A) \[\left( -\infty ,\,\,{{\log }_{e}}\frac{11}{3} \right]\]
B) \[\left[ {{\log }_{e}}\frac{11}{3},\,\,\infty \right)\]
C) \[\left[ -{{\log }_{e}}\frac{11}{3},\,\,{{\log }_{e}}\frac{11}{3} \right]\]
D) None of the above
Correct Answer: B
Solution :
\[f(x)\]is defined, if\[3{{x}^{2}}-4x+5>0\] \[\Rightarrow \] \[3\left[ {{x}^{2}}-\frac{4}{3}x+\frac{5}{3} \right]>0\] \[\Rightarrow \] \[3\left[ {{\left( x-\frac{2}{3} \right)}^{2}}+\frac{11}{9} \right]>0\] which is true for all real\[x\]. \[\therefore \]Domain of\[f=(-\infty ,\,\,\infty )\] Let\[{{\log }_{e}}(3{{x}^{2}}-4x+5)\Rightarrow {{e}^{y}}=3{{x}^{2}}-4x+5\] \[\Rightarrow \]\[3{{x}^{2}}-4x+(5-{{e}^{y}})=0\] For\[x\]to be real, \[16-12(5-{{e}^{y}})\ge 0\] \[\Rightarrow \] \[12{{e}^{y}}\ge 44\] \[\Rightarrow \] \[{{e}^{y}}\ge \frac{11}{3}\] \[\Rightarrow \] \[y\ge {{\log }_{e}}\frac{11}{3}\] Range of\[f=\left[ {{\log }_{e}}\frac{11}{3},\,\,\infty \right)\]
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