1. (a) (i) v(1) = 13 – 4(1)2 + 4(1)
= 1 ms–1 (A1)
(ii) v(0.5) = (0.5)3 – 4(0.5)2 + 4(0.5)
= 1.125 ms–1 accept 1.13 (3 s.f.) (A1) 2
(b) a = v(1.5) = 1.53 – 4(1.5) + 4(1.5)
= 0.375 (A1)
b = v(3) = 33 – 4(32) + 4(3)
= 3 (A1) 2
Table (not required)
v / 0 / 1.125 / 1 / 0.375 / 0 / 0.625 / 3 / 7.875 / 16
(c) (i) = 3t2 – 8t + 4 (A1)
3t2 – 8t + 4 = 0 (M1)
(3t – 2)(t – 2) = 0 (M1)
t = , t = 2 (A1)(A1)
(ii) The function is changing from acceleration to deceleration
or velocity changes from increasing to decreasing
or kite is stationary or velocity is zero (R1)(R1)
Note: Award (R1) for acceleration, (R1) for deceleration.
Gradient = 0 (A1) 8
(d)
(A5) 5
Note: Award (A1) for axes correctly labelled, (A1) if scales correct, (A1) for correct general shape of curve, (A1) for each turning point in approximately the correct place.
(e)
time t / motiont = 0 / stopped
0 < t < / accelerating (increasing in velocity) / (A1)
t = / stopped accelerating
< t < 2 / decelerating (decreasing in velocity) / (A1)
t = 2 / stopped decelerating / (A1) / 3
2 < t £ 4 / accelerating
Note: Stops may be left out
[20]
2. (a) (i) f ¢(x) = 3x2 – 8x – 3 (A1)
(ii) 3x2 – 8x – 3 = 0 (M1)
(3x + l)(x – 3) = 0 (A1)
x = – (A1)
x = 3 (A1)
Note: Alternatively, award (G1) for 1 correct answer,
(G3) for both.
(M1)
= 18.5 (A1)
f(3) = (3)3 – 4(3)2 – 3(3) + 18 (M1)
= 0 (A1)
Points are and (3, 0) (A1) 10
(b) a = 0 (A1)
b = 18 (A1) 2
(c)
(A5) 5
Note: Award (A1) for scales and axes labelled correctly,
(A1)(A1) for maximum and minimum placed correctly, (A1) for smooth curve, (A1) for all points plotted correctly.
(d) (i) or < x < 3 all parts correct (A1)
(ii) or –3 < x < (A1)
and (3, 5) (allow (x > 3)) ft from error in (i). (A1) 3
[20]
3. (a) f (1) = (M1)
Note: (M1) for substituting x =1 into the formula.
= 2 (M1)
Note: (M1) for equating to 2.
k = 4 (AG) 2
(b) q = 2, r = 0.125 (A1)(A1) 2
(c)
(A4) 4
Notes: (A1) for scales and labels
(A1) for accurate smooth curve passing through (0,0)
drawn at least in the given domain
(A1) for asymptotic behaviour (curve must not go up or
cross the x-axis)
(A1) for indicating the position of the maximum point.
(d) M (1.44, 2.12) (G1)(G1) 2
Notes: Brackets required, if missing award (G1)(G0)
Accept x = 1.44 and y = 2.12
(e) y = 0 (A1)(A1) 2
Note: (A1) for “y =” provided the right hand side is
a constant (A1) for 0.
(f) (i) See graph
Note: (A1) for correct line, (A1) for label. (A1)(A1)
(ii) x = 0.3(ft) from candidate’s graph. (A2)(ft) 4
Notes: Accept ±0.1 from their x.
For 0.310 award (G1)(G0)
For other answers taken from the GDC and not given
correct to 3 significant figures award (G0)(AP)(G0)
or (G1)(G0) if (AP) already applied.
[16]
4. (a) y = x + z + z (M1)
Note: Award (M1) for writing a sensible equation.
xz = 162
(M1)
(M1)
(AG) 3
(b) (A1)(A1)(A1) 3
Note: Award (A1) for 1 and no other constant present, (A1) for –324, (A1) for or x–2.
(c)
(M1)
Note: Award (M1) for putting candidate’s derivative equal to zero.
x2 = 324 (A1)(ft)
x = 18 (A1)(ft)(G3) 3
(d) y = 18 + 9 + 9 (M1)
Note: Award (M1) for adding three sides of rectangle.
= 36 (A1)(ft)(G2)
OR
(M1)
= 36 (A1)(ft) 2
(e) a = 36 (A1)
b = 39 (A1) 2
(f)
(A5)(ft) 5
Notes: Award (A1) for correct scales and labels, (A3) for correct points plotted, (A1) for smooth curve with (18, 36) as the minimum value.
Award (A2) for 5 or 6 points correctly plotted, (A1) for 3 or 4 points correctly plotted.
(g) x ³ 18 (A1)(A1)(ft) 2
Notes: Award (A1) for x ³ , (A1) for 18.
Accept x > 18
[20]
5. (a)
(A3) (C3)
Notes: Award (A1) for point (0,5) indicated.
Award (A2) for correct shape.
(b) (1.5, 0.5) (A1)(A1) (C2)
(c) x = 1.5 (A1) (C1)
[6]
6. (a)
(A4) (C4)
Notes: Award (A1) for correct scales.
Award (A1)(A1) for two correct parts to the graph.
Award (A1) if asymptotes are shown.
(b) Horizontal asymptote y = 1. (A1) (C1)
[6]
7. (a)
For correct axes from 0 to 4. (A1)
For correct curve y = x2. (A1)
For correct curve y = 3 – . (A1)
For two intersections. (A1) 4
(b) (0.347, 0.121) or x = 0.347, y = 0.121 (by GDC) (G1)(G1)
(1.53, 2.35) or x = 1.53, y = 2.35. (G1)(G1) 4
(c) (i) for losing the constant. (A1)
For attempting to write as a power (can be implied). (M1)
For correct answer or x–2. (A1)
(ii) 1 (A1) 4
(d) For using y = mx + c or equivalent with their m, to find c. (M1)
c = 1 (A1)
y = x + 1 (A1) 3
[15]
8. (a)
For x-axis from –10 to 10. (A1)
For –4 marked. (A1)
For correct shape of graph. (A1)(A1) 4
(b) Horizontal asymptote (A1)
y = 1 (A1)
Vertical asymptote (A1)
x = 0 (A1) 4
(c) Line drawn on sketch (A2) 2
(d) (2.56, 2.56) (–1.56, –1.56) (A1)(A1)(A1)(A1) 4
(e) Range y Î , y ¹ 1 (A1)(A1) 2
[16]
9. (a)
For labels and scales. (A1)
3 maxima drawn. (A1)
2 minima drawn. (A1)
General shape (A2) 5
(b) (0.827, 4.12) (G2) 2
(c) 0, 1.8, 3.6, 5.4, 7.2, 9 (for any one of these answers). (G1) 1
(d) r = 1 (G2)
Perfect positive correlation. (R1) 3
(e) y = 3x (accept y = 3x + 0.000274) (G2) 2
(f) line on graph (A1) 1
(g) (0, 0) or (1.16, 3.48) (G1) (G1) 2
[16]
10. (a)
cubic function f(x) / graph labelf(x) = x3 + a / B / (A1)
f(x) = (x – a)3 + a / E / (A1)
f(x) = x3 / C / (A1)
f(x) = (x – a)3 / D / (A1) (C4)
Note: If the same letter is written in every box, award (A0) not (A1).
(b) Graph A. (A1) (C1)
(c) Two of them. (Allow “D and C”). (A1) (C1)
[6]
11. (a) With the given domain, the correct answer is
/ (A1)(A1)(ft)
(A1)(ft) (C3)
Notes: Award (A1) for a neat window complying reasonably with the requirements.
The window must clearly have used x values from –3 to 3 and y values at least from 0 to 1. Axes labels are not essential. Some indication of scale must be present but this need not be a formal scale, eg tick marks, a single number on each axis or coordinates of the intersection are all adequate.
Award (A1) for each curve correct and correctly labelled with f and g or the expressions for f and g. Can follow through both curves, for example if curves are incomplete due to a poor window, and penalize only once if both curve labels are missing. Examiners should familiarize themselves with the graph of as this is expected to appear in error. With the correct window, this graph will not be seen at all, but with a larger y interval it might look a little like the correct graph except that it would have asymptotes at x = 0 and y = 1. Award (A0) for this curve.
(b) One solution. (A1)(ft) (C1)
(c) Solution occurs at the point of intersection of the curves,
where x = 0.569840
0.570. (M1)(A1) (C2)
Notes: The (M1) can also be awarded for the intersection point indicated on the sketch.
(0.57 is an (AP))
If a coordinate pair is given as the answer and the x value is correct with no method mentioned, award (C1) or if the method is mentioned, award (M1)(A0).
Can follow through if curve is drawn, answer to (c) is then 1.75.
[6]
12. (a)
(A4) (C4)
Note: Award (A1) for some indication of scale on the y-axis. Award (A1) for at least one asymptote drawn. Award (A1) for each of the two (smooth) branches. The left hand branch must pass through 0. One branch should be above the horizontal asymptote and the other below but if the asymptote is not drawn, then there should be little or no overlap in heights of the branches. If this condition is not fulfilled, award (A1)(A0) for the curve.
(b) (i) Horizontal asymptote y = 3 (A1)
(ii) Vertical asymptote x = 1 (A1)(ft) (C2)
Equations for x and y must be seen, (ft) if reversed.
[6]
13. (a)
(G3) 3
(b) line drawn with –ve gradient and +ve y-intercept (G1)
(2.45, 2.11) (G1)(G1) 3
(c) f¢(1.7) = 3(1.7)2 – 4(1.7) + 1 (M1) 2
Note: Award (M1) for substituting in their f¢(x)
2.87 (A1)(G2)
[8]
14. (a) 6
Notes: (A1) for labels and some idea of scale. (A6)
(A1) for x-intercept seen, (A1) for y-intercept seen in
roughly the correct places (coordinates not required).
(A1) for vertical asymptote seen, (A1) for horizontal
asymptote seen in roughly the correct places (equations
of the lines not required).
(A1) for correct general shape.
(b) x = −4 (A1)(A1)(ft) 2
Note: (A1) for x =, (A1)(ft) for −4
(c) 2
Note: (A1) for correct axis intercepts,
(A1) for straight line (A1)(A1)
(d) (–2.85078, –2.35078) OR (0.35078, 0.85078) (G1)(G1)(A1)(ft) 3
Notes: (A1) for x-coordinate, (A1) for y-coordinate,
(A1)(ft) for correct accuracy. Brackets required.
If brackets not used award (G1)(G0)(A1)(ft).
Accept x = −2.85078, y = −2.35078
or x = 0.35078, y = 0.85078
(e) gradient = 1 (A1) 1
(f) gradient of perpendicular = −1 (A1)(ft)
Note: Can be implied in the next step
y = mx + c
−3 = −1×−2 + c (M1)
c = −5
y = −x − 5 (A1)(ft)(G2)
OR OR
y + 3 = − (x + 2) (M1)(A1)(ft)(G2) 3
Note: Award (G2) for correct answer with no
working at all but (A1)(G1) if the gradient is
mentioned as −1 then correct answer with no
further working.
[17]
1