IB DP Math AA Topic : AHL 3.9 :Definition of the reciprocal trigonometric ratios : IB Style Questions HL Paper 2

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 \(\alpha  = {\rm{A\hat PB}}\) measured in degrees. Assume that the ball travels along the floor.

The maximum for \(\tan \alpha \) gives the maximum for \(\alpha \).

a.Find the value of \(\alpha \) when \(x = 10\).[4]

b.Show that \(\tan \alpha  = \frac{{2x}}{{{x^2} + 35}}\).[4]

c.(i)     Find \(\frac{{\text{d}}}{{{\text{d}}x}}(\tan \alpha )\).

(ii)     Hence or otherwise find the value of \(\alpha \) such that \(\frac{{\text{d}}}{{{\text{d}}x}}(\tan \alpha ) = 0\).

(iii)     Find \(\frac{{{{\text{d}}^2}}}{{{\text{d}}{x^2}}}(\tan \alpha )\) and hence show that the value of \(\alpha \) never exceeds 10°.[11]

d.Find the set of values of \(x\) for which \(\alpha  \geqslant 7^\circ \).[3]
 
▶️Answer/Explanation

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]

 

 

 
 
 
 
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