A) 1
B) 2
C) \[{{2}^{27}}\]
D) None of these
Correct Answer: A
Solution :
\[{{2}^{({{x}^{2}}-3)}}{{^{^{3}}}^{+27}}\]is minimum when \[{{({{x}^{2}}-3)}^{3}}+27\] is minimum Let \[y={{({{x}^{2}}-3)}^{3}}+27\] \[={{x}^{6}}-27-9{{x}^{4}}+27{{x}^{2}}+27\] \[\Rightarrow \] \[y={{x}^{6}}-9{{x}^{4}}+27{{x}^{2}}\] \[\Rightarrow \] \[y={{x}^{2}}({{x}^{4}}-9{{x}^{2}}+27)\] \[\Rightarrow \] \[y={{x}^{2}}\left[ {{x}^{4}}-9{{x}^{2}}+\frac{81}{4}-\frac{81}{4}+27 \right]\] \[\Rightarrow \] \[y={{x}^{2}}\left[ \left( {{x}^{4}}-9{{x}^{2}}+\frac{81}{4} \right)+\frac{27}{4} \right]\] \[\Rightarrow \] \[y={{x}^{2}}\left[ {{\left( {{x}^{2}}-\frac{9}{2} \right)}^{2}}+\frac{27}{2} \right]\ge 0\,\,\forall x\] \[\therefore \] Minimum value of \[{{({{x}^{2}}-3)}^{3}}+27\]is 0. Hence, minimum value of \[{{2}^{{{({{x}^{2}}-3)}^{3}}+27}}\] \[={{2}^{0}}=1\]
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