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Let f: \mathbb{R} \rightarrow \mathbb{R} be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)-$ $1, \forall x, y \in \mathbb{R}. If f^{\prime}(0)=2$, then $|f(-2)| is equal to

 

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{Given\: \: \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})-1 \forall \mathrm{x}, \mathrm{y} \in \operatorname{IR} and \mathrm{f}^{\prime}(0)=2}
Partial differentiate w.r.t \mathrm{x}
\mathrm{\begin{aligned} & \Rightarrow f^{\prime}(x+y) f^{\prime}(x) \\ & \text { for } x=0 \\ \end{aligned} }

\mathrm{\begin{aligned} & f^{\prime}(y)=f^{\prime}(0)=2 \end{aligned} }
on Integrating
\Rightarrow \mathrm{f}(\mathrm{y})=2 \mathrm{y}+\mathrm{c}...............(2)
for \mathrm{y=0}
\Rightarrow \mathrm{f}(0)=\mathrm{C}           ................(3)
Put \mathrm{x=y=0 \: in\: (1)}
\begin{aligned} & \Rightarrow \mathrm{f}(0)=\mathrm{f}(0)+\mathrm{f}(0)-1 \\ & \Rightarrow \mathrm{f}(0)=1 \end{aligned}
from (3) \& (4)
\begin{aligned} & \mathrm{c}=1 \\ \end{aligned}

\begin{aligned} & \Rightarrow \mathrm{f}(\mathrm{y})=2 \mathrm{y}+1 \\ & \Rightarrow \mathrm{f}(-2)=-4+1=-3 \\ & \therefore|\mathrm{f}(-2)|=3 \end{aligned}

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vishal kumar

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