If the fourth term in the binomial expansion | vedclass

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(D) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} b^r$.For the given expression,$n=6$,$a = x^{-\frac{1

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(D) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} b^r$.For the given expression,$n=6$,$a = x^{-\frac{1}{2}(1+\log_{10} x)}$,and $b = x^{\frac{1}{12}}$.The fourth term $(T_4)$ corresponds to $r=3$:$T_4 = ^6C_3 \cdot (x^{-\frac{1}{2}(1+\log_{10} x)})^3 \cdot (x^{\frac{1}{12}})^3 = 200$.$20 \cdot x^{-\frac{3}{2}(1+\log_{10} x)} \cdot x^{\frac{1}{4}} = 200$.$x^{-\frac{3}{2}(1+\log_{10} x) + \frac{1}{4}} = 10$.Taking $\log_{10}$ on both sides,let $t = \log_{10} x$:$-\frac{3}{2}(1+t)t + \frac{1}{4} = 1$.$-6t(1+t) + 1 = 4$.$-6t^2 - 6t - 3 = 0 \Rightarrow 2t^2 + 2t + 1 = 0$.The discriminant $D = 2^2 - 4(2)(1) = 4 - 8 = -4 Since the discriminant is negative,there is no real value of $x$ that satisfies the equation.Thus,the correct option is $D$.

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

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