OCR C1 2012 January — Question 3 4 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2012
SessionJanuary
Marks4
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Mark schemeDownload PDF ↗
TopicSolving quadratics and applications
TypeAlgebraic identity, find constants
DifficultyModerate -0.8 This is a straightforward algebraic identity problem requiring expansion of the right-hand side and coefficient matching. It involves only basic algebraic manipulation with no problem-solving insight needed, making it easier than average but not trivial since students must systematically equate coefficients of x², x, and constants.
Spec1.02e Complete the square: quadratic polynomials and turning points
3 Given that $$5 x ^ { 2 } + p x - 8 = q ( x - 1 ) ^ { 2 } + r$$ for all values of \(x\), find the values of the constants \(p , q\) and \(r\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(5x^2+px-8 = 5(x-1)^2+r = 5(x^2-2x+1)+r = 5x^2-10x+5+r\)B1 \(q=5\) (may be embedded on RHS)
\(p=-10\)B1 \(p=-10\)
M1\(-8 = \pm q + r\) or \(\frac{-p^2}{20}-8=r\)
\(r=-13\)A1 Allow from \(p=10\) 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $5x^2+px-8 = 5(x-1)^2+r = 5(x^2-2x+1)+r = 5x^2-10x+5+r$ | B1 | $q=5$ (may be embedded on RHS) |
| $p=-10$ | B1 | $p=-10$ |
| | M1 | $-8 = \pm q + r$ or $\frac{-p^2}{20}-8=r$ |
| $r=-13$ | A1 | Allow from $p=10$ |

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3 Given that

$$5 x ^ { 2 } + p x - 8 = q ( x - 1 ) ^ { 2 } + r$$

for all values of $x$, find the values of the constants $p , q$ and $r$.

\hfill \mbox{\textit{OCR C1 2012 Q3 [4]}}
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Q1 4 Q2 4 Q3 4 Q4 5 Q5 5 Q6 7 Q7 12 Q8 6 Q9 12 Q10 13