Home / IB DP Math AA Topic: SL 5.6 Derivatives of xn , sinx , cosx , tanx HL Paper 1

IB DP Math AA Topic: SL 5.6 Derivatives of xn , sinx , cosx , tanx HL Paper 1

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]

a.

\(f”(x) = \frac{{{x^2}\frac{{ – 1}}{x} – \left( {1 – \ln x} \right)2x}}{{{x^4}}}\)     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]

b.

     A1A1A1

Note: Award A1 for the maximum and intercept, A1 for a vertical asymptote and A1 for shape (including turning concave up).

[3 marks]

c.

(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]

d.

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.

a.

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.

b.

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.

c.

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.

d.

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]

a.

\(\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = \frac{2}{3}( – 2\sin \theta + 1)\)     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]

b.

\(\frac{{{{\text{d}}^2}t}}{{{\text{d}}{\theta ^2}}} = – \frac{4}{3}\cos \theta < 0\,\,\,\,\left( {{\text{at }}\theta = \frac{\pi }{6}} \right)\)     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]

c.

Examiners report

The fairly easy trigonometry challenged a large number of candidates.

a.

Part (b) was very well done.

b.

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.

c.

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]

a.

\(f'(x) = \frac{1}{2}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{3}{2}}}\)

\( \Rightarrow f”(x) = -\frac{1}{4}{x^{ – \frac{3}{2}}}{(1 – x)^{ – \frac{3}{2}}} + \frac{3}{4}{x^{ – \frac{1}{2}}}{(1 – x)^{ – \frac{5}{2}}}\)     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] 

b.

\(x = {\sin ^2}\theta  \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sin \theta \cos \theta \)     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] 

c.

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.

a.

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.

b.

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.

c.

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]

a.

\(\theta  = \frac{\pi }{4} – \arctan \frac{\pi }{{1 + \pi }}\) (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]

b.

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.

a.

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.

b.

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]

c.Find the coordinates of B, the point of inflexion.[2]
d.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 \(f(x) = g(x)\).[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]

a(i)(ii).

\(f”(x) =  – {{\text{e}}^{ – x}} – {{\text{e}}^{ – x}} + x{{\text{e}}^{ – x}}{\text{ }}\left( { =  – {{\text{e}}^{ – x}}(2 – x)} \right)\)     A1

substituting \(x = 1\) into \(f”(x)\)     M1

\(f”(1){\text{ }}\left( { =  – {{\text{e}}^{ – 1}}} \right) < 0\) hence maximum     R1AG

[3 marks]

b.

\(f”(x) = 0{\text{ (}} \Rightarrow x = 2)\)     M1

coordinates \(\left( {2,{\text{ 2}}{{\text{e}}^{ – 2}}} \right)\)     A1

[2 marks]

c.

(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.

d.     A4

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]

e.

\(A = \int_0^1 {\frac{x}{2}{{\text{e}}^{ – \frac{x}{2}}}{\text{d}}x} \)     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]

f.

Examiners report

Part a) proved to be an easy start for the vast majority of candidates.

a(i)(ii).

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.

b.

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.

c.

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 \(g(x)\).

d.

Part e) was also answered well by those who had earned full marks up to that point.

e.

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).

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]

a.

\(h \circ g(x) = \arctan \left( {\frac{1}{x}} \right)\)     A1

domain of \(h \circ g\) is equal to the domain of \(g:x \in  \circ ,{\text{ }}x \ne 0\)     A1

[2 marks]

b.

(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]

c.

(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]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

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]

a.

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]

b.

\({{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = 0 \Rightarrow 1 + 3x = 0 \Rightarrow x =  – \frac{1}{3}\)     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]

c.

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\)     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]

d.

(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]

e.

Examiners report

Well done.

a.

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.

b.

Good, some forgot to test for min/max, some forgot to give the \(y\) value.

c.

Again quite good, some forgot to check for change in curvature and some forgot the \(y\) value.

d.

Some accurate sketches, some had all the information from earlier parts but could not apply it. The asymptote was often missed.

e.

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]

a.

\(\frac{{\tan x + 1}}{{\tan x – 1}} = \frac{{\frac{{\sin x}}{{\cos x}} + 1}}{{\frac{{\sin x}}{{\cos x}} – 1}}\)     M1A1

\( = \frac{{\sin x + \cos x}}{{\sin x – \cos x}}\)     AG

[2 marks]

b.

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]

c.

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]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

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]

a.

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]

b.

Examiners report

[N/A]

a.

[N/A]

b.

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]

a.

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]

b.

Examiners report

[N/A]

a.

[N/A]

b.

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]

a.

    A3

Note: Award A1 for maximum, A1 for minimum and A1 for general shape.

[3 marks]

b.

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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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]

a.

\(f”'(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\)     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]

b.

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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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]

a.

\(f”'(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\)     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]

b.

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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

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]

a.

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]

b.

\(x = 2y – 1 \Rightarrow y = \frac{{x + 1}}{2}\)     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]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.
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