2}}} + \frac{1}{{1 + \frac{1}{{{x^2}}}}} \times – \frac{1}{{{x^2}}}\) M1A1
\(f'(x) = \frac{1}{{1 + {x^2}}} + \frac{{ – \frac{1}{{{x^2}}}}}{{\frac{{{x^2} + 1}}{{{x^2}}}}}\) (A1)
\( = \frac{1}{{1 + {x^2}}} – \frac{1}{{1 + {x^2}}}\)
\( = 0\) A1
(ii) METHOD 1
f is a constant R1
when \(x > 0\)
\(f(1) = \frac{\pi }{4} + \frac{\pi }{4}\) M1A1
\( = \frac{\pi }{2}\) AG
METHOD 2
from diagram
\(\theta = \arctan \frac{1}{x}\) A1
\(\alpha = \arctan x\) A1
\(\theta + \alpha = \frac{\pi }{2}\) R1
hence \(f(x) = \frac{\pi }{2}\) AG
METHOD 3
\(\tan \left( {f(x)} \right) = \tan \left( {\arctan (x) + \arctan \left( {\frac{1}{x}} \right)} \right)\) M1
\( = \frac{{x + \frac{1}{x}}}{{1 – x\left( {\frac{1}{x}} \right)}}\) A1
denominator = 0, so \(f(x) = \frac{\pi }{2}{\text{ (for }}x > 0)\) R1
[7 marks]
(i) Nigel is correct. A1
METHOD 1
\(\arctan (x)\) is an odd function and \(\frac{1}{x}\) is an odd function
composition of two odd functions is an odd function and sum of two odd functions is an odd function R1
METHOD 2
\(f( – x) = \arctan ( – x) + \arctan \left( { – \frac{1}{x}} \right) = – \arctan (x) – \arctan \left( {\frac{1}{x}} \right) = – f(x)\)
therefore f is an odd function. R1
(ii) \(f(x) = – \frac{\pi }{2}\) A1
[3 marks]
Examiners report
[N/A]
[N/A]
[N/A]
[N/A]
Question
Let \(y(x) = x{e^{3x}},{\text{ }}x \in \mathbb{R}\).
a.Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[2]
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 \(n \in {\mathbb{Z}^ + }\).[7]
b.Find the coordinates of any local maximum and minimum points on the graph of \(y(x)\).
c.Justify whether any such point is a maximum or a minimum.[5]
d.Find the coordinates of any points of inflexion on the graph of \(y(x)\). Justify whether any such point is a point of inflexion.[5]
e.Hence sketch the graph of \(y(x)\), indicating clearly the points found in parts (c) and (d) and any intercepts with the axes.[2]
▶️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
Consider the functions \(f(x) = \tan x,{\text{ }}0 \le \ x < \frac{\pi }{2}\) and \(g(x) = \frac{{x + 1}}{{x – 1}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne 1\).
a.Find an expression for \(g \circ f(x)\), stating its domain.[2]
b.Hence show that \(g \circ f(x) = \frac{{\sin x + \cos x}}{{\sin x – \cos x}}\).[2]
c.Let \(y = g \circ f(x)\), find an exact value for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) at the point on the graph of \(y = g \circ f(x)\) where \(x = \frac{\pi }{6}\), expressing your answer in the form \(a + b\sqrt 3 ,{\text{ }}a,{\text{ }}b \in \mathbb{Z}\).[6]
d.Show that the area bounded by the graph of \(y = g \circ f(x)\), the \(x\)-axis and the lines \(x = 0\) and \(x = \frac{\pi }{6}\) is \(\ln \left( {1 + \sqrt 3 } \right)\).[6]
▶️Answer/Explanation
Markscheme
\(g \circ f(x) = \frac{{\tan x + 1}}{{\tan x – 1}}\) A1
\(x \ne \frac{\pi }{4},{\text{ }}0 \le x < \frac{\pi }{2}\) A1
[2 marks]
\(\frac{{\tan x + 1}}{{\tan x – 1}} = \frac{{\frac{{\sin x}}{{\cos x}} + 1}}{{\frac{{\sin x}}{{\cos x}} – 1}}\) M1A1
\( = \frac{{\sin x + \cos x}}{{\sin x – \cos x}}\) AG
[2 marks]
METHOD 1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(\sin x – \cos x)(\cos x – \sin x) – (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x – \cos x)}^2}}}\) M1(A1)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(2\sin x\cos x – {{\cos }^2}x – {{\sin }^2}x) – (2\sin x\cos x + {{\cos }^2}x + {{\sin }^2}x)}}{{{{\cos }^2}x + {{\sin }^2}x – 2\sin x\cos x}}\)
\( = \frac{{ – 2}}{{1 – \sin 2x}}\)
Substitute \(\frac{\pi }{6}\) into any formula for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1
\(\frac{{ – 2}}{{1 – \sin \frac{\pi }{3}}}\)
\( = \frac{{ – 2}}{{1 – \frac{{\sqrt 3 }}{2}}}\) A1
\( = \frac{{ – 4}}{{2 – \sqrt 3 }}\)
\( = \frac{{ – 4}}{{2 – \sqrt 3 }}\left( {\frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}} \right)\) M1
\( = \frac{{ – 8 – 4\sqrt 3 }}{1} = – 8 – 4\sqrt 3 \) A1
METHOD 2
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(\tan x – 1){{\sec }^2}x – (\tan x + 1){{\sec }^2}x}}{{{{(\tan x – 1)}^2}}}\) M1A1
\( = \frac{{ – 2{{\sec }^2}x}}{{{{(\tan x – 1)}^2}}}\) A1
\( = \frac{{ – 2{{\sec }^2}\frac{\pi }{6}}}{{{{\left( {\tan \frac{\pi }{6} – 1} \right)}^2}}} = \frac{{ – 2\left( {\frac{4}{3}} \right)}}{{{{\left( {\frac{1}{{\sqrt 3 }} – 1} \right)}^2}}} = \frac{{ – 8}}{{{{\left( {1 – \sqrt 3 } \right)}^2}}}\) M1
Note: Award M1 for substitution \(\frac{\pi }{6}\).
\(\frac{{ – 8}}{{{{\left( {1 – \sqrt 3 } \right)}^2}}} = \frac{{ – 8}}{{\left( {4 – 2\sqrt 3 } \right)}}\frac{{\left( {4 + 2\sqrt 3 } \right)}}{{\left( {4 + 2\sqrt 3 } \right)}} = – 8 – 4\sqrt 3 \) M1A1
[6 marks]
Area \(\left| {\int_0^{\frac{\pi }{6}} {\frac{{\sin x + \cos x}}{{\sin x – \cos x}}{\text{d}}x} } \right|\) M1
\( = \left| {\left[ {\ln \left| {\sin x – \cos x} \right|} \right]_0^{\frac{\pi }{6}}} \right|\) A1
Note: Condone absence of limits and absence of modulus signs at this stage.
\( = \left| {\ln \left| {\sin \frac{\pi }{6} – \cos \frac{\pi }{6}} \right| – \ln \left| {\sin 0 – \cos 0} \right|} \right|\) M1
\( = \left| {\ln \left| {\frac{1}{2} – \frac{{\sqrt 3 }}{2}} \right| – 0} \right|\)
\( = \left| {\ln \left( {\frac{{\sqrt 3 – 1}}{2}} \right)} \right|\) A1
\( = – \ln \left( {\frac{{\sqrt 3 – 1}}{2}} \right) = \ln \left( {\frac{2}{{\sqrt 3 – 1}}} \right)\) A1
\( = \ln \left( {\frac{2}{{\sqrt 3 – 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}} \right)\) M1
\( = \ln \left( {\sqrt 3 + 1} \right)\) AG
[6 marks]
Total [16 marks]
Examiners report
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Question
Consider the curve \(y = \frac{1}{{1 – x}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne 1\).
a.Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[2]
b.Determine the equation of the normal to the curve at the point \(x = 3\) in the form \(ax + by + c = 0\) where \(a,{\text{ }}b,{\text{ }}c \in \mathbb{Z}\). [4]
▶️Answer/Explanation
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(1 – x)^{ – 2}}\;\;\;\left( { = \frac{1}{{{{(1 – x)}^2}}}} \right)\) (M1)A1
[2 marks]
gradient of Tangent \( = \frac{1}{4}\) (A1)
gradient of Normal \( = – 4\) (M1)
\(y + \frac{1}{2} = – 4(x – 3)\) or attempt to find \(c\) in \(y = mx + c\) M1
\(8x + 2y – 23 = 0\) A1
[4 marks]
Total [6 marks]
Examiners report
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Question
a.Show that \(\sin \left( {\theta + \frac{\pi }{2}} \right) = \cos \theta \).[1]
b.Consider \(f(x) = \sin (ax)\) where \(a\) is a constant. Prove by mathematical induction that \({f^{(n)}}(x) = {a^n}\sin \left( {ax + \frac{{n\pi }}{2}} \right)\) where \(n \in {\mathbb{Z}^ + }\) and \({f^{(n)}}(x)\) represents the \({{\text{n}}^{{\text{th}}}}\) derivative of \(f(x)\).[7]
▶️Answer/Explanation
Markscheme
\(\sin \left( {\theta + \frac{\pi }{2}} \right) = \sin \theta \cos \frac{\pi }{2} + \cos \theta \sin \frac{\pi }{2}\) M1
\( = \cos \theta \) AG
Note: Accept a transformation/graphical based approach.
[1 mark]
consider \(n = 1,{\text{ }}f'(x) = a\cos (ax)\) M1
since \(\sin \left( {ax + \frac{\pi }{2}} \right) = \cos ax\) then the proposition is true for \(n = 1\) R1
assume that the proposition is true for \(n = k\) so \({f^{(k)}}(x) = {a^k}\sin \left( {ax + \frac{{k\pi }}{2}} \right)\) M1
\({f^{(k + 1)}}(x) = \frac{{{\text{d}}\left( {{f^{(k)}}(x)} \right)}}{{{\text{d}}x}}\;\;\;\left( { = a\left( {{a^k}\cos \left( {ax + \frac{{k\pi }}{2}} \right)} \right)} \right)\) M1
\( = {a^{k + 1}}\sin \left( {ax + \frac{{k\pi }}{2} + \frac{\pi }{2}} \right)\) (using part (a)) A1
\( = {a^{k + 1}}\sin \left( {ax + \frac{{(k + 1)\pi }}{2}} \right)\) A1
given that the proposition is true for \(n = k\) then we have shown that the proposition is true for \(n = k + 1\). Since we have shown that the proposition is true for \(n = 1\) then the proposition is true for all \(n \in {\mathbb{Z}^ + }\) R1
Note: Award final R1 only if all prior M and R marks have been awarded.
[7 marks]
Total [8 marks]
Examiners report
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Question
The function f is defined on the domain \(x \geqslant 0\) by \(f(x) = {{\text{e}}^x} – {x^{\text{e}}}\) .
a.(i) Find an expression for \(f'(x)\) .
(ii) Given that the equation \(f'(x) = 0\) has two roots, state their values.[3]
b.Sketch the graph of f , showing clearly the coordinates of the maximum and minimum.[3]
c.Hence show that \({{\text{e}}^\pi } > {\pi ^{\text{e}}}\) .[1]
▶️Answer/Explanation
Markscheme
(i) \(f'(x) = {{\text{e}}^x} – {\text{e}}{x^{{\text{e}} – 1}}\) A1
(ii) by inspection the two roots are 1, e A1A1
[3 marks]
A3
Note: Award A1 for maximum, A1 for minimum and A1 for general shape.
[3 marks]
from the graph: \({{\text{e}}^x} > {x^{\text{e}}}\) for all \(x > 0\) except x = e R1
putting \(x = \pi \) , conclude that \({{\text{e}}^\pi } > {\pi ^{\text{e}}}\) AG
[1 mark]
Examiners report
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Question
The function f is defined by \(f(x) = {{\text{e}}^x}\sin x\) .
a.Show that \(f”(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) .[3]
b.Obtain a similar expression for \({f^{(4)}}(x)\) .[4]
c.Suggest an expression for \({f^{(2n)}}(x)\), \(n \in {\mathbb{Z}^ + }\), and prove your conjecture using mathematical induction.[8]
▶️Answer/Explanation
Markscheme
\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\) A1
\(f”(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x + {{\text{e}}^x}\cos x – {{\text{e}}^x}\sin x\) A1
\( = 2{{\text{e}}^x}\cos x\) A1
\( = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) AG
[3 marks]
\(f”'(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\) A1
\({f^{(4)}}(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) – 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) A1
\( = 4{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\) A1
\( = 4{{\text{e}}^x}\sin (x + \pi )\) A1
[4 marks]
the conjecture is that
\({f^{(2n)}}(x) = {2^n}{{\text{e}}^x}\sin \left( {x + \frac{{n\pi }}{2}} \right)\) A1
for n = 1, this formula gives
\(f”(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) which is correct A1
let the result be true for n = k , \(\left( {i.e.{\text{ }}{f^{(2k)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)} \right)\) M1
consider \({f^{(2k + 1)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\) M1
\({f^{\left( {2(k + 1)} \right)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) – {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)\) A1
\( = {2^{k + 1}}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\) A1
\( = {2^{k + 1}}{{\text{e}}^x}\sin \left( {x + \frac{{(k + 1)\pi }}{2}} \right)\) A1
therefore true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 1\)
the result is proved by induction. R1
Note: Award the final R1 only if the two M marks have been awarded.
[8 marks]
Examiners report
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Question
The function f is defined by \(f(x) = {{\text{e}}^x}\sin x\) .
a.Show that \(f”(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) .[3]
b.Obtain a similar expression for \({f^{(4)}}(x)\) .[4]
c.Suggest an expression for \({f^{(2n)}}(x)\), \(n \in {\mathbb{Z}^ + }\), and prove your conjecture using mathematical induction.[8]
▶️Answer/Explanation
Markscheme
\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\) A1
\(f”(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x + {{\text{e}}^x}\cos x – {{\text{e}}^x}\sin x\) A1
\( = 2{{\text{e}}^x}\cos x\) A1
\( = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) AG
[3 marks]
\(f”'(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\) A1
\({f^{(4)}}(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) – 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) A1
\( = 4{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\) A1
\( = 4{{\text{e}}^x}\sin (x + \pi )\) A1
[4 marks]
the conjecture is that
\({f^{(2n)}}(x) = {2^n}{{\text{e}}^x}\sin \left( {x + \frac{{n\pi }}{2}} \right)\) A1
for n = 1, this formula gives
\(f”(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) which is correct A1
let the result be true for n = k , \(\left( {i.e.{\text{ }}{f^{(2k)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)} \right)\) M1
consider \({f^{(2k + 1)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\) M1
\({f^{\left( {2(k + 1)} \right)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) – {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)\) A1
\( = {2^{k + 1}}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\) A1
\( = {2^{k + 1}}{{\text{e}}^x}\sin \left( {x + \frac{{(k + 1)\pi }}{2}} \right)\) A1
therefore true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 1\)
the result is proved by induction. R1
Note: Award the final R1 only if the two M marks have been awarded.
[8 marks]
Examiners report
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Question
The function f is defined by
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{2x – 1,}&{x \leqslant 2} \\
{a{x^2} + bx – 5,}&{2 < x < 3}
\end{array}} \right.\]
where a , \(b \in \mathbb{R}\) .
a.Given that f and its derivative, \(f’\) , are continuous for all values in the domain of f , find the values of a and b .[6]
b.Show that f is a one-to-one function.[3]
c.Obtain expressions for the inverse function \({f^{ – 1}}\) and state their domains.[5]
▶️Answer/Explanation
Markscheme
f continuous \( \Rightarrow \mathop {\lim }\limits_{x \to {2^ – }} f(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f(x)\) M1
\(4a + 2b = 8\) A1
\(f'(x) = \left\{ {\begin{array}{*{20}{c}}
{2,}&{x < 2} \\
{2ax + b,}&{2 < x < 3}
\end{array}} \right.\) A1
\(f'{\text{ continuous}} \Rightarrow \mathop {\lim }\limits_{x \to {2^ – }} f'(x) = \mathop {\lim }\limits_{x \to {2^ \div }} f'(x)\)
\(4a + b = 2\) A1
solve simultaneously M1
to obtain a = –1 and b = 6 A1
[6 marks]
for \(x \leqslant 2,{\text{ }}f'(x) = 2 > 0\) A1
for \(2 < x < 3,{\text{ }}f'(x) = – 2x + 6 > 0\) A1
since \(f'(x) > 0\) for all values in the domain of f , f is increasing R1
therefore one-to-one AG
[3 marks]
\(x = 2y – 1 \Rightarrow y = \frac{{x + 1}}{2}\) M1
\(x = – {y^2} + 6y – 5 \Rightarrow {y^2} – 6y + x + 5 = 0\) M1
\(y = 3 \pm \sqrt {4 – x} \)
therefore
\({f^{ – 1}}(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{x + 1}}{2},}&{x \leqslant 3} \\
{3 – \sqrt {4 – x} ,}&{3 < x < 4}
\end{array}} \right.\) A1A1A1
Note: Award A1 for the first line and A1A1 for the second line.
[5 marks]
Examiners report
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