If α, β are the roots of the equation x2−5+3log3⁡5−5log5⁡3x+33log3⁡513−5log5⁡323−1=0 then the equation, whose roots are α+1β and β+1α,

Bonus because ‘x’ is missing the correct will be,x2−5+3log3⁡5−5log5⁡3x+33log3⁡513−5log5⁡323−1=03log3⁡5=3log3⁡5⋅log3⁡5⋅log5⁡3=3log3⁡5⋅log5⁡3=3log3⁡5log5⁡3=5log5⁡33log3⁡53=3log3⁡5⋅log5⁡323=3log3⁡5log5⁡32/3 =5log5⁡32/3So, equation is x2 – 5x – 3 = 0 and roots are α & βα+β=5; αβ=-3New roots are α+1β & β+1αi.e., αβ+1β & αβ+1α i.e., -2β & -2αLet -2α=t ⇒ α=-2tAs α2-5α-3=0⇒−2t2−5−2t−3=0⇒4t2+10t−3=0⇒4+10t−3t2=0⇒3t2−10t−4=0 i.e., 3x2−10x−4=0

If α, β are the roots of the equation x2−5+3log3⁡5−5log5⁡3x+33log3⁡513−5log5⁡323−1=0 then the equation, whose roots are α+1β and β+1α,

Bonus because ‘x’ is missing the correct will be,x2−5+3log3⁡5−5log5⁡3x+33log3⁡513−5log5⁡323−1=03log3⁡5=3log3⁡5⋅log3⁡5⋅log5⁡3=3log3⁡5⋅log5⁡3=3log3⁡5log5⁡3=5log5⁡33log3⁡53=3log3⁡5⋅log5⁡323=3log3⁡5log5⁡32/3 =5log5⁡32/3So, equation is x2 – 5x – 3 = 0 and roots are α & βα+β=5; αβ=-3New roots are α+1β & β+1αi.e., αβ+1β & αβ+1α i.e., -2β & -2αLet -2α=t ⇒ α=-2tAs α2-5α-3=0⇒−2t2−5−2t−3=0⇒4t2+10t−3=0⇒4+10t−3t2=0⇒3t2−10t−4=0 i.e., 3x2−10x−4=0

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If $α, β$ are the roots of the equation $x^{2} - (5 + 3^{\sqrt{\log_{3} 5}} - 5^{\sqrt{\log_{5} 3}}) x + 3 (3^{{(\log_{3} 5)}^{\frac{1}{3}}} - 5^{{(\log_{5} 3)}^{\frac{2}{3}}} - 1) = 0$ then the equation, whose roots are $α + \frac{1}{β} and β + \frac{1}{α}$ ,

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3x² – 10x – 4 = 0

3x² – 10x + 2 = 0

3x² – 20x + 16 = 0

3x² – 20x – 12 = 0

answer is B.

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Detailed Solution

Bonus because ‘x’ is missing the correct will be,

$\begin{array}{l} x^{2} - (5 + 3^{\sqrt{\log_{3} 5}} - 5^{\sqrt{\log_{5} 3}}) x + 3 3^{{(\log_{3} 5)}^{\frac{1}{3}}} - 5^{{(\log_{5} 3)}^{\frac{2}{3}}} - 1) = 0 \\ 3^{\sqrt{\log_{3} 5}} = 3^{\sqrt{\log_{3} 5} \cdot \sqrt{\log_{3} 5} \cdot \sqrt{\log_{5} 3}} = 3^{\log_{3} 5 \cdot \sqrt{\log_{5} 3}} = {(3^{\log_{3} 5})}^{\sqrt{\log_{5} 3}} = 5^{\sqrt{\log_{5} 3}} \\ 3^{\sqrt[3]{\log_{3} 5}} = 3^{\log_{3} 5 \cdot \sqrt[3]{{(\log_{5} 3)}^{2}}} = {(3^{\log_{3} 5})}^{{(\log_{5} 3)}^{2 / 3}} \\ = 5^{{(\log_{5} 3)}^{2 / 3}} \end{array}$

So, equation is x² – 5x – 3 = 0 and roots are $α & β$

$\{α + β = 5; αβ = - 3\}$

New roots are $α + \frac{1}{β} & β + \frac{1}{α}$

$i . e ., \frac{αβ + 1}{β} & \frac{αβ + 1}{α} i . e ., \frac{- 2}{β} & \frac{- 2}{α}$

Let $\frac{- 2}{α} = t \Rightarrow α = \frac{- 2}{t}$

As $α^{2} - 5 α - 3 = 0$

$\begin{array}{l} \Rightarrow {(\frac{- 2}{t})}^{2} - 5 (\frac{- 2}{t}) - 3 = 0 \\ \Rightarrow \frac{4}{t^{2}} + \frac{10}{t} - 3 = 0 \\ \Rightarrow 4 + 10 t - 3 t^{2} = 0 \\ \Rightarrow 3 t^{2} - 10 t - 4 = 0 \\ i.e., 3 x^{2} - 10 x - 4 = 0 \end{array}$