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Date May 2016 Marks available 4 Reference code 16M.1.hl.TZ2.3
Level HL only Paper 1 Time zone TZ2
Command term Find Question number 3 Adapted from N/A

Question

Show that cotα=tan(π2α)cotα=tan(π2α) for 0<α<π20<α<π2.

[1]
a.

Hence find cotαtanα11+x2dx, 0<α<π2cotαtanα11+x2dx, 0<α<π2.

[4]
b.

Markscheme

EITHER

use of a diagram and trig ratios

eg,

M16/5/MATHL/HP1/ENG/TZ2/03/M

tanα=OAcotα=AOtanα=OAcotα=AO

from diagram, tan(π2α)=AOtan(π2α)=AO     R1

OR

use of tan(π2α)=sin(π2α)cos(π2α)=cosαsinαtan(π2α)=sin(π2α)cos(π2α)=cosαsinα     R1

THEN

cotα=tan(π2α)cotα=tan(π2α)     AG

[1 mark]

a.

cotαtanα11+x2dx=[arctanx]cotαtanαcotαtanα11+x2dx=[arctanx]cotαtanα     (A1)

 

Note:     Limits (or absence of such) may be ignored at this stage.

 

=arctan(cotα)arctan(tanα)=arctan(cotα)arctan(tanα)     (M1)

=π2αα=π2αα     (A1)

=π22α=π22α     A1

[4 marks]

b.

Examiners report

This was generally well done.

a.

This was generally well done. Some weaker candidates tried to solve part (b) through use of a substitution, though the standard result arctanxarctanx was well known. A small number used arctanx+carctanx+c and went on to obtain an incorrect final answer.

b.

Syllabus sections

Topic 6 - Core: Calculus » 6.5 » Anti-differentiation with a boundary condition to determine the constant of integration.

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