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 $\mathrm{x}$ is :
Correct Option: , 4
Solution:
$200={ }^{6} C_{3}\left(x^{\frac{1}{x+\log _{10} x}}\right)^{\frac{3}{2}} \times x^{\frac{1}{4}}$
$\Rightarrow 10=x^{\frac{3}{2\left(1+\log _{10} x\right)} \cdot \frac{1}{4}}$
$\Rightarrow 1=\left(\frac{3}{2(1+t)}+\frac{1}{4}\right) t$
where $t=\log _{10} x$
$\Rightarrow \mathrm{t}^{2}+3 \mathrm{t}-4=0$
$\Rightarrow \mathrm{t}=1,-4$
$\Rightarrow \mathrm{x}=10,10^{-4}$
$\Rightarrow \mathrm{x}=10($ As $\mathrm{x}>1)$
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