Pd across $R_{\text {s }}$

$V_1=8-5=3 \mathrm{~V}$

Current through the load resistor

$\mathrm{I}=\frac{5}{1 \times 10^3}=5 \mathrm{~mA}$

Maximum current through Zener diode

$I_{z \max .}=\frac{10}{5}=2 \mathrm{~mA}$

And minimum current through Zener diode

$\quad \mathrm{I}_{z_{\text {min. }}}=0$

$\therefore \mathrm{I}_{5 \text { max. }}=5+2=7 \mathrm{~mA}$

$\text { And } \mathrm{R}_{5 \text { min }}=\frac{\mathrm{V}_1}{\mathrm{I}_{5 \max } \cdot}=\frac{3}{7} \mathrm{k} \Omega$

$\text { Similarly }$

$\mathrm{I}_{5 \text { min. }}=5 \mathrm{~mA}$

$\text { And } \mathrm{R}_{5 \text { max. }}=\frac{\mathrm{V}_1}{\mathrm{I}_{\mathrm{s} \text { min. }}}=\frac{3}{5} \mathrm{k} \Omega$

$\therefore \frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_5<\frac{3}{5} \mathrm{k} \Omega$