Question 3:
--- 3 (a) ---
3 (a) | y =48
Graphical interpretation of
( )
f x = 4 8
One of 8,−8 | B1
Attempts to solve an appropriate equation
E.g.
2 x 2 − 1 0 x = 4 8  x 2 − 5 x − 2 4 = 0  ( x  8 ) ( x  3 ) = 0  x = . . . | M1
x = 8 , − 8 with no additional values | A1
(3)
(b) | Graphical interpretation of
5
f ( x ) … x
2
5
y = x
2
5 2 5
Attempts to solve 2 x 2 − 1 0 x = x  4 x 2 − 2 5 x = 0  x =
2 4
5 1 5
OR attempts to solve (o.e.) 1 0 x − 2 x 2 = x  4 x 2 − 1 5 x = 0  x =
2 4 | M1
5 2 5
Attempts to solve 2 x 2 − 1 0 x = x  4 x 2 − 2 5 x = 0  x =
2 4
5 1 5
AND attempts to solve (o.e.) 1 0 x − 2 x 2 = x  4 x 2 − 1 5 x = 0  x =
2 4 | dM1
15 25
Achieves both critical values x= , x =
4 4 | A1
1 5 2 5
Correct set of values x „ or x …
4 4 | A1
(4)
Total 7
The equation must be given in the form of a correct 3TQ, but not necessarily collected on one side of
the = sign. Solving is allowed by all methods including via a calculator.
All the following score this M mark
Example 1:
2 x 2 − 1 0 x = 4 8 o .e  x = 8 , ( − 3 ) …only the 8 is required
Example 2:
2 x 2 − 1 0 x = 4 8 o .e  x 2 − 5 x − 2 4 = 0  ( x − 8 ) ( x + 3 ) = 0  x = ...
Example 3:
2 x 2 − 1 0 x − 4 8 = 0 o r x 2 − 5 x − 2 4 = 0 leading to x o r x = 8 , ( − 3 )
Example 4:
2 x 2 + 1 0 x − 4 8 = 0  x =
− 1 0  1 0 0
2
−
(
2
4
)
( 2 ) ( − 4 8 )
= ...
Example 5:
2x2 +10x =48o.e  x =−8,(3) …only the −8 is required
A1 x = 8 , − 8 with no additional values unless they have been rejected
(b)
M1: Attempts to solve a correct equation or inequality, leading to a non-zero value for x.
The method must be shown with the terms being collected as a minimum response
Look for 2 x 2 − 1 0 x = 5
2
x  a x 2  b x = 0  x = c , c  0 . Allow with the = as any inequality including <
OR − 2 x 2 + 1 0 x = 5
2
x  a x 2  b x = 0  x = c , c  0 . Allow with the = as any inequality including <
You may see as an alternative to the 2nd equation/inequality 2 x 2 − 1 0 x = − 5
2
x  a x 2  b x = 0  x = c , c  0
It is acceptable to divide through by x, so 2 x 2 − 1 0 x =
5
2
x  2 x − 1 0 =
5
2
 2 x =
2
2
5
 x = 6 . 2 5
dM1: Attempts to solve BOTH correct equations or inequations leading to two non-zero values for x
Allow even if there are extra equations solved.
15
A1: Achieves both critical values x= and
4
x =
2
4
5
and no other values apart from 0 (and perhaps 5)
15
following both equations or inequalities. Allow with incorrect inequalities x… and
4
x „
2
4
5
, you are
looking for the critical values only
A1: Correct range in allowable form. E.g. x „
1 5
4
or x …
2
4
5  15 25 
, −,



, but condone
 4   4 
x „
1 5
4
, x …
2
4
5
PMT
15 25  15 25  25 15
Do NOT incorrect forms such as x„ and x… , −,



, , „ x„
4 4  4   4  4 4
The two aspects must be brought together on a single line. Mark their final answer
Alt Method: You may see attempts via squaring. It can be marked using the above guidelines
( 2x2 −10x )2 = 

5 x 

2  x 2 ( 16x 2 −160x+375 ) =0 x 2(4x−15)(4x−25)=0
2 