the King’s Factor Year 13 partial solutions 7F
This file contains specific solutions to parts of some questions set in Year 13 tKF
sessions, so that you can check that you are on the right track. Some questions,
for instance those where the result is given and a proof is required, do not have
solutions given here.
These brief solutions are NOT model answers and do not indicate how a full solution
should be presented
Updated May 8, 2015
1. [2004 STEP III question 8]
If
g(x)
dy
= f (x)y +
dx
y
and u = y 2 then
du
= 2f (x)u + 2g(x) .
dx
To solve
dy
y 1
= −
dx
x y
start with the above method, using f (x) =
1
x
and g(x) = −1, obtaining
u
du
= 2 − 2.
dx
x
The solution curves of this equation which pass through (1, 1), (2, 2) and (4, 4) are respectively
y 2 + (x − 1)2 = 1, y 2 = 2x and −2y 2 + (x + 2)2 = 4.
All three curves pass through the origin. The first is a circle, the second a parabola and the
third a hyperbola.
2. [2004 STEP III question 5]
(a)
cos x − 7 sin x =
where cos α =
√1
50
√
50 cos(x − α)
and sin α = − √750 .
Also since cos(x − α) =
√1 ,
2
tan(x − α) = ±1.
Using the first part, it can then be shown that either tan x =
Hence x = ω or x = 3π
+ ω.
2
(b) Similar arguments give tan x = 43 or tan x = − 24
.
7
The two solutions are ω and 2ω.
1
4
3
or tan x = − 43 .
the King’s Factor Year 13 partial solutions 7F
3. [2005 STEP III question 7]
Let u = xm . Then du
= mxm−1 and so
dx
R m
R
R
dx = mxmmf (u) du = uf1(u) du = F (u) + c = F (xm ) + c.
xf (xm )
Find
(a) Using f (u) = u − 1 and m = n − 1,
n−1
Z
1
1
x
−1
dx =
ln
+ c;
xn − x
n−1
xn−1
√
(b) Using f (u) = u + 1 and m = n − 2 gives
√
Z
xn−2 + 1 − 1 1
1
√
ln √
dx =
+ c.
n
2
n−2
n
−
2
x +x
x
+ 1 + 1
2