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Date November 2016 Marks available 3 Reference code 16N.1.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number H_11 Adapted from N/A

Question

Let y = e x sin x .

Consider the function f   defined by f ( x ) = e x sin x ,   0 x π .

The curvature at any point ( x ,   y ) on a graph is defined as κ = | d 2 y d x 2 | ( 1 + ( d y d x ) 2 ) 3 2 .

Find an expression for d y d x .

[2]
a.

Show that d 2 y d x 2 = 2 e x cos x .

[2]
b.

Show that the function f has a local maximum value when x = 3 π 4 .

[2]
c.

Find the x -coordinate of the point of inflexion of the graph of f .

[2]
d.

Sketch the graph of f , clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.

[3]
e.

Find the area of the region enclosed by the graph of f and the x -axis.

 

[6]
f.

Find the value of the curvature of the graph of f at the local maximum point.

[3]
g.

Find the value κ for x = π 2 and comment on its meaning with respect to the shape of the graph.

[2]
h.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

d y d x = e x sin x + e x cos x   ( = e x ( sin x + cos x ) )    M1A1

[2 marks]

a.

d 2 y d x 2 = e x ( sin x + cos x ) + e x ( cos x sin x )    M1A1

= 2 e x cos x    AG

[2 marks]

b.

d y d x = e 3 π 4 ( sin 3 π 4 + cos 3 π 4 ) = 0    R1

d 2 y d x 2 = 2 e 3 π 4 cos 3 π 4 < 0    R1

hence maximum at x = 3 π 4      AG

[2 marks]

c.

d 2 y d x 2 = 0 2 e x cos x = 0    M1

x = π 2    A1

 

Note: Award M1A0 if extra zeros are seen.

 

[2 marks]

d.

N16/5/MATHL/HP1/ENG/TZ0/11.e/M

correct shape and correct domain     A1

max at x = 3 π 4 , point of inflexion at x = π 2      A1

zeros at x = 0 and x = π      A1

 

Note: Penalize incorrect domain with first A mark; allow FT from (d) on extra points of inflexion.

 

[3 marks]

e.

EITHER

0 x e x sin x d x = [ e x sin x ] 0 π 0 π e x cos x d x    M1A1

0 π e x sin x d x = [ e x sin x ] 0 π ( [ e x cos x ] 0 x + 0 π e x sin x d x )    A1

OR

0 π e x sin x d x = [ e x cos x ] 0 π + 0 π e x cos x d x    M1A1

0 π e x sin x d x = [ e x cos x ] 0 π + ( [ e x sin x ] 0 π 0 π e x sin x d x )    A1

THEN

0 π e x sin x d x = 1 2 ( [ e x sin x ] 0 x [ e x cos x ] 0 x )    M1A1

0 π e x sin x d x = 1 2 ( e x + 1 )    A1

[6 marks]

f.

d y d x = 0    (A1)

  d 2 y d x 2 = 2 e 3 π 4 cos 3 π 4 = 2 e 3 π 4 (A1)

κ = | 2 e 3 π 4 | 1 = 2 e 3 π 4    A1

[3 marks]

g.

κ = 0    A1

the graph is approximated by a straight line     R1

[2 marks]

h.

Examiners report

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Syllabus sections

Topic 5—Calculus » AHL 5.16—Eulers method for 1st order DEs
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