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Question Number 28406 by Tinkutara last updated on 25/Jan/18
Image not getting attached. Please see the link below.
$${Image}\:{not}\:{getting}\:{attached}.\:{Please} \\ $$$${see}\:{the}\:{link}\:{below}. \\ $$
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 −6p_1 x+2=0; β,γ are roots of x^2 −6p_2 x+3=0; γ ,αare roots of x^2 −6p_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) (1/2),(2/3),(5/6) (2) 1,2,5 (3) 6,1,4 (4) 2,(3/2),(6/5) −.−.−.−.−.− 1. αβ=2 ,βγ=3 , γα=6 Multiplying α^2 β^2 γ^2 =36 αβγ=±6 ((αβγ)/(βγ))=((±6)/3)⇒α=±2 ((αβγ)/(αγ))=((±6)/6)⇒β=±1 ((αβγ)/(αβ))=((±6)/2)=±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=6p_1 ⇒p_1 =((2+1)/6)=(1/2) β+γ=6p_2 ⇒p_2 =((1+3)/6)=(2/3) γ+α=6p_2 ⇒p_3 =((3+2)/6)=(5/6) (p_1 ,p_2 ,p_3 )=((1/2),(2/3),(5/6)) (1) (1/2),(2/3),(5/6) ⋮
$$\:\:\:\:\:\alpha,\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} −\mathrm{6p}_{\mathrm{1}} \mathrm{x}+\mathrm{2}=\mathrm{0}; \\ $$$$\:\:\:\:\:\beta,\gamma\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} −\mathrm{6p}_{\mathrm{2}} \mathrm{x}+\mathrm{3}=\mathrm{0}; \\ $$$$\:\:\:\:\:\gamma\:,\alpha\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} −\mathrm{6p}_{\mathrm{3}} \mathrm{x}+\mathrm{6}=\mathrm{0}; \\ $$$$\:\:\:\:\:\mathrm{p}_{\mathrm{1}} ,\mathrm{p}_{\mathrm{2}} \:\&\mathrm{p}_{\mathrm{3}\:} \mathrm{are}\:\mathrm{positive}\:\mathrm{then}: \\ $$$$\boldsymbol{\mathrm{Choose}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{answer}} \\ $$$$\mathrm{1}.\mathrm{The}\:\mathrm{values}\:\mathrm{of}\:\alpha,\beta,\gamma\:\:\mathrm{respectively}\:\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2},\mathrm{3},\mathrm{1}\:\:\:\:\left(\mathrm{2}\right)\:\mathrm{2},\mathrm{1},\mathrm{3}\:\:\left(\mathrm{3}\right)\:\mathrm{1},\mathrm{2},\mathrm{3}\:\:\left(\mathrm{4}\right)\:\:−\mathrm{1},−\mathrm{2},−\mathrm{3} \\ $$$$\mathrm{2}.\mathrm{The}\:\mathrm{values}\:\mathrm{of}\:\mathrm{p}_{\mathrm{1}} ,\mathrm{p}_{\mathrm{2}} ,\mathrm{p}_{\mathrm{3}} \:\mathrm{respectively}\:\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{6}}\:\:\:\left(\mathrm{2}\right)\:\mathrm{1},\mathrm{2},\mathrm{5}\:\:\left(\mathrm{3}\right)\:\mathrm{6},\mathrm{1},\mathrm{4}\:\:\:\left(\mathrm{4}\right)\:\mathrm{2},\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$−.−.−.−.−.− \\ $$$$\mathrm{1}. \\ $$$$\alpha\beta=\mathrm{2}\:,\beta\gamma=\mathrm{3}\:,\:\gamma\alpha=\mathrm{6} \\ $$$$\mathrm{Multiplying} \\ $$$$\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} =\mathrm{36} \\ $$$$\alpha\beta\gamma=\pm\mathrm{6} \\ $$$$\frac{\alpha\beta\gamma}{\beta\gamma}=\frac{\pm\mathrm{6}}{\mathrm{3}}\Rightarrow\alpha=\pm\mathrm{2} \\ $$$$\frac{\alpha\beta\gamma}{\alpha\gamma}=\frac{\pm\mathrm{6}}{\mathrm{6}}\Rightarrow\beta=\pm\mathrm{1} \\ $$$$\frac{\alpha\beta\gamma}{\alpha\beta}=\frac{\pm\mathrm{6}}{\mathrm{2}}=\pm\mathrm{3} \\ $$$$\left(\alpha,\beta,\gamma\right)=\left(\mathrm{2},\mathrm{1},\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Or} \\ $$$$\left(\alpha,\beta,\gamma\right)=\left(−\mathrm{2},−\mathrm{1},−\mathrm{3}\right)\:\left(\mathrm{Not}\:\mathrm{included}\:\mathrm{in}\:\mathrm{options}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{2},\mathrm{1},\mathrm{3} \\ $$$$−.−.−.−.−.− \\ $$$$\mathrm{2}.\: \\ $$$$\:\:\alpha+\beta=\mathrm{6p}_{\mathrm{1}} \Rightarrow\mathrm{p}_{\mathrm{1}} =\frac{\mathrm{2}+\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\beta+\gamma=\mathrm{6p}_{\mathrm{2}} \Rightarrow\mathrm{p}_{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\gamma+\alpha=\mathrm{6p}_{\mathrm{2}} \Rightarrow\mathrm{p}_{\mathrm{3}} =\frac{\mathrm{3}+\mathrm{2}}{\mathrm{6}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\left(\mathrm{p}_{\mathrm{1}} ,\mathrm{p}_{\mathrm{2}} ,\mathrm{p}_{\mathrm{3}} \right)=\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{6}}\right) \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\vdots \\ $$
Commented by Tinkutara last updated on 26/Jan/18
Thanks Sir! Rest I solved.

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