# IBDP Maths analysis and approaches Topic: SL 5.8 Optimization problems HL Paper 1

## Question

André wants to get from point A located in the sea to point Y located on a straight stretch of beach. P is the point on the beach nearest to A such that AP = 2 km and PY = 2 km. He does this by swimming in a straight line to a point Q located on the beach and then running to Y.

When André swims he covers 1 km in $$5\sqrt 5$$ minutes. When he runs he covers 1 km in 5 minutes.

(a)     If PQ = x km, $$0 \leqslant x \leqslant 2$$ , find an expression for the time T minutes taken by André to reach point Y.

(b)     Show that $$\frac{{{\text{d}}T}}{{{\text{d}}x}} = \frac{{5\sqrt 5 x}}{{\sqrt {{x^2} + 4} }} – 5$$.

(c)     (i)     Solve $$\frac{{{\text{d}}T}}{{{\text{d}}x}} = 0$$.

(ii)     Use the value of x found in part (c) (i) to determine the time, T minutes, taken for André to reach point Y.

(iii)     Show that $$\frac{{{{\text{d}}^2}T}}{{{\text{d}}{x^2}}} = \frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{\frac{3}{2}}}}}$$ and hence show that the time found in part (c) (ii) is a minimum.

## Markscheme

(a)     $${\text{AQ}} = \sqrt {{x^2} + 4} {\text{ (km)}}$$     (A1)

$${\text{QY}} = (2 – x){\text{ (km)}}$$     (A1)

$$T = 5\sqrt 5 {\text{AQ}} + 5{\text{QY}}$$     (M1)

$${\text{ = 5}}\sqrt 5 \sqrt {({x^2} + 4)} + 5(2 – x){\text{ (mins)}}$$     A1

[4 marks]

(b)     Attempting to use the chain rule on $${\text{5}}\sqrt 5 \sqrt {({x^2} + 4)}$$     (M1)

$$\frac{{\text{d}}}{{{\text{d}}x}}\left( {{\text{5}}\sqrt 5 \sqrt {({x^2} + 4)} } \right) = 5\sqrt 5 \times \frac{1}{2}{({x^2} + 4)^{ – \frac{1}{2}}} \times 2x$$     A1

$$\left( { = \frac{{5\sqrt 5 x}}{{\sqrt {{x^2} + 4} }}} \right)$$

$$\frac{{\text{d}}}{{{\text{d}}x}}\left( {5(2 – x)} \right) = – 5$$     A1

$$\frac{{{\text{d}}T}}{{{\text{d}}x}} = \frac{{5\sqrt 5 x}}{{\sqrt {{x^2} + 4} }} – 5$$     AG     N0

[3 marks]

(c)     (i)     $$\sqrt 5 x = \sqrt {{x^2} + 4}$$     A1

Squaring both sides and rearranging to obtain $$5{x^2} = {x^2} + 4$$     M1

x = 1     A1     N1

Note: Do not award the final A1 for stating a negative solution in final answer.

(ii)     $$T = 5\sqrt 5 \sqrt {1 + 4} + 5(2 – 1)$$     M1

= 30 (mins)     A1     N1

Note: Allow FT on incorrect x value.

(iii)     METHOD 1

Attempting to use the quotient rule     M1

$$u = x{\text{ , }}v = \sqrt {{x^2} + 4} {\text{, }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1{\text{ and }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = x{({x^2} + 4)^{ – 1/2}}$$     (A1)

$$\frac{{{{\text{d}}^2}T}}{{{\text{d}}{x^2}}} = 5\sqrt 5 \left[ {\frac{{\sqrt {{x^2} + 4} – \frac{1}{2}{{({x^2} + 4)}^{ – 1/2}} \times 2{x^2}}}{{({x^2} + 4)}}} \right]$$     A1

Attempt to simplify     (M1)

$$= \frac{{5\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}}[{x^2} + 4 – {x^2}]\,\,\,\,\,$$or equivalent     A1

$$= \frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}}$$     AG

When $$x = 1{\text{ , }}\frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}} > 0$$ and hence T = 30 is a minimum     R1     N0

Note: Allow FT on incorrect x value, $$0 \leqslant x \leqslant 2$$.

METHOD 2

Attempting to use the product rule     M1

$$u = x{\text{ , }}v = \sqrt {{x^2} + 4} {\text{, }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = 1{\text{ and }}\frac{{{\text{d}}v}}{{{\text{d}}x}} = x{({x^2} + 4)^{ – 1/2}}$$     (A1)

$$\frac{{{{\text{d}}^2}T}}{{{\text{d}}{x^2}}} = 5\sqrt 5 {({x^2} + 4)^{ – 1/2}} – \frac{{5\sqrt 5 x}}{2}{({x^2} + 4)^{ – 3/2}} \times 2x$$     A1

$$\left( { = \frac{{5\sqrt 5 }}{{{{({x^2} + 4)}^{1/2}}}} – \frac{{5\sqrt 5 {x^2}}}{{{{({x^2} + 4)}^{3/2}}}}} \right)$$

Attempt to simplify     (M1)

$$= \frac{{5\sqrt 5 ({x^2} + 4) – 5\sqrt 5 {x^2}}}{{{{({x^2} + 4)}^{3/2}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { = \frac{{5\sqrt 5 ({x^2} + 4 – {x^2})}}{{{{({x^2} + 4)}^{3/2}}}}} \right)$$     A1

$$= \frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}}$$     AG

When $$x = 1{\text{ , }}\frac{{20\sqrt 5 }}{{{{({x^2} + 4)}^{3/2}}}} > 0$$ and hence T = 30 is a minimum     R1     N0

Note: Allow FT on incorrect x value, $$0 \leqslant x \leqslant 2$$.

[11 marks]

Total [18 marks]

## Examiners report

Most candidates scored well on this question. The question tested their competence at algebraic manipulation and differentiation. A few candidates failed to extract from the context the correct relationship between velocity, distance and time.

## Question

A packaging company makes boxes for chocolates. An example of a box is shown below. This box is closed and the top and bottom of the box are identical regular hexagons of side x cm.

(a)     Show that the area of each hexagon is $$\frac{{3\sqrt 3 {x^2}}}{2}{\text{c}}{{\text{m}}^2}$$ .

(b)     Given that the volume of the box is $${\text{90 c}}{{\text{m}}^2}$$ , show that when $$x = \sqrt[3]{{20}}$$ the total surface area of the box is a minimum, justifying that this value gives a minimum.

## Markscheme

(a)     Area of hexagon $$= 6 \times \frac{1}{2} \times x \times x \times \sin 60^\circ$$     M1

$$= \frac{{3\sqrt 3 {x^2}}}{2}$$     AG

(b)     Let the height of the box be h

Volume $$= \frac{{3\sqrt 3 h{x^2}}}{2} = 90$$     M1

Hence $$h = \frac{{60}}{{\sqrt 3 {x^2}}}$$     A1

Surface area, $$A = 3\sqrt 3 {x^2} + 6hx$$     M1

$$= 3\sqrt 3 {x^2} + \frac{{360}}{{\sqrt 3 }}{x^{ – 1}}$$     A1

$$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 6\sqrt 3 x – \frac{{360}}{{\sqrt 3 }}{x^{ – 2}}$$     A1

$$\left( {\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0} \right)$$

$$6\sqrt 3 {x^3} = \frac{{360}}{{\sqrt 3 }}$$     M1

$${x^3} = 20$$

$$x = \sqrt[3]{{20}}$$     AG

$$\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = 6\sqrt 3 + \frac{{720{x^{ – 3}}}}{{\sqrt 3 }}$$

which is positive when $$x = \sqrt[3]{{20}}$$, and hence gives a minimum value.     R1

[8 marks]

## Examiners report

There were a number of wholly correct answers seen and the best candidates tackled the question well. However, many candidates did not seem to understand what was expected in such a problem. It was disappointing that a significant number of candidates were unable to find the area of the hexagon.

## 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.

Show that $$t = \frac{2}{3}(2\cos \theta + \theta )$$.

[3]
a.

Find the value of $$\theta$$ for which $$\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = 0$$.

[2]
b.

What route should Jorg take to travel from A to B in the least amount of time?

[3]
c.

## 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

At 12:00 a boat is 20 km due south of a freighter. The boat is travelling due east at $$20{\text{ km}}\,{{\text{h}}^{ – 1}}$$, and the freighter is travelling due south at $$40{\text{ km}}\,{{\text{h}}^{ – 1}}$$.

Determine the time at which the two ships are closest to one another, and justify your answer.

[8]
a.

If the visibility at sea is 9 km, determine whether or not the captains of the two ships can ever see each other’s ship.

[3]
b.

## Markscheme

(M1)

$${s^2} = {(20t)^2} + {(20 – 40t)^2}$$     M1

$${s^2} = 2000{t^2} – 1600t + 400$$     A1

to minimize s it is enough to minimize $${s^2}$$

$$f'(t) = 4000t – 1600$$     A1

setting $$f'(t)$$ equal to 0     M1

$$4000t – 1600 = 0 \Rightarrow t = \frac{2}{5}$$ or 24 minutes     A1

$$f”(t) = 4000 > 0$$     M1

$$\Rightarrow$$ at $$t = \frac{2}{5},{\text{ }}f(t)$$ is minimized

hence, the ships are closest at 12:24     A1

Note: accept solution based on s.

[8 marks]

a.

$$f\left( {\frac{2}{5}} \right) = \sqrt {80}$$     M1A1

since $$\sqrt {80} < 9$$, the captains can see one another     R1

[3 marks]

b.

## Examiners report

This was, disappointingly, a poorly answered question. Some tried to talk their way through the question without introducing the time variable. Even those who did use the distance as a function of time often did not check for a minimum.

a.

This was, disappointingly, a poorly answered question. Some tried to talk their way through the question without introducing the time variable. Even those who did use the distance as a function of time often did not check for a minimum.

b.

## Question

At 12:00 a boat is 20 km due south of a freighter. The boat is travelling due east at $$20{\text{ km}}\,{{\text{h}}^{ – 1}}$$, and the freighter is travelling due south at $$40{\text{ km}}\,{{\text{h}}^{ – 1}}$$.

Determine the time at which the two ships are closest to one another, and justify your answer.

[8]
a.

If the visibility at sea is 9 km, determine whether or not the captains of the two ships can ever see each other’s ship.

[3]
b.

## Markscheme

(M1)

$${s^2} = {(20t)^2} + {(20 – 40t)^2}$$     M1

$${s^2} = 2000{t^2} – 1600t + 400$$     A1

to minimize s it is enough to minimize $${s^2}$$

$$f'(t) = 4000t – 1600$$     A1

setting $$f'(t)$$ equal to 0     M1

$$4000t – 1600 = 0 \Rightarrow t = \frac{2}{5}$$ or 24 minutes     A1

$$f”(t) = 4000 > 0$$     M1

$$\Rightarrow$$ at $$t = \frac{2}{5},{\text{ }}f(t)$$ is minimized

hence, the ships are closest at 12:24     A1

Note: accept solution based on s.

[8 marks]

a.

$$f\left( {\frac{2}{5}} \right) = \sqrt {80}$$     M1A1

since $$\sqrt {80} < 9$$, the captains can see one another     R1

[3 marks]

b.

## Examiners report

This was, disappointingly, a poorly answered question. Some tried to talk their way through the question without introducing the time variable. Even those who did use the distance as a function of time often did not check for a minimum.

a.

This was, disappointingly, a poorly answered question. Some tried to talk their way through the question without introducing the time variable. Even those who did use the distance as a function of time often did not check for a minimum.

b.

## 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 $$f'(x) = \frac{{1 – \ln x}}{{{x^2}}}$$.

[2]
a.

Find the coordinates of B, at which the curve reaches its maximum value.

[3]
b.

Find the coordinates of C, the point of inflexion on the curve.

[5]
c.

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.

[4]
d.

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 $$x = {\text{e}}$$.

[7]
e.

## Markscheme

$$f'(x) = \frac{{x \times \frac{1}{x} – \ln x}}{{{x^2}}}$$     M1A1

$$= \frac{{1 – \ln x}}{{{x^2}}}$$     AG

[2 marks]

a.

$$\frac{{1 – \ln x}}{{{x^2}}} = 0$$ 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]

b.

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

c.

$$f(1) = 0$$     A1

$$f'(1) = 1$$     (A1)

$$y = x + c$$     (M1)

through (1, 0)

equation is $$y = x – 1$$     A1

[4 marks]

d.

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]

e.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

## 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 $$f'(x)$$.

[2]
a.

Hence find the $$x$$-coordinates of the points where the gradient of the graph of $$f$$ is zero.

[1]
b.

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.

[3]
c.

Find the $$x$$-coordinates of the other two points of inflexion.

[4]
d.

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.

[6]
e.

## 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]

a.

$$\frac{{ – {x^2} – 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0$$

$$x = – 1 \pm \sqrt 2$$     A1

[1 mark]

b.

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

c.

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]

d.

$$\int_{ – 1}^0 {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x}$$     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]

e.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

## 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 $$f'(x)$$.

[2]
a.

Hence find the $$x$$-coordinates of the points where the gradient of the graph of $$f$$ is zero.

[1]
b.

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.

[3]
c.

Find the $$x$$-coordinates of the other two points of inflexion.

[4]
d.

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.

[6]
e.

## 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]

a.

$$\frac{{ – {x^2} – 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0$$

$$x = – 1 \pm \sqrt 2$$     A1

[1 mark]

b.

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

c.

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]

d.

$$\int_{ – 1}^0 {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x}$$     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]

e.

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

## Question

A window is made in the shape of a rectangle with a semicircle of radius $$r$$ metres on top, as shown in the diagram. The perimeter of the window is a constant P metres.

Find the area of the window in terms of P and $$r$$.

[4]
a.i.

Find the width of the window in terms of P when the area is a maximum, justifying that this is a maximum.

[5]
a.ii.

Show that in this case the height of the rectangle is equal to the radius of the semicircle.

[2]
b.

## Markscheme

the width of the rectangle is $$2r$$ and let the height of the rectangle be $$h$$

$$P = 2r + 2h + \pi r$$     (A1)

$$A = 2rh + \frac{{\pi {r^2}}}{2}$$     (A1)

$$h = \frac{{{\text{P}} – 2r – \pi r}}{2}$$

$$A = 2r\left( {\frac{{{\text{P}} – 2r – \pi r}}{2}} \right) + \frac{{\pi {r^2}}}{2}\,\,\,\left( { = \operatorname{P} r – 2{r^2} – \frac{{\pi {r^2}}}{2}} \right)$$     M1A1

[4 marks]

a.i.

$$\frac{{{\text{d}}A}}{{{\text{d}}r}} = {\text{P}} – 4r – \pi r$$     A1

$$\frac{{{\text{d}}A}}{{{\text{d}}r}} = 0$$     M1

$$\Rightarrow r = \frac{{\text{P}}}{{4 + \pi }}$$     (A1)

hence the width is $$\frac{{2{\text{P}}}}{{4 + \pi }}$$     A1

$$\frac{{{{\text{d}}^2}A}}{{{\text{d}}{r^2}}} = – 4 – \pi < 0$$     R1

hence maximum     AG

[5 marks]

a.ii.

EITHER

$$h = \frac{{{\text{P}} – 2r – \pi r}}{2}$$

$$h = \frac{{{\text{P}} – \frac{{2{\text{P}}}}{{4 + \pi }} – \frac{{{\text{P}}\pi }}{{4 + \pi }}}}{2}$$     M1

$$h = \frac{{4{\text{P}} + \pi {\text{P}} – 2{\text{P}} – \pi {\text{P}}}}{{2(4 + \pi )}}$$    A1

$$h = \frac{{\text{P}}}{{(4 + \pi )}} = r$$     AG

OR

$$h = \frac{{{\text{P}} – 2r – \pi r}}{2}$$

$$P = r(4 + \pi )$$     M1

$$h = \frac{{r(4 + \pi ) – 2r – \pi r}}{2}$$     A1

$$h = \frac{{4r + \pi r – 2r – \pi r}}{2} = r$$     AG

[2 marks]

b.

## Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

### Question

Consider all closed cylinders of constant volume V.
(a) Find the cylinder of minimum total surface area (i.e. its dimensions in terms of V); justify your answer.
(b) Do there exist cylinders of maximum total surface area?

Ans
(only guidelines and the final answers)
Let us call r the radius of the rectangle. The height will be $$h=\frac{V}{\pi r^{2}}$$
Then the total surface area is given by $$S=\frac{2V}{r}+2\pi r^{2}$$
(a) It is the cylinder with radius $$r=\sqrt[3]{\frac{V}{2\pi }}$$ (and height $$h=\frac{V}{\pi r^{2}}$$)
(b) The domain is $$r> 0$$. There is no cylinder of maximum total surface area.

### Question

Consider all closed cylinders of constant total surface area S.
(a) Find the cylinder of maximum volume (i.e. its dimensions in terms of S); justify your answer.
(b) Do there exist cylinders of minimum volume?

Ans
(only guidelines and the final answers)
Let us call r the radius of the rectangle. The height will be $$h=\frac{S-2\, \pi\, r^{2}}{2\, \pi\, r}$$
(a) It is the cylinder with radius $$r=\sqrt{\frac{S}{6\pi }}$$ (and height $$h=\frac{S-2\, \pi \, r^{2}}{2\, \pi \, r}$$)
(b) The domain is $$0< r\leq \sqrt{\frac{S}{2\pi }}$$. At the endpoint $$r=\sqrt{\frac{S}{2\pi }}$$, the volume is $$0$$.

### Question

Consider the graph G of the function $$y=\sqrt{x}$$ and its reflection G’ about the vertical line $$x=4$$.
(a) Sketch the graphs of G and G’; indicate the coordinates of the intersection point of the two graphs.
A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices on the curves G and G’ respectively. The area of this rectangle is denoted by S.

Ans
$$S=(8-2x)\sqrt{x},\, S_{max}=\frac{32\sqrt{3}}{9}$$

### Question

A point P$$(x,x^{2})$$ lies on the curve $$y=x^{2}$$. Calculate the minimum distance from the point $$A\left ( 2,-\frac{1}{2} \right )$$ to the point P . (Total 6 marks)

Ans
METHOD 1
Let $$S=\textup{AP}^{2}=(x-2)^{2}+(x^{2}+\frac{1}{2})^{2}$$                            (M1)
The graph of S is as follows:

The minimum value of S is $$2.6686$$.                                                                                                                         (G3)
Therefore the minimum distance $$=\sqrt{2.6686}=1.63\, (3\, \textup{s.f.})$$                                                            (A2)
OR
The minimum point is $$(0.682,1.63)$$                                                                                                                      (G3)
The minimum distance is $$1.63\, (3\, \textup{s.f.})$$                                                                                                                  (G2)  (C6)
METHOD 2
Let $$S=\textup{AP}^{2}=(x-2)^{2}+(x^{2}+\frac{1}{2})^{2}$$.                                                                                                            (M1)
$$\frac{\textup{d}S}{\textup{d}x}=2(x-2)+4x(x^{2}+\frac{1}{2})=4(x^{3}+x^{2}+1)$$                                                                                   (A2)
Solving $$x^{3}+x-1=0$$ gives $$x=0.68233$$                                                                                                    (G2)
Therefore, minimum distance $$=\sqrt{(0.68233-2)^{2}+(0.68233^{2}+0.5)^{2}}=1.63$$              (A1)  (C6)                                 [6]

### Question

The point $$B(a,b)$$ b) is on the curve $$f(x)=x^{2}$$ such that B is the point which is closest to A(6, 0). Calculate the value of a. (Total 6 marks)

Ans

$$b=a^{2}$$
$$AB^{2}=(a-6)^{2}+a^{4}$$                                                                                          (M1)(A1)
Minimum value of $$(x-6)^{2}+x^{4}$$ occurs at $$x=1.33$$                       (G3)
$$=> a=1.33$$                                                                                                             (G1)  (C6)                                                  [6]

### Question

A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve $$y=\textup{sin}\, x$$ where $$0\leq x\leq \pi$$.
(a) Write down an expression for the area of the rectangle.
(b) Find the maximum area of the rectangle.  (Total 3 marks)

Ans

(a) Area $$=(\pi -2x)\, \textup{sin}\, x$$.                                                         (M1)(A1)   (C2)
(b) Maximum Area $$=1.12$$ units2                                           (A1)  (C1)                                          [3]

### Question

A rectangle is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve $$y=e^{-x^{2}}$$. The area of this rectangle is denoted by A.
(a) Write down an expression for A in terms of x.
(b) Find the maximum value of A. (Total 6 marks)

(a)  $$A=2x \times e^{-x^{2}}=2xe^{-x^{2}}$$                          (M1)(A1)   (C2)
(b)  $$\frac{\textup{d}A}{\textup{d}x}=2(1-2x^{2})e^{-x^{2}}$$                                 (A2)
$$\frac{\textup{d}A}{\textup{d}x}=0\, \textup{when}\, x=\frac{1}{\sqrt{2}}$$                                       (A1)
$$A_{max}=\sqrt{2}e^{-\frac{1}{2}}\: \: (\textup{or}\, 0.858)$$                            (A1)   (C4)                                               [6]