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Question Number 28406 by Tinkutara last updated on 25/Jan/18

I m a g e n o t g e t t i n g a t t a c h e d . P l e a s e

s e e t h e l i n k b e l o w .

Commented by Tinkutara last updated on 25/Jan/18

http://ibb.co/iaU0Jb Any method other than looking options?

Answered by Rasheed.Sindhi last updated on 26/Jan/18

α, β are roots of x^{2} - {6 p}_{1} x + 2 = 0;

β, γ are roots of x^{2} - {6 p}_{2} x + 3 = 0;

γ, α are roots of x^{2} - {6 p}_{3} x + 6 = 0;

p_{1}, p_{2} & p_{3} are positive then :

Choose the correct answer

1 . The values of α, β, γ respectively are

(1) 2, 3, 1 (2) 2, 1, 3 (3) 1, 2, 3 (4) - 1, - 2, - 3

2 . The values of p_{1}, p_{2}, p_{3} respectively are

(1) \frac{1}{2}, \frac{2}{3}, \frac{5}{6} (2) 1, 2, 5 (3) 6, 1, 4 (4) 2, \frac{3}{2}, \frac{6}{5}

- . - . - . - . - . -

1 .

α β = 2, β γ = 3, γ α = 6

Multiplying

α^{2} β^{2} γ^{2} = 36

α β γ = \pm 6

\frac{α β γ}{β γ} = \frac{\pm 6}{3} \Rightarrow α = \pm 2

\frac{α β γ}{α γ} = \frac{\pm 6}{6} \Rightarrow β = \pm 1

\frac{α β γ}{α β} = \frac{\pm 6}{2} = \pm 3

(α, β, γ) = (2, 1, 3)

Or

(α, β, γ) = (- 2, - 1, - 3) (Not included in options)

(2) 2, 1, 3

- . - . - . - . - . -

2 .

α + β = {6 p}_{1} \Rightarrow p_{1} = \frac{2 + 1}{6} = \frac{1}{2}

β + γ = {6 p}_{2} \Rightarrow p_{2} = \frac{1 + 3}{6} = \frac{2}{3}

γ + α = {6 p}_{2} \Rightarrow p_{3} = \frac{3 + 2}{6} = \frac{5}{6}

(p_{1}, p_{2}, p_{3}) = (\frac{1}{2}, \frac{2}{3}, \frac{5}{6})

(1) \frac{1}{2}, \frac{2}{3}, \frac{5}{6}

⋮

Commented by Tinkutara last updated on 26/Jan/18

Thanks Sir! Rest I solved.