OCR MEI C1 2010 January — Question 9 5 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Year2010
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimultaneous equations
TypeIntersection existence or conditions
DifficultyModerate -0.5 This is a straightforward application of the discriminant method: substitute the line equation into the curve, rearrange to get a quadratic, and show b²-4ac < 0. While it requires understanding of what 'no intersection' means algebraically, it's a standard technique taught early in C1 with minimal computational complexity, making it slightly easier than average.
Spec1.01a Proof: structure of mathematical proof and logical steps1.02d Quadratic functions: graphs and discriminant conditions

9 Prove that the line \(y = 3 x - 10\) does not intersect the curve \(y = x ^ { 2 } - 5 x + 7\).

Question 9:
AnswerMarks Guidance
AnswerMark Guidance
\(x^2 - 5x + 7 = 3x - 10\)M1 Or attempt to substitute \(\frac{y+10}{3}\) for \(x\)
\(x^2 - 8x + 17 [= 0]\) o.e. or \(y^2 - 4y + 13 [= 0]\) o.e.M1 Condone one error; allow M1 for \(x^2 - 8x = -17\) [oe for \(y\)] only if they go on to completing the square method
Use of \(b^2 - 4ac\) with numbers substituted (condone one error in substitution) (may be in quadratic formula)M1 Or \((x-4)^2 = 16 - 17\) or \((x-4)^2 + 1 = 0\) (condone one error)
\(b^2 - 4ac = 64 - 68\) or \(-4\) cao [or \(16 - 52\) or \(-36\) if \(y\) used]A1 Or \((x-4)^2 = -1\) or \(x = 4 \pm \sqrt{-1}\) [or \((y-2)^2 = -9\) or \(y = 2 \pm \sqrt{-9}\)]
\([< 0]\) so no [real] roots [so line and curve do not intersect]A1 Or conclusion from completing square; needs to be explicit correct conclusion and correct ft; allow '\(< 0\) so no intersection' o.e.; allow '\(-4\) so no roots' etc; allow A2 for full argument from sum of two squares \(= 0\); A1 for weaker correct conclusion; some may use the condition \(b^2 < 4ac\) for no real roots; allow equivalent marks, with first A1 for \(64 < 68\) o.e.
## Question 9:

| Answer | Mark | Guidance |
|--------|------|----------|
| $x^2 - 5x + 7 = 3x - 10$ | M1 | Or attempt to substitute $\frac{y+10}{3}$ for $x$ |
| $x^2 - 8x + 17 [= 0]$ o.e. or $y^2 - 4y + 13 [= 0]$ o.e. | M1 | Condone one error; allow M1 for $x^2 - 8x = -17$ [oe for $y$] only if they go on to completing the square method |
| Use of $b^2 - 4ac$ with numbers substituted (condone one error in substitution) (may be in quadratic formula) | M1 | Or $(x-4)^2 = 16 - 17$ or $(x-4)^2 + 1 = 0$ (condone one error) |
| $b^2 - 4ac = 64 - 68$ or $-4$ cao [or $16 - 52$ or $-36$ if $y$ used] | A1 | Or $(x-4)^2 = -1$ or $x = 4 \pm \sqrt{-1}$ [or $(y-2)^2 = -9$ or $y = 2 \pm \sqrt{-9}$] |
| $[< 0]$ so no [real] roots [so line and curve do not intersect] | A1 | Or conclusion from completing square; needs to be explicit correct conclusion and correct ft; allow '$< 0$ so no intersection' o.e.; allow '$-4$ so no roots' etc; allow A2 for full argument from sum of two squares $= 0$; A1 for weaker correct conclusion; some may use the condition $b^2 < 4ac$ for no real roots; allow equivalent marks, with first A1 for $64 < 68$ o.e. |
9 Prove that the line $y = 3 x - 10$ does not intersect the curve $y = x ^ { 2 } - 5 x + 7$.

\hfill \mbox{\textit{OCR MEI C1 2010 Q9 [5]}}