A) \[k=2\]
B) \[k=-3\]
C) \[k\ne \frac{-15}{2}\]
D) \[k\ne \frac{-1}{2}\]
Correct Answer: D
Solution :
| [d] Given, \[x+ky=0\] and \[2x-y=0\] |
| Here, \[{{a}_{1}}=1,\,{{b}_{1}}=k,\,{{c}_{1}}=0\] |
| and \[{{a}_{2}}=2,\,{{b}_{2}}=-1,\,{{c}_{2}}=0\] |
| System of equations has unique solution |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\frac{1}{2}\ne \frac{k}{-1}\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,k\ne -\frac{1}{2}\] |
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