Question

If the $9^{th}$ term in the expansion of ${\left(\frac{1}{x^2}+\frac{x}{2}lo{g}_{2}x\right)}^{12}$ is equal to 495, then the value of x can be

• A. 1
• B. An integer greater than 1
• C. a fraction
• D. an irrational number

Answer 1

Answer: (B), (C)

The correct options are
B An integer greater than 1
C a fraction

From the given question, $T_9{=}^{12}{C}_4\frac{1}{x^8}\frac{x^8}{2^8}({\log }_{2}x{)}^8=495$
$\Rightarrow \frac{12\times 1\times 10\times 4}{24}\times \frac{1}{2^8}({\log }_{2}x{)}^8=495$
$\Rightarrow {\log }_{2}x=\pm 2\Rightarrow x=\frac{1}{4}$ or $4$.