If f(1)=0 and dfdx>f(x)∀x≥1 , then

dfdx−f(x)>0⇒e−xdfdx−f(x)>e⇒ddxe−x⋅f(x)>0⇒e−x⋅f(x) is on increasing function f(1)=0→e−x⋅f(x)>0→f(x)>0

If f(1)=0 and dfdx>f(x)∀x≥1 , then

dfdx−f(x)>0⇒e−xdfdx−f(x)>e⇒ddxe−x⋅f(x)>0⇒e−x⋅f(x) is on increasing function f(1)=0→e−x⋅f(x)>0→f(x)>0

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$If f (1) = 0 and \frac{d f}{d x} > f (x) \forall x \geq 1, then$

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$f (x) > 0$

$f (x) > 1$

$f (x) > e$

$f (x) > \frac{1}{e}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{d f}{d x} - f (x) > 0 \Rightarrow e^{- x} (\frac{d f}{d x} - f (x)) > e \\ \Rightarrow \frac{d}{d x} (e^{- x} \cdot f (x)) > 0 \Rightarrow e^{- x} \cdot f (x) is on increasing function \\ f (1) = 0 \to e^{- x} \cdot f (x) > 0 \to f (x) > 0 \end{array}$