Standard +0.3 This is a standard tangent-to-curve problem requiring students to use the discriminant condition (b²-4ac=0) after substituting the line equation into the curve. It involves algebraic manipulation and solving a quadratic equation, but follows a well-practiced procedure with no novel insight required. Slightly above average difficulty due to the parameter k appearing in multiple terms, requiring careful algebraic handling.
6 The equation of a curve is \(y = ( 2 k - 3 ) x ^ { 2 } - k x - ( k - 2 )\), where \(k\) is a constant. The line \(y = 3 x - 4\) is a tangent to the curve.
Find the value of \(k\).
Use of discriminant (dependent on both previous M marks)
\(9k^2 - 54k + 81[=0]\) leading to \(k^2 - 6k + 9 = 0\)
M1
Simplifying and solving *their* 3-term quadratic in \(k\)
\(k = 3\)
A1
Alternative method:
\((2k-3)x^2 - kx - (k-2) = 3x-4\)
\*M1
Equating curve and line
\(2(2k-3)x - k = 3 \Rightarrow x = \dfrac{k+3}{4k-6}\) or \(k = \dfrac{3+6x}{4x-1}\)
DM1
Differentiating and solving for \(x\) or \(k\)
Either \((2k-3)\!\left(\dfrac{k+3}{4k-6}\right)^2 - k\!\left(\dfrac{k+3}{4k-6}\right) - (k-2) = 3\!\left(\dfrac{k+3}{4k-6}\right) - 4\) or \(4x\!\left(\dfrac{3x^2+3x-6}{2x^2-x-1}\right) - 6x - \!\left(\dfrac{3x^2+3x-6}{2x^2-x-1}\right) = 3\)
DM1
Substituting *their* \(x\) into equation or *their* \(k = \dfrac{3x^2+3x-6}{2x^2-x-1}\) or \(k = \dfrac{3x+6}{2x+1}\) into derivative equation (dependent on both previous M marks)
\(9k^2 - 54k + 81[=0]\) leading to \(k^2 - 6k + 9 = 0\)
M1
Simplifying and solving *their* 3-term quadratic in \(k\) (or solving for \(x\))
\(k = 3\)
A1
SC If M0, B1 for differentiating, equating to 3 and solving for \(x\) or \(k\)
5
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2k-3)x^2 - kx - (k-2) = 3x - 4$ | \*M1 | Equating curve and line |
| $(2k-3)x^2 - (k+3)x - (k-6)[=0]$ | DM1 | Forming a 3-term quadratic |
| $(k+3)^2 + 4(2k-3)(k-6)[=0]$ | DM1 | Use of discriminant (dependent on **both** previous M marks) |
| $9k^2 - 54k + 81[=0]$ leading to $k^2 - 6k + 9 = 0$ | M1 | Simplifying and solving *their* 3-term quadratic in $k$ |
| $k = 3$ | A1 | |
| **Alternative method:** | | |
| $(2k-3)x^2 - kx - (k-2) = 3x-4$ | \*M1 | Equating curve and line |
| $2(2k-3)x - k = 3 \Rightarrow x = \dfrac{k+3}{4k-6}$ or $k = \dfrac{3+6x}{4x-1}$ | DM1 | Differentiating and solving for $x$ or $k$ |
| Either $(2k-3)\!\left(\dfrac{k+3}{4k-6}\right)^2 - k\!\left(\dfrac{k+3}{4k-6}\right) - (k-2) = 3\!\left(\dfrac{k+3}{4k-6}\right) - 4$ or $4x\!\left(\dfrac{3x^2+3x-6}{2x^2-x-1}\right) - 6x - \!\left(\dfrac{3x^2+3x-6}{2x^2-x-1}\right) = 3$ | DM1 | Substituting *their* $x$ into equation or *their* $k = \dfrac{3x^2+3x-6}{2x^2-x-1}$ or $k = \dfrac{3x+6}{2x+1}$ into derivative equation (dependent on **both** previous M marks) |
| $9k^2 - 54k + 81[=0]$ leading to $k^2 - 6k + 9 = 0$ | M1 | Simplifying and solving *their* 3-term quadratic in $k$ (or solving for $x$) |
| $k = 3$ | A1 | SC If M0, B1 for differentiating, equating to 3 and solving for $x$ or $k$ |
| | **5** | |
6 The equation of a curve is $y = ( 2 k - 3 ) x ^ { 2 } - k x - ( k - 2 )$, where $k$ is a constant. The line $y = 3 x - 4$ is a tangent to the curve.
Find the value of $k$.\\
\hfill \mbox{\textit{CAIE P1 2021 Q6 [5]}}