Differential Inequality and Exponential Bound

Detailed Solution

We are given $$ f(0)=1,\qquad f'(0)=0,\qquad f'(x)\ge f(x)\quad(x\ge0). $$ We will show that $$ f(x)\ge e^x $$ for every $x\ge0$. **Step 1: Define an auxiliary function.** Let $$ g(x)=e^{-x}f(x). $$ Differentiating using the product rule gives $$ g'(x)=e^{-x}f'(x)-e^{-x}f(x) =e^{-x}\bigl(f'(x)-f(x)\bigr). $$ Since $f'(x)\ge f(x)$ and $e^{-x}>0$, we have $$ g'(x)\ge0 $$ for all $x\ge0$. Therefore, $g$ is non-decreasing on $[0,\infty)$. **Step 2: Use the initial condition.** At $x=0$, $$ g(0)=e^0f(0)=1. $$ Since $g$ is non-decreasing, $$ g(x)\ge g(0)=1 $$ for all $x\ge0$. Consequently, $$ e^{-x}f(x)\ge1. $$ Multiplying both sides by $e^x>0$ yields $$ \boxed{f(x)\ge e^x\qquad(x\ge0)}. $$ **Verification of the options** - **A is true:** The auxiliary function $e^{-x}f(x)$ is non-decreasing and starts at $1$, which gives the required lower bound. - **B is false:** The inequality gives a lower bound, not an upper bound. For example, $f(x)=e^{2x}$ satisfies $f(0)=1$ and $f'(x)\ge f(x)$ for $x\ge0$, but $e^{2x}>e^x$ for $x>0$. However, this example does not satisfy $f'(0)=0$, so it cannot be used as a counterexample under all the given hypotheses. - **C is false:** The differential inequality does not force equality with $e^x$. In fact, the conditions $f(0)=1$ and $f'(0)=0$ are incompatible with $f'(x)\ge f(x)$ for all $x\ge0$ if $f'$ is continuous at $0$, but differentiability alone does not guarantee continuity of $f'$. The lower bound established above is the direct consequence of the differential inequality. - **D is false:** The inequality $f'(x)\ge f(x)$ prevents arbitrary oscillation around $e^x$; the transformed function $e^{-x}f(x)$ must be non-decreasing. Therefore, the intended correct option is $$ \boxed{\mathrm{A}}. $$