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Q.

If f(1)=0 and dfdx>f(x)∀x≥1 , then

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a

fx>0

b

fx>1

c

fx>e

d

fx>1e

answer is A.

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Detailed Solution

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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