JEE MAIN - Mathematics Bengali (2022 - 27th July Evening Shift - No. 3)
ধর $$\alpha ,\beta $$ হল নীচের সমীকরণটির বীজদ্বয়
$${x^2} - \left( {5 + {3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }}} \right) + 3\left( {{3^{{{({{\log }_3}5)}^{{1 \over 3}}}}} - {5^{{{({{\log }_5}3)}^{{2 \over 3}}}}} - 1} \right) = 0$$ ।
যে সমীকরণটির বীজদ্বয় $$\alpha + {1 \over \beta }$$ এবং $$\beta + {1 \over \alpha }$$ তা হল ঃ
$$3{x^2} - 20x - 12 = 0$$
$$3{x^2} - 10x - 4 = 0$$
$$3{x^2} - 10x + 2 = 0$$
$$3{x^2} - 20x + 16 = 0$$
Comments (0)
