A) \[1<m<5\]
B) \[-1<m<1\]
C) \[-5<m<\frac{11}{24}\]
D) \[m<\frac{71}{24}\]
Correct Answer: D
Solution :
| \[\because \]\[{{x}^{2}}+2x+2={{(x+1)}^{2}}+1>0\,\forall \,x\in \mathbf{R}\] |
| \[\therefore \frac{m{{x}^{2}}+3x+4}{{{x}^{2}}+2x+2}<5\] |
| \[\Rightarrow \]\[(m-5){{x}^{2}}-7x-6<0\,\forall \,x\in \mathbf{R}\] |
| \[\Rightarrow \]\[m-5<0\]and\[D<0\Rightarrow m<5\] |
| \[\operatorname{and}49+24\left( m-5 \right)<0\Rightarrow m<\frac{71}{24}\] |
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