Two point charges Q and -Q/4 are separated by a distance x. Then

At a point on the line joining the charges at a distance which is $\frac{\mathrm{x}}{3}$on the right side of the charge  $\frac{-\mathrm{Q}}{4}$:    $\text{V}=\frac{\text{KQ}}{\left(\frac{\text{4x}}{\text{3}}\right)}\text{+}\frac{\text{K}\left(\frac{-\text{Q}}{\text{4}}\right)}{\left(\frac{\text{x}}{\text{3}}\right)}=\frac{\text{3KQ}}{\text{4x}}-\frac{\text{3KQ}}{\text{4x}}=0$

Option (A) is correct 

At a point on the line joining the charges at a distance which is $\frac{\mathrm{x}}{5}$on the left side of the charge  $\frac{-\mathrm{Q}}{4}$: $\text{V}=\frac{\text{KQ}}{\left(\frac{\text{4x}}{\text{5}}\right)}\text{+}\frac{\text{K}\left(\frac{-\text{Q}}{\text{4}}\right)}{\left(\frac{\text{x}}{\text{5}}\right)}=\frac{\text{5KQ}}{\text{4x}}-\frac{\text{5KQ}}{\text{4x}}=0$

Option (B) is correct 

At a point on the line joining the charges which is at a distance x on the right side of the charge  $\frac{-\mathrm{Q}}{4}$: $\text{E}=\frac{\text{KQ}}{{\left(\text{2x}\right)}^{\text{2}}}-\frac{\text{K}\left(\frac{\text{Q}}{\text{4}}\right)}{{\text{x}}^{\text{2}}}=0$

Option (C) is correct

There is only one point along the line joining the charges where the electric field is zero. It is located at a distance x on the right side of the charge $\frac{-\text{Q}}{\text{4}}$
Option (D) is wrong