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Date	November 2021	Marks available	2	Reference code	21N.1.SL.TZ0.13
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Hence	Question number	13	Adapted from	N/A

Question

Irina uses a set of coordinate axes to draw her design of a window. The base of the window is on the $x$ -axis, the upper part of the window is in the form of a quadratic curve and the sides are vertical lines, as shown on the diagram. The curve has end points $(0, 10)$ and $(8, 10)$ and its vertex is $(4, 12)$ . Distances are measured in centimetres.

The quadratic curve can be expressed in the form $y = a x^{2} + b x + c$ for $0 \leq x \leq 8$ .

Write down the value of $c$ .

[1]

a.i.

Hence form two equations in terms of $a$ and $b$ .

[2]

a.ii.

Hence find the equation of the quadratic curve.

[2]

a.iii.

Find the area of the shaded region in Irina’s design.

[3]

Markscheme

$c = 10$ A1

[1 mark]

a.i.

$64 a + 8 b + 10 = 10$ A1

$16 a + 4 b + 10 = 12$ A1

Note: Award A1 for each equivalent expression or A1 for the use of the axis of symmetry formula to find $4 = \frac{- b}{2 a}$ or from use of derivative. Award A0A1 for $64 a + 8 b + c = 10$ and $16 a + 4 b + c = 12$ .

[2 marks]

a.ii.

$y = - \frac{1}{8} x^{2} + x + 10$ A1A1

Note: Award A1A0 if one term is incorrect, A0A0 if two or more terms are incorrect. Award at most A1A0 if correct $a, b$ and $c$ values are seen but answer not expressed as an equation.

[2 marks]

a.iii.

recognizing the need to integrate their expression (M1)

$\int_{0}^{8} - \frac{1}{8} x^{2} + x + 10 d x$ (A1)

Note: Award (A1) for correct integral, including limits. Condone absence of $d x$ .

$90.7 {cm}^{2} (\frac{272}{3}, 90.6666 \dots)$ A1

[3 marks]

Examiners report

Generally, the responses were good for this last question on the paper. The main issue here was to not give the two equations in part (a)(ii) with simplified coefficients of $a$ and $b$ . Several candidates understood what was required but left their answers with ${(8)}^{2} a$ and ${(4)}^{2} a$ un-simplified and lost marks. Some candidates used the coordinates $(0, 0)$ to substitute in the equation with an incorrect equation of $a + b = 0$ . Candidates were successful at writing the equations in part (a)(iii). In part (b), most candidates realized that they had to use integration to find the area of the shaded region and, for the most part, were able to find a correct value for the area using either the correct equation or their obtained equation from the previous part. A common error was to integrate between $0$ and $10$ instead of $0$ and $8$ .

a.i.

a.ii.

a.iii.

Syllabus sections

Topic 5—Calculus » SL 5.5—Integration introduction, areas between curve and x axis

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