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Differentiation

Practice exam-style IB Math AI questions for Differentiation, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

The table shows values of a function ff for values of xx close to 22. The value of the function at x=2x=2 is also shown.

x

f(x)

1.8

4.8

1.9

4.9

1.99

4.99

2

1

2.01

5.01

2.1

5.1

2.2

5.2

A

lim⁡x→2f(x) \lim_{x\to 2}f(x).

[1]
B

the value of f(2)f(2).

[1]
C

why the answers to parts (a) and (b) are different.

[2]
Question 2
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

The depth, DD centimetres, of water in a container is modelled by

D(t)=0.02t3−0.6t2+4t+10D(t)=0.02t^3-0.6t^2+4t+10

where tt is the time in hours after measurements begin, for 0≤t≤120\leq t\leq12.

A

an expression for dDdt\dfrac{dD}{dt}.

[2]
B

the value of dDdt\dfrac{dD}{dt} when t=5t=5.

[1]
C

your answer to part (b) in context.

[2]
Question 3
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

Let

f(x)=x3−3x2−9x+5f(x)=x^3-3x^2-9x+5

where x∈Rx\in\mathbb{R}.

A

f′(x)f'(x).

[2]
B

the intervals on which ff is increasing and the interval on which ff is decreasing.

[3]
Question 4
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The curve CC has equation

y=x3−4x+1y=x^3-4x+1

The point PP on CC has xx-coordinate 22.

A

dydx\dfrac{dy}{dx} and the gradient of CC at PP.

[2]
B

the equation of the tangent to CC at PP.

[2]
C

the equation of the normal to CC at PP.

[2]

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Question 5
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A farmer uses 40 m40\ \text{m} of fencing to enclose a rectangular field beside a straight river. No fencing is required along the river. The width perpendicular to the river is x mx\ \text{m}.

A

the area of the field is given by A(x)=40x−2x2A(x)=40x-2x^2 and the feasible domain of xx.

[2]
B

the value of xx that maximizes the area.

[2]
C

the other dimension of the field and the maximum area.

[2]
Question 6
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The total cost, C(x)C(x) dollars, of producing xx items is modelled by

C(x)=0.02x3−1.2x2+30x+500C(x)=0.02x^3-1.2x^2+30x+500

for 0≤x≤400\leq x\leq40. The marginal cost is given by C′(x)C'(x).

A

an expression for the marginal cost.

[2]
B

the production levels at which the marginal cost is 1212 dollars per item.

[2]
C

the two answers from part (b) in context.

[1]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A function is defined by

f(x)=e0.3xsin⁡(2x)f(x)=e^{0.3x}\sin(2x)

where angles are measured in radians.

A

f′(x)f'(x).

[3]
B

the gradient of the graph of ff at x=1x=1.

[2]
Question 8
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

Let

f(x)=x4−8x2+3f(x)=x^4-8x^2+3
A

the values of xx for which the gradient of the graph of ff is zero.

[3]
B

the coordinates of the stationary points.

[1]
C

Classify each stationary point as a local maximum or a local minimum.

[2]

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Question 9
HL • Paper 1
Medium
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HL • Paper 1
Medium
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The function ff is defined by

f(x)=ln⁡(x2+1)x,x>0f(x)=\frac{\ln(x^2+1)}{x},\qquad x>0
A

f′(x)f'(x).

[3]
B

the equation of the tangent to the graph of ff at x=1x=1.

[3]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Let

f(x)=xe−x2/2,x∈Rf(x)=xe^{-x^2/2},\qquad x\in\mathbb{R}
A

f′(x)f'(x).

[2]
B

the coordinates of the stationary points.

[2]
C

Classify the stationary points.

[2]
Question 11
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A spherical balloon is being inflated. Its volume is increasing at a constant rate of 20 cm3 s−120\ \text{cm}^3\text{ s}^{-1}. At a particular instant, its radius is 1.5 cm1.5\ \text{cm}.

The volume and surface area of a sphere are given by

V=43πr3,A=4πr2V=\frac43\pi r^3,\qquad A=4\pi r^2
A

the rate at which the radius is increasing at this instant.

[4]
B

the rate at which the surface area is increasing at this instant.

[2]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Water is poured into an inverted conical tank at a rate of 0.400 m3 min−10.400\ \text{m}^3\text{ min}^{-1}. The tank has height 4 m4\ \text{m} and upper radius 2 m2\ \text{m}. At time tt, the water has depth h mh\ \text{m} and surface radius r mr\ \text{m}.

A

the volume of water is V=πh312V=\dfrac{\pi h^3}{12}.

[2]
B

the rate at which the water depth is increasing when h=1.20 mh=1.20\ \text{m}.

[3]

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Question 13
HL • Paper 1
Medium
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HL • Paper 1
Medium
Calculator Permitted

Let

f(x)=x5−5x4f(x)=x^5-5x^4
A

f′′(x)f''(x).

[2]
B

the intervals on which the graph is concave-up and concave-down.

[2]
C

all points of inflexion of the graph.

[2]
Question 14
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A closed cylindrical container has volume 500 cm3500\ \text{cm}^3. Its radius is r cmr\ \text{cm} and its height is h cmh\ \text{cm}. The container is made using the least possible surface area.

A

the surface area can be written as

S(r)=2πr2+1000rS(r)=2\pi r^2+\frac{1000}{r}
[2]
B

the radius and height of the container.

[4]
C

that these dimensions give a minimum surface area.

[1]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The function ff is defined by

f(x)=x4−4x3f(x)=x^4-4x^3
A

the stationary points of the graph of ff.

[2]
B

f′′(x)f''(x) and use it to classify the stationary point at x=3x=3.

[2]
C

why the second derivative test is inconclusive at x=0x=0, and classify the stationary point (0,0)(0,0).

[2]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A company models its monthly revenue, R(x)R(x) thousand dollars, by

R(x)=5xe−0.04xR(x)=5xe^{-0.04x}

where x>0x>0 is the number of hundreds of units sold.

A

R′(x)R'(x).

[2]
B

the value of xx at which the revenue is stationary.

[2]
C

that this stationary value is a maximum.

[2]
D

the maximum monthly revenue, in dollars.

[1]

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Question 17
SL • Paper 2
Medium
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SL • Paper 2
Medium
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A sensor is calibrated using the function

f(x)={x2−9x−3,x≠3,10,x=3.f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x\ne3,\\10,&x=3.\end{cases}

A table gives values of f(x)f(x) for values of xx close to 33.

x

f(x)

2.9

5.9

2.99

5.99

2.999

5.999

3

10

3.001

6.001

3.01

6.01

3.1

6.1

A
I.

Use the table to lim⁡x→3f(x)\displaystyle\lim_{x\to3}f(x).

[1]
II.

whether ff is continuous at x=3x=3. Give a reason.

[1]
B
I.

For h≠0h\ne0, simplify the secant gradient

f(3+h)−6h\frac{f(3+h)-6}{h}
[2]
II.

the gradient approached by the curve as xx approaches 33.

[1]
C

the equation of the tangent to the branch y=x2−9x−3y=\dfrac{x^2-9}{x-3} at x=4x=4.

[3]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The mass of algae, MM kilograms, in an experimental pond is modelled by

M(t)=−t3+9t2+24M(t)=-t^3+9t^2+24

where tt is the number of weeks after the experiment begins and 0≤t≤80\le t\le8.

A
I.

M′(t)M'(t).

[2]
II.

M′(4)M'(4) and the answer in context.

[2]
B

the stationary point of the model in the interior of the given domain.

[2]
C

the global maximum mass of algae during the eight weeks. your answer.

[2]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The function ff is defined by

f(x)=x3+ax2−4x+2f(x)=x^3+ax^2-4x+2

where aa is a constant. At x=1x=1, the tangent to the graph has gradient 55.

A
I.

f′(x)f'(x) in terms of aa.

[1]
II.

the value of aa.

[2]
B

the equation of the tangent at x=1x=1.

[2]
C

The normal at x=1x=1 meets the yy-axis at QQ. the coordinates of QQ.

[3]
Question 20
HL • Paper 3
Medium
Calculator Permitted
HL • Paper 3
Medium
Calculator Permitted

A sensor records a quantity using the function

F(t)={t2−25t−5,t≠5,13,t=5.F(t)=\begin{cases}\dfrac{t^2-25}{t-5},&t\ne5,\\13,&t=5.\end{cases}

The manufacturer considers recalibrating the sensor at t=5t=5.

A
I.

lim⁡t→5F(t)\displaystyle\lim_{t\to5}F(t).

[1]
II.

why FF is not continuous at t=5t=5.

[1]
B

Use the difference quotient to F′(5)F'(5) does not exist.

[2]
C

The recalibrated function GG has the same values as FF except that G(5)=10G(5)=10. the equation of the tangent to the graph of GG at t=5t=5.

[2]
D

the tt-intercept of the normal to the graph of GG at t=5t=5.

[2]

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Question 21
HL • Paper 3
Medium
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HL • Paper 3
Medium
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The flow of vehicles along a road is modelled by

q(v)=2.4v−0.03v2q(v)=2.4v-0.03v^2

where vv is the mean speed in km h−1\text{km h}^{-1} and q(v)q(v) is the flow in hundreds of vehicles per hour.

A
I.

q′(v)q'(v).

[1]
II.

the speed at which the traffic flow is stationary.

[2]
B

For the operational domain 10≤v≤7010\le v\le70, the intervals on which the flow is increasing and decreasing.

[2]
C

the maximum modelled flow, in vehicles per hour.

[1]
D

A generalized road model is qa(v)=av−0.03v2q_a(v)=av-0.03v^2, where a>0a>0. , in terms of aa, the stationary speed and the corresponding maximum value of qaq_a.

[2]
Question 22
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

An open box is made from a rectangular sheet of card measuring 30 cm30\ \text{cm} by 24 cm24\ \text{cm}. Squares of side x cmx\ \text{cm} are removed from each corner and the remaining sides are folded upwards.

A
I.

the volume of the box is

V(x)=x(30−2x)(24−2x)V(x)=x(30-2x)(24-2x)
[1]
II.

the feasible domain for xx.

[2]
B

the value of xx that gives the maximum volume.

[3]
C

the maximum volume of the box, correct to three significant figures.

[2]
Question 23
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
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A theatre charges pp dollars for a ticket. The number of tickets expected to be sold is modelled by

q=600−20pq=600-20p

where 6≤p≤306\le p\le30. The variable cost is $6 per ticket and the fixed cost is $1200.

A
I.

an expression for the revenue in terms of pp.

[1]
II.

the profit is

P(p)=−20p2+720p−4800P(p)=-20p^2+720p-4800
[2]
B

the ticket price that maximizes the profit.

[3]
C

the expected number of tickets sold and the resulting maximum profit.

[2]
Question 24
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The energy generated by a prototype solar panel during a test is modelled by

E(t)=−12t4+8t3−36t2+64tE(t)=-\frac12t^4+8t^3-36t^2+64t

where EE is measured in kilojoules, tt is measured in hours and 0≤t≤80\le t\le8.

A
I.

E′(t)E'(t).

[2]
II.

E′(t)=−2(t−2)2(t−8)E'(t)=-2(t-2)^2(t-8).

[1]
B

why the stationary point at t=2t=2 is not a local maximum or a local minimum.

[2]
C

the global maximum value of EE during the test and when it occurs.

[3]

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Question 25
SL • Paper 2
Hard
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SL • Paper 2
Hard
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The function ff is defined on −2≤x≤5-2\le x\le5 by

f(x)=x4−4x3−8x2+10f(x)=x^4-4x^3-8x^2+10
A
I.

f′(x)=4x(x−4)(x+1)f'(x)=4x(x-4)(x+1)
[1]
II.

the coordinates of all stationary points.

[2]
B

Classify each stationary point.

[3]
C

the global maximum and global minimum values of ff on the stated domain.

[2]
Question 26
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
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The height of a water jet above the ground is modelled by

h(x)=−0.02x3+0.18x2+0.6x+1h(x)=-0.02x^3+0.18x^2+0.6x+1

where hh and the horizontal distance xx are measured in metres, and 0≤x≤120\le x\le12. The polynomial is treated as a mathematical model on this interval, including where its extrapolated value is below ground.

A
I.

h′(x)h'(x).

[1]
II.

the gradient of the jet at x=4x=4.

[2]
B

the equation of the tangent to the path of the jet at x=4x=4.

[2]
C

the maximum height reached by the jet within the stated domain.

[3]
Question 27
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The concentration of a substance in a sample is modelled by

C(t)=t2e−t/2C(t)=t^2e^{-t/2}

where CC is measured in milligrams per litre, tt is measured in hours and 0≤t≤120\le t\le12.

A
I.

C′(t)C'(t).

[2]
II.

the time at which the concentration is greatest.

[1]
B

the maximum concentration.

[2]
C

the times at which the graph of CC changes concavity.

[3]
Question 28
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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The function ff is defined by

f(x)=ln⁡xx,x>0f(x)=\frac{\ln x}{x},\qquad x>0
A
I.

f′(x)f'(x).

[2]
II.

the coordinates of the stationary point.

[1]
B

f′′(x)f''(x) and classify the stationary point.

[3]
C

the point of inflexion of the graph.

[2]

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Question 29
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A cube of ice melts while retaining its cubic shape. Its volume decreases at a constant rate of 18 cm3 min−118\ \text{cm}^3\text{ min}^{-1}. Let the side length be s cms\ \text{cm}, the surface area be A cm2A\ \text{cm}^2 and the space diagonal be d cmd\ \text{cm}.

A
I.

expressions for the volume VV, surface area AA and diagonal dd in terms of ss.

[2]
II.

the value of dVdt\dfrac{dV}{dt}.

[1]
B

the rate of change of ss when s=6s=6.

[2]
C

the rates of change of the surface area and the space diagonal when s=6s=6.

[3]
Question 30
HL • Paper 2
Hard
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HL • Paper 2
Hard
Calculator Permitted

The monthly energy output of a system is modelled by

E(t)=40+10sin⁡(πt6)E(t)=40+10\sin\left(\frac{\pi t}{6}\right)

where EE is measured in megawatt-hours, tt is measured in months and 0≤t≤120\le t\le12.

A
I.

E′(t)E'(t).

[2]
II.

the maximum and minimum energy outputs during the year.

[1]
B

E′′(t)E''(t) and the point of inflexion in the interior of the domain.

[3]
C

the equation of the tangent to the graph at the point of inflexion.

[2]
Question 31
HL • Paper 3
Hard
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HL • Paper 3
Hard
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The electrical power produced by a tilting solar panel is modelled by

P(θ)=900e−0.2θsin⁡θP(\theta)=900e^{-0.2\theta}\sin\theta

where 0≤θ≤π20\le\theta\le\dfrac{\pi}{2} is measured in radians and PP is measured in watts.

A
I.

P′(θ)P'(\theta).

[2]
II.

the stationary value of θ\theta.

[2]
B

the power produced at this angle.

[2]
C

that the value found in part (a)(ii) gives the global maximum power on the stated domain.

[2]
Question 32
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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The concentration of a medicine in a patient's blood is modelled by

C(t)=12te−0.5t,t≥0C(t)=12te^{-0.5t},\qquad t\ge0

where CC is measured in mg L−1\text{mg L}^{-1} and tt is measured in hours.

A
I.

C′(t)C'(t).

[2]
II.

when the concentration is greatest.

[1]
B

the greatest concentration.

[1]
C

the graph has a point of inflexion at t=4t=4.

[2]
D

the point of inflexion in the context of the medicine concentration.

[2]

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Question 33
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A lamp is mounted 6 m6\ \text{m} above level ground. A person of height 1.8 m1.8\ \text{m} walks directly away from the lamp. At time tt, the person is x mx\ \text{m} from the lamp and the length of the person's shadow is y my\ \text{m}.

A
I.

Use similar triangles to y=37xy=\dfrac37x.

[2]
II.

The person walks at 1.40 m s−11.40\ \text{m s}^{-1}. the rate at which the shadow length is increasing.

[1]
B

the speed of the tip of the shadow.

[2]
C

For a lamp of height HH and a person of height hh, where H>hH>h, derive the speed of the shadow tip when the person walks at constant speed uu.

[2]
D

A different lamp is used with the same person. its height if the tip of the shadow always moves at twice the person's speed.

[1]
Question 34
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A cube of ice melts while retaining its cubic shape. Its side length is s cms\ \text{cm}, its volume is V cm3V\ \text{cm}^3 and its surface area is A cm2A\ \text{cm}^2. At one instant, s=4 cms=4\ \text{cm} and dVdt=−24 cm3 min−1\dfrac{dV}{dt}=-24\ \text{cm}^3\text{ min}^{-1}.

A
I.

dsdt\dfrac{ds}{dt} at this instant.

[2]
II.

dAdt\dfrac{dA}{dt} at this instant.

[2]
B

, for any positive value of ss, the instantaneous rates satisfy

dAdt=4sdVdt\frac{dA}{dt}=\frac4s\frac{dV}{dt}
[2]
C

When s=5 cms=5\ \text{cm}, the surface area is decreasing at 30 cm2 min−130\ \text{cm}^2\text{ min}^{-1}. the rate of change of the volume.

[2]
Question 35
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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An open box is made from a square sheet of card of side 30 cm30\ \text{cm} by cutting a square of side x cmx\ \text{cm} from each corner and folding up the sides. The feasible domain is 0<x<150<x<15.

A
I.

the volume is V(x)=x(30−2x)2V(x)=x(30-2x)^2.

[1]
II.

V′(x)=(30−2x)(30−6x)V'(x)=(30-2x)(30-6x).

[2]
B

the dimensions of the box with maximum volume and this volume.

[2]
C

A manufacturer instead maximizes the score

J(x)=V(x)−kx2J(x)=V(x)-kx^2

where k>0k>0 represents a waste penalty. For dimensional consistency, treat JJ as volume-valued: VV is measured in cm3\text{cm}^3 and xx in cm\text{cm}, so kk has units of cm\text{cm}. kk if the optimal cut size is x=4 cmx=4\ \text{cm}.

[2]
D

that x=4x=4 gives a local maximum of JJ for this value of kk.

[1]
Question 36
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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The daily demand for a product is modelled by

D(p)=1000e−0.05pD(p)=1000e^{-0.05p}

where pp is the selling price in dollars. The production cost is $8 per item, so the daily profit is

P(p)=(p−8)D(p),p≥8P(p)=(p-8)D(p),\qquad p\ge8
A
I.

P′(p)P'(p).

[2]
II.

the price that maximizes daily profit.

[1]
B

the corresponding maximum daily profit.

[2]
C

A tax of τ\tau dollars is imposed on each item sold. the new profit-maximizing price is p=28+τp=28+\tau.

[2]
D

the result from part (c) in terms of how the model allocates the tax.

[1]

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Question 37
HL • Paper 3
Hard
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HL • Paper 3
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A rotating beacon is located 2 km2\ \text{km} from the nearest point OO on a straight coastline. Its beam makes an angle θ\theta with the line perpendicular to the coast at OO. The illuminated point is x kmx\ \text{km} from OO along the coast. The beacon rotates so that θ\theta is increasing at a constant rate of 0.300 rad s−10.300\ \text{rad s}^{-1}.

A
I.

x=2tan⁡θx=2\tan\theta.

[1]
II.

dxdt\dfrac{dx}{dt} when θ=π4\theta=\dfrac\pi4.

[2]
B

d2xdt2\dfrac{d^2x}{dt^2} when θ=π4\theta=\dfrac\pi4.

[2]
C

For a beacon at distance dd from the coast rotating so that θ\theta increases at a constant angular speed ω>0\omega>0, derive the speed of the illuminated point.

[2]
D

the positive angle at which the illuminated point moves at twice its minimum speed.

[1]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A spotlight is located 30 m30\ \text{m} from a straight wall. Its beam rotates so that the angle θ\theta between the beam and the perpendicular to the wall increases at a constant rate of 0.080 rad s−10.080\ \text{rad s}^{-1}. The illuminated point is x mx\ \text{m} from the point on the wall nearest the spotlight.

A
I.

x=30tan⁡θx=30\tan\theta.

[1]
II.

an expression for dxdt\dfrac{dx}{dt} in terms of θ\theta.

[2]
B

the speed of the illuminated point when θ=0.600\theta=0.600 radians.

[2]
C

the acceleration d2xdt2\dfrac{d^2x}{dt^2} of the illuminated point when θ=0.600\theta=0.600 radians.

[3]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The population of bacteria in a culture is modelled by

P(t)=12001+5e−0.4tP(t)=\frac{1200}{1+5e^{-0.4t}}

where tt is measured in hours and PP is the number of bacteria.

A
I.

P′(t)P'(t).

[2]
II.

why the population is increasing for all t≥0t\ge0.

[1]
B

The rate of growth is greatest when P=600P=600. the corresponding value of tt.

[2]
C

the greatest rate of population growth and the significance of this time on the graph of PP.

[3]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function ff is defined for x≥0x\ge0 by

f(x)=(x2+1)e−xf(x)=(x^2+1)e^{-x}
A
I.

f′(x)=−(x−1)2e−xf'(x)=-(x-1)^2e^{-x}
[2]
II.

the stationary point.

[1]
B

why this stationary point is not a local maximum or local minimum.

[2]
C

all points of inflexion on the domain.

[3]

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Question 41
HL • Paper 2
Hard
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HL • Paper 2
Hard
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A rectangular page must contain a printed area of 384 cm2384\ \text{cm}^2. The side margins are each 2 cm2\ \text{cm} wide and the top and bottom margins are each 3 cm3\ \text{cm} wide. The width of the printed area is x cmx\ \text{cm}.

A
I.

the height of the printed area in terms of xx.

[1]
II.

the total area of the page is

A(x)=408+6x+1536x,x>0A(x)=408+6x+\frac{1536}{x},\qquad x>0
[2]
B

the value of xx that minimizes the total area of the page.

[3]
C

the dimensions of the page and that they give a minimum area.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The vertical profile of a designed track is modelled by

f(x)=(x2+100)e−0.1x,0≤x≤40f(x)=(x^2+100)e^{-0.1x},\qquad 0\le x\le40

where xx and f(x)f(x) denote numerical values of distances measured in metres.

A
I.

f′(x)=−0.1e−0.1x(x−10)2f'(x)=-0.1e^{-0.1x}(x-10)^2.

[2]
II.

the stationary point.

[1]
B

why this stationary point is not a local maximum or minimum.

[1]
C

f′′(x)f''(x) and the points of inflexion.

[3]
D

the equation of the tangent at the stationary point.

[1]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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The displacement of a damped vibrating component is modelled by

A(t)=e−0.4tcos⁡(2t),t≥0A(t)=e^{-0.4t}\cos(2t),\qquad t\ge0

where tt is measured in seconds.

A
I.

A′(t)A'(t).

[2]
II.

the first stationary time after t=0t=0.

[1]
B

the displacement at this stationary time.

[1]
C

Use the second derivative to classify this stationary point.

[2]
D

the first positive time at which the graph has a point of inflexion.

[2]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The mass of algae in a culture is modelled by the Gompertz function

A(t)=500exp⁡(−4e−0.3t),t≥0A(t)=500\exp\left(-4e^{-0.3t}\right),\qquad t\ge0

where AA is measured in grams and tt in days.

A
I.

A′(t)A'(t).

[2]
II.

A′′(t)A''(t) in factored form.

[1]
B

the time and mass at the point of inflexion.

[2]
C

the maximum growth rate of the algae culture.

[1]
D

For the generalized model

M(t)=Lexp⁡(−be−kt)M(t)=L\exp\left(-be^{-kt}\right)

where L,b,k>0L,b,k>0 and tt is any real number, the time, mass and growth rate at its point of inflexion.

[2]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A lifeguard is 30 m30\ \text{m} offshore from point AA on a straight beach. A swimmer is at point BB, 100 m100\ \text{m} along the beach from AA. The lifeguard swims to a landing point XX, where AX=xAX=x, and then runs from XX to BB. The swimming speed is 2 m s−12\ \text{m s}^{-1} and the running speed is 5 m s−15\ \text{m s}^{-1}.

A
I.

the total travel time is

T(x)=x2+9002+100−x5,0≤x≤100T(x)=\frac{\sqrt{x^2+900}}{2}+\frac{100-x}{5},\qquad 0\le x\le100
[1]
II.

T′(x)T'(x).

[2]
B

the value of xx that minimizes the travel time.

[2]
C

the minimum travel time and that it is a global minimum on the stated domain.

[2]
D

If the optimal landing point is interior, and the swimming and running speeds are vsv_s and vrv_r, respectively, derive the condition satisfied by the optimal landing point.

[1]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For the curve y=ln⁡xy=\ln x, where x>0x>0, let P(a,ln⁡a)P(a,\ln a) be a point on the curve. The tangent and normal at PP meet the coordinate axes.

A
I.

the equation of the tangent at PP.

[2]
II.

the xx-coordinate of the tangent's xx-intercept is X(a)=a(1−ln⁡a)X(a)=a(1-\ln a).

[1]
B

the greatest possible value of X(a)X(a) and the corresponding value of aa.

[2]
C

the yy-intercept of the normal at PP is N(a)=ln⁡a+a2N(a)=\ln a+a^2.

[1]
D

the value of aa for which the normal passes through the origin, and that this value is unique.

[2]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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A conical drinking cup has fixed slant height 12 cm12\ \text{cm}. Its circular opening has radius r cmr\ \text{cm} and its vertical height is h cmh\ \text{cm}.

A
I.

h=144−r2h=\sqrt{144-r^2}.

[1]
II.

the volume is

V(r)=π3r2144−r2,0<r<12V(r)=\frac{\pi}{3}r^2\sqrt{144-r^2},\qquad 0<r<12
[2]
B

the value of rr that maximizes the volume.

[2]
C

the corresponding height and maximum volume.

[2]
D

that this stationary value is the global maximum for 0<r<120<r<12.

[1]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Consider the family of functions

fa(x)=x4−2ax2f_a(x)=x^4-2ax^2

where aa is a real parameter. The shape of the graph changes as aa passes through zero.

A
I.

fa′(x)f_a'(x) and fa′′(x)f_a''(x).

[2]
II.

For a>0a>0, the stationary points.

[1]
B

For a>0a>0, classify each stationary point using the second derivative.

[2]
C

For a>0a>0, the points of inflexion and the intervals of concavity.

[2]
D

and how the stationary-point structure changes when a=0a=0 and when a<0a<0.

[1]

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Differential Equations

Integration