A) 1
B) 2
C) 0
D) none of these
Correct Answer: A
Solution :
| \[{{x}^{2}}-(k-2)\,\,x+{{k}^{2}}=0\] |
| \[{{x}^{2}}+kx-2k-1=0\] should have both roots. Common or each should have equal roots. |
| Case. (i) \[\frac{1}{1}=\frac{-\,(k-2)}{k}=\frac{{{k}^{2}}}{2k-1}\] |
| \[\Rightarrow k=-\,k+2\] and \[2k-1={{k}^{2}}\] |
| \[\Rightarrow k=1\] |
| Case. (ii) \[{{(k-2)}^{2}}-4{{k}^{2}}=0\] and \[{{k}^{2}}-4(2k-1)=0\] |
| \[(3k-2)(-\,k-2)=0\] and \[{{k}^{2}}-8k+4=0\] have no common value, \[k=1\] is the only solution. |
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