A) 2
B) 4
C) \[-4\]
D) None of these
Correct Answer: C
Solution :
If the points\[(x+1,2),(1,x+2)\]and \[\left( \frac{1}{x+1},\frac{2}{x+1} \right)\]are collinear, then \[\left| \begin{matrix} x+1 & 2 & 1 \\ 1 & x+2 & 1 \\ \frac{1}{x+1} & \frac{2}{x+1} & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} x+1 & 2 & 1 \\ -x & x & 0 \\ \frac{1}{x+1}-(x+1) & \frac{2}{x+1}-2 & 0 \\ \end{matrix} \right|=0\] \[\left[ \begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{align} \right]\] \[\Rightarrow \] \[1\left[ -x\left( \frac{2}{x+1}-2 \right)-x\left( \frac{1}{x+1}-(x+1) \right) \right]=0\] \[\Rightarrow \] \[x\left[ -\frac{2}{x+}+2-\frac{1}{x+1}+(x+1) \right]=0\] \[\Rightarrow \] \[[-2+2(x+1)-1+{{(x+1)}^{2}}]=0\] \[\Rightarrow \] \[-2+2x+2-1+{{x}^{2}}+1+2x=0\] \[\Rightarrow \] \[{{x}^{2}}+4x=0\] \[\Rightarrow \] \[x(x+4)=0\] \[\Rightarrow \] \[x=0\] \[\Rightarrow \] \[x=-4\]
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