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Differential Calculus

Practice exam-style IB Math AA questions for Differential Calculus, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Non Calculator
SL • Paper 1
Easy
Non Calculator

The curve CC has equation y=x3−3x2+2x+4y=x^3-3x^2+2x+4.

A

dydx\frac{dy}{dx}.

[1]
B

the coordinates of the point on CC where x=1x=1, and the gradient of the tangent to CC at this point.

[2]
C

the equation of the normal to CC at this point.

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

Let f(x)=sin⁡xf(x)=\sin x. The table shows values of

q(h)=f(2+h)−f(2)hq(h)=\frac{f(2+h)-f(2)}{h}

for values of hh close to 00.

h

q(h)

-0.10

-0.370

-0.05

-0.393

-0.02

-0.407

-0.01

-0.412

0.01

-0.421

0.02

-0.425

0.05

-0.439

0.10

-0.461

A

lim⁡h→0q(h)\displaystyle \lim_{h\to 0}q(h).

[2]
B

the answer to part (a) in terms of the graph of ff.

[1]
C

the equation of the tangent to the graph of ff at x=2x=2.

[2]
Question 3
SL • Paper 2
Easy
Calculator Permitted
SL • Paper 2
Easy
Calculator Permitted

The curve CC has equation

y=x3−4x2+2x+5y=x^3-4x^2+2x+5

The point PP on CC has xx-coordinate 33.

A

the coordinates of PP.

[1]
B

the equation of the tangent to CC at PP.

[2]
C

the equation of the normal to CC at PP.

[2]
Question 4
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let f(x)=x3+32x2−6x+1f(x)=x^3+\frac{3}{2}x^2-6x+1, for x∈Rx\in\mathbb{R}.

A

f′(x)f'(x) in factorized form.

[2]
B

the intervals on which ff is increasing and decreasing.

[2]
C

the coordinates and nature of each local extremum of ff.

[2]

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

The function ff is defined by f(x)=xe2x+1f(x)=xe^{2x}+1.

A

f′(x)f'(x).

[2]
B

the equation of the tangent to the graph of ff at the point where x=0x=0.

[3]
Question 6
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let f(x)=x4−4x3f(x)=x^4-4x^3, for x∈Rx\in\mathbb{R}.

A

f′′(x)f''(x).

[2]
B

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

[2]
C

the coordinates of the points of inflexion of the graph of ff.

[2]
Question 7
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The temperature TT of a liquid, in degrees Celsius, is modelled by T(t)=t2+4tT(t)=t^2+4t, where tt is the time in minutes.

A

the average rate of change of TT between t=1t=1 and t=1+ht=1+h, where h≠0h\ne0.

[2]
B

the instantaneous rate of change of TT when t=1t=1.

[1]
C

the equation of the tangent to the graph of TT at t=1t=1.

[2]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A spherical balloon has radius rr cm, volume VV cm3^3 and surface area SS cm2^2. At an instant when r=2r=2, the volume is increasing at a rate of 12π12\pi cm3^3 s−1^{-1}.

A

drdt\frac{dr}{dt} at this instant.

[3]
B

dSdt\frac{dS}{dt} at this instant.

[2]

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Question 9
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Let

f(x)=xe−0.4x,g(x)=ln⁡(x+2),x≥0f(x)=xe^{-0.4x},\qquad g(x)=\ln(x+2),\qquad x\geq 0
A

f′(x)f'(x).

[3]
B

the value of x≥0x\geq 0 for which the tangent to ff at xx is parallel to the tangent to gg at xx.

[3]
Question 10
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

Let

f(x)=xe−x24,x∈Rf(x)=xe^{-\frac{x^2}{4}},\qquad x\in\mathbb{R}
A

f′(x)f'(x).

[2]
B

f′′(x)=e−x24(x34−3x2)f''(x)=e^{-\frac{x^2}{4}}\left(\dfrac{x^3}{4}-\dfrac{3x}{2}\right).

[2]
C

the positive xx-coordinate of the point of inflexion of the graph of ff.

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

The function ff is defined by

f(x)={x2+kx,x<1,ax+b,x≥1.f(x)= \begin{cases} x^2+kx, & x<1,\\ ax+b, & x\geq 1. \end{cases}

The function is differentiable at x=1x=1, and f(3)=8f(3)=8.

A

three equations involving aa, bb and kk.

[3]
B

the values of aa, bb and kk.

[2]
Question 12
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The diagram shows the curve y=6−x2y=6-x^2, for 0≤x≤60\le x\le \sqrt{6}. A rectangle has one vertex at the origin, sides parallel to the axes, and its upper right vertex on the curve. The width of the rectangle is xx and its area is AA.

A

Express AA in terms of xx.

[2]
B

the value of xx for which AA is a maximum.

[3]
C

the maximum area of the rectangle.

[1]

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

Let f(x)=x3−2xf(x)=x^3-2x.

A

Use first principles to f′(x)f'(x).

[4]
B

the equation of the tangent to the graph of ff at the point where x=−1x=-1.

[2]
Question 14
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Consider the limit

lim⁡x→0ex−1−xx2\lim_{x\to0}\frac{e^x-1-x}{x^2}
A

direct substitution gives an indeterminate form.

[1]
B

Use l'Hopital's rule to evaluate the limit.

[4]
Question 15
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The curve CC is defined implicitly by

x2+xy+y2=7x^2+xy+y^2=7

The point P(1,2)P(1,2) lies on CC.

A

dydx\frac{dy}{dx} in terms of xx and yy.

[3]
B

the equation of the tangent to CC at PP.

[2]
C

the equation of the normal to CC at PP.

[1]
Question 16
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let

f(x)=3x+arcsin⁡(2x−1)f(x)=3^x+\arcsin(2x-1)

where 12≤x≤1\frac{1}{2}\le x\le1.

A

f′(x)f'(x).

[3]
B

the equation of the tangent to the graph of ff at x=12x=\frac{1}{2}, using the right-hand derivative at this endpoint.

[2]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function ff is defined by

f(x)=x3−3x,−2≤x≤1f(x)=x^3-3x,\qquad -2\le x\le1
A

the stationary points of ff in the interval −2≤x≤1-2\le x\le1.

[2]
B

the absolute maximum and absolute minimum values of ff on this interval.

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

Consider the function

h(x)=x4−6x2+3x+2,−3≤x≤3h(x)=x^4-6x^2+3x+2,\qquad -3\leq x\leq 3
A

the xx-coordinates of the stationary points of hh.

[3]
B

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

[2]
C

the absolute maximum value of hh on the interval [−3,3][-3,3].

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

A rectangular printed region has area 600 cm2600\ \text{cm}^2. It is surrounded by margins of 2 cm2\ \text{cm} on the left and right, and 3 cm3\ \text{cm} at the top and bottom. Let xx be the width, in cm, of the printed region.

A

the total area A cm2A\ \text{cm}^2 of the paper is

A=624+6x+2400xA=624+6x+\frac{2400}{x}
[2]
B

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

[3]
C

the minimum total area of the paper.

[1]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let

p(x)=x3+2xp(x)=x^3+2x

and

r(x)=p′(x)e2xr(x)=p'(x)e^{2x}
A

Use first principles to p′(x)=3x2+2p'(x)=3x^2+2.

[4]
B

r′′(0)r''(0).

[2]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the following limits.

A

Evaluate

lim⁡x→0ex−1−x−x22x3\lim_{x\to0}\frac{e^x-1-x-\frac{x^2}{2}}{x^3}
[4]
B

the horizontal asymptote of

y=5x+ex2ex−xy=\frac{5x+e^x}{2e^x-x}

as x→∞x\to\infty.

[2]
Question 22
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Water is poured into a right circular cone. At all times the radius rr of the water surface and the depth hh of the water satisfy

r=25hr=\frac{2}{5}h

The volume of water is increasing at a rate of 0.080 m3 min−10.080\ \text{m}^3\text{ min}^{-1}.

A

the volume of water is

V=4π75h3V=\frac{4\pi}{75}h^3
[1]
B

an expression for dVdt\dfrac{dV}{dt} in terms of hh and dhdt\dfrac{dh}{dt}.

[2]
C

dhdt\dfrac{dh}{dt} when h=6h=6.

[2]
D

drdt\dfrac{dr}{dt} when h=6h=6.

[1]
Question 23
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let

f(x)=arctan⁡(2x−1)+3x,0≤x≤2f(x)=\arctan(2x-1)+3^x,\qquad 0\leq x\leq 2
A

f′(x)f'(x).

[3]
B

the value of xx for which the tangent to the graph of ff has gradient 55.

[1]
C

the equation of this tangent.

[2]
Question 24
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let f(x)=x3−6x2+9x+1f(x)=x^3-6x^2+9x+1, where x∈Rx\in\mathbb{R}.

A

Part (a)

I.

f′(x)f'(x).

[2]
II.

the xx-coordinates of the stationary points of the graph of ff.

[2]
B

Part (b)

I.

the coordinates and nature of each stationary point.

[4]
II.

the intervals on which ff is increasing and decreasing.

[2]
C

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

[3]

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Question 25
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=(x+1)e−xf(x)=(x+1)e^{-x}, for x≥−1x\geq -1.

A
I.

f′(x)=−xe−xf'(x)=-xe^{-x}.

[3]
II.

f′′(x)f''(x).

[2]
B

the coordinates and nature of the stationary point of the graph of ff.

[3]
C

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

[2]
Question 26
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

Let f(x)=x2ln⁡xf(x)=x^2\ln x, where x>0x>0.

A
I.

f′(x)f'(x).

[3]
B
I.

the xx-coordinate of the stationary point of the graph of ff.

[2]
II.

the coordinates and nature of this stationary point.

[4]
C

the equation of the tangent to the graph of ff at the point where x=ex=e.

[2]
Question 27
SL • Paper 1
Medium
Non Calculator
SL • Paper 1
Medium
Non Calculator

The curve CC has equation y=x+4xy=x+\frac{4}{x}, where x>0x>0.

A
I.

dydx\frac{dy}{dx}.

[2]
II.

the coordinates and nature of the stationary point of CC.

[2]
B

A tangent to CC is parallel to the line y=−3x+5y=-3x+5. the equation of this tangent.

[3]
C

the range of yy for points on CC.

[2]
D

the equation of the tangent to CC at the point where x=4x=4.

[2]
Question 28
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The rate R(t)R(t), in litres per hour, at which water flows through a filter is modelled by R(t)=0.04t3−0.9t2+5t+12R(t)=0.04t^3-0.9t^2+5t+12, where tt is the time in hours after the filter is switched on and 0≤t≤180\le t\le18.

A
I.

R′(t)R'(t).

[1]
II.

R′(6)R'(6) in the context of the model.

[2]
B

the times at which RR has stationary points, and the nature of each stationary point.

[3]
C

the maximum flow rate predicted by the model for 0≤t≤180\le t\le18.

[2]

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Question 29
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A closed cylindrical container has radius r cmr\ \text{cm} and height h cmh\ \text{cm}. Its volume is fixed at 750 cm3750\ \text{cm}^3. The total surface area is denoted by S cm2S\ \text{cm}^2.

A

S=2πr2+1500rS=2\pi r^2+\frac{1500}{r}
[2]
B
I.

dSdr\dfrac{dS}{dr}.

[1]
II.

the value of rr which minimizes SS.

[3]
C

the height and the minimum total surface area of the container.

[2]
Question 30
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A company models its weekly profit, in hundreds of dollars, by P(x)=−0.02x3+1.2x2+15x−200P(x)=-0.02x^3+1.2x^2+15x-200 where xx is the number of units produced in one week, in hundreds, and 0≤x≤700\le x\le70.

A
I.

P′(x)P'(x).

[1]
II.

the meaning of P′(30)=33P'(30)=33.

[1]
B

the interval on which the profit is increasing.

[3]
C

the maximum weekly profit predicted by the model.

[3]
Question 31
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The curve CC is defined implicitly by

x2+y2+xy+ey=5x^2+y^2+xy+e^y=5

The point P(2,0)P(2,0) lies on CC.

A

dydx=−2x+y2y+x+ey\frac{dy}{dx}=-\frac{2x+y}{2y+x+e^y}
[3]
B

the equation of the normal to CC at PP.

[2]
C

the coordinates of the second point where this normal intersects CC.

[2]
Question 32
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

A square sheet of card has side length 12 cm12\ \text{cm}. Squares of side length x cmx\ \text{cm} are cut from each corner, and the remaining card is folded to make an open rectangular box. The volume of the box is V cm3V\ \text{cm}^3.

A
I.

V=x(12−2x)2V=x(12-2x)^2.

[3]
II.

the possible values of xx.

[1]
B
I.

dVdx\frac{dV}{dx} in factorized form.

[3]
II.

the value of xx for which the volume is a maximum. your answer.

[3]
C

the maximum volume of the box.

[1]

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Question 33
SL • Paper 1
Hard
Non Calculator
SL • Paper 1
Hard
Non Calculator

The function ff is defined by f(x)=x+2cos⁡xf(x)=x+2\cos x, for 0≤x≤π0\leq x\leq \pi.

A
I.

f′(x)f'(x) and f′′(x)f''(x).

[3]
B

the xx-coordinates of the stationary points of the graph of ff.

[2]
C

Classify the stationary points as local maxima or local minima.

[2]
D

the absolute maximum and absolute minimum values of ff on 0≤x≤π0\leq x\leq \pi.

[3]
Question 34
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let p(x)=x3+3x2−1p(x)=x^3+3x^2-1, where x∈Rx\in\mathbb{R}.

A
I.

Use first principles to p′(x)=3x2+6xp'(x)=3x^2+6x.

[4]
II.

the equation of the tangent to the graph of pp at the point where x=−1x=-1.

[2]
B
I.

p′′(x)p''(x) and p(3)(x)p^{(3)}(x).

[2]
II.

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

[2]
C

the coordinates and nature of the stationary points of the graph of pp.

[2]
Question 35
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For a∈Ra\in\mathbb{R}, consider L(a)=lim⁡x→0e2x−1−2x−ax2x2L(a)=\lim_{x\to0}\frac{e^{2x}-1-2x-ax^2}{x^2}

A
I.

direct substitution gives an indeterminate form.

[1]
II.

Use l'Hopital's rule to L(a)=2−aL(a)=2-a.

[4]
B

the value of aa for which L(a)=0L(a)=0.

[1]
C
I.

Evaluate lim⁡x→∞x2+ex3ex−2x\lim_{x\to\infty}\frac{x^2+e^x}{3e^x-2x}

[2]
II.

the horizontal asymptote of y=x2+ex3ex−2xy=\frac{x^2+e^x}{3e^x-2x} as x→∞x\to\infty.

[2]
Question 36
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let h(x)=arctan⁡x−x2h(x)=\arctan x-\frac{x}{2}, where x∈Rx\in\mathbb{R}.

A
I.

h′(x)h'(x).

[2]
II.

the xx-coordinates of the stationary points of the graph of hh.

[2]
B

the coordinates and nature of each stationary point.

[3]
C

the intervals on which hh is increasing and decreasing.

[1]
D

the equation of the normal to the graph of hh at the origin.

[2]

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Question 37
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A sensor reading is modelled by f(x)=xe−0.3x+0.5sin⁡xf(x)=xe^{-0.3x}+0.5\sin x for 0≤x≤100\le x\le10, where xx is measured in seconds.

A
I.

f′(x)=e−0.3x(1−0.3x)+0.5cos⁡xf'(x)=e^{-0.3x}(1-0.3x)+0.5\cos x.

[2]
II.

the equation of the tangent to the graph of ff at x=2x=2.

[3]
B

all values of xx in the interval 0≤x≤100\le x\le10 for which the tangent to the graph of ff is horizontal.

[2]
C

the intervals on which ff is increasing.

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

A function ff has derivative f′(x)=0.1(x+3)(x−1)2(5−x)f'(x)=0.1(x+3)(x-1)^2(5-x) for −4≤x≤6-4\le x\le6.

A
I.

the xx-values at which ff has stationary points.

[1]
II.

the intervals on which ff is increasing.

[2]
B

the nature of each stationary point of ff.

[2]
C

the xx-coordinates of the points of inflexion of ff.

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

The function ff is defined by f(x)=ln⁡(x+1)x+1,x>0f(x)=\frac{\ln(x+1)}{x+1},\qquad x>0

A
I.

f′(x)f'(x).

[3]
B

the equation of the tangent to the graph of ff at x=3x=3.

[3]
C

the coordinates of the stationary point of the graph of ff.

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

A drone is moving in a vertical plane. At time tt, its height above level ground is h mh\ \text{m} and its horizontal distance from an observer is x mx\ \text{m}. The angle of elevation of the drone from the observer is θ\theta, where

tan⁡θ=hx\tan\theta=\frac{h}{x}

At a particular instant, h=120h=120, x=160x=160, dhdt=3\dfrac{dh}{dt}=3 and dxdt=−4\dfrac{dx}{dt}=-4, with distances measured in metres and time in seconds.

A
I.

sec⁡2θdθdt=xdhdt−hdxdtx2\sec^2\theta\frac{d\theta}{dt}=\frac{x\frac{dh}{dt}-h\frac{dx}{dt}}{x^2}
[3]
II.

dθdt\dfrac{d\theta}{dt} at this instant.

[1]
B

Let ss be the distance from the observer to the drone. dsdt\dfrac{ds}{dt} at this instant.

[3]
C

the sign of your answer to part (b).

[1]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a>0a>0, consider the family of functions

fa(x)=x4−4ax3+2a2x2,x∈Rf_a(x)=x^4-4ax^3+2a^2x^2,\qquad x\in\mathbb{R}

The diagram shows the graph of faf_a for one positive value of aa, with its three stationary points indicated.

A
I.

Use first principles to if p(x)=x3p(x)=x^3, then p′(x)=3x2p'(x)=3x^2.

[3]
II.

fa′(x)f_a'(x) and fa′′(x)f_a''(x).

[2]
B
I.

For a=1a=1, the xx-coordinates of the stationary points of f1f_1.

[2]
II.

Classify the three stationary points for a=1a=1.

[2]
C
I.

By writing x=atx=at, fa(at)=a4(t4−4t3+2t2)f_a(at)=a^4(t^4-4t^3+2t^2).

[1]
II.

the geometrical transformation that maps the graph of f1f_1 onto the graph of faf_a.

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

A student investigates limits of quotients near x=0x=0 using a GDC. For real mm, define

Lm=lim⁡x→0ln⁡(1+x)−x+mx21−cos⁡xL_m=\lim_{x\to0}\frac{\ln(1+x)-x+mx^2}{1-\cos x}

The table generated by the student suggests that Lm=3L_m=3 for one value of mm.

x

Quotient

-0.100

2.9304

-0.050

2.9660

-0.010

2.9933

-0.001

2.9993

0.001

3.0007

0.010

3.0066

0.050

3.0328

0.100

3.0646

A
I.

direct substitution in the quotient defining LmL_m gives an indeterminate form.

[1]
II.

Using Maclaurin series, Lm=2m−1L_m=2m-1.

[3]
B

the value of mm suggested by the table.

[2]
C
I.

For m=2m=2, Lm=3L_m=3 using l'Hopital's rule.

[3]
II.

For k>0k>0, define

Mk=lim⁡x→0ekx−1−kxln⁡(1+x)−xM_k=\lim_{x\to0}\frac{e^{kx}-1-kx}{\ln(1+x)-x}

kk if Mk=−9M_k=-9.

[3]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A differentiable function FF is defined on R\mathbb{R} and satisfies

F′(x)=e−x(x2−4),F(0)=1F'(x)=e^{-x}(x^2-4),\qquad F(0)=1

The graph of F′F' is shown.

A
I.

the intervals on which FF is increasing and decreasing.

[2]
II.

Classify the stationary points of FF.

[2]
B

points of inflexion of FF occur where x=1±5x=1\pm\sqrt5. You are not required to the corresponding yy-coordinates.

[4]
C
I.

For c>0c>0, suppose Gc′(x)=e−x(x2−c2)G_c'(x)=e^{-x}(x^2-c^2). the xx-coordinates of the stationary points of GcG_c.

[1]
II.

the horizontal separation between the two points of inflexion of GcG_c is always 2c2+12\sqrt{c^2+1}.

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

Two autonomous vehicles move on straight roads that meet at right angles at a junction OO. Vehicle AA moves east along the horizontal road and vehicle BB moves south along the vertical road. At time tt seconds after noon,

x=40+6t,y=150−8tx=40+6t,\qquad y=150-8t

where xx and yy are the distances in metres of AA and BB from OO, respectively. The model is used while 0≤t≤180\le t\le18.

A
I.

an expression for the square of the distance DD between the vehicles.

[1]
II.

dDdt\dfrac{dD}{dt} in terms of xx, yy and DD.

[3]
B

the rate of change of the distance between the vehicles when t=10t=10.

[3]
C
I.

the vehicles are closest when t=9.60t=9.60, correct to three significant figures.

[3]
II.

the minimum distance between the vehicles.

[2]

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Question 45
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The curve CC is defined implicitly by x2+2y2−xy=8x^2+2y^2-xy=8 The point P(2,2)P(2,2) lies on CC.

A
I.

(a)(i) dydx=y−2x4y−x\frac{dy}{dx}=\frac{y-2x}{4y-x}

[3]
II.

(a)(ii) the equation of the tangent to CC at PP.

[2]
B
I.

(b)(i) the points on CC where the tangent is horizontal.

[3]
II.

(b)(ii) the points on CC where the tangent is vertical.

[3]
C

why no point on CC has both a horizontal and a vertical tangent.

[1]
Question 46
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A point (x,y)(x,y) moves in the first quadrant on the ellipse x2+4y2=100x^2+4y^2=100 At a certain instant, x=6x=6, y=4y=4 and dxdt=−2 cm s−1\frac{dx}{dt}=-2\ \text{cm s}^{-1}. A rectangle is formed with vertices (±x,±y)(\pm x,\pm y), so its area is A cm2A\ \text{cm}^2.

A
I.

dydx\frac{dy}{dx} in terms of xx and yy.

[2]
II.

dydt\frac{dy}{dt} at the instant when x=6x=6 and y=4y=4.

[3]
B
I.

Express AA in terms of xx only.

[2]
II.

the maximum possible area of the rectangle.

[4]
C

dAdt\frac{dA}{dt} at the instant when x=6x=6 and y=4y=4.

[1]
Question 47
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The function ff is defined by

f(x)={ax2+bx+1,x<1,cln⁡x+x,x≥1,f(x)=\begin{cases} ax^2+bx+1, & x<1,\\ c\ln x+x, & x\geq 1,\end{cases}

where aa, bb and cc are constants. The function is differentiable at x=1x=1, and f′(2)=2f'(2)=2.

A
I.

three equations involving aa, bb and cc.

[3]
II.

the values of aa, bb and cc.

[3]
B

Using the values found in part (a), the coordinates and nature of any stationary point of the graph of ff.

[3]
C

whether the graph of ff has a point of inflexion at x=1x=1. your answer.

[2]
Question 48
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let f(x)=x4−3x2f(x)=x^4-3x^2

A
I.

Use first principles to f′(x)=4x3−6xf'(x)=4x^3-6x.

[4]
B

f(3)(x)f^{(3)}(x) and f(4)(x)f^{(4)}(x).

[2]
C

Let p(x)=x4−3x2+ae2xp(x)=x^4-3x^2+ae^{2x} the value of aa for which p(4)(0)=p′′(0)p^{(4)}(0)=p''(0).

[2]

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Question 49
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For x≠0x\ne0, define

F(x)=sin⁡x−xcos⁡xx3F(x)=\frac{\sin x-x\cos x}{x^3}
A
I.

direct substitution in lim⁡x→0F(x)\displaystyle\lim_{x\to0}F(x) gives an indeterminate form.

[1]
II.

Use l'Hopital's rule to evaluate lim⁡x→0F(x)\displaystyle\lim_{x\to0}F(x).

[3]
B

A new function GG is defined by G(x)=F(x)G(x)=F(x) for x≠0x\ne0 and G(0)=kG(0)=k. the value of kk for which GG is continuous at x=0x=0.

[1]
C

Using G(0)=13G(0)=\dfrac13, the absolute minimum value of GG on −2≤x≤2-2\le x\le2.

[3]
Question 50
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The curve CC is defined implicitly by x2y+y3=10x^2y+y^3=10 The point P(1,2)P(1,2) lies on CC.

A
I.

dydx=−2xyx2+3y2\frac{dy}{dx}=-\frac{2xy}{x^2+3y^2}

[3]
B

the equation of the tangent to CC at PP.

[2]
C

d2ydx2\dfrac{d^2y}{dx^2} at PP and whether the curve is concave-up or concave-down at PP.

[3]
Question 51
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function ff is defined by f(x)=2xarctan⁡x+log⁡3(x+2),x>−2f(x)=2^x\arctan x+\log_3(x+2),\qquad x>-2

A
I.

f′(x)f'(x).

[3]
B

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

[2]
C

The tangent to the graph of ff at x=ax=a has gradient 22, where 0<a<10<a<1. the equation of this tangent.

[3]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The function ff is defined by

f(x)={x2+px+q,x<2,rex−2+s(x−2),x≥2.f(x)=\begin{cases}x^2+px+q, & x<2,\\ re^{x-2}+s(x-2), & x\ge2.\end{cases}

The function is differentiable at x=2x=2, and f(0)=5f(0)=5, f(4)=10f(4)=10.

A
I.

an equation involving qq.

[1]
II.

Use continuity and differentiability at x=2x=2 to two further equations.

[2]
III.

the values of pp, qq, rr and ss.

[2]
B

f′(x)f'(x).

[2]
C

whether ff has any local minimum points.

[3]

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Question 53
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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The curve CC is given in the first quadrant by

x23+y23=a23,a>0x^{\frac23}+y^{\frac23}=a^{\frac23},\qquad a>0

A point PP on CC may be written as P(acos⁡3t,asin⁡3t)P(a\cos^3 t,a\sin^3 t), where 0<t<π20<t<\frac{\pi}{2}.

A

Use implicit differentiation to

dydx=−(yx)13\frac{dy}{dx}=-\left(\frac{y}{x}\right)^{\frac13}
[3]
B
I.

the tangent to CC at P(acos⁡3t,asin⁡3t)P(a\cos^3 t,a\sin^3 t) has equation

xcos⁡t+ysin⁡t=a\frac{x}{\cos t}+\frac{y}{\sin t}=a
[3]
II.

the intercepts made by this tangent with the coordinate axes.

[1]
C
I.

A rectangle with sides parallel to the coordinate axes has opposite vertices at the origin and at PP. its area is A(t)=a2cos⁡3tsin⁡3tA(t)=a^2\cos^3t\sin^3t.

[1]
II.

the maximum possible area of the rectangle, giving the corresponding coordinates of PP.

[3]
D

In the case a=8a=8, a particle moves along CC in the first quadrant. At an instant when x=33x=3\sqrt3 and y=1y=1, xx is decreasing at 0.060 m s−10.060\ \text{m s}^{-1}. dydt\dfrac{dy}{dt} at this instant.

[3]
Question 54
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a>0a>0, define

fa(x)=arctan⁡(ax)−ln⁡(x+1),x>0f_a(x)=\arctan(ax)-\ln(x+1),\qquad x>0

This question investigates how the number of stationary points of faf_a depends on aa.

A
I.

fa′(x)f_a'(x).

[2]
II.

stationary points satisfy

a2x2−ax+1−a=0a^2x^2-ax+1-a=0
[2]
B

For a=2a=2, the exact xx-coordinate of the stationary point in the domain x>0x>0.

[3]
C
I.

the discriminant of the quadratic in part (a)(ii) is a2(4a−3)a^2(4a-3).

[2]
II.

the number of stationary points of faf_a in the domain x>0x>0 for all a>0a>0.

[4]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a positive integer nn, define

fn(x)=xne−x,x>0f_n(x)=x^ne^{-x},\qquad x>0

These curves are used to model quantities that rise from zero and then decay.

A
I.

fn′(x)f_n'(x).

[2]
II.

the xx-coordinate of the maximum point of fnf_n.

[2]
B
I.

fn′′(x)=e−xxn−2(x2−2nx+n(n−1))f_n''(x)=e^{-x}x^{n-2}\left(x^2-2nx+n(n-1)\right)
[3]
II.

the xx-coordinates of the points of inflexion of fnf_n, taking into account the domain x>0x>0.

[2]
C

For n=9n=9, the interval on which f9f_9 is concave-down and its position relative to the maximum point.

[4]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For c>0c>0, define

sc(x)=x2+c2,x∈Rs_c(x)=\sqrt{x^2+c^2},\qquad x\in\mathbb{R}

These functions are sometimes used as smooth approximations to ∣x∣|x|.

A
I.

sc′(x)s_c'(x).

[2]
II.

sc′(0)s_c'(0) and its meaning geometrically.

[2]
B
I.

For fixed x≠0x\ne0, lim⁡c→0+sc′(x)\displaystyle \lim_{c\to0^+}s_c'(x).

[2]
II.

why this does not that ∣x∣|x| is differentiable at x=0x=0.

[2]
C

sc′′(x)s_c''(x) and the value of xx at which the concavity of scs_c is greatest.

[4]

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Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let

f(x)=x+ex,x∈Rf(x)=x+e^x,\qquad x\in\mathbb{R}

The inverse function of ff is denoted by gg. This question investigates derivatives of gg without finding an explicit formula for gg.

A
I.

ff has an inverse function.

[2]
II.

the value of g(1)g(1).

[1]
B
I.

g′(1)g'(1).

[2]
II.

g′′(1)=−18g''(1)=-\dfrac18.

[2]
C

Use the tangent line to y=g(x)y=g(x) at x=1x=1 to g(1.1)g(1.1).

[2]
D

that, for any twice differentiable function ff with inverse gg and with f′(x)≠0f'(x)\ne0,

g′′(f(x))=−f′′(x)(f′(x))3g''(f(x))=-\frac{f''(x)}{(f'(x))^3}
[4]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For real aa, consider

Qa(x)=cos⁡x−1+x22+ax4x4,x≠0Q_a(x)=\frac{\cos x-1+\frac{x^2}{2}+ax^4}{x^4},\qquad x\ne0

A GDC graph of QaQ_a near x=0x=0 is used to whether QaQ_a has a removable discontinuity at x=0x=0.

A
I.

Use a Maclaurin series to

lim⁡x→0Qa(x)=a+124\lim_{x\to0}Q_a(x)=a+\frac{1}{24}
[3]
II.

the value of aa for which the limit is 00.

[2]
B

For a=0a=0, evaluate lim⁡x→0Q0(x)\displaystyle \lim_{x\to0}Q_0(x) using repeated l'Hopital's rule.

[4]
C
I.

Define Qa(0)Q_a(0) so that QaQ_a is continuous at x=0x=0.

[1]
II.

why this definition removes the discontinuity at x=0x=0.

[2]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For 0≤x≤30\le x\le3, let CC be the curve

y=arctan⁡xy=\arctan x

For a fixed real number aa, the point P(a,0)P(a,0) lies on the xx-axis. A point QQ on CC has coordinates (x,arctan⁡x)(x,\arctan x). The distance PQPQ is to be minimized.

A
I.

minimizing PQPQ is equivalent to minimizing

D(x)=(x−a)2+(arctan⁡x)2D(x)=(x-a)^2+(\arctan x)^2
[1]
II.

an interior stationary point satisfies

a=x+arctan⁡x1+x2a=x+\frac{\arctan x}{1+x^2}
[3]
B

For a=2a=2, the coordinates of the point QQ on CC closest to PP.

[4]
C
I.

Let

ϕ(x)=x+arctan⁡x1+x2\phi(x)=x+\frac{\arctan x}{1+x^2}

ϕ\phi is increasing on 0≤x≤30\le x\le3.

[3]
II.

the values of aa for which the closest point occurs at the endpoint x=3x=3.

[3]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

Let

f(x)=e2xsin⁡xf(x)=e^{2x}\sin x

Repeated differentiation of ff produces expressions of the form Rne2xsin⁡(x+αn)R_ne^{2x}\sin(x+\alpha_n). This question investigates this pattern.

A

(a)

I.

f′(x)f'(x).

[2]
II.

f′′(x)f''(x).

[2]
B

(b)

I.

Let θ=arctan⁡(12)\theta=\arctan\left(\dfrac12\right).

f′(x)=5e2xsin⁡(x+θ)f'(x)=\sqrt5e^{2x}\sin(x+\theta)
[2]
II.

by induction that, for all integers n≥0n\ge0,

f(n)(x)=5n2e2xsin⁡(x+nθ)f^{(n)}(x)=5^{\frac n2}e^{2x}\sin(x+n\theta)
[3]
C

Use your result to f(6)(0)f^{(6)}(0) in exact form.

[2]
D

Using a GDC or otherwise, the least positive integer nn for which ∣f(n)(0)∣>1000|f^{(n)}(0)|>1000.

[2]

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Differential Equations