If the third term in the binomial expansion of

Question:

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:

  1. $2 \sqrt{2}$

  2. $\frac{1}{8}$

  3. $4 \sqrt{2}$

  4. $\frac{1}{4}$


Correct Option: , 4

Solution:

$\left(1+x^{\log _{2} x}\right)^{5}$

$\mathrm{T}_{3}={ }^{5} \mathrm{C}_{2} \cdot\left(\mathrm{x}^{\log _{2} \mathrm{x}}\right)^{2}=2560$

$\Rightarrow 10 \cdot x^{2 \log _{2} x}=2560$

$\Rightarrow x^{2 \log _{2} x}=256$

$\Rightarrow 2\left(\log _{2} x\right)^{2}=\log _{2} 256$

$\Rightarrow 2\left(\log _{2} x\right)^{2}=8$

$\Rightarrow\left(\log _{2} x\right)^{2}=4$

$\Rightarrow \quad \log _{2} x=2$ or $-2$

$x=4$ or $\frac{1}{4}$

 

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