Let f:mathbbR→ mathbbR be a differentiable function that satisfies the relation f(x+y)=f(x)+f(y)-1,forall x, y∈ mathbbR. If f^'(0)=2, then |f(-2)| is…

Given,

Functional equation $f(x+y)=f(x)+f(y)-1$f(x+y)=f(x)+f(y)−1

Now taking x=0&y=0 in above equation we get,

$$f(0+0)=f(0)+f(0)-1\Rightarrow f(0)=1$$f(0+0)=f(0)+f(0)−1⇒f(0)=1

Now we know that,

$${f}^{'}(x)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(x+h)-f(x)}{h}$$f′(x)=h→0lim​hf(x+h)−f(x)​

$$\Rightarrow {f}^{'}(x)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(x)+f(h)-1-f(x)}{h}$$⇒f′(x)=h→0lim​hf(x)+f(h)−1−f(x)​

$$\Rightarrow {f}^{'}(x)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(h)-1}{h}$$⇒f′(x)=h→0lim​hf(h)−1​

Now let $\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(h)-1}{h}=k$h→0lim​hf(h)−1​=k

So, the equation becomes ${f}^{'}(x)=k$f′(x)=k

Now putting $x=0$x=0 in above equation we get,

${f}^{'}(0)=k\Rightarrow k=2$f′(0)=k⇒k=2 ${\text{as given }{f}^{'}(0)=2}$as given f′(0)=2

So, ${f}^{'}(x)=2$f′(x)=2

Now integrate both side we get,

$$f(x)=2x+c$$f(x)=2x+c

Now again taking $x=0$x=0 we get,

$$f(0)=2\times 0+c\Rightarrow c=1{\text{as }f(0)=1}$$f(0)=2×0+c⇒c=1as f(0)=1

So, $f(x)=2x+1$f(x)=2x+1

And $f(-2)=2\times (-2)+1\Rightarrow f(-2)=-3$f(−2)=2×(−2)+1⇒f(−2)=−3

Hence, $|f(-2)|=3$∣f(−2)∣=3