QJEEM19Ma9c02: JEE Main 2019 Mathematics - Class 11 JEEM19Ma9c02 Solution: If the third term in the binomial expansion of equals 2560,…

Question

Verified Solution • 2025–26 Session moderate

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 -

Accepted Answer

{\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}

Step-by-Step Verified Solution