CAIE P1 2007 June — Question 1 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2007
SessionJune
Marks4
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Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypeTangent with given gradient or condition
DifficultyStandard +0.3 This is a straightforward tangent problem requiring substitution of the line equation into the curve equation, forming a quadratic, and applying the discriminant condition (b²-4ac=0) for tangency. It's slightly above average difficulty due to the squared y term requiring careful algebraic manipulation, but follows a standard method taught explicitly in P1 courses.
Spec1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations
1 Find the value of the constant \(c\) for which the line \(y = 2 x + c\) is a tangent to the curve \(y ^ { 2 } = 4 x\).
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Eliminates \(x\) or \(y\) completelyM1 Aims to make \(x\) or \(y\) subject + substitute
\(y^2 - 2y + 2c\) or \(4x^2 + x(4c-4) + c^2 = 0\)M1 Correct quadratic – not necessarily \(= 0\)
Use of \(b^2 - 4ac = 0\)A1 Used correctly on "quadratic = 0"
\(\rightarrow c = \frac{1}{2}\)A1 [4] co
[or gradients equal \(2 = \frac{1}{\sqrt{x}}\) M1A1 \(\rightarrow\) value for \(x, y\) and \(c\). M1A1] 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Eliminates $x$ or $y$ completely | M1 | Aims to make $x$ or $y$ subject + substitute |
| $y^2 - 2y + 2c$ or $4x^2 + x(4c-4) + c^2 = 0$ | M1 | Correct quadratic – not necessarily $= 0$ |
| Use of $b^2 - 4ac = 0$ | A1 | Used correctly on "quadratic = 0" |
| $\rightarrow c = \frac{1}{2}$ | A1 [4] | co |
| [or gradients equal $2 = \frac{1}{\sqrt{x}}$ M1A1 $\rightarrow$ value for $x, y$ and $c$. M1A1] | | |

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1 Find the value of the constant $c$ for which the line $y = 2 x + c$ is a tangent to the curve $y ^ { 2 } = 4 x$.

\hfill \mbox{\textit{CAIE P1 2007 Q1 [4]}}
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Q1 4 Q2 4 Q3 4 Q4 4 Q5 5 Q6 7 Q7 7 Q8 8 Q9 8 Q10 12 Q11 12