Binomial Theorem 1 Question 12
13.
If the third term in the binomial expansion of $\left(1+x^{\log _{2} x}\right)^{5}$ equals 2560 , then a possible value of $x$ is
(a) $4 \sqrt{2}$
(b) $\frac{1}{4}$
(c) $\frac{1}{8}$
(d) $2 \sqrt{2}$
(2019 Main, 10 Jan I)
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Answer:
Correct Answer: 13. (b)
Solution:
- The $(r+1)$ th term in the expansion of $(a+x)^{n}$ is given by $T_{r+1}={ }^{n} C_{r} a^{n-r} x^{r}$
$\therefore 3^{\text {rd }}$ term in the expansion of $\left(1+x^{\log _{2} x}\right)^{5}$ is
${ }^{5} C_{2}(1)^{5-2}\left(x^{\log _{2} x}\right)^{2}$
$\Rightarrow{ }^{5} C_{2}(1)^{5-2}\left(x^{\log _{2} x}\right)^{2}=2560$ (given)
$\Rightarrow \quad 10\left(x^{\log _{2} x}\right)^{2}=2560$
$\Rightarrow \quad x^{\left(2 \log _{2} x\right)}=256$
$\Rightarrow \quad \log _{2} x^{2 \log _{2} x}=\log _{2} 256$
taking $ \log _{2} \text { on both sides } $ $ \left(\because \log _{2} 256=\log _{2} 2^{8}=8\right) $
$\Rightarrow 2\left(\log _{2} x\right)\left(\log _{2} x\right)=8 $
$\left(\log _{2} x\right)^{2}=4 $
$\Rightarrow \log _{2} x= \pm 2 $
$\Rightarrow \log _{2} x=2 \text { or } \quad \log _{2} x=-2 $
$\Rightarrow x=4 \text { or } x=2^{-2}=\frac{1}{4}$
