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Date May 2016 Marks available 4 Reference code 16M.2.hl.TZ2.11
Level HL only Paper 2 Time zone TZ2
Command term Show that Question number 11 Adapted from N/A

Question

Points A , B and T lie on a line on an indoor soccer field. The goal, [AB] , is 2 metres wide. A player situated at point P kicks a ball at the goal. [PT] is perpendicular to (AB) and is 6 metres from a parallel line through the centre of [AB] . Let PT be x metros and let α=AˆPB measured in degrees. Assume that the ball travels along the floor.

M16/5/MATHL/HP2/ENG/TZ2/11

The maximum for tanα gives the maximum for α.

Find the value of α when x=10.

[4]
a.

Show that tanα=2xx2+35.

[4]
b.

(i)     Find ddx(tanα).

(ii)     Hence or otherwise find the value of α such that ddx(tanα)=0.

(iii)     Find d2dx2(tanα) and hence show that the value of α never exceeds 10°.

[11]
c.

Find the set of values of x for which α.

[3]
d.

Markscheme

EITHER

\alpha  = \arctan \frac{7}{{10}} - \arctan \frac{5}{{10}}{\text{ }}( = 34.992 \ldots ^\circ  - 26.5651 \ldots ^\circ )     (M1)(A1)(A1)

Note:     Award (M1) for \alpha  = {\rm{A\hat PT}} - {\rm{B\hat PT}}, (A1) for a correct {\rm{A\hat PT}} and (A1) for a correct {\rm{B\hat PT}}.

OR

\alpha  = \arctan {\text{ }}2 - \arctan \frac{{10}}{7}{\text{ }}( = 63.434 \ldots ^\circ  - 55.008 \ldots ^\circ )     (M1)(A1)(A1)

Note:     Award (M1) for \alpha  = {\rm{P\hat BT}} - {\rm{P\hat AT}}, (A1) for a correct {\rm{P\hat BT}} and (A1) for a correct {\rm{P\hat AT}}.

OR

\alpha  = \arccos \left( {\frac{{125 + 149 - 4}}{{2 \times \sqrt {125}  \times \sqrt {149} }}} \right)     (M1)(A1)(A1)

Note:     Award (M1) for use of cosine rule, (A1) for a correct numerator and (A1) for a correct denominator.

THEN

= 8.43^\circ    A1

[4 marks]

a.

EITHER

\tan \alpha  = \frac{{\frac{7}{x} - \frac{5}{x}}}{{1 + \left( {\frac{7}{x}} \right)\left( {\frac{5}{x}} \right)}}    M1A1A1

Note:     Award M1 for use of \tan (A - B), A1 for a correct numerator and A1 for a correct denominator.

= \frac{{\frac{2}{x}}}{{1 + \frac{{35}}{{{x^2}}}}}    M1

OR

\tan \alpha  = \frac{{\frac{x}{5} - \frac{x}{7}}}{{1 + \left( {\frac{x}{5}} \right)\left( {\frac{x}{7}} \right)}}    M1A1A1

Note:     Award M1 for use of xxx, A1 for a correct numerator and A1 for a correct denominator.

= \frac{{\frac{{2x}}{{35}}}}{{1 + \frac{{{x^2}}}{{35}}}}       M1

OR

\cos \alpha  = \frac{{{x^2} + 35}}{{\sqrt {({x^2} + 25)({x^2} + 49)} }}    M1A1

Note:     Award M1 for either use of the cosine rule or use of \cos (A - B).

\sin \alpha \frac{{2x}}{{\sqrt {({x^2} + 25)({x^2} + 49)} }}    A1

\tan \alpha  = \frac{{\frac{{2x}}{{\sqrt {({x^2} + 25)({x^2} + 49)} }}}}{{\frac{{{x^2} + 35}}{{\sqrt {({x^2} + 25)({x^2} + 49)} }}}}    M1

THEN

\tan \alpha  = \frac{{2x}}{{{x^2} + 35}}    AG

[4 marks]

b.

(i)     \frac{{\text{d}}}{{{\text{d}}x}}(\tan \alpha ) = \frac{{2({x^2} + 35) - (2x)(2x)}}{{{{({x^2} + 35)}^2}}}{\text{ }}\left( { = \frac{{70 - 2{x^2}}}{{{{({x^2} + 35)}^2}}}} \right)     M1A1A1

Note:     Award M1 for attempting product or quotient rule differentiation, A1 for a correct numerator and A1 for a correct denominator.

(ii)     METHOD 1

EITHER

\frac{{\text{d}}}{{{\text{d}}x}}(\tan \alpha ) = 0 \Rightarrow 70 - 2{x^2} = 0    (M1)

x = \sqrt {35} {\text{ (m) }}\left( { = 5.9161 \ldots {\text{ (m)}}} \right)    A1

\tan \alpha  = \frac{1}{{\sqrt {35} }}{\text{ }}( = 0.16903 \ldots )    (A1)

OR

attempting to locate the stationary point on the graph of

\tan \alpha  = \frac{{2x}}{{{x^2} + 35}}    (M1)

x = 5.9161 \ldots {\text{ (m) }}\left( { = \sqrt {35} {\text{ (m)}}} \right)    A1

\tan \alpha  = 0.16903 \ldots {\text{ }}\left( { = \frac{1}{{\sqrt {35} }}} \right)    (A1)

THEN

\alpha  = 9.59^\circ    A1

METHOD 2

EITHER

\alpha  = \arctan \left( {\frac{{2x}}{{{x^2} + 35}}} \right) \Rightarrow \frac{{{\text{d}}\alpha }}{{{\text{d}}x}} = \frac{{70 - 2{x^2}}}{{{{({x^2} + 35)}^2} + 4{x^2}}}    M1

\frac{{{\text{d}}\alpha }}{{{\text{d}}x}} = 0 \Rightarrow x = \sqrt {35} {\text{ (m) }}\left( { = 5.9161{\text{ (m)}}} \right)    A1

OR

attempting to locate the stationary point on the graph of

\alpha  = \arctan \left( {\frac{{2x}}{{{x^2} + 35}}} \right)    (M1)

x = 5.9161 \ldots {\text{ (m) }}\left( { = \sqrt {35} {\text{ (m)}}} \right)    A1

THEN

\alpha  = 0.1674 \ldots {\text{ }}\left( { = \arctan \frac{1}{{\sqrt {35} }}} \right)    (A1)

= 9.59^\circ    A1

(iii)     \frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}(\tan \alpha ) = \frac{{{{({x^2} + 25)}^2}( - 4x) - (2)(2x)({x^2} + 35)(70 - 2{x^2})}}{{{{({x^2} + 35)}^4}}}{\text{ }}\left( { = \frac{{4x({x^2} - 105)}}{{{{({x^2} + 35)}^3}}}} \right)     M1A1

substituting x = \sqrt {35} {\text{ }}( = 5.9161 \ldots ) into \frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}(\tan \alpha )     M1

\frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}(\tan \alpha ) < 0{\text{ }}( =- 0.004829 \ldots ) and so \alpha  = 9.59^\circ is the maximum value of \alpha      R1

\alpha never exceeds 10°     AG

[11 marks]

c.

attempting to solve \frac{{2x}}{{{x^2} + 35}} \geqslant \tan 7^\circ      (M1)

Note:     Award (M1) for attempting to solve \frac{{2x}}{{{x^2} + 35}} = \tan 7^\circ .

x = 2.55 and x = 13.7     (A1)

2.55 \leqslant x \leqslant 13.7{\text{ (m)}}    A1

[3 marks]

d.

Examiners report

This question was generally accessible to a large majority of candidates. It was pleasing to see a number of different (and quite clever) trigonometric methods successfully employed to answer part (a) and part (b).

a.

This question was generally accessible to a large majority of candidates. It was pleasing to see a number of different (and quite clever) trigonometric methods successfully employed to answer part (a) and part (b).

b.

The early parts of part (c) were generally well done. In part (c) (i), a few candidates correctly found \frac{{\text{d}}}{{{\text{d}}x}}(\tan \alpha ) in unsimplified form but then committed an algebraic error when endeavouring to simplify further. A few candidates merely stated that \frac{{\text{d}}}{{{\text{d}}x}}(\tan \alpha ) = {\sec ^2}\alpha .

Part (c) (ii) was reasonably well done with a large number of candidates understanding what was required to find the correct value of \alpha in degrees. In part (c)(iii), a reasonable number of candidates were able to successfully find \frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}(\tan \alpha ) in unsimplified form. Some however attempted to solve \frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}(\tan \alpha ) = 0 for \chi rather than examine the value of \frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}(\tan \alpha ) at x = \sqrt {35} .

c.

Part (d), which required use of a GCD to determine an inequality, was surprisingly often omitted by candidates. Of the candidates who attempted this part, a number stated that x \geqslant 2.55. Quite a sizeable proportion of candidates who obtained the correct inequality did not express their answer to 3 significant figures.

d.

Syllabus sections

Topic 3 - Core: Circular functions and trigonometry » 3.2 » Exact values of \sin, \cos and \tan of 0, \frac{\pi }{6}, \frac{\pi }{4}, \frac{\pi }{3}, \frac{\pi }{2} and their multiples.

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