Question

If the third term in the binomial expansion of $(1+x^{{\log }_{2}x}{)}^5$ equals $2560$, then a possible value of $x$ is :

• A. $2\sqrt{2}$
• B. $\frac{1}{8}$
• C. $4\sqrt{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (D)

${\left(1+x^{{\log }_{2}x}\right)}^5$
$T_3={}^5{C}_2\cdot {\left(x^{{\log }_{2}x}\right)}^2=2560$
$\Rightarrow 10\cdot x^{2{\log }_{2}x}=2560$
$\Rightarrow x^{2{\log }_{2}x}=256$
$\Rightarrow 2{\left({\log }_{2}x\right)}^2={\log }_{2}256$
$\Rightarrow 2{\left({\log }_{2}x\right)}^2=8$
$\Rightarrow {\left({\log }_{2}x\right)}^2=4\Rightarrow {\log }_{2}x=2$ or $-2$
$x=4$ or $\frac{1}{4}$