Let f:[0,1] rightarrow mathbb{R} be a functio | vedclass

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(D) $1.$ Given $f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x$. Multiplying by $e^{-x}$,we get $e^{-x}f^{\prime \prime}(x) - 2e^{-x}f^{\prime}(x)

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$(B, C)$

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(D) $1.$ Given $f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x$. Multiplying by $e^{-x}$,we get $e^{-x}f^{\prime \prime}(x) - 2e^{-x}f^{\prime}(x) + e^{-x}f(x) \geq 1$.This can be written as $\frac{d^2}{dx^2}(e^{-x}f(x)) \geq 1$. Let $g(x) = e^{-x}f(x)$. Then $g^{\prime \prime}(x) \geq 1$.Since $f(0)=f(1)=0$,we have $g(0)=0$ and $g(1)=0$. Since $g^{\prime \prime}(x) > 0$,$g(x)$ is strictly convex. $A$ convex function with $g(0)=g(1)=0$ must be negative in $(0,1)$. Thus $g(x) $2.$ Since $g(x) = e^{-x}f(x)$ has a minimum at $x=1/4$,$g^{\prime}(1/4) = 0$. For $x  1/4$,$g^{\prime}(x) > 0$.$g^{\prime}(x) = e^{-x}f^{\prime}(x) - e^{-x}f(x) = e^{-x}(f^{\prime}(x) - f(x))$.For $x \in (0, 1/4)$,$g^{\prime}(x) For $x \in (1/4, 1)$,$g^{\prime}(x) > 0 \Rightarrow f^{\prime}(x) - f(x) > 0 \Rightarrow f^{\prime}(x) > f(x)$.Thus,$(B)$ and $(C)$ are true.

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