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Question:
If the fourth term in the binomial expansion of
$\left(\sqrt{\frac{1}{x^{1+\log _{10} x}}}+x^{\frac{1}{12}}\right)^{6}$ is equal to 200 , and $x>1$,
then the value of x is :
Correct Option: , 4
Solution:
$200={ }^{6} \mathrm{C}_{3}\left(\mathrm{x}^{\frac{1}{x+\log _{10} x}}\right)^{\frac{3}{2}} \times \mathrm{x}^{\frac{1}{4}}$
$\Rightarrow 10=x^{\frac{3}{2\left(1+\log _{10} x\right)}+\frac{1}{4}}$
$\Rightarrow 1=\left(\frac{3}{2(1+t)}+\frac{1}{4}\right) t$
where $t=\log _{10} x$
$\Rightarrow t^{2}+3 t-4=0$
$\Rightarrow t=1,-4$
$\Rightarrow x=10,10^{-4}$
$\Rightarrow x=10($ As $x>1)$