First slide

Binomial theorem for positive integral Index

Question

If the third term in the binomial expansion of ${(1 + x^{\log_{2} x})}^{5}$ equals 2560, then a possible value of x is

Moderate

A

$4 \sqrt{2}$
B

$\frac{1}{2}$
C

$\frac{1}{8}$
D

$2 \sqrt{2}$

Solution

The (r + 1)th term in the expansion of (a + x)ⁿ is Given by $T_{r + 1} =^{n} C_{r} a^{n - r} x^{r}$
.'. 3^rd term in the expansion of ${(1 + x^{\log_{2} x})}^{5}$ is
$\begin{aligned} ^{5} C_{2} (1)^{5 - 2} {(x^{\log_{2} x})}^{2} \\ \Rightarrow^{5} C_{2} (1)^{5 - 2} {(x^{\log_{2} x})}^{2} \\ \Rightarrow 2560 \\ \Rightarrow 10 {(x^{\log_{2} x})}^{2} = 2560 \\ \Rightarrow x^{(2 \log_{2} x)} = 256 \\ \Rightarrow \log_{2} x^{2 \log_{2} x} = \log_{2} 256 \end{aligned}$ (given)
(taking Log₂ on both sides)
$\begin{aligned} \Rightarrow 2 (\log_{2} x) (\log_{2} x) = 8 [∵ \log_{2} 256 = \log_{2} 2^{8} = 8] \\ {(\log_{2} x)}^{2} = 4 \\ \Rightarrow \log_{2} x = \pm 2 \\ \Rightarrow \log_{2} x = 2 or \log_{2} x = - 2 \\ \Rightarrow x = 4 \\ or x = 2^{- 2} = \frac{1}{4} \end{aligned}$

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