Question

If the fourth term in the binomial expansion of ${(\sqrt{\frac{1}{x^{1+{\log }_{10}x}}}+x^{1/12})}^6$ is equal to $200$, and $x>1$, then the value of $x$ is

• A. $100$
• B. $10$
• C. ${10}^4$
• D. ${10}^3$
• E. None of these

Answer 1

The correct option is B $10$

$NOTE$: This is a BONUS question as no option is correct.
General term, $T_{r+1}={}^6{C}_r{\left(\frac{1}{x^{1+{\log }_{10}x}}\right)}^{(6-r)/2}{\left(x\right)}^{r/12}$

$\Rightarrow T_{r+1}={}^6{C}_r{\left(x^{1+{\log }_{10}x}\right)}^{(r-6)/2}{\left(x\right)}^{r/12}$

$\Rightarrow T_4={}^6{C}_3{\left(x^{1+{\log }_{10}x}\right)}^{-3/2}{\left(x\right)}^{1/4}$

$\Rightarrow 20\times x^{-5/4-3\left({\log }_{10}x\right)/2}=200$

$\Rightarrow x^{-5/4-3\left({\log }_{10}x\right)/2}=10$
$\Rightarrow (\frac{-5}{4}-\frac{3}{2}{\log }_{10}x){\log }_{10}x=1$
$\Rightarrow \frac{3}{2}({\log }_{10}x{)}^2+\frac{5}{4}{\log }_{10}x+1=0$

Let ${\log }_{10}x=t$
Then $6t^2+5t+4=0\cdots \left(1\right)$
Discriminant, $D=25-96<0$
Hence, Eq $\left(1\right)$ has imaginary roots.