| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show discriminant inequality, then solve |
| Difficulty | Standard +0.3 This is a standard discriminant problem requiring algebraic manipulation to form a quadratic in x, then applying b²-4ac > 0 for distinct real roots. The algebra involves clearing fractions and collecting terms, which is routine for P1. Part (b) requires solving a quadratic inequality, which is also standard curriculum content. While multi-step, each step follows predictable techniques without requiring novel insight. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Scheme | Marks | Guidance |
| \(4(p-2x) = \frac{12+15p}{x+p}\) \(8x^2+4px+12+15p-4p^2 = (=0)\) e.g. \(a=8, b=4p, c=12+15p-4p^2\) | M1 | Multiplies by \((x+p)\), multiplies out the brackets and collects terms on one side. Condone slips and the omission of \(=0\), but there must be an ...x² term. May be implied by further work e.g. their attempt at the discriminant. |
| \((4p)^2-4 \times 8 \times (12+15p-4p^2) (>0)\) | dM1 | Attempts \(b^2-4ac\) for their quadratic in \(x\) with their \(b\) and their \(c\) both being in terms of \(p\). The values/expressions for \(a\), \(b\) and \(c\) for their quadratic in \(x\) must be embedded in the correct positions in the expression but condone invisible brackets and sign slips miscopying. It is dependent on the previous method mark. Condone for this mark if they work in terms of \(x\) instead of \(p\) provided the coefficients were actually those in terms of \(p\) from their original quadratic. i.e. \(8x^2+4px+12+15p-4p^2 = 0 \Rightarrow a=8, b=4p, c=12+15p-4p^2\) \(\Rightarrow (4x)^2-4 \times 8 \times (12+15x-4x^2) (>0)\) (they have just written \(x\) instead of \(p\) which is fine). |
| \(3p^2-10p-8 > 0\) * | A1* | Achieves \(3p^2-10p-8 > 0\) with all stages of working shown: Multiplying out \((p-2x)(x+p)\), Collecting terms on one side of an equation (\(=0\) may be implied), Finding the discriminant. At least one intermediate stage of working between the discriminant and the final answer. There must be no errors seen including invisible brackets but condone a missing trailing bracket if it does not alter the processing. e.g. \((4p)^2-4 \times 8 \times (12+15p-4p^2\) is A1*. Condone the \(>0\) appearing for the first time in their final answer, provided an incorrect inequality has not been used in earlier working. Do not withhold this mark if they state e.g. "b² – 4ac = 0" as part of their working. |
| (a) Total: 3 marks | ||
| e.g. \((p-4)(3p+2) (=0) \Rightarrow 4, -\frac{2}{3}\) | M1 | Attempts to find critical values by solving the given quadratic by either: factorising (do not accept \((p-4)(p+\frac{3}{2})\) unless they have divided both sides of the inequality/equation by 3 first), completing the square \(3(p-\frac{5}{3})^2 \pm ... \Rightarrow (p=)...\), quadratic formula. Usual rules apply (see general marking principles for guidance). May be in another variable e.g. \(x\). This mark cannot be awarded from directly using a calculator and stating the roots. |
| \(p < -\frac{2}{3}\) or \(p > 4\) | M1A1 | M1: Attempts to find the outside region for their critical values. May use another variable e.g. \(x\). May be implied by \(p < "-\frac{2}{3}", p > "4"\) or incorrect use of inequalities e.g. \(4 < p < -\frac{2}{3}\). This is not dependent on the first method mark so if a calculator has been used then this mark can still be scored. A1: \(p < -\frac{2}{3}\) or \(p > 4\) (or equivalent). (may appear on separate lines). Isw once a correct answer is seen, provided there is no contradiction and no part of the range is rejected. This can only be scored provided both previous method marks have been awarded. Must be in terms of \(p\). Accept e.g. \(p < -\frac{2}{3}, p > 4, p < -\frac{2}{3}\) or \(p > 4\), \(\left\{p: p < -\frac{2}{3} \cup p > 4\right\}\), or variations of these. Accept \(\left(-\infty, -\frac{2}{3}\right), (4, \infty)\). Do not accept \(-\infty, -\frac{2}{3}, [4, \infty)\), \(p \leq -\frac{2}{3}, p \geq 4\), \(4 < p < -\frac{2}{3}\). |
| (b) Total: 3 marks | ||
| Total for Question 6: 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Scheme | Marks | Guidance |
| (i) \(f'(4) = \frac{4(4)^2+10-7(4)^{\frac{1}{2}}}{4(4)^2} = \frac{15}{2}\) | B1 | \(\frac{15}{2}\) oe stated (as the gradient of \(f(x)\) at \(P\)). |
| (i) Total: 1 mark | ||
| (ii) "\(\frac{15}{2} \to -\frac{2}{15}\)" | M1 | Finds the negative reciprocal of their gradient in part (i). If they do not have a gradient in (i) then only allow \(-\frac{2}{15}\) or an attempt at \(\frac{1}{f'(4)}\). |
| \(y+1 = "-\frac{2}{15}"(x-4)\) | M1 | Attempts to find the equation of the normal using a changed gradient to that found in (i) and (4, −1) with the coordinates in the correct positions. If they do not have a gradient in (i) then allow any gradient \(\neq \frac{15}{2}\). If they use \(y = mx+c\) they must proceed as far as \(c = ...\). |
| \(2x+15y+7 = 0\) | A1 | \(2x+15y+7 = 0\) or any multiple of this where all the coefficients are integers and all terms are on the same side of the equation. e.g. \(30y+14+4x = 0\) scores A1. |
| (ii) Total: 4 marks | ||
| 7a Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Scheme | Marks | Guidance |
| \(\frac{4x^2+10-7x^{\frac{1}{2}}}{4x^2} = \pm ...x^2 \pm ...x^{-\frac{3}{2}} \pm ...\) | M1 | Splits into three separate terms with at least one term with the correct index. The index does not need to be processed. e.g. \(x^{-\frac{1}{2}}\). |
| Two of the terms of \(x^2 + \frac{5}{2}x^{-2} - \frac{7}{4}x^{-\frac{1}{2}}\) | A1 | Two of \(x^2 + \frac{5}{2}x^{-2} - \frac{7}{4}x^{-\frac{1}{2}}\) (unsimplified but the indices must be processed). May appear as a list of terms on different lines. May be implied by a correctly integrated expression. |
| dM1A1ft | dM1: Attempts to increase the power by one on at least one term. It is dependent on the previous method mark. The index does not need to be processed for this mark. It cannot be scored for attempting to integrate individual terms on the numerator or denominator. \(\int\left(x^2+\frac{5}{2}x^{-2}-\frac{7}{4}x^{-\frac{1}{2}}\right) dx = \frac{2}{5}x^2+5x^{-2}-\frac{7}{4}x(+c)\) | dM1A1ft |
| \(\frac{2}{5}(4)^{\frac{5}{2}}+5(4)^{\frac{1}{2}}-\frac{7}{4}(4)-c = -1 \Rightarrow c = ...\) | ddM1 | Attempts to substitute \(x=4\) into their changed expression (condone one slip in substituting in), sets equal to −1 and attempts to find \(c\). This mark cannot be scored if they do not have a constant of integration. It is dependent on both of the previous method marks. Do not be concerned by the mechanics of the rearrangement and condone arithmetical slips, but they must achieve a value. |
| \((f(x) =) \frac{2}{5}x^2+5x^{-\frac{1}{2}}-\frac{7}{4}x-\frac{84}{5}\) | A1 | \(\left(f(x)=\right) \frac{2}{5}x^2+5x^{-\frac{1}{2}}-\frac{7}{4}x-\frac{84}{5}\) or any equivalent expression e.g. \(0.4x^2+5x^{-\frac{1}{2}}-1.75x-16.8\). Accept \(x^{\frac{1}{2}}\) for \(x\). Condone equivalent fractions for the coefficients provided both numerator and denominator are integers. e.g. \(\left(f(x)=\right) \frac{x^2}{2.5}+5x^{-\frac{1}{2}}-\frac{7}{4}x-\frac{84}{5}\) is A0. Withhold the final mark if there is still integration notation around the answer or if it is set equal to 0. |
| (b) Total: 6 marks | ||
| Total for Question 7: 10 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Scheme | Marks | Guidance |
| \(x\left(x^3-\frac{7}{2}x-5\right) = \frac{15}{2}-5x\) \(\Rightarrow 2x^4-7x^2-15 = 0\) * | M1A1* | M1: Sets up a correct equation in \(x\). Invisible brackets may be implied by further work. e.g. \(x^4-\frac{7}{2}x^2-5x = \frac{15}{2}-5x\) is acceptable as they can use the equation of the asymptote. A1*: Multiplies out and rearranges the equation to achieve the given answer with no errors seen including invisible brackets. Condone the implied \(=0\) in their working, provided it is present in their final answer. There must be at least one stage of intermediate working between their starting equation and proceeding to the given answer. \(x^4-\frac{7}{2}x^2-5x = \frac{15}{2}-5x \Rightarrow 2x^4-7x^2-15 = 0\) is M1A0*. |
| (a) Total: 2 marks | ||
| \((2x^2+3)(x^2-5) = 0 \Rightarrow (x^2=)...\) | M1 | Attempts to solve the quadratic equation in \(x^2\) by either: factorising, completing the square, quadratic formula. Usual rules apply. Do not be too concerned by the labelling e.g. it is acceptable to use a different variable as well as condoning \(y=x^2\) and \(x=x^2\). This mark cannot be awarded from directly using a calculator and stating the roots. |
| \((x=)(\pm)\sqrt{5}\) | B1 | At least one of \(\sqrt{5}\) and \(-\sqrt{5}\) (and no others). |
| (b) Work for (b) may be seen in (a) which can score | ||
| Note that if they do not square root their solution to the quadratic in \(x^2\) then this mark cannot be scored (and no further marks). Attempts to substitute in $\pm n^{\ |
| Answer/Scheme | Marks | Guidance |
|---|---|---|
| $4(p-2x) = \frac{12+15p}{x+p}$ $8x^2+4px+12+15p-4p^2 = (=0)$ e.g. $a=8, b=4p, c=12+15p-4p^2$ | M1 | Multiplies by $(x+p)$, multiplies out the brackets and collects terms on one side. Condone slips and the omission of $=0$, but there must be an ...x² term. May be implied by further work e.g. their attempt at the discriminant. |
| $(4p)^2-4 \times 8 \times (12+15p-4p^2) (>0)$ | dM1 | Attempts $b^2-4ac$ for their quadratic in $x$ with their $b$ and their $c$ both being in terms of $p$. The values/expressions for $a$, $b$ and $c$ for their quadratic in $x$ must be embedded in the correct positions in the expression but condone invisible brackets and sign slips miscopying. It is dependent on the previous method mark. Condone for this mark if they work in terms of $x$ instead of $p$ provided the coefficients were actually those in terms of $p$ from their original quadratic. i.e. $8x^2+4px+12+15p-4p^2 = 0 \Rightarrow a=8, b=4p, c=12+15p-4p^2$ $\Rightarrow (4x)^2-4 \times 8 \times (12+15x-4x^2) (>0)$ (they have just written $x$ instead of $p$ which is fine). |
| $3p^2-10p-8 > 0$ * | A1* | Achieves $3p^2-10p-8 > 0$ with all stages of working shown: Multiplying out $(p-2x)(x+p)$, Collecting terms on one side of an equation ($=0$ may be implied), Finding the discriminant. At least one intermediate stage of working between the discriminant and the final answer. There must be no errors seen including invisible brackets but condone a missing trailing bracket if it does not alter the processing. e.g. $(4p)^2-4 \times 8 \times (12+15p-4p^2$ is A1*. Condone the $>0$ appearing for the first time in their final answer, provided an incorrect inequality has not been used in earlier working. Do not withhold this mark if they state e.g. "b² – 4ac = 0" as part of their working. |
| **(a) Total: 3 marks** | | |
| e.g. $(p-4)(3p+2) (=0) \Rightarrow 4, -\frac{2}{3}$ | M1 | Attempts to find critical values by solving the given quadratic by either: factorising (do not accept $(p-4)(p+\frac{3}{2})$ unless they have divided both sides of the inequality/equation by 3 first), completing the square $3(p-\frac{5}{3})^2 \pm ... \Rightarrow (p=)...$, quadratic formula. Usual rules apply (see general marking principles for guidance). May be in another variable e.g. $x$. **This mark cannot be awarded from directly using a calculator and stating the roots.** |
| $p < -\frac{2}{3}$ or $p > 4$ | M1A1 | M1: Attempts to find the outside region for their critical values. May use another variable e.g. $x$. May be implied by $p < "-\frac{2}{3}", p > "4"$ or incorrect use of inequalities e.g. $4 < p < -\frac{2}{3}$. **This is not dependent on the first method mark so if a calculator has been used then this mark can still be scored.** A1: $p < -\frac{2}{3}$ or $p > 4$ (or equivalent). (may appear on separate lines). Isw once a correct answer is seen, provided there is no contradiction and no part of the range is rejected. **This can only be scored provided both previous method marks have been awarded. Must be in terms of $p$.** Accept e.g. $p < -\frac{2}{3}, p > 4, p < -\frac{2}{3}$ or $p > 4$, $\left\{p: p < -\frac{2}{3} \cup p > 4\right\}$, or variations of these. Accept $\left(-\infty, -\frac{2}{3}\right), (4, \infty)$. Do not accept $-\infty, -\frac{2}{3}, [4, \infty)$, $p \leq -\frac{2}{3}, p \geq 4$, $4 < p < -\frac{2}{3}$. |
| **(b) Total: 3 marks** | | |
| **Total for Question 6: 6 marks** | | |
**Allow part (b) to follow on from (a) without any labelling of specific parts.**
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# Question 7a:
| Answer/Scheme | Marks | Guidance |
|---|---|---|
| (i) $f'(4) = \frac{4(4)^2+10-7(4)^{\frac{1}{2}}}{4(4)^2} = \frac{15}{2}$ | B1 | $\frac{15}{2}$ oe stated (as the gradient of $f(x)$ at $P$). |
| **(i) Total: 1 mark** | | |
| (ii) "$\frac{15}{2} \to -\frac{2}{15}$" | M1 | Finds the negative reciprocal of their gradient in part (i). If they do not have a gradient in (i) then only allow $-\frac{2}{15}$ or an attempt at $\frac{1}{f'(4)}$. |
| $y+1 = "-\frac{2}{15}"(x-4)$ | M1 | Attempts to find the equation of the normal using a changed gradient to that found in (i) and (4, −1) with the coordinates in the correct positions. If they do not have a gradient in (i) then allow any gradient $\neq \frac{15}{2}$. If they use $y = mx+c$ they must proceed as far as $c = ...$. |
| $2x+15y+7 = 0$ | A1 | $2x+15y+7 = 0$ or any multiple of this where all the coefficients are integers and all terms are on the same side of the equation. e.g. $30y+14+4x = 0$ scores A1. |
| **(ii) Total: 4 marks** | | |
| **7a Total: 4 marks** | | |
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# Question 7b:
| Answer/Scheme | Marks | Guidance |
|---|---|---|
| $\frac{4x^2+10-7x^{\frac{1}{2}}}{4x^2} = \pm ...x^2 \pm ...x^{-\frac{3}{2}} \pm ...$ | M1 | Splits into **three separate terms** with at least one term with the correct index. The index does not need to be processed. e.g. $x^{-\frac{1}{2}}$. |
| Two of the terms of $x^2 + \frac{5}{2}x^{-2} - \frac{7}{4}x^{-\frac{1}{2}}$ | A1 | Two of $x^2 + \frac{5}{2}x^{-2} - \frac{7}{4}x^{-\frac{1}{2}}$ (unsimplified but the indices must be processed). May appear as a list of terms on different lines. May be implied by a correctly integrated expression. |
| dM1A1ft | dM1: Attempts to increase the power by one on at least one term. It is dependent on the previous method mark. The index does not need to be processed for this mark. It cannot be scored for attempting to integrate individual terms on the numerator or denominator. $\int\left(x^2+\frac{5}{2}x^{-2}-\frac{7}{4}x^{-\frac{1}{2}}\right) dx = \frac{2}{5}x^2+5x^{-2}-\frac{7}{4}x(+c)$ | dM1A1ft |
| $\frac{2}{5}(4)^{\frac{5}{2}}+5(4)^{\frac{1}{2}}-\frac{7}{4}(4)-c = -1 \Rightarrow c = ...$ | ddM1 | Attempts to substitute $x=4$ into their changed expression (condone one slip in substituting in), sets equal to −1 and attempts to find $c$. **This mark cannot be scored if they do not have a constant of integration.** It is dependent on both of the previous method marks. Do not be concerned by the mechanics of the rearrangement and condone arithmetical slips, but they must achieve a value. |
| $(f(x) =) \frac{2}{5}x^2+5x^{-\frac{1}{2}}-\frac{7}{4}x-\frac{84}{5}$ | A1 | $\left(f(x)=\right) \frac{2}{5}x^2+5x^{-\frac{1}{2}}-\frac{7}{4}x-\frac{84}{5}$ or any equivalent expression e.g. $0.4x^2+5x^{-\frac{1}{2}}-1.75x-16.8$. Accept $x^{\frac{1}{2}}$ for $x$. Condone equivalent fractions for the coefficients provided both numerator and denominator are integers. e.g. $\left(f(x)=\right) \frac{x^2}{2.5}+5x^{-\frac{1}{2}}-\frac{7}{4}x-\frac{84}{5}$ is A0. Withhold the final mark if there is still integration notation around the answer or if it is set equal to 0. |
| **(b) Total: 6 marks** | | |
| **Total for Question 7: 10 marks** | | |
**Mark (a) and (b) together so do not be concerned with labelling of the parts.**
---
# Question 8a:
| Answer/Scheme | Marks | Guidance |
|---|---|---|
| $x\left(x^3-\frac{7}{2}x-5\right) = \frac{15}{2}-5x$ $\Rightarrow 2x^4-7x^2-15 = 0$ * | M1A1* | M1: Sets up a correct equation in $x$. Invisible brackets may be implied by further work. e.g. $x^4-\frac{7}{2}x^2-5x = \frac{15}{2}-5x$ is acceptable as they can use the equation of the asymptote. A1*: Multiplies out and rearranges the equation to achieve the given answer with no errors seen including invisible brackets. Condone the implied $=0$ in their working, provided it is present in their final answer. There must be at least one stage of intermediate working between their starting equation and proceeding to the given answer. $x^4-\frac{7}{2}x^2-5x = \frac{15}{2}-5x \Rightarrow 2x^4-7x^2-15 = 0$ is M1A0*. |
| **(a) Total: 2 marks** | | |
| $(2x^2+3)(x^2-5) = 0 \Rightarrow (x^2=)...$ | M1 | Attempts to solve the quadratic equation in $x^2$ by either: factorising, completing the square, quadratic formula. Usual rules apply. Do not be too concerned by the labelling e.g. it is acceptable to use a different variable as well as condoning $y=x^2$ and $x=x^2$. **This mark cannot be awarded from directly using a calculator and stating the roots.** |
| $(x=)(\pm)\sqrt{5}$ | B1 | At least one of $\sqrt{5}$ and $-\sqrt{5}$ (and no others). |
| **(b) Work for (b) may be seen in (a) which can score** | | |
| Note that if they do not square root their solution to the quadratic in $x^2$ then this mark cannot be scored (and no further marks). Attempts to substitute in $\pm n^{\\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying on calculator technology are not acceptable.}
The equation
$$4 ( p - 2 x ) = \frac { 12 + 15 p } { x + p } \quad x \neq - p$$
where $p$ is a constant, has two distinct real roots.\\
(a) Show that
$$3 p ^ { 2 } - 10 p - 8 > 0$$
(b) Hence, using algebra, find the range of possible values of $p$
\hfill \mbox{\textit{Edexcel P1 2023 Q6 [6]}}