A) \[a<2\]
B) \[2\le a\le 3\]
C) \[3<a\le 4\]
D) \[a>4\]
Correct Answer: A
Solution :
Let\[f(x)={{x}^{2}}-2ax+{{a}^{2}}+a-3=0\]. Since, \[f(x)\]has real roots both less than 3. Therefore,\[D>0\]and\[f(3)>0\] \[\Rightarrow \] \[4{{a}^{2}}-4({{a}^{2}}+a-3)>0\] and \[{{a}^{2}}-5a+6>0\] \[\Rightarrow \] \[a<3\]and\[a<2\]or\[a>3\] \[\Rightarrow \] \[a<2\]You need to login to perform this action.
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