Let Φ be a function satisfying Φ(x+y)=Φx+Φy and Φx=xψx for all x,y ∈ R. If ψx is continuous then

∅x+y=∅x+∅y and ∅x=xψx ∅1x=limh→0∅x+h-∅xh =limh→0∅x+∅h-∅xh =limh→0∅hh =limh→0hψhh =limh→0ψh ∅1x=ψ0

Let Φ be a function satisfying Φ(x+y)=Φx+Φy and Φx=xψx for all x,y ∈ R. If ψx is continuous then

∅x+y=∅x+∅y and ∅x=xψx ∅1x=limh→0∅x+h-∅xh =limh→0∅x+∅h-∅xh =limh→0∅hh =limh→0hψhh =limh→0ψh ∅1x=ψ0

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Let $Φ$ be a function satisfying $Φ (x + y) = Φ (x) + Φ (y)$ and $Φ (x) = xψ (x)$ for
all $x, y \in R .$ If $ψ (x)$ is continuous then

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$Φ^{'} (x) = ψ (0)$

$Φ^{'} (x) = ψ (x)$

$Φ^{'} (x) = ψ^{'} (x)$

$Φ^{'} (x) = ψ (x) + ψ^{'} (x)$

answer is A.

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Detailed Solution

$\emptyset (x + y) = \emptyset (x) + \emptyset (y) and \emptyset (x) = xψ (x) \emptyset^{1} (x) = \lim_{h \to 0} \frac{\emptyset (x + h) - \emptyset (x)}{h} = \lim_{h \to 0} \frac{\emptyset (x) + \emptyset (h) - \emptyset (x)}{h} = \lim_{h \to 0} \frac{\emptyset (h)}{h} = \lim_{h \to 0} \frac{hψ (h)}{h} = \lim_{h \to 0} ψ (h) \emptyset^{1} (x) = ψ (0)$