JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 25)
Let $$f:\mathbb{R}\to\mathbb{R}$$ be a differentiable function that satisfies the relation $$f(x+y)=f(x)+f(y)-1,\forall x,y\in\mathbb{R}$$. If $$f'(0)=2$$, then $$|f(-2)|$$ is equal to ___________.
Answer
3
Explanation
$f(x+y)=f(x)+f(y)-1$
$$ \begin{aligned} & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\\\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2 \\\\ & f^{\prime}(x)=2 \Rightarrow d y=2 d x \\\\ & y=2 x+C \\\\ & \mathrm{x}=0, \mathrm{y}=1, \mathrm{c}=1 \\\\ & \mathrm{y}=2 \mathrm{x}+1 \\\\ & |f(-2)|=|-4+1|=|-3|=3 \end{aligned} $$
$$ \begin{aligned} & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\\\ & f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2 \\\\ & f^{\prime}(x)=2 \Rightarrow d y=2 d x \\\\ & y=2 x+C \\\\ & \mathrm{x}=0, \mathrm{y}=1, \mathrm{c}=1 \\\\ & \mathrm{y}=2 \mathrm{x}+1 \\\\ & |f(-2)|=|-4+1|=|-3|=3 \end{aligned} $$
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