Let f:[0,10] rightarrow [1,20] be a function | vedclass

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(C) Given the function $f(x) = \begin{cases} \frac{60-5x}{3}, & 0 \leq x \leq 6 \ 10, & 6 \leq x \leq 7 \ 31-3x, & 7 \leq x \leq 10 \end{cases}$.For

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onto but not one-one function

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(C) Given the function $f(x) = \begin{cases} \frac{60-5x}{3}, & 0 \leq x \leq 6 \ 10, & 6 \leq x \leq 7 \ 31-3x, & 7 \leq x \leq 10 \end{cases}$.For $x \in [6, 7]$, $f(x) = 10$. Since the function takes the same value for all $x$ in the interval $[6, 7]$, it is not one-one.Now, let us find the range of $f(x)$:For $0 \leq x \leq 6$, $f(x) = \frac{60-5x}{3}$. As $x$ goes from $0$ to $6$, $f(x)$ goes from $\frac{60}{3} = 20$ to $\frac{60-30}{3} = 10$. So, the range is $[10, 20]$.For $6 \leq x \leq 7$, $f(x) = 10$. So, the range is ${10}$.For $7 \leq x \leq 10$, $f(x) = 31-3x$. As $x$ goes from $7$ to $10$, $f(x)$ goes from $31-21 = 10$ to $31-30 = 1$. So, the range is $[1, 10]$.The union of these ranges is $[1, 10] \cup {10} \cup [10, 20] = [1, 20]$.Since the range $[1, 20]$ is equal to the co-domain $[1, 20]$, the function is onto.Thus, $f(x)$ is onto but not one-one.

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

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