Sketch the graphs of f and g on the same set of axes for x ∈ [− 150°; 180°] Clearly show all intercepts with the axes and the coordinates of turning points. (7) Use your graph to answer the following questions:
Write down the period of f. (1)
For which values of x is f(x) = g(x)? (2)
For which values of x is f(x) > 0? (2)
For which values of x is g(x) increasing? (2)
Determine one value of x for which f(x) − g(x) = 1,5. (1)
If the curve of f is moved down one unit, write down the new equation of f. (2)
If the curve of g is moved 45° to the left, write down the new equation of g. (2)
Sketch below are the graphs of g(x) = asin(x+p) and f(x) = bcosqx for x ∈ [0°; 180°]
Determine the numerical values of a, p, b and q. (5)
If the graph of g(x) is shifted two units down:
Write down the amplitude of the new graph (1)
Write down the equation of the new graph (2)
If the graph of f(x) is shifted 60 to the left, write down two possible equations of the new graph. (2) [29]
Solutions
for g(x) = 2 cos x and for f(x) = sin(x + 30°)
period = 360° (1)
x = –120° or 60° (2)
for f(x) >0; x ∈ (–90°; 90°) (2)
g(x) increasing when x ∈ (–120°; 60°) (2)
x = 0° (1)
New f(x) = 2cos x –1 (2)
Original equation: g(x) = sin (x + 30°), with 45° shift to the left: g(x) = sin (x + 30° + 45°) so g(x) = sin (x + 75°) (2)
a = 2 (amplitude of the f(x)) f(x) = 2sin(x+p)… Substitute 60° ∴ 2 = 2sin(60° + p) ÷ 2 ∴ 1 = sin(60° + p) Pressing shift sin −1 (1) = 90° ∴ 60° + p = 90° ∴ p = 30°∴ f(x) = 2sin(x + 30°) b=3 (amplitude of the g(x)) period = 720 720° = 360°∴ q = ½ ∴ g(x) = 3 cos ½ x (5)
(1) Amplitude = 3 (shift up or down has no effect on the amplitude) (1) (2) g(x) = 3 cos ½ x − 2 (2)