Question

If the sum of a the roots of the equation ${\left(\frac{1}{2x}-1\right)}^{109}=-1$ is $s$,then $\left[s\right]$ where $\left[\right]$ denotes integral part is

• A. $28$
• B. $27$
• C. $26$
• D. $25$

Answer 1

Answer: (B)

$27$

$(\frac{1}{2x}-1{)}^{109}=-1$

$(1-2x{)}^{109}=-2^{109}{x}^{109}$
$(2x-1{)}^{109}=2^{109}{x}^{109}$
$2^{109}{x}^{109}-{1092}^{108}{x}^{108}+...=2^{109}{x}^{109}$
$-{1092}^{108}{x}^{108}+...=0$
Sum of roots=$\frac{109\left(54\right)}{109\left(2\right)}$
$=27$