 ```Algebraic Fractions
1.
Express
x
x  12
+ 2
as a single fraction in its simplest form.
( x  1)( x  3) x  9
(6)
2.
f(x) =
x2  x  6
, x  0, x  3.
x 2  3x
(a) Express f(x) in its simplest form.
(3)
(b) Hence, or otherwise, find the exact solutions of f(x) = x + 1.
(3)
3.
The function f is defined by
f: x 
(a) Show that f(x) =
5x  1
3
–
, x > 1.
x  x2 x2
2
2
, x > 1.
x 1
(4)
(b) Find f –1(x).
(3)
The function g is defined by
g: x  x2 + 5, x  ℝ.
(c) Solve fg(x) =
1
4
.
(3)
4.
Express
6
2 x 2  3x
– 2
(2 x  3)( x  2) x  x  2
as a single fraction in its simplest form.
(7)
5.
(a) Simplify
3x 2  x  2
.
x2 1
(3)
(b) Hence, or otherwise, express
3x 2  x  2
1
–
as a single fraction in its simplest form.
2
x( x  1)
x 1
(3)
6. really (cant edit the number!)
7. Given that:
f ( x) 
2x  3
9  2x
 2
x  2 2 x  3x  2
Show that f ( x) 
4x  6
2x 1
7 Marks
8. Given that:
2 x 4  3x 2  x  1
dx  e
 ax 2  bx  c  2
2
x 1
x 1
find the values of the constants a, b, c, d , e.
Question
Scheme
Marks
number
x2 – 9 = (x – 3)(x + 3) seen
1.
B1
Attempt at forming single fraction
M1; A1
x( x  3)  ( x  12)( x  1)
2 x 2  10 x  12
;=
( x  1)( x  3)( x  3)
( x  1)( x  3)( x  3)
Factorising numerator =
2( x  2)( x  3)
2( x  2 )
or equivalent =
( x  1)( x  3)( x  3)
( x  1)( x  3)
M1 M1 A1
(6)
(6 marks)
( x  3)( x  2) ( x  2)
2
;
or 1 
x( x  3)
x
x
2a)
B1 numerator, B1 denominator ;
B1,B1,B1
M1 for equating f(x) to x + 1 and
(3)
( x  2)
 x  1  x2  2
x
M1 A1√
2b)
3
(a)
x 2
A1
5x  1
3

( x  2)( x  1) x  2
B1
5 x  1  3( x  1)
( x  2)( x  1)
=
M1
M1 for combining fractions even if the denominator is not lowest common
2x  4
( x  2)( x  1) =
=
M1 A1 cso
2( x  2)
2
( x  2)( x  1) = x  1 *
(4)
M1 must have linear numerator
(b)
2
x 1 
y=
M1A1
xy  y  2  xy = 2 + y
A1
2 x
f –1(x) = x
2
x 4
fg(x) =
(3)
o.e.
2
(attempt)
2
2
x 4 =
Setting
2
" g"  1 ]
[
1
4 and finding x2 = …;
M1
M1; A1
x= 2
(3)

Question
Scheme
Marks
Number
x2  x  2   x  2 x  1
4.
At any
B1
stage
x  2 x  3
2 x 2  3x
x


 2 x  3 x  2   2 x  3 x  2  x  2
B1
x  x  1  6
2 x 2  3x
6
 2

 2 x  3 x  2  x  x  2  x  2  x  1
M1
x2  x  6

 x  2  x  1
A1

 x  3 x  2 
 x  2  x  1
M1
A1

x3
x 1
A1
(7)

Alternative method
x2  x  2   x  2 x  1
At any
B1
stage
 2 x  3 appearing as a factor of the numerator at any stage
2 x 2  3x   x  1  6  2 x  3

2 x 2  3x
6


 2 x  3 x  2   x  2  x  1
 2 x  3 x  2  x  1

2 x3  5 x 2  9 x  18
 2 x  3 x  2  x  1
B1
M1
can be
A1
implied
M1
 x  2  2x2  9x  9

 2 x  3 x  2  x  1
or
 2 x  3  x 2  x  6 
 2 x  3 x  2  x  1
or
 x  3  2 x 2  x  6 
 2 x  3 x  2  x  1
Any one linear factor ×

 2 x  3 x  2  x  3
 2 x  3 x  2  x  1

x3
x 1
Complete
A1
factors
Question
number
5.
(a)
A1
(7)
Scheme
3x  2x  1 ,
x  1x  1

3x  2
x 1
Marks
M1B1, A1 (3)
Notes
M1 attempt to factorise numerator, usual rules
B1 factorising denominator seen anywhere in (a),
If factorisation of denom. not seen, correct answer implies B1
(b) Expressing over common denominator
3x  2
1
x3x  2  1


x 1
x x  1
x x  1
(3x 2  x  2) x  ( x  1)
[Or “Otherwise” :
]
x( x 2  1)
M1
Multiplying out numerator and attempt to factorise
3x
2
M1

 2 x  1  3x  1x  1