If the middle term in the expansion of 1x+xlog2⁡x5 is 40 then x equal

The middle term of 1x+xlog2⁡x5 is 5C21x3xlog2⁡x2=10x2log2⁡x−3=40∴ x2log2⁡x−3=4=22Taking logarithm to the base 2 of both the sides, we get(2t−3)t=2 where t=log2⁡x⇒ 2t2−3t−2=0 or (2t+1)(t−2)=0⇒ t=−1/2 or 2⇒x=1/2,4

If the middle term in the expansion of 1x+xlog2⁡x5 is 40 then x equal

The middle term of 1x+xlog2⁡x5 is 5C21x3xlog2⁡x2=10x2log2⁡x−3=40∴ x2log2⁡x−3=4=22Taking logarithm to the base 2 of both the sides, we get(2t−3)t=2 where t=log2⁡x⇒ 2t2−3t−2=0 or (2t+1)(t−2)=0⇒ t=−1/2 or 2⇒x=1/2,4

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If the middle term in the expansion of ${(\frac{1}{x} + x^{\log_{2} x})}^{5}$ is 40 then $x$ equal

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$1 / \sqrt{2}, 2$

$\sqrt{2}, 4$

$1 / \sqrt{2}, 4$

$\sqrt{2}, 1 / \sqrt{2}$

answer is C.

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Detailed Solution

The middle term of ${(\frac{1}{x} + x^{\log_{2} x})}^{5}$ is

$\begin{array}{l} ^{5} C_{2} {(\frac{1}{x})}^{3} {(x^{\log_{2} x})}^{2} = 10 x^{2 \log_{2} x - 3} = 40 \\ ∴ x^{2 \log_{2} x - 3} = 4 = 2^{2} \end{array}$

Taking logarithm to the base 2 of both the sides, we get

$\begin{aligned} (2 t - 3) t = 2 where t = \log_{2} x \\ \Rightarrow 2 t^{2} - 3 t - 2 = 0 or (2 t + 1) (t - 2) = 0 \\ \Rightarrow t = - 1 / 2 or 2 \Rightarrow x = 1 / \sqrt{2}, 4 \end{aligned}$