| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Prove/show always positive |
| Difficulty | Moderate -0.8 Part (i) is a standard completing-the-square exercise to show a quadratic is always positive (discriminant negative or complete the square to get (x-4)²+1>0). Part (ii) requires translating words to algebra ((x+3)²>x²) and simplifying to 6x+9>0, then recognizing this fails for negative x values. Both parts are routine AS-level techniques with minimal problem-solving demand. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x^2 - 8x + 17 = (x-4)^2 - 16 + 17\) | M1 | Attempt to complete the square, accept \((x-4)^2\ldots\) |
| \(= (x-4)^2 + 1\) with comment | A1 | With either \((x-4)^2 \geq 0\), \((x-4)^2+1 \geq 1\) or min at \((4,1)\) |
| As \((x-4)^2 \geq 0 \Rightarrow (x-4)^2 + 1 \geq 1\), hence \(x^2 - 8x + 17 > 0\) for all \(x\) | A1 | Fully written solution with correct statements, valid reason and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Explanation that statement may not always be true; tests e.g. \(x=-5\): \((-5+3)^2 = 4\) whereas \((-5)^2 = 25\) | M1 | Attempts a value where not true and shows/implies it is not true |
| States "sometimes true" AND gives reason e.g. when \(x=5\): \((5+3)^2 = 64 > 25\) ✓; when \(x=-5\): \((-5+3)^2 = 4 < 25\) ✗ | A1 | Shows true for one value AND not true for another, states "sometimes true" |
# Question 2(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x^2 - 8x + 17 = (x-4)^2 - 16 + 17$ | M1 | Attempt to complete the square, accept $(x-4)^2\ldots$ |
| $= (x-4)^2 + 1$ with comment | A1 | With either $(x-4)^2 \geq 0$, $(x-4)^2+1 \geq 1$ or min at $(4,1)$ |
| As $(x-4)^2 \geq 0 \Rightarrow (x-4)^2 + 1 \geq 1$, hence $x^2 - 8x + 17 > 0$ for all $x$ | A1 | Fully written solution with correct statements, valid reason and conclusion |
Alternative methods accepted:
- **Discriminant:** $b^2 - 4ac = -4$, no real roots, U-shaped curve
- **Differentiation:** $\frac{dy}{dx} = 2x - 8 \Rightarrow$ turning point $(4,1)$, show minimum $> 0$
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# Question 2(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Explanation that statement may not always be true; tests e.g. $x=-5$: $(-5+3)^2 = 4$ whereas $(-5)^2 = 25$ | M1 | Attempts a value where not true and shows/implies it is not true |
| States "sometimes true" AND gives reason e.g. when $x=5$: $(5+3)^2 = 64 > 25$ ✓; when $x=-5$: $(-5+3)^2 = 4 < 25$ ✗ | A1 | Shows true for one value AND not true for another, states "sometimes true" |
---\begin{enumerate}
\item (i) Show that $x ^ { 2 } - 8 x + 17 > 0$ for all real values of $x$\\
(ii) "If I add 3 to a number and square the sum, the result is greater than the square of the original number."
\end{enumerate}
State, giving a reason, if the above statement is always true, sometimes true or never true.
\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q2 [5]}}