Chapter(^8)
307
P1^
(iv) (a)
(b) y = sin x
(v) (a)
(b) y = −cos x
4 (i) y = tan x + 4
(ii) y = tan (x + 30°)
(iii) y = tan (0.5x)
5 (i) y = 4 sin x
(ii) –2^3
6 (i) a = 3, b = − 4
(ii) x = 0.361 or x = 2.78
(iii)
7 (i) a = 4, b = 6
(ii) x = 48.2 or x = 311.8
(iii)
8 (i) a = 6, b = 2, c = 3
(ii) 7
12
9 (i) 2 f(x) 8
(ii)
(iii) No, it is a many-to-one
function.
10 (i) x = 0.730 or x = 2.41
(ii)
(iii) k 1, k 7
(iv) 32
(v) 2.80
Chapter 8
●?^ (Page^ 254)
To find the distance between the
vapour trails you need two pieces of
information for each of them: either
two points that it goes through,
or else one point and its direction.
All of these need to be in three
dimensions. However, if you want
to find the closest approach of the
aircraft you also need to know, for
each of them, the time at which it was
at a given point on its trail and the
speed at which it was travelling. (This
answer assumes constant speeds and
directions.)
(^) ● (Page 261)
The vector a 1 i + a 2 j + a 3 k is shown
in the diagram.
x
y
1
–1
O
1 3
x
y
1
–1
O
90 270
180 360 450 x
y
y = tanx + 4
0 90 270
4
150 330 x
y
0 60 240
y = tan (x + 30)
y
0 x
y = tan (0.5x)
180 360 540
x
f(x)
O
–1
7 y = 3 – 4 cos^2 x
π 4 π 2 34 π π
x
f(x)
O
10
–2
90° 180° 270° 360°
y = 4 – 6 cosx
x
f(x)
O
8
5
2
π 2 π
y = 5 – 3 sin 2 x
x
f(x)
O
7
4
1
π 2 π 32 π 2 π
y = 4 – 3 sinx
O
P
Q
a 3
a 1
a 2
z
y
x