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 :
\[f(x)=ax{{e}^{b{{x}^{2}}}};f(2)=2\] and \[f'(2)=0;\]\[2a{{e}^{4b}}=1\] ?. (1) also\[f'(x)=a[x{{e}^{b{{x}^{2}}}}\cdot 2bx+{{e}^{b{{x}^{2}}}}]\] \[=ax{{e}^{b{{x}^{2}}}}[2b{{x}^{2}}+1]\] \[f'(2)=a{{e}^{4b}}(8b+1)=0;\] \[a=0\]or\[b=-1/8\]but\[a\ne 0;a=e\text{/}2\] Hence \[a=\frac{\sqrt{e}}{2}\]and\[b=\frac{-1}{8}\].
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