A) \[g(x)>0\]
B) \[g(x)\ge 0\]
C) \[g(x)<0\]
D) \[g(x)\le 0\]
Correct Answer: A
Solution :
Clearly the equation \[f'(x)=0\] must also have repeated roots. So, \[f'(x)\ge 0\,\forall \,x.\] |
Let \[f'(x)={{a}_{1}}{{x}^{2}}+{{b}_{1}}x+{{c}_{1}},\] |
Where \[{{a}_{1}}>0\] and \[b_{1}^{2}-4{{a}_{1}}{{c}_{1}}=0,\] then \[g(x)={{a}_{1}}{{x}_{2}}+({{b}_{1}}-2{{a}_{1}})x+2{{a}_{1}}-{{b}_{1}}+{{c}_{1}}\] |
Its discriminant \[b_{1}^{2}-4{{a}_{1}}{{c}_{1}}-4a_{1}^{2}<0\] \[\Rightarrow \]\[g(x)>0\,\forall \,x\in R\] |
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