A) \[-2<a<8\]
B) \[2<a<8\]
C) \[-2\le a\le 8\]
D) \[2\le a\le 8\]
Correct Answer: C
Solution :
Roots of \[{{x}^{2}}-8x+({{a}^{2}}-6a)=0\] are real. So\[D\ge 0\] Þ \[64-4({{a}^{2}}-6a)\ge 0\]Þ \[16-{{a}^{2}}+6a\ge 0\] Þ \[{{a}^{2}}-6a-16\le 0\]Þ \[(a-8)(a+2)\le 0\] Now we have two cases: Case I : \[(a-8)\le 0\]and \[(a+2)\ge 0\] Þ \[a\le 8\]and \[a\ge -2\] Case II : \[(a-8)\ge 0\]and \[(a+2)\le 0\] Þ \[a\ge 8\]and \[a\le -2\]but it is impossible Therefore, we get \[-2\le a\le 8\] Aliter : Students should note that the expression \[(x-a)(x-b)\{a<b\}\] will be less than or equal to zero if \[x\in [a,b]\] or otherwise \[x\notin [a,b]\]. Therefore\[(a-8)(a+2)\le 0\] i.e., \[\{a-(-2)\}(a-8)\le 0\Rightarrow a\in [-2,\ 8]\].
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