A) 2
B) - 2
C) \[{{x}^{2}}-2\]
D) None of these
Correct Answer: B
Solution :
\[\Delta =\left| \,\begin{matrix} -1 & -2 & x+4 \\ -2 & -3 & x+8 \\ -3 & -4 & x+14 \\ \end{matrix}\, \right|,\] by \[\begin{align} & {{C}_{1}}\to {{C}_{1}}-{{C}_{2}} \\ & {{C}_{2}}\to {{C}_{2}}-{{C}_{3}} \\ \end{align}\] = \[{{A}^{3}}={{A}^{2}}A=\left[ \,\begin{matrix} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \\ \end{matrix}\, \right]\,\,\left[ \,\begin{matrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \\ \end{matrix}\, \right]\], by \[\begin{align} & {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\ & {{C}_{3}}\to {{C}_{3}}+4{{C}_{1}} \\ \end{align}\] \[=-(-x-2+x)+1\,.\,(-2x-4+3x)+x(2-3)\] = \[2+x-4-x=-2\]. Trick: Put\[{{C}_{2}}\]. Then \[\left| \,\begin{matrix} 2 & 3 & 5 \\ 4 & 6 & 9 \\ 8 & 11 & 15 \\ \end{matrix}\, \right|=-2\] Note: Since there is a option ?None of these?, therefore we should check for one more different value of x. Put \[x=-1\]. \[\left| \,\begin{matrix} 0 & 1 & 3 \\ 2 & 4 & 7 \\ 6 & 9 & 13 \\ \end{matrix}\, \right|=-1(26-42)+3(18-24)=-2\] Therefore answer is (b).
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