User interface language: English | Español

Date November 2017 Marks available 5 Reference code 17N.2.AHL.TZ0.H_10
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Show that Question number H_10 Adapted from N/A

Question

Consider the function f ( x ) = x sin x ,   0 < x < π .

Consider the region bounded by the curve y = f ( x ) , the x -axis and the lines x = π 6 ,   x = π 3 .

Show that the x -coordinate of the minimum point on the curve y = f ( x ) satisfies the equation tan x = 2 x .

[5]
a.i.

Determine the values of x for which f ( x ) is a decreasing function.

[2]
a.ii.

Sketch the graph of y = f ( x ) showing clearly the minimum point and any asymptotic behaviour.

[3]
b.

Find the coordinates of the point on the graph of f where the normal to the graph is parallel to the line y = x .

[4]
c.

This region is now rotated through 2 π radians about the x -axis. Find the volume of revolution.

[3]
d.

Markscheme

attempt to use quotient rule or product rule     M1

f ( x ) = sin x ( 1 2 x 1 2 ) x cos x sin 2 x   ( = 1 2 x sin x x cos x sin 2 x )     A1A1

 

Note:     Award A1 for 1 2 x sin x or equivalent and A1 for x cos x sin 2 x or equivalent.

 

setting f ( x ) = 0     M1

sin x 2 x x cos x = 0

sin x 2 x = x cos x or equivalent     A1

tan x = 2 x     AG

[5 marks]

a.i.

x = 1.17

0 < x 1.17     A1A1

 

Note:     Award A1 for 0 < x and A1 for x 1.17 . Accept x < 1.17 .

 

[2 marks]

a.ii.

N17/5/MATHL/HP2/ENG/TZ0/10.b/M

concave up curve over correct domain with one minimum point above the x -axis.     A1

approaches x = 0 asymptotically     A1

approaches x = π asymptotically     A1

 

Note:     For the final A1 an asymptote must be seen, and π must be seen on the x -axis or in an equation.

 

[3 marks]

b.

f ( x )   ( = sin x ( 1 2 x 1 2 ) x cos x sin 2 x ) = 1     (A1)

attempt to solve for x     (M1)

x = 1.96     A1

y = f ( 1.96 )

= 1.51     A1

[4 marks]

c.

V = π π 6 π 3 x d x sin 2 x     (M1)(A1)

 

Note:     M1 is for an integral of the correct squared function (with or without limits and/or π ).

 

= 2.68   ( = 0.852 π )     A1

[3 marks]

d.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.
[N/A]
d.

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