A) Contains at least four elements
B) Is a singleton
C) Contains exactly two elements
D) Is an empty set
Correct Answer: B
Solution :
| \[{{3}^{x}}\left( {{3}^{x}}-1 \right)+2=\left| {{3}^{x}}-1 \right|+\left| {{3}^{x}}-2 \right|\] |
| Case I: \[0<{{3}^{x}}<1\] \[\Rightarrow \,\,\,-\infty <x<0\] |
| \[\Rightarrow \,\,\,\,\,\,{{\left( {{3}^{x}} \right)}^{2}}-{{3}^{x}}+2=1-{{3}^{x}}+2-{{3}^{x}}\] |
| \[\Rightarrow \,\,\,\,\,\,{{\left( {{3}^{x}} \right)}^{2}}+{{3}^{x}}-1=0\] \[\Rightarrow \,\,\,\,\,\,{{3}^{x}}=\frac{-1+\sqrt{5}}{2}<1\] |
| One real solution |
| Case II: \[1<{{3}^{x}}<2\] \[\Rightarrow \text{ }0<x<lo{{g}_{2}}3\] |
| \[\Rightarrow \,\,\,\,\,{{\left( {{3}^{x}} \right)}^{2}}-{{3}^{x}}+2={{3}^{x}}-1+2-{{3}^{x}}\] |
| \[\Rightarrow \,\,\,\,\,{{\left( {{3}^{x}} \right)}^{2}}-{{3}^{X}}+1=0\] |
| \[\Rightarrow \] No solution |
| \[\therefore \] Discriminant is negative |
| Case III: \[2<{{3}^{x}}<\infty \] |
| \[\Rightarrow \,\,\,\,\,{{\left( {{3}^{x}} \right)}^{2}}-{{3}^{x}}+2={{2.3}^{x}}-3\] |
| \[\Rightarrow \text{ }{{\left( {{3}^{x}} \right)}^{2}}-3.\left( {{3}^{x}} \right)+5=0\] |
| \[\Rightarrow \] No solution \[\therefore \] Discriminant is negative |
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