Question

If $2^{-m}\times \frac{1}{2^m}=\frac{1}{4}$, then $\frac{1}{14}\left[\right(4^m{)}^{\frac{1}{2}}+{\left(\frac{1}{5^m}\right)}^{-1}]$, is equal to

Answer 1

$2^{-m}\times \frac{1}{2^m}=\frac{1}{4}$
$\Rightarrow \frac{2^{-m}}{2^m}=\frac{1}{2^2}$
$\Rightarrow \frac{1}{2^{m+m}}=\frac{1}{2^2}$
$\Rightarrow \frac{1}{2^{2m}}=\frac{1}{2^2}$
On comparing both the sides , we get
$2m=2$
$\Rightarrow m=1$
$\therefore \frac{1}{14}\left[\right(4^m{)}^{\frac{1}{2}}+{\left(\frac{1}{5^m}\right)}^{-1}]$

$=\frac{1}{14}\left[\right(4^1{)}^{\frac{1}{2}}+{\left(\frac{1}{5^1}\right)}^{-1}]$ $\because m=1$
$=\frac{1}{14}\left[\right(2^2{)}^{\frac{1}{2}}+5]$
$=\frac{1}{14}[2+5]$
$=\frac{1}{14}\left(7\right)$
$=\frac{1}{2}$

Hence, option (a) is correct.