OCR Further Pure Core AS 2023 June — Question 5 4 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2023
SessionJune
Marks4
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Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeReciprocal sum of roots
DifficultyModerate -0.3 This is a straightforward application of symmetric functions of roots using Vieta's formulas. Students need to recognize that 1/α² + 1/β² = (α² + β²)/(αβ)², then express α² + β² as (α+β)² - 2αβ. While it requires algebraic manipulation across multiple steps, it's a standard textbook exercise with a well-established method that Further Maths students practice regularly. Slightly easier than average due to its routine nature.
Spec4.05a Roots and coefficients: symmetric functions
5 In this question you must show detailed reasoning. The roots of the equation \(5 x ^ { 2 } - 3 x + 12 = 0\) are \(\alpha\) and \(\beta\). By considering the symmetric functions of the roots, \(\alpha + \beta\) and \(\alpha \beta\), determine the exact value of \(\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } }\).
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
\(\alpha + \beta = \frac{3}{5}\)B1
\(\alpha\beta = \frac{12}{5}\)B1
\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2\beta^2} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}\)M1 Rewrite expression in terms of standard symmetric functions. Need to see \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\)
\(= \frac{\left(\frac{3}{5}\right)^2 - 2 \times \frac{12}{5}}{\left(\frac{12}{5}\right)^2} = -\frac{37}{48}\)A1 SC B1 for correct answer if B0B0M0. Accept any equivalent fraction 
# Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha + \beta = \frac{3}{5}$ | **B1** | |
| $\alpha\beta = \frac{12}{5}$ | **B1** | |
| $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2\beta^2} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{(\alpha\beta)^2}$ | **M1** | Rewrite expression in terms of standard symmetric functions. Need to see $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ |
| $= \frac{\left(\frac{3}{5}\right)^2 - 2 \times \frac{12}{5}}{\left(\frac{12}{5}\right)^2} = -\frac{37}{48}$ | **A1** | SC B1 for correct answer if B0B0M0. Accept any equivalent fraction |

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5 In this question you must show detailed reasoning.
The roots of the equation $5 x ^ { 2 } - 3 x + 12 = 0$ are $\alpha$ and $\beta$.

By considering the symmetric functions of the roots, $\alpha + \beta$ and $\alpha \beta$, determine the exact value of $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } }$.

\hfill \mbox{\textit{OCR Further Pure Core AS 2023 Q5 [4]}}
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