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

  2. (2) $4 \sqrt{2}$

  3. (3) $\frac{1}{8}$

  4. (4) $2 \sqrt{2}$


Correct Option: 1

Solution:

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 $\frac{1}{4}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now