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:
Correct Option: 1
Third term of $\left(1+x^{\log _{2} x}\right)^{5}={ }^{5} C_{2}\left(x^{\log _{2} x}\right)^{5-3}$
$={ }^{5} C_{2}\left(x^{\log _{2} x}\right)^{2}$
Given, ${ }^{5} \mathrm{C}_{2}\left(x^{\log _{2} x}\right)^{2}=2560$
$\Rightarrow\left(x^{\log _{2} x}\right)^{2}=256=(\pm 16)^{2}$
$\Rightarrow x^{\log _{2} x}=16$ or $x^{\log _{2} x}=-16$ (rejected)
$\Rightarrow x^{\log _{2} x}=16 \Rightarrow \log _{2} x \log _{2} x=\log _{2} 16=4$
$\Rightarrow \log _{2} x=\pm 2 \Rightarrow x=2^{2}$ or $2^{-2}$
$\Rightarrow x=4$ or $\frac{1}{4}$
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