# IBDP Maths analysis and approaches Topic: SL 5.6 Differentiation of sums and multiples of functions 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]

## 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

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]

## 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]

## 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]

## 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.
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