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Date May 2008 Marks available 7 Reference code 08M.2.hl.TZ2.6
Level HL only Paper 2 Time zone TZ2
Command term Find and Justify Question number 6 Adapted from N/A

Question

Consider the curve with equation f(x)=e2x2 for x<0f(x)=e2x2 for x<0 .

Find the coordinates of the point of inflexion and justify that it is a point of inflexion.

Markscheme

METHOD 1

EITHER

Using the graph of y=f(x)     (M1)

     A1

 

The maximum of f(x) occurs at x = −0.5 .     A1

OR

Using the graph of y=f(x).     (M1)

     A1

 

The zero of f(x) occurs at x = −0.5 .     A1

THEN

Note: Do not award this A1 for stating x = ±0.5 as the final answer for x .

 

f(0.5)=0.607(=e0.5)     A2

Note: Do not award this A1 for also stating (0.5, 0.607) as a coordinate.

 

EITHER

Correctly labelled graph of f(x) for x<0 denoting the maximum f(x)     R1

(e.g. f(0.6)=1.17 and f(0.4)=1.16 stated)     A1     N2

OR

Correctly labelled graph of f(x) for x<0 denoting the maximum f(x)     R1

(e.g. f(0.6)=0.857 and f(0.4)=1.05 stated)     A1     N2

OR

f(0.5)1.21. f(x)<1.21 just to the left of x=12

and f(x)<1.21 just to the right of x=12     R1

(e.g. f(0.6)=1.17 and f(0.4)=1.16 stated)     A1     N2

OR

f(x)>0 just to the left of x=12 and f(x)<0 just to the right of x=12     R1

(e.g. f(0.6)=0.857 and f(0.4)=1.05 stated)     A1     N2

[7 marks] 

METHOD 2

f(x)=4xe2x2     A1

f(x)=4e2x2+16x2e2x2(=(16x24)e2x2)     A1

Attempting to solve f(x)=0     (M1)

x=12     A1

Note: Do not award this A1 for stating x=±12 as the final answer for x .

 

f(12)=1e (=0.607)     A1

Note: Do not award this A1 for also stating (12,1e) as a coordinate.

 

EITHER

Correctly labelled graph of f(x) for x<0 denoting the maximum f(x)     R1

(e.g. f(0.6)=1.17 and f(0.4)=1.16 stated)     A1     N2

OR

Correctly labelled graph of f(x) for x<0 denoting the maximum f(x)     R1

(e.g. f(0.6)=0.857 and f(0.4)=1.05 stated)     A1     N2

OR

f(0.5)1.21. f(x)<1.21 just to the left of x=12

and f(x)<1.21 just to the right of x=12     R1

(e.g. f(0.6)=1.17 and f(0.4)=1.16 stated)     A1     N2

OR

f(x)>0 just to the left of x=12 and f(x)<0 just to the right of x=12     R1

(e.g. f(0.6)=0.857 and f(0.4)=1.05 stated)     A1     N2

[7 marks]

Examiners report

Most candidates adopted an algebraic approach rather than a graphical approach. Most candidates found f(x) correctly, however when attempting to find f(x), a surprisingly large number either made algebraic errors using the product rule or seemingly used an incorrect form of the product rule. A large number ignored the domain restriction and either expressed x=±12 as the x-coordinates of the point of inflection or identified x=12 rather than x=12. Most candidates were unsuccessful in their attempts to justify the existence of the point of inflection.

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Points of inflexion with zero and non-zero gradients.

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