The coordinates of point A are (75, 450). Determine whether point A is on the bicycle track. Give a reason for your answer.

Find the derivative of \(y = \frac{{ - {x^2}}}{{10}} + \frac{{27}}{2}x\).

Use the answer in part (b) to determine if A (75, 450) is the point furthest north on the track between O and B. Give a reason for your answer.

(i) Write down the midpoint of the line segment OB.

(ii) Find the gradient of the line segment OB.

Scott starts from a point C(0,150) . He hikes along a straight road towards the bicycle track, parallel to the line segment OB.

Find the equation of Scott’s road. Express your answer in the form \(ax + by = c\), where \(a, b {\text{ and }} c \in \mathbb{R}\).

Use your graphic display calculator to find the coordinates of the point where Scott first crosses the bicycle track.

\(y =  - \frac{{{{75}^2}}}{{10}} + \frac{{27}}{2} \times 75\)     (M1)

Note: Award (M1) for substitution of 75 in the formula of the function.

= 450     (A1)

Yes, point A is on the bike track. (A1)

Note: Do not award the final (A1) if correct working is not seen.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{{2x}}{{10}} + \frac{{27}}{2}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 0.2x + 13.5} \right)\)     (A1)(A1)

Notes: Award (A1) for each correct term. If extra terms are seen award at most (A1)(A0). Accept equivalent forms.

\( - \frac{{2x}}{{10}} + \frac{{27}}{2} = 0\)     (M1)

Note: Award (M1) for equating their derivative from part (b) to zero.

\(x = 67.5\)     (A1)(ft)

Note: Follow through from their derivative from part (b).

\( {\text{(Their) }} 67.5 \ne 75\)      (R1)

Note: Award (R1) for a comparison of their 67.5 with 75. Comparison may be implied (eg 67.5 is the x-coordinate of the furthest north point).

OR

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{{2 \times (75)}}{{10}} + \frac{{27}}{2}\)     (M1)

Note: Award (M1) for substitution of 75 into their derivative from part (b).

\(= -1.5\)     (A1)(ft)

Note: Follow through from their derivative from part (b).

\({\text{(Their)}} -1.5 \ne 0\)     (R1)

Note: Award (R1) for a comparison of their –1.5 with 0. Comparison may be implied (eg The gradient of the parabola at the furthest north point (vertex) is 0).

Hence A is not the furthest north point.     (A1)(ft)   

Note: Do not award (R0)(A1)(ft). Follow through from their derivative from part (b).

(i) M(50,175)     (A1)

Note: If parentheses are omitted award (A0). Accept x = 50, y = 175.

(ii) \(\frac{{350 - 0}}{{100 - 0}}\)     (M1)

Note: Award (M1) for correct substitution in gradient formula.

\( = 3.5\left( {\frac{{350}}{{100}},\frac{7}{2}} \right)\)     (A1)(ft)(G2)

Note: Follow through from (d)(i) if midpoint is used to calculate gradient. Award (G1)(G0) for answer 3.5x without working.

\(y = 3.5x + 150\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for using their gradient from part (d), (A1)(ft) for correct equation of line.

\(3.5x - y = -150\) or \(7x - 2y = -300\) (or equivalent)     (A1)(ft)

Note: Award (A1)(ft) for expressing their equation in the form \(ax + by = c\).

(18.4, 214) (18.3772..., 214.320...)     (A1)(ft)(A1)(ft)(G2)(ft)

Notes: Follow through from their equation in (e). Coordinates must be positive for follow through marks to be awarded. If parentheses are omitted and not already penalized in (d)(i) award at most (A0)(A1)(ft). If coordinates of the two intersection points are given award (A0)(A1)(ft). Accept x = 18.4, y = 214.