Moderate -0.5 This is a straightforward discriminant problem requiring students to set the equations equal, rearrange to a quadratic, and show b²-4ac ≥ 0 for all k. It's slightly easier than average because it's a direct application of a standard technique with clear signposting ('show that... for all values'), requiring only algebraic manipulation without problem-solving insight.
1 A line has equation \(y = 3 x - 2 k\) and a curve has equation \(y = x ^ { 2 } - k x + 2\), where \(k\) is a constant. Show that the line and the curve meet for all values of \(k\).
\(x^2 - kx + 2 = 3x - 2k\) leading to \(x^2 - x(k+3) + (2+2k) [= 0]\)
M1
3-term quadratic, may be implied in the discriminant
\(b^2 - 4ac = (k+3)^2 - 8(1+k)\) (ignore '\(= 0\)' at this stage)
DM1
Cannot just be seen in the quadratic formula
\(= (k-1)^2\) accept \((k-1)(k-1)\)
A1
Or use of calculus to show minimum of zero at \(k = 1\) or sketch of \(f(k) = k^2 - 2k + 1\)
\(\geqslant 0\) Hence will meet for all values of \(k\)
A1
Clear conclusion
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - kx + 2 = 3x - 2k$ leading to $x^2 - x(k+3) + (2+2k) [= 0]$ | **M1** | 3-term quadratic, may be implied in the discriminant |
| $b^2 - 4ac = (k+3)^2 - 8(1+k)$ (ignore '$= 0$' at this stage) | **DM1** | Cannot just be seen in the quadratic formula |
| $= (k-1)^2$ accept $(k-1)(k-1)$ | **A1** | Or use of calculus to show minimum of zero at $k = 1$ or sketch of $f(k) = k^2 - 2k + 1$ |
| $\geqslant 0$ Hence will meet for all values of $k$ | **A1** | Clear conclusion |
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1 A line has equation $y = 3 x - 2 k$ and a curve has equation $y = x ^ { 2 } - k x + 2$, where $k$ is a constant. Show that the line and the curve meet for all values of $k$.\\
\hfill \mbox{\textit{CAIE P1 2023 Q1 [4]}}