Step-by-step explanation:
[tex] \red{\large\underline{\sf{Solution-}}}[/tex]
We know that,
[tex]\rm :\longmapsto\:Wronkian \: of \: the \: set \: {\left \{f_{1}(x),f_{2}(x),f_{3}(x) \right \}} \: is \: [/tex]
[tex]W{\left(f_{1},f_{2},f_{3} \right)}\left(x\right) = \left|\begin{array}{ccc}f_{1}\left(x\right) & f_{2}\left(x\right) & f_{3}\left(x\right)\\f_{1}^{\prime}\left(x\right) & f_{2}^{\prime}\left(x\right) & f_{3}^{\prime}\left(x\right)\\f_{1}^{\prime\prime}\left(x\right) & f_{2}^{\prime\prime}\left(x\right) & f_{3}^{\prime\prime}\left(x\right)\end{array}\right|[/tex]
Here,
[tex]\rm :\longmapsto\:f_1(x) = 1 - x[/tex]
[tex]\rm :\longmapsto\:f_2(x) = 1 + x[/tex]
[tex]\rm :\longmapsto\:f_3(x) = 1 - 3x[/tex]
So, there respective differential coefficients are as follow
[tex]\rm :\longmapsto\:f_1'(x) = 0 - 1 = - 1[/tex]
[tex]\rm :\longmapsto\:f_2'(x) = 0 + 1 = 1[/tex]
[tex]\rm :\longmapsto\:f_3'(x) = 0 - 3 = - 3[/tex]
Also,
[tex]\rm :\longmapsto\:f_1''(x) = 0[/tex]
[tex]\rm :\longmapsto\:f_2''(x) = 0[/tex]
[tex]\rm :\longmapsto\:f_3''(x) = 0[/tex]
So, on substituting the values in
[tex]W{\left(f_{1},f_{2},f_{3} \right)}\left(x\right) = \left|\begin{array}{ccc}f_{1}\left(x\right) & f_{2}\left(x\right) & f_{3}\left(x\right)\\f_{1}^{\prime}\left(x\right) & f_{2}^{\prime}\left(x\right) & f_{3}^{\prime}\left(x\right)\\f_{1}^{\prime\prime}\left(x\right) & f_{2}^{\prime\prime}\left(x\right) & f_{3}^{\prime\prime}\left(x\right)\end{array}\right|.[/tex]
We get
[tex]\rm :\longmapsto\:W= \: \ \: \begin{gathered}\sf \left | \begin{array}{ccc}1 - x&1 + x&1 - 3x\\ - 1&1& - 3\\0&0&0\end{array}\right | \end{gathered} \\ [/tex]
[tex]\rm \: = \: 0[/tex]
[ If elements of any row all are 0, its determinant value is zero]
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
