A) \[-2<k<2\]
B) \[-5<k<3\]
C) \[-3<k<5\]
D) \[-1<k<3\]
Correct Answer: D
Solution :
\[{{\operatorname{x}}^{2}}-2kx+{{k}^{2}}-4=0\] \[\Rightarrow \,\,\,{{(x-k)}^{2}}-{{2}^{2}}=0\] \[\Rightarrow \,\,\,\left( x-k\,-\,2 \right)\left( x-k+2 \right)=0\] \[\Rightarrow \,\,\,\,x=k+2,\,\,k-2\] \[\Rightarrow \,\,\,\operatorname{k}+2<5\,\,\,\And \,\,k-2>-\,3\] \[\Rightarrow \,\,\,\,\operatorname{k}<3\,\,\,\And \,\,\,k>-1\] \[\Rightarrow \,\,\,\,-1<k<3\]
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