IB Mathematics AI SL 5.1 Calculus Question Bank

Practise IB Mathematics AI SL 5.1 by interpreting derivatives, integrals, rates of change, area and optimisation in contextual models with technology.

Exam points

Use derivative rules and graph meaning to analyse gradients, rates of change and stationary points.
Apply definite and indefinite integration to accumulation, area and simple motion or finance models.
Use technology for roots, intersections and numerical checks while respecting domains and units.

Covered syllabus points

SL 5.1—Limits and derivative concept SL 5.2—Increasing and decreasing functions SL 5.3—Derivative of powers SL 5.4—Tangents and normals SL 5.5—Integration as anti-differentiation SL 5.6—Stationary points SL 5.7—Optimization SL 5.8—Trapezoidal rule

[Maximum number: 4]

Consider the graph of the cubic function $f(x)=x^{3}+2 x^{2}-4 x-2$ . Part of the graph of y=f(x) is shown in the following diagram.

Figure for Question 5.1 Calculus - SL content question 1 — IB Maths AI SL

(a)

Write down

[ 4 ]

(i)

the gradient of the tangent.

[ 2 ]

(ii)

the equation of the tangent.

[ 2 ]

$\quad y+3=3(x-1)$ OR y=3 x-6

[Maximum number: 5]

Consider the graph of the following function:

g(x)=\frac{8}{x}+\frac{x^{2}}{2}, \text { for } x \neq 0

(a)

Find $g^{\prime}(x)$ .

[ 3 ]

$\quad\left(g^{\prime}(x)=\right)-8 x^{-2}+x$

A1A1A1
Note: Award A1 for -8 seen, A1 for $x^{-2}$ (or $\frac{1}{x^{2}}$ ) and A1 for second term being x. Award at most A1A1AO if additional terms are seen.

(b)

Write down the interval in which g(x) is increasing.

[ 2 ]

$x>2 \mathbf{OR}(2, \infty) \mathbf{OR} \quad 2<x<\infty$

A1A1
Note: Award A1 for 2 seen and award A1 for correct inequality.

[Maximum number: 7]

A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is $h \mathrm{~cm}$ , and the top and base of the prism have sides of length $x \mathrm{~cm}$ .

Figure for Question 5.1 Calculus - SL content question 3 — IB Maths AI SL

(a)

Find an expression for $\frac{\mathrm{d} V}{\mathrm{~d} x}$ .

[ 2 ]

$\frac{\mathrm{d} V}{\mathrm{~d} x}=300 \sqrt{3}-\frac{27}{4} x^{2}$

Note: Award A1 for a correct term.

(b)

Find the value of x which maximizes the volume of the box.

[ 2 ]

from the graph of V or $\frac{\mathrm{d} V}{\mathrm{~d} x}$ OR solving $\frac{\mathrm{d} V}{\mathrm{~d} x}=0$

x=8.77(8.77382 \ldots)

(c)

Hence, or otherwise, find the maximum possible volume of the box.

The box will contain spherical chocolates. The production manager assumes that they can calculate the exact number of chocolates in each box by dividing the volume of the box by the volume of a single chocolate and then rounding down to the nearest integer.

[ 2 ]

from the graph of V OR substituting their value for x into V
$V_{\text {max }}=3040 \mathrm{~cm}^{3}(3039.34 \ldots)$

(d)

Explain why the production manager is incorrect.

[ 1 ]

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Covered syllabus points

5.1 Calculus - SL content question 1

Question (a)

Question (i)

Question (ii)

5.1 Calculus - SL content question 2

Question (a)

Question (b)

5.1 Calculus - SL content question 3

Question (a)

Question (b)

Question (c)

Question (d)