A) 4
B) 5
C) 6
D) 7
Correct Answer: B
Solution :
[b] Equation \[{{x}^{2}}+2(a+b)x+(a-b+8)=0\] has unequal real roots \[\forall \,\,x\in R\] \[\Rightarrow \,\,D>0\,\,\forall a\in R\] \[\Rightarrow \,\,4{{(a+b)}^{2}}-4(a-b+8)>0,\,\,\forall \,a\in R\] \[\Rightarrow \,\,{{a}^{2}}+a(2b-1)+({{b}^{2}}+b-8)>0,\,\,\forall a\in R\] \[\Rightarrow \,\,D<0\] \[\Rightarrow \,\,{{(2b-1)}^{2}}-4({{b}^{2}}+b-8)<0\] \[\Rightarrow \,\,\,\,b>\frac{33}{8}\]You need to login to perform this action.
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