| Exam Board | Edexcel |
|---|---|
| Module | PURE |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof of inequality |
| Difficulty | Standard +0.8 This is a proof by contradiction requiring students to set equations equal, form a quartic, substitute to create a quadratic in x², and show it has no real solutions using the discriminant. While the technique is standard, it requires careful algebraic manipulation across multiple steps and proper proof structure, making it moderately challenging but within reach of a well-prepared Further Maths student. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks |
|---|---|
| 2 | Assume that C and C do intersect so that x 4 + 1 0 x 2 + 8 = 2 x 2 − 7 |
| Answer | Marks |
|---|---|
| (and has real solutions). | B1 |
| Answer | Marks |
|---|---|
| x4 +8x2 +15=0 ( x2 +4 )2 −1=0 x2 =... | M1 |
| x2 = −5, −3 | A1 |
| Answer | Marks |
|---|---|
| 1 2 | A1 |
Total 4
Question 2:
2 | Assume that C and C do intersect so that x 4 + 1 0 x 2 + 8 = 2 x 2 − 7
1 2
(and has real solutions). | B1
( )( )
x4 +8x2 +15=0 x2 +5 x2 +3 =0 x2 =...
or
−8 82 −415
x4 +8x2 +15=0x2 =
2
or
x4 +8x2 +15=0 ( x2 +4 )2 −1=0 x2 =... | M1
x2 = −5, −3 | A1
e.g. The two values of x2 < 0 (so x 4 + 8 x 2 + 1 5 = 0 has no real roots) which is not
possible so C and C do not intersect
1 2 | A1
(4)
Total 4
Alternative considering the sign of each term / the expression
B1: As above in main scheme notes
M1: Either one of:
• considers the sign of one side of an equation or inequality which includes either an x2 or x4
term. e.g. x 4 + 8 x 2 … 0 e.g. x4 +8x2 +150 Allow to compare with 0 e.g. x2 +3 0
• considers the sign of x 2 or x 4 e.g. x 2 … 0
• considers separately at least one part of the product of two terms. e.g. x2(x2 +8)>0 or
(x2 +8)0
• considers the range of values using their completed square form e.g.
(
x 2 + 4
) 2
… 1 6 or
x2 +4…4
Do not be concerned by > or …
A1: Correct three-term quadratic in x2 and correct deductions that e.g. x 4 … 0 and x 2 … 0
depending on their expression (condone > 0)
This mark cannot be scored if the expression they use to make their deductions is incorrect.
e.g. x 2 ( x 2 + 8 ) is positive (or zero) since x 2 ( ) and x2 +8 are both greater than (or equal to)
zero is A1
e.g. x 4 + 8 x 2 … 0 since x 2 and x 4 are both … 0 is A1
e.g. x2 +4…4 since x2…0 is A1
e.g. x 4 + 8 x 2 … 0 is A0
A1: This mark can only be scored provided the previous M1A1 has been scored.
Acceptable conclusion which refers to
• the equation x 4 + 8 x 2 + 1 5 = 0 gives no solutions o.e. e.g. x4 +8x2 +15cannot equal zero,
e.g. x 4 + 8 x 2 cannot equal − 1 5 e.g. x2 +4 cannot equal − 1
• C and C not intersecting
1 2
• Correct use of inequalities throughout e.g. x 2 … 0
PMT
(not just x2 >0).
Do not accept e.g. −3„ 0\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
The curve $C _ { 1 }$ has equation
$$y = x ^ { 4 } + 10 x ^ { 2 } + 8 \quad x \in \mathbb { R }$$
The curve $C _ { 2 }$ has equation
$$y = 2 x ^ { 2 } - 7 \quad x \in \mathbb { R }$$
Use algebra to prove by contradiction that $C _ { 1 }$ and $C _ { 2 }$ do not intersect.
\hfill \mbox{\textit{Edexcel PURE 2024 Q2}}