Question

If the third term in the expansion of
${(\frac{1}{x}+x^{{\log }_{2}x})}^5$ is $40$ then $x$ equals

• A. $\frac{1}{\sqrt{2}},2$
• B. $\sqrt{2},4$
• C. $\frac{1}{\sqrt{2}},4$
• D. $\sqrt{2},\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}},4$

The middle term of ${(\frac{1}{x}+x{\log }_{2}x)}^5$ is ${}^5{C}_2{\left(\frac{1}{x}\right)}^3{\left(x^{{\log }_{2}x}\right)}^2=10x^{2{\log }_{2}x-3}=40$

$\therefore x^{2{\log }_{2}x-3}=4=2^2$

Taking logarithm to the base $2$ of both the sides, we get $(2t-3)t=2$ where $t={\log }_{2}x$

$\Rightarrow 2t^2-3t-2=0$ or $(2t+1)(t-2)=0$

$\Rightarrow t=-\frac{1}{2},2\Rightarrow x=\frac{1}{\sqrt{2}},4$