117097 If \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{N}\) is defined by
\(f(x)=\left\{\begin{array}{ccl}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0 \\ -2 n-1, & \text { if } & n\lt 0\end{array}\right.\)then \(f\) is

117098 Given that for any \(n \in N\) there exist an odd integer \(q\) and a non-negative integer \(r\) such that, \(n\) can be written uniquely as \(n=q \times 2^r\). Let \(\mathbf{f}: \mathbf{N} \rightarrow \mathbf{N} \times \mathbf{N}\) be function defined by
\(\mathbf{f}(\mathbf{n})=\left(\mathbf{r}+\mathbf{1}, \frac{\mathbf{q}+\mathbf{1}}{\mathbf{2}}\right)\). Then,

117099 If \(f:[0, \infty) \rightarrow[0, \infty)\) is defined by \(f(x)=\frac{x}{1+x}\), then is

117100 Let \(f:[0,10] \rightarrow[1,20]\) be a function defined
\(\text { as } f(x)= \begin{cases}\frac{60-5 x}{3}, 0 \leq x \leq 6 \\ 10, 6 \leq x \leq 7, \text { then } f \text { is } \\ 31-3 x, 7 \leq x \leq 10\end{cases}\)

117097 If \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{N}\) is defined by
\(f(x)=\left\{\begin{array}{ccl}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0 \\ -2 n-1, & \text { if } & n\lt 0\end{array}\right.\)then \(f\) is

117098 Given that for any \(n \in N\) there exist an odd integer \(q\) and a non-negative integer \(r\) such that, \(n\) can be written uniquely as \(n=q \times 2^r\). Let \(\mathbf{f}: \mathbf{N} \rightarrow \mathbf{N} \times \mathbf{N}\) be function defined by
\(\mathbf{f}(\mathbf{n})=\left(\mathbf{r}+\mathbf{1}, \frac{\mathbf{q}+\mathbf{1}}{\mathbf{2}}\right)\). Then,

117099 If \(f:[0, \infty) \rightarrow[0, \infty)\) is defined by \(f(x)=\frac{x}{1+x}\), then is

117100 Let \(f:[0,10] \rightarrow[1,20]\) be a function defined
\(\text { as } f(x)= \begin{cases}\frac{60-5 x}{3}, 0 \leq x \leq 6 \\ 10, 6 \leq x \leq 7, \text { then } f \text { is } \\ 31-3 x, 7 \leq x \leq 10\end{cases}\)

117097 If \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{N}\) is defined by
\(f(x)=\left\{\begin{array}{ccl}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0 \\ -2 n-1, & \text { if } & n\lt 0\end{array}\right.\)then \(f\) is

117098 Given that for any \(n \in N\) there exist an odd integer \(q\) and a non-negative integer \(r\) such that, \(n\) can be written uniquely as \(n=q \times 2^r\). Let \(\mathbf{f}: \mathbf{N} \rightarrow \mathbf{N} \times \mathbf{N}\) be function defined by
\(\mathbf{f}(\mathbf{n})=\left(\mathbf{r}+\mathbf{1}, \frac{\mathbf{q}+\mathbf{1}}{\mathbf{2}}\right)\). Then,

117099 If \(f:[0, \infty) \rightarrow[0, \infty)\) is defined by \(f(x)=\frac{x}{1+x}\), then is

117100 Let \(f:[0,10] \rightarrow[1,20]\) be a function defined
\(\text { as } f(x)= \begin{cases}\frac{60-5 x}{3}, 0 \leq x \leq 6 \\ 10, 6 \leq x \leq 7, \text { then } f \text { is } \\ 31-3 x, 7 \leq x \leq 10\end{cases}\)

117097 If \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{N}\) is defined by
\(f(x)=\left\{\begin{array}{ccl}2 n, & \text { if } & n>0 \\ 1, & \text { if } & n=0 \\ -2 n-1, & \text { if } & n\lt 0\end{array}\right.\)then \(f\) is

117098 Given that for any \(n \in N\) there exist an odd integer \(q\) and a non-negative integer \(r\) such that, \(n\) can be written uniquely as \(n=q \times 2^r\). Let \(\mathbf{f}: \mathbf{N} \rightarrow \mathbf{N} \times \mathbf{N}\) be function defined by
\(\mathbf{f}(\mathbf{n})=\left(\mathbf{r}+\mathbf{1}, \frac{\mathbf{q}+\mathbf{1}}{\mathbf{2}}\right)\). Then,

117099 If \(f:[0, \infty) \rightarrow[0, \infty)\) is defined by \(f(x)=\frac{x}{1+x}\), then is

117100 Let \(f:[0,10] \rightarrow[1,20]\) be a function defined
\(\text { as } f(x)= \begin{cases}\frac{60-5 x}{3}, 0 \leq x \leq 6 \\ 10, 6 \leq x \leq 7, \text { then } f \text { is } \\ 31-3 x, 7 \leq x \leq 10\end{cases}\)