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

Question

Consider the graph of the semicircle given by f(x)=6xx2, for 0. 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 » 6.1 » Derivative interpreted as gradient function and as rate of change.
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