A) \[a=\frac{\sqrt{e}}{2}\] and \[b=\frac{-1}{8}\]
B) \[a=\frac{-\sqrt{e}}{2}\] and \[b=\frac{1}{8}\]
C) \[a=\frac{1}{8}\] and \[b=\frac{\sqrt{e}}{2}\]
D) \[a=\frac{-\sqrt{e}}{2}\] and \[b=\frac{-1}{8}\]
Correct Answer: A
Solution :
\[f(x)=ax{{e}^{b{{x}^{2}}}};\,\,f(2)=1\] \[f'(2)=0;\,\,2a{{e}^{4b}}=1\] also \[f'(x)=a[x{{e}^{b{{x}^{2}}}}.2bx+{{e}^{b{{x}^{2}}}}]=ax{{e}^{b{{x}^{2}}}}[2b{{x}^{2}}+1]\] \[f'(2)=a{{e}^{4b}}(8b+1)=0\] \[a=0\]or\[b=-1/8\]; but\[a\ne 0;\,\,a=e/2\] hence \[a=\frac{\sqrt{e}}{2}\]and\[b=\frac{-1}{8}\]
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