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(D) Given the inequality $f(x) + f'(x) \le 1$.Multiply both sides by the integrating factor $e^x$:$e^x f'(x) + e^x f(x) \le e^x$This can be written as

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$\frac{e-1}{e}$

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(D) Given the inequality $f(x) + f'(x) \le 1$.Multiply both sides by the integrating factor $e^x$:$e^x f'(x) + e^x f(x) \le e^x$This can be written as the derivative of a product:$\frac{d}{dx} (f(x) e^x) \le e^x$Now,integrate both sides from $0$ to $1$:$\int_{0}^{1} \frac{d}{dx} (f(x) e^x) dx \le \int_{0}^{1} e^x dx$$[f(x) e^x]_{0}^{1} \le [e^x]_{0}^{1}$$f(1) e^1 - f(0) e^0 \le e^1 - e^0$Since $f(0) = 0$,we have:$f(1) e \le e - 1$$f(1) \le \frac{e-1}{e}$Thus,the largest possible value of $f(1)$ is $\frac{e-1}{e}$.

Suppose $f(x)$ is a differentiable real function such that $f(x) + f'(x) \le 1$ for all $x$ and $f(0)=0$. The largest possible value of $f(1)$ is

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