CAIE P1 2021 June — Question 3 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscriminant and conditions for roots
TypeLine tangent to curve, find constant
DifficultyModerate -0.3 This is a standard tangent-to-curve problem requiring students to set the line equal to the curve, use the discriminant condition (b²-4ac=0) for tangency, solve for m, then find coordinates. It's slightly easier than average as it follows a well-practiced procedure with straightforward algebra, though it does require understanding the tangency condition and completing multiple steps.
Spec1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations

3 A line with equation \(y = m x - 6\) is a tangent to the curve with equation \(y = x ^ { 2 } - 4 x + 3\).
Find the possible values of the constant \(m\), and the corresponding coordinates of the points at which the line touches the curve.

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(x^2 - 4x + 3 = mx - 6\) leading to \(x^2 - x(4+m) + 9\)*M1 Equating and gathering terms. May be implied on the next line.
\(b^2 - 4ac\) leading to \((4+m)^2 - 4 \times 9\)DM1 SOI. Use of the discriminant with their \(a\), \(b\) and \(c\)
\(4 + m = \pm 6\) or \((m-2)(m+10) = 0\) leading to \(m = 2\) or \(-10\)A1 Must come from \(b^2 - 4ac = 0\) SOI
Substitute both their \(m\) values into their equation in line 1DM1
\(m = 2\) leading to \(x = 3\); \(m = -10\) leading to \(x = -3\)A1
\((3, 0), (-3, 24)\)A1 Accept 'when \(x=3, y=0\); when \(x=-3, y=24\)'. If final A0A0 scored, SC B1 for one point correct WWW
Alternative method:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dy}{dx} = 2x - 4 \rightarrow 2x - 4 = m\)*M1
\(x^2 - 4x + 3 = (2x-4)x - 6\)DM1
\(x^2 - 4x + 3 = 2x^2 - 4x - 6 \rightarrow 9 = x^2 \rightarrow x = \pm 3\)A1
\(y = 0\), \(24\) or \((3, 0)\), \((-3, 24)\)A1
Substitute both their \(x\) values into their equation in line 1DM1 Or substitute both their \((x, y)\) into \(y = mx - 6\)
When \(x = 3\), \(m = 2\); when \(x = -3\), \(m = -10\)A1 If A0, DM1, A0 scored, SC B1 for one point correct WWW
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - 4x + 3 = mx - 6$ leading to $x^2 - x(4+m) + 9$ | *M1 | Equating and gathering terms. May be implied on the next line. |
| $b^2 - 4ac$ leading to $(4+m)^2 - 4 \times 9$ | DM1 | SOI. Use of the discriminant with their $a$, $b$ and $c$ |
| $4 + m = \pm 6$ or $(m-2)(m+10) = 0$ leading to $m = 2$ or $-10$ | A1 | Must come from $b^2 - 4ac = 0$ SOI |
| Substitute both their $m$ values into their equation in line 1 | DM1 | |
| $m = 2$ leading to $x = 3$; $m = -10$ leading to $x = -3$ | A1 | |
| $(3, 0), (-3, 24)$ | A1 | Accept 'when $x=3, y=0$; when $x=-3, y=24$'. If final A0A0 scored, **SC B1** for one point correct WWW |

**Alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 2x - 4 \rightarrow 2x - 4 = m$ | *M1 | |
| $x^2 - 4x + 3 = (2x-4)x - 6$ | DM1 | |
| $x^2 - 4x + 3 = 2x^2 - 4x - 6 \rightarrow 9 = x^2 \rightarrow x = \pm 3$ | A1 | |
| $y = 0$, $24$ or $(3, 0)$, $(-3, 24)$ | A1 | |
| Substitute both their $x$ values into their equation in line 1 | DM1 | Or substitute both their $(x, y)$ into $y = mx - 6$ |
| When $x = 3$, $m = 2$; when $x = -3$, $m = -10$ | A1 | If A0, DM1, A0 scored, **SC B1** for one point correct WWW |

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3 A line with equation $y = m x - 6$ is a tangent to the curve with equation $y = x ^ { 2 } - 4 x + 3$.\\
Find the possible values of the constant $m$, and the corresponding coordinates of the points at which the line touches the curve.\\

\hfill \mbox{\textit{CAIE P1 2021 Q3 [6]}}