OCR PURE — Question 9 6 marks

Exam BoardOCR
ModulePURE
Marks6
PaperDownload PDF ↗
TopicTangents, normals and gradients
TypeTangent parallel to given line
DifficultyStandard +0.3 This is a slightly above-routine tangent problem requiring students to set up simultaneous conditions (line passes through a given point AND is tangent to the curve), leading to a quadratic discriminant condition. While it involves multiple steps, the technique is standard for A-level and requires no novel insight beyond applying the tangent condition Δ=0.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07m Tangents and normals: gradient and equations
9 In this question you must show detailed reasoning.
Find the equation of the straight line with positive gradient that passes through \(( 0,2 )\) and is a tangent to the curve \(y = x ^ { 2 } - x + 6\).
Question 9:
AnswerMarks Guidance
\(y = mx + 2\)M1 seen or implied
\(mx + 2 = x^2 - x + 6\)M1 Attempt solve equations of line and curve simultaneously
\(x^2 - (m+1)x + 4 = 0\)A1
Equal roots, hence \((m+1)^2 - 16 = 0\)M1 Attempt use \(b^2 - 4ac = 0\)
\(m = -5\) inadmissibleB1
\(m = 3\), Equation is \(y = 3x+2\)A1
Alternative Method:
AnswerMarks Guidance
grad of tangent \(= 2x-1\)M1 Must be correct
Where tgt touches curve \(y = (2x-1)x + 2\)M1 Allow M1 for just \(y=(2x-1)x+2\) oe
Hence \((2x-1)x + 2 = x^2 - x + 6\)M1
\(x^2 = 4\)A1
\(x = -2\) gives gradient \(= -5\), so rejectB1
\(x = 2\) gives gradient \(= 3\), Equation is \(y = 3x+2\)A1
[6] 
# Question 9:
| $y = mx + 2$ | M1 | seen or implied |
|---|---|---|
| $mx + 2 = x^2 - x + 6$ | M1 | Attempt solve equations of line and curve simultaneously |
| $x^2 - (m+1)x + 4 = 0$ | A1 | |
| Equal roots, hence $(m+1)^2 - 16 = 0$ | M1 | Attempt use $b^2 - 4ac = 0$ |
| $m = -5$ inadmissible | B1 | |
| $m = 3$, Equation is $y = 3x+2$ | A1 | |

**Alternative Method:**
| grad of tangent $= 2x-1$ | M1 | Must be correct |
|---|---|---|
| Where tgt touches curve $y = (2x-1)x + 2$ | M1 | Allow M1 for just $y=(2x-1)x+2$ oe |
| Hence $(2x-1)x + 2 = x^2 - x + 6$ | M1 | |
| $x^2 = 4$ | A1 | |
| $x = -2$ gives gradient $= -5$, so reject | B1 | |
| $x = 2$ gives gradient $= 3$, Equation is $y = 3x+2$ | A1 | |
| | [6] | |
9 In this question you must show detailed reasoning.\\
Find the equation of the straight line with positive gradient that passes through $( 0,2 )$ and is a tangent to the curve $y = x ^ { 2 } - x + 6$.

\hfill \mbox{\textit{OCR PURE  Q9 [6]}}
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