Loading [MathJax]/jax/element/mml/optable/BasicLatin.js

User interface language: English | Español

Date November 2016 Marks available 3 Reference code 16N.1.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number H_11 Adapted from N/A

Question

Consider the function f  defined by f(x)=exsinx, 0xπ.

The curvature at any point (x, y) on a graph is defined as κ=|d2ydx2|(1+(dydx)2)32.

Show that the function f has a local maximum value when x=3π4.

[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.

The curvature at any point (x, y) on a graph is defined as κ=|d2ydx2|(1+(dydx)2)32.

[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=π2 and comment on its meaning with respect to the shape of the graph.

[2]
h.

Markscheme

dydx=e3π4(sin3π4+cos3π4)=0    R1

d2ydx2=2e3π4cos3π4<0    R1

hence maximum at x=3π4     AG

[2 marks]

c.

d2ydx2=02excosx=0    M1

x=π2    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=3π4, point of inflexion at x=π2     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

x0exsinxdx=[exsinx]π0π0excosxdx    M1A1

π0exsinxdx=[exsinx]π0([excosx]x0+π0exsinxdx)    A1

OR

π0exsinxdx=[excosx]π0+π0excosxdx    M1A1

π0exsinxdx=[excosx]π0+([exsinx]π0π0exsinxdx)    A1

THEN

π0exsinxdx=12([exsinx]x0[excosx]x0)    M1A1

π0exsinxdx=12(ex+1)    A1

[6 marks]

f.

dydx=0    (A1)

 d2ydx2=2e3π4cos3π4=2e3π4 (A1)

κ=|2e3π4|1=2e3π4    A1

[3 marks]

g.

κ=0    A1

the graph is approximated by a straight line     R1

[2 marks]

h.

Examiners report

[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.
[N/A]
g.
[N/A]
h.

Syllabus sections

Topic 5 —Calculus » SL 5.8—Testing for max and min, optimisation. Points of inflexion
Show 202 related questions
Topic 5 —Calculus

View options