| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Line-curve intersection conditions |
| Difficulty | Standard +0.3 This is a standard discriminant problem requiring substitution of the line into the curve equation, rearranging to a quadratic, and applying b²-4ac > 0 for two distinct intersections. It's slightly easier than average as it follows a well-practiced procedure with straightforward algebra and no conceptual surprises. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Eliminate \(y\) and form quadratic equation \(= 0\) or quadratic expression \(\{=0\}\) | M1 | AO3.1a |
| \(\{3x - 2(2x^2-5) = k \Rightarrow\}\ -4x^2 + 3x + 10 - k = 0\) | A1 | AO1.1b |
| \(\{"b^2-4ac">0 \Rightarrow\}\ 3^2 - 4(-4)(10-k) > 0\) | dM1 | AO2.1 |
| \(9 + 16(10-k) > 0 \Rightarrow 169 - 16k > 0\) | ||
| Critical value of \(\frac{169}{16}\) obtained | B1 | AO1.1b |
| \(k < \frac{169}{16}\) | A1 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Eliminate \(y\), form quadratic \(= 0\): \(8y^2 + (8k-9)y + 2k^2 - 45 = 0\) | M1, A1 | AO3.1a, AO1.1b |
| \(\{"b^2-4ac">0\}\ (8k-9)^2 - 4(8)(2k^2-45) > 0\) | dM1 | AO2.1 |
| \(64k^2 - 144k + 81 - 64k^2 + 1440 > 0 \Rightarrow -144k + 1521 > 0\) | ||
| Critical value \(\frac{169}{16}\) | B1 | AO1.1b |
| \(k < \frac{169}{16}\) | A1 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 4x,\ m_l = \frac{3}{2} \Rightarrow 4x = \frac{3}{2} \Rightarrow x = \frac{3}{8}\). So \(y = 2\left(\frac{3}{8}\right)^2 - 5 = -\frac{151}{32}\) | M1, A1 | AO3.1a, AO1.1b |
| \(k = 3\left(\frac{3}{8}\right) - 2\left(-\frac{151}{32}\right) \Rightarrow k = \ldots\) | dM1 | AO2.1 |
| Critical value \(\frac{169}{16}\) | B1 | AO1.1b |
| \(k < \frac{169}{16}\) | A1 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy of eliminating \(x\) or \(y\) and manipulating to form a quadratic equation \(= 0\) | M1 | Must form quadratic equation or quadratic expression |
| \(-4x^2 + 3x + 10 - k = 0\) or \(4x^2 - 3x - 10 + k = 0\) or one-sided quadratic \(-4x^2 + 3x + 10 - k\) or \(4x^2 - 3x - 10 + k\); OR \(8y^2 + (8k-9)y + 2k^2 - 45 = 0\) or one-sided equivalent | A1 | Correct algebra leading to either form |
| Interprets \(3x - 2y = k\) intersecting \(y = 2x^2 - 5\) at two distinct points by applying \(b^2 - 4ac > 0\) to their quadratic | dM1 | Depends on previous M mark |
| See scheme | B1 | |
| \(k < \dfrac{169}{16}\) or \(\left\{k : k < \dfrac{169}{16}\right\}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy of using differentiation to find values of \(x\) and \(y\) where \(3x - 2y = k\) is a tangent to \(y = 2x^2 - 5\) | M1 | |
| Correct algebra leading to \(x = \dfrac{3}{8}\), \(y = -\dfrac{151}{32}\) | A1 | |
| Full method of substituting \(x = \dfrac{3}{8}\), \(y = -\dfrac{151}{32}\) into \(l\) and attempting to find value of \(k\) | dM1 | Depends on previous M mark |
| See scheme | B1 | |
| \(k < \dfrac{169}{16}\) or \(\left\{k : k < \dfrac{169}{16}\right\}\) | A1 | Deduces correct answer |
## Question 5:
### Main Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Eliminate $y$ and form quadratic equation $= 0$ or quadratic expression $\{=0\}$ | M1 | AO3.1a |
| $\{3x - 2(2x^2-5) = k \Rightarrow\}\ -4x^2 + 3x + 10 - k = 0$ | A1 | AO1.1b |
| $\{"b^2-4ac">0 \Rightarrow\}\ 3^2 - 4(-4)(10-k) > 0$ | dM1 | AO2.1 |
| $9 + 16(10-k) > 0 \Rightarrow 169 - 16k > 0$ | | |
| Critical value of $\frac{169}{16}$ obtained | B1 | AO1.1b |
| $k < \frac{169}{16}$ | A1 | AO1.1b |
### Alternative Method 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Eliminate $y$, form quadratic $= 0$: $8y^2 + (8k-9)y + 2k^2 - 45 = 0$ | M1, A1 | AO3.1a, AO1.1b |
| $\{"b^2-4ac">0\}\ (8k-9)^2 - 4(8)(2k^2-45) > 0$ | dM1 | AO2.1 |
| $64k^2 - 144k + 81 - 64k^2 + 1440 > 0 \Rightarrow -144k + 1521 > 0$ | | |
| Critical value $\frac{169}{16}$ | B1 | AO1.1b |
| $k < \frac{169}{16}$ | A1 | AO1.1b |
### Alternative Method 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 4x,\ m_l = \frac{3}{2} \Rightarrow 4x = \frac{3}{2} \Rightarrow x = \frac{3}{8}$. So $y = 2\left(\frac{3}{8}\right)^2 - 5 = -\frac{151}{32}$ | M1, A1 | AO3.1a, AO1.1b |
| $k = 3\left(\frac{3}{8}\right) - 2\left(-\frac{151}{32}\right) \Rightarrow k = \ldots$ | dM1 | AO2.1 |
| Critical value $\frac{169}{16}$ | B1 | AO1.1b |
| $k < \frac{169}{16}$ | A1 | AO1.1b |
# Question 5:
## Part (main):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy of eliminating $x$ or $y$ and manipulating to form a quadratic equation $= 0$ | M1 | Must form quadratic equation or quadratic expression |
| $-4x^2 + 3x + 10 - k = 0$ or $4x^2 - 3x - 10 + k = 0$ or one-sided quadratic $-4x^2 + 3x + 10 - k$ or $4x^2 - 3x - 10 + k$; OR $8y^2 + (8k-9)y + 2k^2 - 45 = 0$ or one-sided equivalent | A1 | Correct algebra leading to either form |
| Interprets $3x - 2y = k$ intersecting $y = 2x^2 - 5$ at two distinct points by applying $b^2 - 4ac > 0$ to their quadratic | dM1 | Depends on previous M mark |
| See scheme | B1 | |
| $k < \dfrac{169}{16}$ or $\left\{k : k < \dfrac{169}{16}\right\}$ | A1 | Correct answer |
## Alt 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy of using differentiation to find values of $x$ and $y$ where $3x - 2y = k$ is a tangent to $y = 2x^2 - 5$ | M1 | |
| Correct algebra leading to $x = \dfrac{3}{8}$, $y = -\dfrac{151}{32}$ | A1 | |
| Full method of substituting $x = \dfrac{3}{8}$, $y = -\dfrac{151}{32}$ into $l$ and attempting to find value of $k$ | dM1 | Depends on previous M mark |
| See scheme | B1 | |
| $k < \dfrac{169}{16}$ or $\left\{k : k < \dfrac{169}{16}\right\}$ | A1 | Deduces correct answer |
---\begin{enumerate}
\item The line $l$ has equation
\end{enumerate}
$$3 x - 2 y = k$$
where $k$ is a real constant.\\
Given that the line $l$ intersects the curve with equation
$$y = 2 x ^ { 2 } - 5$$
at two distinct points, find the range of possible values for $k$.
\hfill \mbox{\textit{Edexcel Paper 2 Q5 [5]}}