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Date November 2014 Marks available 7 Reference code 14N.2.hl.TZ0.10
Level HL only Paper 2 Time zone TZ0
Command term Find, Justify, and State Question number 10 Adapted from N/A

Question

Consider the triangle PQR where QˆPR=30, PQ=(x+2) cm and PR=(5x)2 cm, where 2<x<5.

Show that the area, Acm2, of the triangle is given by A=14(x38x2+5x+50).

[2]
a.

(i)     State dAdx.

(ii)     Verify that dAdx=0 when x=13.

[3]
b.

(i)     Find d2Adx2 and hence justify that x=13 gives the maximum area of triangle PQR.

(ii)     State the maximum area of triangle PQR.

(iii)     Find QR when the area of triangle PQR is a maximum.

[7]
c.

Markscheme

use of A=12qrsinθ to obtain A=12(x+2)(5x)2sin30     M1

=14(x+2)(2510x+x2)     A1

A=14(x38x2+5x+50)     AG

[2 marks]

a.

(i)     dAdx=14(3x216x+5)=14(3x1)(x5)     A1

(ii)     METHOD 1

EITHER

dAdx=14(3(13)216(13)+5)=0     M1A1

OR

dAdx=14(3(13)1)((13)5)=0     M1A1

THEN

so dAdx=0 when x=13     AG

METHOD 2

solving dAdx=0 for x     M1

2<x<5x=13     A1

so dAdx=0 when x=13     AG

METHOD 3

a correct graph of dAdx versus x     M1

the graph clearly showing that dAdx=0 when x=13     A1

so dAdx=0 when x=13     AG

[3 marks]

b.

(i)     d2Adx2=12(3x8)     A1

for x=13, d2Adx2=3.5 (<0)     R1

so x=13 gives the maximum area of triangle PQR     AG

(ii)     Amax=34327 (=12.7) (cm2)     A1

(iii)     PQ=73 (cm) and PR=(143)2 (cm)     (A1)

QR2=(73)2+(143)42(73)(143)2cos30     (M1)(A1)

=391.702

QR = 19.8 (cm)     A1

[7 marks]

Total [12 marks]

c.

Examiners report

This question was generally well done. Parts (a) and (b) were straightforward and well answered.

a.

This question was generally well done. Parts (a) and (b) were straightforward and well answered.

b.

This question was generally well done. Parts (c) (i) and (ii) were also well answered with most candidates correctly applying the second derivative test and displaying sound reasoning skills.

Part (c) (iii) required the use of the cosine rule and was reasonably well done. The most common error committed by candidates in attempting to find the value of QR was to use PR=143 (cm) rather than PR=(143)2 (cm). The occasional candidate used cos30=12.

c.

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Local maximum and minimum values.
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