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)
$\Rightarrow \quad 10\left(x^{\log _2 x}\right)^{2}=2560$
$\Rightarrow \quad x^{\left(2 \log _2 x\right)}=256$
$\Rightarrow \quad \log _2 x^{2 \log _2 x}=\log _2 256$
$$ \begin{array}{ccc} \Rightarrow & 2\left(\log _2 x\right)\left(\log _2 x\right)=8 & \left(\text { taking } \log _2\right. \text { on both sides } \\ \left(\log _2 x\right)^{2}=4 & \left(\because \log _2 256=\log _2 2^{8}=8\right) \\ \Rightarrow & \log _2 x= \pm 2 \\ \Rightarrow & \log _2 x=2 \text { or } \quad \log _2 x=-2 \\ \Rightarrow & x=4 \text { or } x=2^{-2}=\frac{1}{4} \end{array} $$
