A) \[f(1)\]
B) f (3)
C) \[f(1)+f(3)\]
D) \[f(1)+f(5)\]
E) (e) \[f(1)+f(3)+f(5)\]
Correct Answer: B
Solution :
\[f(x)=2(x-3)(x-5)\]; \[\left| \,\begin{matrix} 1 & x+3 & 3({{x}^{2}}+3x+9) \\ 1 & x+5 & 4({{x}^{2}}+5x+25) \\ 1 & 1 & 3 \\ \end{matrix}\, \right|\] (Taking out \[(x-3),(x-5)\]and 2 from Ist row, IInd row and IIrd column respectively) \[f(x)=2(x-3)(x-5)\] \[\left| \,\begin{matrix} 0 & (x+2) & 3({{x}^{2}}+3x+8) \\ 0 & 2 & {{x}^{2}}+11x+73 \\ 1 & 1 & 3 \\ \end{matrix}\, \right|,\,\,\]\[\]\[({{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\]and\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]) \[=2(x-3)(x-5)[1(x+2)\]\[({{x}^{2}}+11x+73)-6({{x}^{2}}+3x+8)]\] \[=2({{x}^{2}}-8x+15)({{x}^{3}}+13{{x}^{2}}+95x\]\[+146-6{{x}^{2}}-18x-48)\] \[=2({{x}^{2}}-8x+15)({{x}^{3}}+7{{x}^{2}}+77x+98)\] \[=2({{x}^{5}}-{{x}^{4}}+36{{x}^{3}}-413{{x}^{2}}\]\[+371x+1470)\] \[f(1)=2928\],\[f(3)=0\],\[f(5)=0\] \ \[f(1).f(3)+f(3).f(5)+f(5).f(1)\]\[=0+0+0\]\[=0=f(3)\].
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