Pre-U Pre-U 9795/1 2016 June — Question 9 10 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2016
SessionJune
Marks10
TopicRoots of polynomials
TypeSymmetric functions of roots
DifficultyChallenging +1.2 Part (i) is direct recall of Vieta's formulas. Part (ii) requires standard algebraic manipulation using symmetric function identities (α²+β²+γ² = (α+β+γ)² - 2(αβ+βγ+γα)). Part (iii) involves expanding a symmetric product and substituting known values—tedious but methodical. Part (iv) is a straightforward logical deduction from (iii). While this is a Further Maths question with multiple parts requiring careful algebra, it follows well-established techniques without requiring novel insight, making it moderately above average difficulty.
Spec4.05a Roots and coefficients: symmetric functions
9 The cubic equation \(x ^ { 3 } - a x ^ { 2 } + b x - c = 0\) has roots \(\alpha , \beta\) and \(\gamma\).
  1. State, in terms of \(a , b\) and \(c\), the values of \(\alpha + \beta + \gamma , \alpha \beta + \beta \gamma + \gamma \alpha\) and \(\alpha \beta \gamma\).
  2. Find, in terms of \(a , b\) and \(c\), the values of \(\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }\) and \(\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }\).
  3. Show that \(( \alpha - 2 \beta \gamma ) ( \beta - 2 \gamma \alpha ) ( \gamma - 2 \alpha \beta ) = c ( 2 a + 1 ) ^ { 2 } - 2 ( b + 2 c ) ^ { 2 }\).
  4. Deduce that one root of the equation \(x ^ { 3 } - a x ^ { 2 } + b x - c = 0\) is twice the product of the other two roots if and only if \(c ( 2 a + 1 ) ^ { 2 } = 2 ( b + 2 c ) ^ { 2 }\).
Question 9 [2+4+4+1 marks]
(i)
\(\alpha+\beta+\gamma = a\), \(\alpha\beta+\beta\gamma+\gamma\alpha = b\) and \(\alpha\beta\gamma = c\) B1B1 (B1 any 2 correct; +B1 all 3 correct)
[2]
(ii)
\(\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\) M1
\(= a^2 - 2b\) A1
\(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma)\) M1
\(= b^2 - 2ac\) A1
[4]
(iii)
\((\alpha-2\beta\gamma)(\beta-2\gamma\alpha)(\gamma-2\alpha\beta)\)
\(= \left(\alpha\beta - 2\beta^2\gamma - 2\alpha^2\gamma + 4\gamma^2\alpha\beta\right)(\gamma-2\alpha\beta)\) M1
\(= \alpha\beta\gamma - 2(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2) + 4\alpha\beta\gamma(\alpha^2+\beta^2+\gamma^2) - 8(\alpha\beta\gamma)^2\) M1 (Collecting up in terms of the symmetric fns.)
\(= c - 2(b^2-2ac) + 4c(a^2-2b) - 8c^2\) M1 (Use of (i)'s and (ii)'s results)
\(= c(1+4a+4a^2) - 2(b^2+4bc+4c^2)\)
\(= c(2a+1)^2 - 2(b+2c)^2\) A1 legitimately
[4]
Alternative method: Using \(\alpha\beta\gamma=c\),
\((\alpha-2\beta\gamma)(\beta-2\gamma\alpha)(\gamma-2\alpha\beta) = \left(\alpha-\frac{2c}{\alpha}\right)\left(\beta-\frac{2c}{\beta}\right)\left(\gamma-\frac{2c}{\gamma}\right)\)
\(= \frac{1}{\alpha\beta\gamma}\left(\alpha^2-2c\right)\left(\beta^2-2c\right)\left(\gamma^2-2c\right)\)
\(= \frac{1}{c}\left((\alpha\beta\gamma)^2 - 2c\sum\alpha^2\beta^2 + 4c^2\sum\alpha^2 - 8c^3\right)\)
\(= \frac{1}{c}\left(c^2 - 2c[b^2-2ac] + 4c^2[a^2-2b] - 8c^3\right) =\) etc. as above
(iv)
One root is the product of the other two
\(\Leftrightarrow (\alpha-2\beta\gamma)(\beta-2\gamma\alpha)(\gamma-2\alpha\beta)=0\)
\(\Leftrightarrow c(2a+1)^2 = 2(b+2c)^2\) B1 legitimately
Must reason \(\Rightarrow\) and \(\Leftarrow\) explicitly (or together)
[1]
**Question 9** [2+4+4+1 marks]

**(i)**
$\alpha+\beta+\gamma = a$, $\alpha\beta+\beta\gamma+\gamma\alpha = b$ and $\alpha\beta\gamma = c$ B1B1 (B1 any 2 correct; +B1 all 3 correct)

**[2]**

**(ii)**
$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$ M1

$= a^2 - 2b$ A1

$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma)$ M1

$= b^2 - 2ac$ A1

**[4]**

**(iii)**
$(\alpha-2\beta\gamma)(\beta-2\gamma\alpha)(\gamma-2\alpha\beta)$

$= \left(\alpha\beta - 2\beta^2\gamma - 2\alpha^2\gamma + 4\gamma^2\alpha\beta\right)(\gamma-2\alpha\beta)$ M1

$= \alpha\beta\gamma - 2(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2) + 4\alpha\beta\gamma(\alpha^2+\beta^2+\gamma^2) - 8(\alpha\beta\gamma)^2$ M1 (Collecting up in terms of the symmetric fns.)

$= c - 2(b^2-2ac) + 4c(a^2-2b) - 8c^2$ M1 (Use of **(i)**'s and **(ii)**'s results)

$= c(1+4a+4a^2) - 2(b^2+4bc+4c^2)$

$= c(2a+1)^2 - 2(b+2c)^2$ A1 **legitimately**

**[4]**

Alternative method: Using $\alpha\beta\gamma=c$,

$(\alpha-2\beta\gamma)(\beta-2\gamma\alpha)(\gamma-2\alpha\beta) = \left(\alpha-\frac{2c}{\alpha}\right)\left(\beta-\frac{2c}{\beta}\right)\left(\gamma-\frac{2c}{\gamma}\right)$

$= \frac{1}{\alpha\beta\gamma}\left(\alpha^2-2c\right)\left(\beta^2-2c\right)\left(\gamma^2-2c\right)$

$= \frac{1}{c}\left((\alpha\beta\gamma)^2 - 2c\sum\alpha^2\beta^2 + 4c^2\sum\alpha^2 - 8c^3\right)$

$= \frac{1}{c}\left(c^2 - 2c[b^2-2ac] + 4c^2[a^2-2b] - 8c^3\right) =$ etc. as above

**(iv)**
One root is the product of the other two

$\Leftrightarrow (\alpha-2\beta\gamma)(\beta-2\gamma\alpha)(\gamma-2\alpha\beta)=0$

$\Leftrightarrow c(2a+1)^2 = 2(b+2c)^2$ B1 **legitimately**

Must reason $\Rightarrow$ and $\Leftarrow$ explicitly (or together)

**[1]**
9 The cubic equation $x ^ { 3 } - a x ^ { 2 } + b x - c = 0$ has roots $\alpha , \beta$ and $\gamma$.\\
(i) State, in terms of $a , b$ and $c$, the values of $\alpha + \beta + \gamma , \alpha \beta + \beta \gamma + \gamma \alpha$ and $\alpha \beta \gamma$.\\
(ii) Find, in terms of $a , b$ and $c$, the values of $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ and $\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }$.\\
(iii) Show that $( \alpha - 2 \beta \gamma ) ( \beta - 2 \gamma \alpha ) ( \gamma - 2 \alpha \beta ) = c ( 2 a + 1 ) ^ { 2 } - 2 ( b + 2 c ) ^ { 2 }$.\\
(iv) Deduce that one root of the equation $x ^ { 3 } - a x ^ { 2 } + b x - c = 0$ is twice the product of the other two roots if and only if $c ( 2 a + 1 ) ^ { 2 } = 2 ( b + 2 c ) ^ { 2 }$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q9 [10]}}
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