A) \[{{e}^{-2}}\]
B) \[{{e}^{-1}}\]
C) \[1-{{e}^{-1}}\]
D) \[1-{{e}^{-2}}\]
Correct Answer: C
Solution :
Given\[f'(x)+f(x)\le 1\] Multiplying by \[{{e}^{x}},\] we get \[f'(x){{e}^{x}}+f(x){{e}^{x}}\le {{e}^{x}}\] \[\frac{d}{dx}({{e}^{x}}.f(x))\le {{e}^{x}}\] Integrating between 0 and 1 \[\int\limits_{0}^{1}{\frac{d}{dx}}({{e}^{x}}f(x))dx\le \int\limits_{0}^{1}{{{e}^{x}}dx}\] \[[{{e}^{x}}f(x)]_{0}^{1}\le [{{e}^{x}}]_{0}^{1}\] \[e.f(1)-{{e}^{0}}.f(0)\le e-1\] \[f(1)\le \frac{e-1}{e}\]
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