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Date November 2013 Marks available [N/A] Reference code 13N.2.sl.TZ0.7
Level SL only Paper 2 Time zone TZ0
Command term Find, Hence, and Write down Question number 7 Adapted from N/A

Question

Consider the graph of the semicircle given by \(f(x) = \sqrt {6x - {x^2}} \), for \(0 \leqslant x \leqslant 6\). A rectangle \(\rm{PQRS}\) is drawn with upper vertices \(\rm{R}\) and \(\rm{S}\) on the graph of \(f\), and \(\rm{PQ}\) on the \(x\)-axis, as shown in the following diagram.

Let \({\text{OP}} = x\).

(i)     Find \({\text{PQ}}\), giving your answer in terms of \(x\).

(ii)     Hence, write down an expression for the area of the rectangle, giving your answer in terms of \(x\).

[[N/A]]
a.

Find the rate of change of area when \(x = 2\).

[2]
b(i).

The area is decreasing for \(a < x < b\). Find the value of \(a\) and of \(b\).

[2]
b(ii).

Markscheme

(i)     valid approach (may be seen on diagram)     (M1)

eg     \({\text{Q}}\) to \(6\) is \(x\)

\({\text{PQ}} = 6 - 2x\)     A1     N2

(ii)     \(A = (6 - 2x)\sqrt {6x - {x^2}} \)     A1     N1

[3 marks]

a.

recognising \(\frac{{{\text{d}}A}}{{{\text{d}}x}}\) at \(x = 2\) needed (must be the derivative of area)     (M1)

\(\frac{{{\text{d}}A}}{{{\text{d}}x}} =  - \frac{{7\sqrt 2 }}{2},{\text{ }} - 4.95\)     A1     N2

[2 marks]

b(i).

\(a = 0.879{\text{  }}b = 3\)     A1A1     N2

[4 marks]

b(ii).

Examiners report

[N/A]
a.
[N/A]
b(i).
[N/A]
b(ii).

Syllabus sections

Topic 6 - Calculus » 6.3 » Applications.

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