JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 9)
If the third term in the binomial expansion
of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is -
of $${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$ equals 2560, then a possible value of x is -
$$2\sqrt 2 $$
$$4\sqrt 2 $$
$${1 \over 8}$$
$${1 \over 4}$$
Explanation
$${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$$
$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$
$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$
$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$
$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$
$$ \Rightarrow 2{({\log _2}x)^2} = 8$$
$$ \Rightarrow \,\,{({\log _2}x)^2} = 4$$
$$ \Rightarrow \,\,{\log _2}x = 2$$ or $$-$$ 2
$$x = 4$$ or $${1 \over 4}$$
$${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$$
$$ \Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$$
$$ \Rightarrow \,\,{x^{2\log 2x}} = 256$$
$$ \Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$$
$$ \Rightarrow 2{({\log _2}x)^2} = 8$$
$$ \Rightarrow \,\,{({\log _2}x)^2} = 4$$
$$ \Rightarrow \,\,{\log _2}x = 2$$ or $$-$$ 2
$$x = 4$$ or $${1 \over 4}$$
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