Question

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

• A. bijective function
• B. one-one but not on to function
• C. onto but not one-one function
• D. neither one -one nor onto function

Answer 1

Answer: (C)

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,
$0 \leq x \leq 6 $
$\Rightarrow -30 \leq-5 x \leq 0 $
$\Rightarrow 60-30 \leq 60-5 x \leq 60 $
$\Rightarrow 10 \leq \frac{60-5 x}{3} \leq 20 $ and $ 7 \leq x \leq 10$
$\Rightarrow -30 \leq-3 x \leq-21 $
$\Rightarrow 1 \leq 31-3 x \leq 10 $
$\therefore $ Range of $f(x)=[1,20]$
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.