If - - - are roots of x3-2x2-3x-1-0- then value of - - - - - - - - - - - is less than

If α, β, γ are roots of x^3 + 2x^2 3x + 1 = 0 Then 1α, 1β , 1γ are roots of x^3 3x^2 + 2x + 1 = 0 Now, αβα + β⇒1(1α + 1β ) We know , {1α + 1β + 1γ = 3, 1αβ ...

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Standard XII

Mathematics

Relation between Roots and Coefficients for Higher Order Equations

If α, β, γ ar...

Question

If $α, β, γ$ are roots of $x^{3} + 2 x^{2} - 3 x + 1 = 0,$ then value of $\frac{α β}{α + β} + \frac{α γ}{α + γ} + \frac{β γ}{β + γ}$ is less than

3

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2

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4

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5

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Solution

The correct option is D $5$
If $α, β, γ$ are roots of $x^{3} + 2 x^{2} - 3 x + 1 = 0$
Then $\frac{1}{α}, \frac{1}{β}, \frac{1}{γ}$ are roots of $x^{3} - 3 x^{2} + 2 x + 1 = 0$
Now, $\frac{α β}{α + β} \Rightarrow \frac{1}{(\frac{1}{α} + \frac{1}{β})}$
We know , ${\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = 3, \frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α} = 2, \frac{1}{α β γ} = - 1}$
$\frac{1}{α} + \frac{1}{β} = (3 - \frac{1}{γ})$
Similarly $\frac{1}{γ} + \frac{1}{β} = (3 - \frac{1}{α}) & (\frac{1}{α} + \frac{1}{γ}) = (3 - \frac{1}{β})$
Put values in given expression,
$⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1}{3 - \frac{1}{γ}} + \frac{1}{3 - \frac{1}{β}} + \frac{1}{3 + \frac{1}{α}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$= \frac{\sum (3 - \frac{1}{γ}) (3 - \frac{1}{β})}{(3 - \frac{1}{α}) (3 - \frac{1}{β}) (3 - \frac{1}{γ})}$
$= \frac{27 - 6 [\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}] + (\frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α})}{(3 - \frac{1}{γ}) (3 - \frac{1}{β}) (3 - \frac{1}{γ})}$
$= \frac{27 - 6 \times 3 + 2}{27 - 9 \times 3 + 3 \times 2 - (- 1)}$
$= \frac{29 - 18}{28 - 27 + 6} = \frac{11}{7}$

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If α,β,γ are roots of x3+2x2−3x+1=0, then value of αβα+β+αγα+γ+βγβ+γ is less than

If $α, β, γ$ are roots of $x^{3} + 2 x^{2} - 3 x + 1 = 0,$ then value of $\frac{α β}{α + β} + \frac{α γ}{α + γ} + \frac{β γ}{β + γ}$ is less than