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7.Binomial Theorem
hard
यदि $\left(1+x^{\log _{2} x}\right)^{5}$ के द्विपद प्रसार में तीसरा पद $2560$ के बराबर है, तो $x$ का एक संभव मान है
A
$\frac{1}{4}$
B
$4\sqrt 2 $
C
$\frac{1}{8}$
D
$2\sqrt 2 $
(JEE MAIN-2019)
Solution
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
$\Rightarrow 2\left(\log _{2} x\right)^{2}=8 \Rightarrow\left(\log _{2} x\right)=\pm 2$
$\Rightarrow x=4, \frac{1}{4}$
Here $x=\frac{1}{4}$
Standard 11
Mathematics
