Binomial Theorem 1 Question 6
7. If the fourth term in the binomial expansion of $\sqrt{x^{\frac{1}{1+\log _{10} x}}}+x^{\frac{1}{12}}$ is equal to 200 , and $x>1$, then the
value of $x$ is
(a) 100
(b) $10^{4}$
(c) 10
(d) $10^{3}$
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Answer:
Correct Answer: 7. (a)
Solution:
- Given binomial is $\sqrt{x^{\frac{1}{1+\log _{10} x}}}+x^{\frac{1}{12}}$
Since, the fourth term in the given expansion is 200.
$\therefore{ }^{6} C _3 x^{x^{1+\log _{10} x}}{ }^{\frac{3}{2}} x^{\frac{1}{12}}{ }^{3}=200$
$$ \begin{array}{lc} \Rightarrow & 20 \times x^{\frac{3}{2\left(1+\log _{10} x\right)}+\frac{1}{4}}=200 \\ \Rightarrow & x^{\frac{3}{2\left(1+\log _{10} x\right)}}+\frac{1}{4}=10 \\ \Rightarrow & \frac{3}{2\left(1+\log _{10} x\right)}+\frac{1}{4} \log _{10} x=1 \end{array} $$
[applying $\log _{10}$ both sides]
$$ \begin{aligned} & \Rightarrow\left[6+\left(1+\log _{10} x\right)\right] \log _{10} x=4\left(1+\log _{10} x\right) \\ & \Rightarrow \quad\left(7+\log _{10} x\right) \log _{10} x=4+4 \log _{10} x \\ & \Rightarrow \quad t^{2}+7 t=4+4 t \quad\left[\text { let } \log _{10} x=t\right] \\ & \Rightarrow \quad t^{2}+3 t-4=0 \\ & \Rightarrow \quad t=1,-4=\log _{10} x \\ & \Rightarrow \quad x=10,10^{-4} \\ & \text { Since, } \quad x>1 \quad x=10 \end{aligned} $$
