Question 6 - A-Level Maths

CAIE P1 2019 November — Question 6 7 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2019
Session	November
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chain Rule
Type	Tangent with specified gradient
Difficulty	Standard +0.3 This is a standard tangent problem requiring equating line and curve equations, using the discriminant condition (b²-4ac=0) for tangency, then finding contact points. It involves routine algebraic manipulation and applying a well-known technique, making it slightly easier than average for A-level.
Spec	1.02f Solve quadratic equations: including in a function of unknown 1.07m Tangents and normals: gradient and equations

6 A straight line has gradient \(m\) and passes through the point ( \(0 , - 2\) ). Find the two values of \(m\) for which the line is a tangent to the curve \(y = x ^ { 2 } - 2 x + 7\) and, for each value of \(m\), find the coordinates of the point where the line touches the curve.

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Question 6:

Answer	Marks	Guidance
Answer	Marks	Guidance
Equation of line is \(y = mx - 2\)	B1	OR
\(x^2 - 2x + 7 = mx - 2 \rightarrow x^2 - x(2+m) + 9 = 0\)	M1
Apply \(b^2 - 4ac(=0) \rightarrow (2+m)^2 - 4\times 9\ (=0)\)	*M1
\(m = 4\) or \(-8\)	A1
\(m=4 \rightarrow x^2 - 6x + 9 = 0 \rightarrow x = 3\) ; \(m=-8 \rightarrow x^2 + 6x + 9 = 0 \rightarrow x = -3\)	DM1
\((3, 10),\ (-3, 22)\)	A1A1
Alternative method:
\(\frac{dy}{dx} = 2x - 2\)	B1
\(2x - 2 = m\)	M1
\(x^2 - 2x + 7 = (2x-2)x - 2 = 2x^2 - 2x - 2\)	M1
\(x^2 - 9 = 0 \rightarrow x = \pm 3\)	A1
\((3, 10),\ (-3, 22)\)	A1A1
When \(x=3,\ m=4\); when \(x=-3,\ m=-8\)	A1
7

## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Equation of line is $y = mx - 2$ | B1 | OR |
| $x^2 - 2x + 7 = mx - 2 \rightarrow x^2 - x(2+m) + 9 = 0$ | M1 | |
| Apply $b^2 - 4ac(=0) \rightarrow (2+m)^2 - 4\times 9\ (=0)$ | *M1 | |
| $m = 4$ or $-8$ | A1 | |
| $m=4 \rightarrow x^2 - 6x + 9 = 0 \rightarrow x = 3$ ; $m=-8 \rightarrow x^2 + 6x + 9 = 0 \rightarrow x = -3$ | DM1 | |
| $(3, 10),\ (-3, 22)$ | A1A1 | |
| **Alternative method:** | | |
| $\frac{dy}{dx} = 2x - 2$ | B1 | |
| $2x - 2 = m$ | M1 | |
| $x^2 - 2x + 7 = (2x-2)x - 2 = 2x^2 - 2x - 2$ | M1 | |
| $x^2 - 9 = 0 \rightarrow x = \pm 3$ | A1 | |
| $(3, 10),\ (-3, 22)$ | A1A1 | |
| When $x=3,\ m=4$; when $x=-3,\ m=-8$ | A1 | |
| | **7** | |

Show LaTeX source

6 A straight line has gradient $m$ and passes through the point ( $0 , - 2$ ). Find the two values of $m$ for which the line is a tangent to the curve $y = x ^ { 2 } - 2 x + 7$ and, for each value of $m$, find the coordinates of the point where the line touches the curve.\\

\hfill \mbox{\textit{CAIE P1 2019 Q6 [7]}}

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