Standard +0.3 This is a straightforward discriminant problem requiring students to set the equations equal, rearrange to standard quadratic form, and show b²-4ac > 0 for all m. The algebra is routine and the 'for all m' condition makes it easier since students just need to show the discriminant is a positive constant or has no real roots that make it non-positive. Slightly easier than average due to its mechanical nature.
3 The equation of a curve is \(y = 2 x ^ { 2 } + m ( 2 x + 1 )\), where \(m\) is a constant, and the equation of a line is \(y = 6 x + 4\).
Show that, for all values of \(m\), the line intersects the curve at two distinct points.
\(y\) eliminated and all terms on one side with correct algebraic steps. Condone \(\pm\) errors
Using \(b^2 - 4ac\) on \(2x^2 + x(2m-6) + m - 4\ (= 0)\)
DM1
Any use of discriminant with their \(a\), \(b\) and \(c\) identified correctly.
\(4m^2 - 32m + 68\) or \(2m^2 - 16m + 34\) or \(m^2 - 8m + 17\)
A1
\((2m-8)^2 + k\) or \((m-4)^2 + k\) or minimum point \((4, k)\) or finds \(b^2 - 4ac\ (= -4, -16, -64)\)
DM1
OE. Any valid method attempted on their 3-term quadratic
\((m-4)^2 + 1\) oe + always \(> 0 \rightarrow 2\) solutions for all values of \(m\), or Minimum point \((4,1)\) + (fn) always \(> 0 \rightarrow 2\) solutions for all values of \(m\), or \(b^2 - 4ac < 0\) + no solutions \(\rightarrow 2\) solutions for the original equation for all values of \(m\)
A1
Clear and correct reasoning and conclusion without wrong working.
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| $2x^2 + m(2x+1) - 6x - 4 (= 0)$ | *M1 | $y$ eliminated and all terms on one side with correct algebraic steps. Condone $\pm$ errors |
| Using $b^2 - 4ac$ on $2x^2 + x(2m-6) + m - 4\ (= 0)$ | DM1 | Any use of discriminant with their $a$, $b$ and $c$ identified correctly. |
| $4m^2 - 32m + 68$ **or** $2m^2 - 16m + 34$ **or** $m^2 - 8m + 17$ | A1 | |
| $(2m-8)^2 + k$ **or** $(m-4)^2 + k$ **or** minimum point $(4, k)$ **or** finds $b^2 - 4ac\ (= -4, -16, -64)$ | DM1 | OE. Any valid method attempted on their 3-term quadratic |
| $(m-4)^2 + 1$ oe + always $> 0 \rightarrow 2$ solutions for all values of $m$, **or** Minimum point $(4,1)$ + (fn) always $> 0 \rightarrow 2$ solutions for all values of $m$, **or** $b^2 - 4ac < 0$ + no solutions $\rightarrow 2$ solutions for the original equation for all values of $m$ | A1 | Clear and correct reasoning and conclusion without wrong working. |
---
3 The equation of a curve is $y = 2 x ^ { 2 } + m ( 2 x + 1 )$, where $m$ is a constant, and the equation of a line is $y = 6 x + 4$.
Show that, for all values of $m$, the line intersects the curve at two distinct points.\\
\hfill \mbox{\textit{CAIE P1 2020 Q3 [5]}}