OCR Further Pure Core 2 2018 March — Question 4 4 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2018
SessionMarch
Marks4
TopicRoots of polynomials
TypeReciprocal sum of roots
DifficultyStandard +0.8 This is a Further Maths question requiring knowledge of transforming cubic equations and applying relationships between roots and coefficients. While the technique (substituting y=1/x) is standard for Further Pure, it requires insight to recognize the appropriate substitution and careful algebraic manipulation. The question tests understanding beyond routine application of sum/product formulas, placing it moderately above average difficulty.
Spec4.05a Roots and coefficients: symmetric functions
You are given that the cubic equation \(2x^3 - 3x^2 + x + 4 = 0\) has three roots, \(\alpha\), \(\beta\) and \(\gamma\). By making a suitable substitution to obtain a related cubic equation, determine the value of \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\). [4]
AnswerMarks
Substitute \(y = \frac{1}{x}\)M1
\(\Rightarrow \frac{2}{y^3} - \frac{3}{y^2} + \frac{1}{y} + 4 = 0\)A1
\(\Rightarrow 4y^3 + y^2 - 3y + 2 = 0\)M1
has roots \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\)M1
\(\Rightarrow \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = -\frac{1}{4}\)A1 [4] 
Substitute $y = \frac{1}{x}$ | M1

$\Rightarrow \frac{2}{y^3} - \frac{3}{y^2} + \frac{1}{y} + 4 = 0$ | A1

$\Rightarrow 4y^3 + y^2 - 3y + 2 = 0$ | M1

has roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ | M1

$\Rightarrow \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = -\frac{1}{4}$ | A1 [4]

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You are given that the cubic equation $2x^3 - 3x^2 + x + 4 = 0$ has three roots, $\alpha$, $\beta$ and $\gamma$.

By making a suitable substitution to obtain a related cubic equation, determine the value of $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$. [4]

\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q4 [4]}}
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