| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Prove/show always positive |
| Difficulty | Moderate -0.8 This is a straightforward question testing basic understanding of discriminants and completing the square. Part (i) requires recognizing that x²-6x+10 has discriminant 36-40=-4<0 with positive leading coefficient, making it always positive. This is a standard textbook exercise requiring only routine application of discriminant knowledge, making it easier than average A-level questions. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Statement | Always True | Sometimes True | Never True | Reason | |||
| The quadratic equation \(a x ^ { 2 } + b x + c = 0 , \quad ( a \neq 0 )\) has 2 real roots. | ✓ | It only has 2 real roots when \(b ^ { 2 } - 4 a c > 0\). When \(b ^ { 2 } - 4 a c = 0\) it has 1 real root and when \(b ^ { 2 } - 4 a c < 0\) it has 0 real roots. | |||||
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| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 - 6x + 10 = (x-3)^2 + 1\) | M1 | Complete the square or differentiation to find minimum |
| Deduces "always true": \((x-3)^2 \geq 0 \Rightarrow (x-3)^2 + 1 \geq 1\), always positive | A1 | State always true with valid reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Explanation that it need not always be true; e.g. if \(a < 0\) then \(ax > b \Rightarrow x < \frac{b}{a}\) | M1 | Show it is "sometimes" true |
| States "sometimes true": if \(a>0\) then \(ax>b \Rightarrow x>\frac{b}{a}\); if \(a<0\) then \(ax>b \Rightarrow x<\frac{b}{a}\) | A1 | Correct full statement with explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Difference \(= (n+1)^2 - n^2 = 2n+1\) | M1 | Set up proof algebraically |
| Deduces "Always true" as \(2n+1 = (\text{even}+1) = \text{odd}\) | A1 | State always true with reason and proof |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 - 6x + 10 = (x-3)^2 + 1$ | M1 | Complete the square or differentiation to find minimum |
| Deduces "always true": $(x-3)^2 \geq 0 \Rightarrow (x-3)^2 + 1 \geq 1$, always positive | A1 | State always true with valid reason |
## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Explanation that it need not always be true; e.g. if $a < 0$ then $ax > b \Rightarrow x < \frac{b}{a}$ | M1 | Show it is "sometimes" true |
| States "sometimes true": if $a>0$ then $ax>b \Rightarrow x>\frac{b}{a}$; if $a<0$ then $ax>b \Rightarrow x<\frac{b}{a}$ | A1 | Correct full statement with explanation |
## Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Difference $= (n+1)^2 - n^2 = 2n+1$ | M1 | Set up proof algebraically |
| Deduces "Always true" as $2n+1 = (\text{even}+1) = \text{odd}$ | A1 | State always true with reason and proof |
---6. Complete the table below. The first one has been done for you.
For each statement you must state if it is always true, sometimes true or never true, giving a reason in each case.\\
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Statement & Always True & Sometimes True & Never True & Reason \\
\hline
The quadratic equation $a x ^ { 2 } + b x + c = 0 , \quad ( a \neq 0 )$ has 2 real roots. & & ✓ & & It only has 2 real roots when $b ^ { 2 } - 4 a c > 0$. When $b ^ { 2 } - 4 a c = 0$ it has 1 real root and when $b ^ { 2 } - 4 a c < 0$ it has 0 real roots. \\
\hline
\begin{tabular}{l}
(i) \\
When a real value of $x$ is substituted into $x ^ { 2 } - 6 x + 10$ the result is positive. \\
\end{tabular} & & & & \\
\hline
\begin{tabular}{l}
(ii) \\
If $a x > b$ then $x > \frac { b } { a }$ \\
(2) \\
\end{tabular} & & & & \\
\hline
\begin{tabular}{l}
(iii) \\
The difference between consecutive square numbers is odd. \\
\end{tabular} & & & & \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{Edexcel Paper 2 Q6 [6]}}