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If $f:R\rightarrow R$ be twice continuously differentiable function such that $2f(x+1)=f(x)+f(2x)$.

then value of $\bigg(\lfloor\frac{f(2)+f(7)}{f(9)}\rfloor\bigg)$, where $\lfloor x \rfloor$ is a floor of $x$

Attempt: Using hit and trial , $f(x)=c$

So $f(2)=f(7) = f(9) = c.$ So $\bigg(\lfloor\frac{f(2)+f(7)}{f(9)}\rfloor\bigg) = \lfloor 2 \rfloor = 2$

could some help me how to solve it , thanks

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  • $\begingroup$ A constant would fit. By plugging in $x=0$ and $x=1$ you can see that even if $f$ is not constant, $f(2)=f(1)=f(0)$. $\endgroup$ – gt6989b May 7 '17 at 12:24
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It can be seen from that given statement that $f(x) , f(x+1)$ and $f(2x)$ are in A.P.

Therefore, let $f(x+1)-f(x) = h(x)$

Now by first principle of differentiation,

$f'(1) = \lim_{x \to 0} \frac{f(1+x)-f(1)}{x}$

$f'(1) = \lim_{x \to 0} \frac{h(x) + f(x)-f(1)}{x}$

For limit to be defined,

$h(0) + f(0)-f(1)=0$ ......(1)

Now putting $x=0$ in the orignal statement

$2f(1)= f(0) + f(0)$

$f(1)=f(0)$

This proves that $h(0)=0$. Therefore it is a constant A.P. with $f(0)=f(1)=f(2)=....f(n)$ where $n \in R$.

Answer to your question will be $2$.

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