Date | November 2016 | Marks available | 3 | Reference code | 16N.2.sl.TZ0.10 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find and Write down | Question number | 10 | Adapted from | N/A |
Question
The following diagram shows the graph of f(x)=asinbx+c, for 0⩽x⩽12.
The graph of f has a minimum point at (3, 5) and a maximum point at (9, 17).
The graph of g is obtained from the graph of f by a translation of (k0). The maximum point on the graph of g has coordinates (11.5, 17).
The graph of g changes from concave-up to concave-down when x=w.
(i) Find the value of c.
(ii) Show that b=π6.
(iii) Find the value of a.
(i) Write down the value of k.
(ii) Find g(x).
(i) Find w.
(ii) Hence or otherwise, find the maximum positive rate of change of g.
Markscheme
(i) valid approach (M1)
eg5+172
c=11 A1 N2
(ii) valid approach (M1)
egperiod is 12, per =2πb, 9−3
b=2π12 A1
b=π6 AG N0
(iii) METHOD 1
valid approach (M1)
eg5=asin(π6×3)+11, substitution of points
a=−6 A1 N2
METHOD 2
valid approach (M1)
eg17−52, amplitude is 6
a=−6 A1 N2
[6 marks]
(i) k=2.5 A1 N1
(ii) g(x)=−6sin(π6(x−2.5))+11 A2 N2
[3 marks]
(i) METHOD 1 Using g
recognizing that a point of inflexion is required M1
egsketch, recognizing change in concavity
evidence of valid approach (M1)
egg″(x)=0, sketch, coordinates of max/min on g′
w=8.5 (exact) A1 N2
METHOD 2 Using f
recognizing that a point of inflexion is required M1
egsketch, recognizing change in concavity
evidence of valid approach involving translation (M1)
egx=w−k, sketch, 6+2.5
w=8.5 (exact) A1 N2
(ii) valid approach involving the derivative of g or f (seen anywhere) (M1)
egg′(w), −πcos(π6x), max on derivative, sketch of derivative
attempt to find max value on derivative M1
eg−πcos(π6(8.5−2.5)), f′(6), dot on max of sketch
3.14159
max rate of change =π (exact), 3.14 A1 N2
[6 marks]