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Let $f:[0,10] \rightarrow[1,20]$ be a function defined
as $f(x)=\left\{\begin{array}{ll}\frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10\end{array}\right.$ then $f$ is
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as $f(x)=\left\{\begin{array}{ll}\frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10\end{array}\right.$ then $f$ is
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Verified Answer
The correct answer is:
onto but not one-one function
We have, $f(x)= \begin{cases}\frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10\end{cases}$
Now, $f(x)=10 \forall x \in[6,7]$
So, $f(x)$ is not one-one.
Again,
$\begin{aligned}
& 0 \leq x \leq 6 \\
& \Rightarrow \quad-30 \leq-5 x \leq 0 \\
& \Rightarrow \quad 60-30 \leq 60-5 x \leq 60 \\
& \Rightarrow \quad \frac{30}{3} \leq \frac{60-5 x}{3} \leq \frac{60}{3} \\
& \Rightarrow \quad 10 \leq \frac{60-5 x}{3} \leq 20 \text { and } 7 \leq x \leq 10 \\
& \Rightarrow \quad-30 \leq-3 x \leq-21 \Rightarrow 1 \leq 31-3 x \leq 10 \\
& \therefore \text { Range of } f(x)=[1,20] \\
&
\end{aligned}$
It is given that co-domain of $f(x)=[1,20]$
$\therefore$ Range of $f(x)=$ co-domain of $f(x)$
So, $f(x)$ is onto.
Now, $f(x)=10 \forall x \in[6,7]$
So, $f(x)$ is not one-one.
Again,
$\begin{aligned}
& 0 \leq x \leq 6 \\
& \Rightarrow \quad-30 \leq-5 x \leq 0 \\
& \Rightarrow \quad 60-30 \leq 60-5 x \leq 60 \\
& \Rightarrow \quad \frac{30}{3} \leq \frac{60-5 x}{3} \leq \frac{60}{3} \\
& \Rightarrow \quad 10 \leq \frac{60-5 x}{3} \leq 20 \text { and } 7 \leq x \leq 10 \\
& \Rightarrow \quad-30 \leq-3 x \leq-21 \Rightarrow 1 \leq 31-3 x \leq 10 \\
& \therefore \text { Range of } f(x)=[1,20] \\
&
\end{aligned}$
It is given that co-domain of $f(x)=[1,20]$
$\therefore$ Range of $f(x)=$ co-domain of $f(x)$
So, $f(x)$ is onto.
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