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If $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(xy+1) = f(x)f(y)-f(y)-x+2\;\forall x, y\in \mathbb{R}$ and $f(0) = 1\;,$ Then $f(x) $ is

$\bf{My\; Try::}$ Put $x=y=0\;,$ We get $f(1) = (f(0))^2-f(0)-0+2 = 1-1+2=2$

Similarly put $x=y=1\;,$ We get $f(2) = (f(1))^2-f(1)-1+2 = 4-2-1+2 = 3$

So from above values function must be in the form of $f(x) = x+1$

But i did not understand how can i calculate it.

Help required, Thanks

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$$f(xy+1) = f(x)f(y)-f(y)-x+2$$

Let $y=0$. Then $$f(1)=f(x)f(0)-x+2$$ Let $f(1)=a$. Then $$f(x)=x+a-2$$

Check:

$xy+1+a-2=(x+a-2)(y+a-2)-(y+a-2)-x+2$

$xy+a-1=xy+ax-2x+ay+a^2-2a-2y-2a+4-y-a+2-x+2$

$(a-2-1)x+(a-2-1)y+(a^2-6a+9)=0$

$(a-3)x+(a-3)y+(a-3)^2=0$ $(\forall x,y \in \mathbb R)$

Then $a=3$

Answer: $$f(x)=x+1$$

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$$\displaystyle f(xy + 1) = f(x)f(y) - f(y) - x + 2 $$

Taking derivative w.r.t. $ x $ both sides,

$ y f'(xy + 1) = f(y)f'(x) - 1 $

Put $ y = 0 $,

$ f'(x) = 1 $

$ \displaystyle \int_1^{f(x)}{d(f(x))} = \int_0^x { dx} $

$$ \displaystyle f(x) = x + 1 $$

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  • 2
    $\begingroup$ Why should $f$ be differentiable? $\endgroup$ – Mr. Chip Aug 11 '16 at 10:11
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    $\begingroup$ But $f$ not necessarily differentiable $\endgroup$ – Roman83 Aug 11 '16 at 10:12
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    $\begingroup$ Oops, yeah. Should I delete it? $\endgroup$ – Kartik Sharma Aug 11 '16 at 16:07

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