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: , 2
$\therefore$ fourth term is equal to 200 .
$T_{4}={ }^{6} C_{3}\left(\sqrt{x\left(\frac{1}{1+\log _{10} x}\right)}\right)^{3}\left(x^{\frac{1}{12}}\right)^{3}=200$
$\Rightarrow 20 x^{\frac{3}{2\left(1+\log _{10} x\right)}} \cdot x^{\frac{1}{4}}=200$
$x^{\frac{1}{4}+\frac{3}{2\left(1+\log _{10} x\right)}}=10$
Taking $\log _{10}$ on both sides and putting $\log _{10} x=t$
$\left(\frac{1}{4}+\frac{3}{2(1+t)}\right) t=1 \Rightarrow t^{2}+3 t-4=0$
$\Rightarrow t^{2}+4 t-t-4=0 \Rightarrow t(t+4)-1(t+4)=0$
$\Rightarrow t=1$ or $t=-4$
$\log _{10} x=1 \Rightarrow x=10$
or $\log _{10} x=-4 \Rightarrow x=10^{-4}$
According to the question $x>1, \therefore x=10$.
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