The value of constant 'k' will be [tex]\frac{1}{3}[/tex].
Given equation in the question is,
[tex]\frac{1}{3}\times (x-k)=kx[/tex]
Further solve the expression for the solution of the equation.
[tex]\frac{3}{3}\times (x-k)=3kx[/tex]
[tex](x-k)=3kx[/tex]
[tex]x-3kx-k=3kx-3kx[/tex]
[tex]x-3kx-k=0[/tex]
[tex]x(1-3k)=k[/tex]
[tex]\frac{x(1-3k)}{(1-3k)}=\frac{3kx}{1-3k}[/tex]
[tex]x=\frac{3kx}{(1-3k)}[/tex]
Since, equation has no value, denominator will be zero.
[tex]1-3k=0[/tex]
[tex]1=3k[/tex]
[tex]k=\frac{1}{3}[/tex]
Therefore, Option (D) will be the correct option.
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