Binomial Theorem 1 Question 5
6.
If the fourth term in the binomial expansion of $(\frac{2}{x}+x^{\log _{8} x )^{6}}(x>0)$ is $20 \times 8^{7}$, then the value of $x$ is
(a) $8^{-2}$
(b) $8^{3}$
(c) 8
(d) $8^{2}$
(2019 Main, 9 April I)
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Answer:
Correct Answer: 6. (d)
Solution:
- Given binomial is $(\frac{2}{x}+x^{\log _{8} x)^{6}}$
Since, general term in the expansion of $(x+a)^{n}$ is $T_{r+1} = {}^{n} C_{r} x^{n-r} a^{r}$
$\therefore \quad T_{4}=T_{3+1}={ }^{6} C_{3} (\frac{2}{x}){ }^{6-3} \quad\left(x^{\log _{8} x}\right)^{3}=20 \times 8^{7} \text { (given) } $
$\Rightarrow 20 (\frac{2^{3}}{x}) x^{3 \log _{8} x}=20 \times 8^{7} $
$\because [{}^{6} C_{3}=20]$
$ \Rightarrow 2^{3} x^{[3(\log_{8} x)-3]}$ $=(2^{3})^{7} \Rightarrow x^({\frac{3}{3}\log_{2} x-3}) $ = $(2^{3})^6$
$\because [\log _{a^{n}}(x)=\frac{1}{n} \log _{a} x \text { for } x>0 ; a>0, \neq 1 ]$
$\Rightarrow x^{\log_{2} x-3}=2^{18}$
On taking $\log _{2}$ both sides, we get
$ \begin{aligned} & \quad\left(\log _{2} x-3\right) \log _{2} x=18 \\ & \Rightarrow \quad\left(\log _{2} x\right)^{2}-3 \log _{2} x-18=0 \\ & \Rightarrow \quad\left(\log _{2} x\right)^{2}-6 \log _{2} x+3 \log _{2} x-18=0 \\ & \Rightarrow \quad \log _{2} x\left(\log _{2} x-6\right)+3\left(\log _{2} x-6\right)=0 \\ & \Rightarrow \quad\left(\log _{2} x-6\right)\left(\log _{2} x+3\right)=0 \\ & \Rightarrow \quad \log _{2} x=-3,6 \\ & \Rightarrow \quad x=2^{-3}, 2^{6} \Rightarrow x=\frac{1}{8}, 8^{2} \end{aligned} $
