Let $f: R \rightarrow R$ be a continuous function satisfying $f(0)=1$ and $f(2x)-f(x)=x$ for all $x \in R$. If $\lim_{n \rightarrow \infty} \{f(x)-f(\frac{x}{2^n})\} = G(x)$,then $\sum_{r=1}^{10} G(r^2)$ is equal to

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then:

Let $f(x) = [x^2 - x] + |-x + [x]|$,where $x \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is

If $a$ is the point of discontinuity of the function $f(x) = \begin{cases} \cos 2 x, & \text{for } -\infty < x < 0 \\ e^{3 x}, & \text{for } 0 \leq x < 3 \\ x^2-4 x+3, & \text{for } 3 \leq x \leq 6 \\ \frac{\log (15 x-89)}{x-6}, & \text{for } x>6 \end{cases}$ Then, $\lim _{x \rightarrow a} \frac{x^2-9}{x^3-5 x^2+9 x-9} =$

If $f(x) = \begin{cases} k, & \text{for } x = 1 \\ \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}, & \text{for } x \neq 1 \end{cases}$ is continuous on $[0, \infty)$, then $k =$

If $f(x) = \lim_{n \rightarrow \infty} \left( \frac{\log(2+x) - x^{2n} \sin x}{1+x^{2n}} \right)$ for $0 \leq x \leq \frac{\pi}{2}$, then at $x=1$, $f(x)$ is