CAIE P1 2023 November — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscriminant and conditions for roots
TypeLine tangent to curve, find constant
DifficultyStandard +0.3 This is a standard tangent problem requiring students to use the condition that tangent means one point of intersection (discriminant = 0), then solve a quadratic in c and find corresponding coordinates. It's slightly above average difficulty due to the parameter c appearing in both equations and requiring careful algebraic manipulation, but follows a well-practiced technique with no novel insight needed.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.07m Tangents and normals: gradient and equations
6 A line has equation \(y = 6 x - c\) and a curve has equation \(y = c x ^ { 2 } + 2 x - 3\), where \(c\) is a constant. The line is a tangent to the curve at point \(P\). Find the possible values of \(c\) and the corresponding coordinates of \(P\).
Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(cx^2 + 2x - 3 = 6x - c\) leading to \(cx^2 - 4x + (c-3) = 0\)B1 3-term quadratic
\(16 - 4c(c-3) = 0\)\*M1 Apply \(b^2 - 4ac = 0\) ('\(= 0\)' may be implied in subsequent work). *Their* coefficients must be substituted correctly
\(4c^2 - 12c - 16 = 0\) leading to \([4](c-4)(c+1) = 0\) leading to \(c = 4\) and \(-1\)A1 Dependent on factorisation oe
When \(c = 4\), \(4x^2 - 4x + 1 = 0\), \([(2x-1)^2 = 0]\)DM1 OE. Substituting *their* \(c = 4\) into *their* quadratic equation
\(x = \frac{1}{2}\), \(y = -1\)A1 Both required
When \(c = -1\), \(x^2 + 4x + 4 = 0\), \([(x+2)^2 = 0]\)DM1 OE. Substituting *their* \(c = -1\) into *their* quadratic equation
\(x = -2\), \(y = -11\)A1 Both required
Alternative method:
\(\frac{dy}{dx} = 2cx + 2\)B1
\(2cx + 2 = 6\)M1 Equating *their* curve gradient and 6
\(c = \frac{2}{x}\)A1 SOI
\(2x^2 + 3x - 2 = 0\)DM1 Substitute \(c = \frac{2}{x}\) into \(cx^2 + 2x - 3 = 6x - c\). Simplify to 3-term quadratic
\((2x-1)(x+2) = 0 \rightarrow x = \frac{1}{2}\) or \(-2\)A1 Dependent on factorisation. Both required
\(c = 4\) and \(-1\)A1 Both required, if DM0 given SC B1 for both
\(y = -1\) and \(-11\)A1 Both required, if DM0 given SC B1 for both. SC one correct \((x, y)\). A1 only
7 
## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $cx^2 + 2x - 3 = 6x - c$ leading to $cx^2 - 4x + (c-3) = 0$ | B1 | 3-term quadratic |
| $16 - 4c(c-3) = 0$ | \*M1 | Apply $b^2 - 4ac = 0$ ('$= 0$' may be implied in subsequent work). *Their* coefficients must be substituted correctly |
| $4c^2 - 12c - 16 = 0$ leading to $[4](c-4)(c+1) = 0$ leading to $c = 4$ and $-1$ | A1 | Dependent on factorisation oe |
| When $c = 4$, $4x^2 - 4x + 1 = 0$, $[(2x-1)^2 = 0]$ | DM1 | OE. Substituting *their* $c = 4$ into *their* quadratic equation |
| $x = \frac{1}{2}$, $y = -1$ | A1 | Both required |
| When $c = -1$, $x^2 + 4x + 4 = 0$, $[(x+2)^2 = 0]$ | DM1 | OE. Substituting *their* $c = -1$ into *their* quadratic equation |
| $x = -2$, $y = -11$ | A1 | Both required |
| **Alternative method:** | | |
| $\frac{dy}{dx} = 2cx + 2$ | B1 | |
| $2cx + 2 = 6$ | M1 | Equating *their* curve gradient and 6 |
| $c = \frac{2}{x}$ | A1 | SOI |
| $2x^2 + 3x - 2 = 0$ | DM1 | Substitute $c = \frac{2}{x}$ into $cx^2 + 2x - 3 = 6x - c$. Simplify to 3-term quadratic |
| $(2x-1)(x+2) = 0 \rightarrow x = \frac{1}{2}$ or $-2$ | A1 | Dependent on factorisation. Both required |
| $c = 4$ and $-1$ | A1 | Both required, if DM0 given **SC B1** for both |
| $y = -1$ and $-11$ | A1 | Both required, if DM0 given **SC B1** for both. **SC** one correct $(x, y)$. A1 only |
| | **7** | |
6 A line has equation $y = 6 x - c$ and a curve has equation $y = c x ^ { 2 } + 2 x - 3$, where $c$ is a constant. The line is a tangent to the curve at point $P$.

Find the possible values of $c$ and the corresponding coordinates of $P$.\\

\hfill \mbox{\textit{CAIE P1 2023 Q6 [7]}}
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