Question:

If the third term in the binomial expansion of \(\left(1+x^{\log_2x}\right)^5\) equals 2560, then a possible value of \(x\) is -

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The third term is \(T_3=\binom52\left(x^{\log_2x}\right)^2\); take \(\log_2\) after equating it to 2560.
Updated On: Nov 5, 2026
  • \(2\sqrt2\)
  • \(4\sqrt2\)
  • \(\frac18\)
  • \(\frac14\)
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The Correct Option is D

Solution and Explanation

\(T_3=\binom52\left(x^{\log_2x}\right)^2=10\,x^{2\log_2x}=2560\Rightarrow x^{2\log_2x}=256\).
Take \(\log_2\): \(2(\log_2x)^2=8\Rightarrow(\log_2x)^2=4\Rightarrow\log_2x=\pm2\).
So \(x=4\) or \(x=\frac14\). Among the options only \(\frac14\) is present.
Answer: (4) \(\frac14\)
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