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If  by using properties of determinants, find the value of \[f(2x)-f(x).\]

Answer:

Let

On multiplying and dividing first column by a, we get

Applying \[{{C}_{1}}\to {{C}_{1}}+b{{C}_{2}}+c{{C}_{3}},\] we get

On taking \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\] common from \[{{C}_{1}},\] we get

Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}},\] we get

Now, expanding along \[{{C}_{1}},\] we get

\[A=\frac{1}{a}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\times 1\times \left| \begin{matrix}    c & -\,a-b  \\    a+c & -\,b  \\ \end{matrix} \right|\]

\[=\frac{1}{a}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})(-bc+{{a}^{2}}+ac+ba+bc)\]

\[=\frac{1}{a}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})({{a}^{2}}+ac+ba)\]

\[=\frac{1}{a}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\times a(a+b+c)\]

\[=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]             Hence proved.

OR

Given,

Taking a common from \[{{C}_{1}},\] we get

Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}},\] we get

Now, on expanding through \[{{R}_{1}},\] we get

\[f(x)=a\,[1\{a\,(x+a)+1({{x}^{2}}+ax)\}]\]

\[=a\,[ax+{{a}^{2}}+{{x}^{2}}+ax]\]

\[=a\,[{{x}^{2}}+2ax+{{a}^{2}}]\]

\[=a\,{{(x+a)}^{2}}\]

\[\therefore \]      \[f(2x)=a{{(2x+a)}^{2}}\]

Now, \[f(2x)-f(x)=a{{(2x+a)}^{2}}-a{{(x\,+a)}^{2}}\]

\[=a\,\,[{{(2x+a)}^{2}}-{{(x+a)}^{2}}]\]

\[=a\,\,[(2x+a+x+a)(2x+a-x-a)]\]

\[[\because \,\,(a\,+b)(a\,-b)={{a}^{2}}-{{b}^{2}}]\]

\[=a\,\,[(3x+2a)(x)]=x(3x+2a)a\]