A) \[\frac{1}{2}\]
B) \[2\]
C) \[-\frac{1}{2}\]
D) \[-2\]
E) \[\frac{3}{2}\]
Correct Answer: C
Solution :
Given, let\[z=\frac{2+i}{ai-1}\]and\[(z)=0,a\in R\] ...(i) \[z=\frac{(2+i)(ai+1)}{(ai-1)(ai+1)}\] \[z=\frac{2ai+a{{i}^{2}}+2+i}{{{a}^{2}}{{i}^{2}}-1}\] \[z=\frac{(2-a)+i(2a+1)}{(-{{a}^{2}}-1)}\] \[(\because {{i}^{2}}=-1)\] \[z=\frac{(a-2)}{({{a}^{2}}+1)}-i\frac{(2a+1)}{({{a}^{2}}+1)}\] From Eq. (i), \[-\frac{(2a+1)}{({{a}^{2}}+1)}=0\] \[\Rightarrow \] \[2a+1=0\] \[\Rightarrow \] \[a=-\frac{1}{2}\]
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