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Date May 2011 Marks available 8 Reference code 11M.1.hl.TZ2.13
Level HL only Paper 1 Time zone TZ2
Command term Find Question number 13 Adapted from N/A

Question

(i)     Sketch the graphs of y=sinx and y=sin2x , on the same set of axes, for 0xπ2 .

(ii)     Find the x-coordinates of the points of intersection of the graphs in the domain 0xπ2 .

(iii)     Find the area enclosed by the graphs.

[9]
a.

Find the value of 10x4xdx using the substitution x=4sin2θ .

[8]
b.

The increasing function f satisfies f(0)=0 and f(a)=b , where a>0 and b>0 .

(i)     By reference to a sketch, show that a0f(x)dx=abb0f1(x)dx .

(ii)     Hence find the value of 20arcsin(x4)dx .

[8]
c.

Markscheme

(i)

    A2

 

Note: Award A1 for correct sinx , A1 for correct sin2x .

 

Note: Award A1A0 for two correct shapes with π2 and/or 1 missing.

 

Note: Condone graph outside the domain.

 

(ii)     sin2x=sinx , 0xπ2

2sinxcosxsinx=0     M1

sinx(2cosx1)=0

x=0,π3     A1A1     N1N1

 

(iii)     area =π30(sin2xsinx)dx     M1

Note: Award M1 for an integral that contains limits, not necessarily correct, with sinx and sin2x subtracted in either order.

 

=[12cos2x+cosx]π30     A1

=(12cos2π3+cosπ3)(12cos0+cos0)     (M1)

=3412

=14     A1

[9 marks]

a.

10x4xdx=π604sin2θ44sin2θ×8sinθcosθdθ     M1A1A1 

Note: Award M1 for substitution and reasonable attempt at finding expression for dx in terms of dθ , first A1 for correct limits, second A1 for correct substitution for dx .

 

π608sin2θdθ     A1

π6044cos2θdθ     M1

=[4θ2sin2θ]π60     A1

=(2π32sinπ3)0     (M1)

=2π33     A1

[8 marks]

b.

(i)     

     M1

from the diagram above

the shaded area =a0f(x)dx=abb0f1(y)dy     R1

=abb0f1(x)dx     AG

 

(ii)     f(x)=arcsinx4f1(x)=4sinx     A1

20arcsin(x4)dx=π3π604sinxdx     M1A1A1

Note: Award A1 for the limit π6 seen anywhere, A1 for all else correct.

 

=π3[4cosx]π60     A1

=π34+23     A1

Note: Award no marks for methods using integration by parts.

 

[8 marks]

c.

Examiners report

A significant number of candidates did not seem to have the time required to attempt this question satisfactorily.

Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes.

Part (ii) was done well by most the most common error being to divide the equation by sinx and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution.

The final part was done well by many candidates.

Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully.

Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well.

a.

A significant number of candidates did not seem to have the time required to attempt this question satisfactorily.

Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes.

Part (ii) was done well by most the most common error being to divide the equation by sinx and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution.

The final part was done well by many candidates.

Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully.

Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well.

b.

A significant number of candidates did not seem to have the time required to attempt this question satisfactorily.

Part (a) was done quite well by most but a number found sketching the functions difficult, the most common error being poor labelling of the axes.

Part (ii) was done well by most the most common error being to divide the equation by sinx and so omit the x = 0 value. Many recognised the value from the graph and corrected this in their final solution.

The final part was done well by many candidates.

Many candidates found (b) challenging. Few were able to substitute the dx expression correctly and many did not even seem to recognise the need for this term. Those that did tended to be able to find the integral correctly. Most saw the need for the double angle expression although many did not change the limits successfully.

Few candidates attempted part c). Those who did get this far managed the sketch well and were able to explain the relationship required. Among those who gave a response to this many were able to get the result although a number made errors in giving the inverse function. On the whole those who got this far did it well.

c.

Syllabus sections

Topic 6 - Core: Calculus » 6.7 » Integration by substitution.
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