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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, \[f(2x)-f(x)=a{{(2x+a)}^{2}}-a{{(x\,+a)}^{2}}\] \[=a\,\,[{{(2x+a)}^{2}}-{{(x+a)}^{2}}]\] \[=a\,\,[(2x+a+x+a)(2x+a-x-a)]\] \[[\because \,\,(a\,+b)(a\,-b)={{a}^{2}}-{{b}^{2}}]\] \[=a\,\,[(3x+2a)(x)]=x(3x+2a)a\]
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