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7.Binomial Theorem
hard
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
A
$10^4$
B
$100$
C
$10^3$
D
None of these
(JEE MAIN-2019)
Solution
$^{6} \mathrm{C}_{3} \times x^{- \frac{3}{2}(1+\operatorname{log} x)} \cdot \mathrm{x}^{\frac{1}{4}}=200$
$x^{\frac{1}{4}- \frac{3}{2}(1+\operatorname{log} x)}=10$
$\Rightarrow \frac{1}{4}-\frac{3}{2}\left(1+\log _{10} x\right) \cdot \log _{10} x=1$
$\Rightarrow 6 t^{2}+5 t+4=0, t=\log _{10} x$
$D<0$
So no real solution
All options are incorrect
Standard 11
Mathematics
