IB Questionbank

User interface language: English | Español

Date	November Example question	Marks available	4	Reference code	EXN.2.SL.TZ0.6
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that and Justify	Question number	6	Adapted from	N/A

Question

A theatre set designer is designing a piece of flat scenery in the shape of a hill. The scenery is formed by a curve between two vertical edges of unequal height. One edge is $2$ metres high and the other is $1$ metre high. The width of the scenery is $6$ metres.

A coordinate system is formed with the origin at the foot of the $2$ metres high edge. In this coordinate system the highest point of the cross‐section is at $(2, 3.5)$ .

A set designer wishes to work out an approximate value for the area of the scenery $(A m^{2})$ .

In order to obtain a more accurate measure for the area the designer decides to model the curved edge with the polynomial $h (x) = a x^{3} + b x^{2} + c x + d a, b, c, d \in ℝ$ where $h$ metres is the height of the curved edge a horizontal distance $x m$ from the origin.

Explain why $A < 21$ .

[1]

By dividing the area between the curve and the $x$ ‐axis into two trapezoids of unequal width show that $A > 14.5$ , justifying the direction of the inequality.

[4]

Write down the value of $d$ .

[1]

Use differentiation to show that $12 a + 4 b + c = 0$ .

[2]

Determine two other linear equations in $a$ , $b$ and $c$ .

[3]

Hence find an expression for $h (x)$ .

[3]

Use the expression found in (f) to calculate a value for $A$ .

[2]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

The area $A$ is less than the rectangle containing the cross-section which is equal to $6 \times 3.5 = 21$ R1

Note: $6 \times 3.5 = 21$ is not sufficient for R1.

[1 mark]

$\frac{1}{2} \times 2 \times (2 + 3.5) + \frac{1}{2} \times 4 \times (3.5 + 1)$ (M1)(A1)

$= 14.5$ A1

This is an underestimate as the trapezoids are enclosed by (are under) the curve. R1

Note: This can be shown in a diagram.

[4 marks]

$h (0) = 2 \Rightarrow d = 2$ A1

[1 mark]

$h' (x) = 3 a x^{2} + 2 b x + c$ A1

$h' (2) = 0$ M1

hence $12 a + 4 b + c = 0$ AG

[2 marks]

Substitute the points $(2, 3.5)$ and $(6, 1)$ (M1)

$8 a + 4 b + 2 c + 2 = 3.5 (8 a + 4 b + 2 c = 1.5)$

and

$216 a + 36 b + 6 c + 2 = 1 (216 a + 36 b + 6 c = - 1)$ A1A1

[3 marks]

Solve on a GDC (M1)

$h (x) = 0.0365 x^{3} - 0.521 x^{2} + 1.65 x + 2$ A2

$(h (x) = 0.0364583 \dots x^{3} - 0.520833 \dots x^{2} + 1.64583 \dots x + 2)$

[3 marks]

$\int_{0}^{6} 0.0364583 \dots x^{3} - 0.520833 \dots x^{2} + 1.64583 \dots x + 2 d x$

$= 15.9 (15.9374 \dots) (m^{2})$ (M1)A1

Note: Accept $16.0 (16.014)$ from the three significant figure answer to part (g).

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » SL 5.8—Trapezoid rule

Show 21 related questions

EXM.3.AHL.TZ0.7a.ii:
With reference to the shape of the graph, explain whether your answer to part (a)(i) will be an over-estimate or an underestimate of the area.
EXM.2.AHL.TZ0.12c.ii:
Find the value of this area.
EXM.3.AHL.TZ0.7b.ii:
Write down the coefficient of determination.
22M.1.AHL.TZ2.14:
The shape of a vase is formed by rotating a curve about the $y$ -axis.

The vase is $10 cm$ high. The internal radius of the vase is measured at $2 cm$ intervals along the height:

Use the trapezoidal rule to estimate the volume of water that the vase can hold.
22M.1.SL.TZ1.6a:
Use the trapezoidal rule, with the values given in the following table, to approximate the area of the shaded region.
EXM.2.AHL.TZ0.12b.ii:
Write down the coefficient of determination.
EXM.3.AHL.TZ0.7c.ii:
Find the value of this area.
EXM.3.AHL.TZ0.7d.ii:
Hence explain how a straight line graph could be drawn using the coordinates in the table.
EXM.2.AHL.TZ0.12b.i:
Use all the coordinates in the table to find the equation of the least squares cubic regression curve.
EXM.2.AHL.TZ0.12a:
Use the trapezoidal rule to find an estimate for the area.
EXM.3.AHL.TZ0.7a.i:
Use the trapezoidal rule to find an estimate for the area.
EXM.3.AHL.TZ0.7b.i:
Use all the coordinates in the table to find the equation of the least squares cubic regression curve.
EXM.2.AHL.TZ0.12c.i:
Write down an expression for the area enclosed by the cubic regression curve, the $x$ -axis, the $y$ -axis and the line $x = 10$ .
EXM.3.AHL.TZ0.7d.i:
Show that ${\text{ln}}\,y = qx + {\text{ln}}\,p$ .
EXM.3.AHL.TZ0.7d.iii:
By finding the equation of a suitable regression line, show that $p = 1.83$ and $q = 0.986$ .
EXM.3.AHL.TZ0.7d.iv:
Hence find the area enclosed by the exponential function, the $x$ -axis, the $y$ -axis and the line $x = 4.4$ .
21M.2.AHL.TZ1.2c:
Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.
21M.2.SL.TZ1.5b:
Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.
EXM.3.AHL.TZ0.7c.i:
Write down an expression for the area enclosed by the cubic function, the $x$ -axis, the $y$ -axis and the line $x = 4.4$ .
21N.1.SL.TZ0.6a:
Use the trapezoidal rule with an interval width of $0.4$ to estimate the total amount of carbon dioxide emitted during these two hours.
21N.1.SL.TZ0.6b:
The real amount of carbon dioxide emitted during these two hours was $72$ tonnes.

Find the percentage error of the estimate found in part (a).

Hide related questions

Topic 5—Calculus

Question

Markscheme

Examiners report

Syllabus sections

View options