JEE MAIN - Mathematics (2022 - 27th July Evening Shift - No. 3)
If $$\alpha, \beta$$ are the roots of the equation
$$ x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0 $$,
then the equation, whose roots are $$\alpha+\frac{1}{\beta}$$ and $$\beta+\frac{1}{\alpha}$$, is :
Explanation
$${3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }} = {3^{\sqrt {{{\log }_3}5} }} - {\left( {{3^{{{\log }_3}5}}} \right)^{\sqrt {{{\log }_5}3} }}$$
$${3^{{{\left( {{{\log }_3}5} \right)}^{{1 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = 0$$
Note : In the given equation 'x' is missing.
So _27th_July_Evening_Shift_en_3_1.png)
$$\alpha + \beta + {1 \over \alpha } + {1 \over \beta } = (\alpha + \beta ) + {{\alpha + \beta } \over {\alpha \beta }}$$
$$ = 5 - {5 \over 3} = {{10} \over 3}$$
$$\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = 2 + \alpha \beta + {1 \over {\alpha \beta }} = 2 - 3 - {1 \over 3} = {{ - 4} \over 3}$$
So Equation must be option (B).
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