Explanation

(d) Given $R = 1\ k\Omega$
Or $\ R = 1 \times 10^{3}$
$C = 1\mu F$
$\ \ \ \ \ \ \ C = 1 \times 10^{- 6}\ F$
In this condition frequency of carrier signal
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{R_{c}} < \ < f_{c}$
$\frac{1}{1 \times 10^{3} \times 10^{- 6}} < \ < f_{c}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore f_{c} > \ > 1\ kHz$
Because frequency greater than 1
So $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f_{c} = 10\ kHz$