is
A) 0
B) 2
C) -1
D) 1
Correct Answer: C
Solution :
\[\left| \begin{matrix} x+3 & x & x+2 \\ x & x+1 & x-1 \\ x+2 & 2x & 3x+1 \\ \end{matrix} \right|=f(x)\] (say) Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[f(x)=\left| \begin{matrix} x+3 & x & x+2 \\ -3 & 1 & -3 \\ -1 & x & 2x-1 \\ \end{matrix} \right|\] Applying \[{{C}_{1}}\to {{C}_{1}}-{{C}_{3}}\] and \[{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\] \[f(x)=\left| \begin{matrix} 1 & -2 & x+2 \\ 0 & 4 & -3 \\ -2x & 1-x & 2x-1 \\ \end{matrix} \right|\] Expand w.r.t. \['{{C}_{1}}'\] \[f(x)=[4(2x-1)+3(1-x)]\] \[+(-2x)[6-4(x+2)]\] \[f(x)=[8x-4+3-3x]+[-2x][-4x-2]\] \[f(x)=(5x-1)+(8{{x}^{2}}+4x)\] \[f(x)=8{{x}^{2}}+9x-1\] Hence, the constant term of quadratic equation is \[-1.\]
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