JEE Mains · Maths · STD 11 - 7. binomial theoram
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
- A \(\frac{1}{4}\)
- B \(4\sqrt 2 \)
- C \(\frac{1}{8}\)
- D \(2\sqrt 2 \)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
In the expansion of \(\left(1+x^{\log _{2} x}\right)^{5}\) third term say \(\mathrm{T}_{3}=^{5} \mathrm{C}_{2}\left(\mathrm{x}^{\log _{2} \mathrm{x}}\right)^{2}=2560\) \(\Rightarrow\left(x^{\log x}\right)^{2}=256\) taking lograthium to the base \(2\) on both sides…
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