Let f:[0,10]arrow[1,20] be a function defined as f(x)={cases}60-5x/3, & 0≤ x≤ 6 \\ 10, & 6≤ x≤ 7 \\ 31-3x, & 7≤ x≤ 10{cases} then f is

Correct Answer is : onto but not one-one function

neither one -one nor onto function

Correct Answer is : onto but not one-one function

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TS EAMCET 10-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 1 of 160

Marks: +1, -0

f:[0,10]\rightarrow[1,20]

f(x)=\begin{cases}\frac{60-5x}{3}, & 0\le x\le 6 \\ 10, & 6\le x\le 7 \\ 31-3x, & 7\le x\le 10\end{cases}

f

Solution: 👈: Video Solution

We have,

f(x)=\begin{cases}\frac{60-5x}{3}, & 0\le x\le 6 \\ 10, & 6\le x\le 7 \\ 31-3x, & 7\le x\le 10.\end{cases}

Now,

f(x)=10 \forall x \in [6,7]

So,

f(x)

is not one-one.

Again,

0 \le x \le 6

\Rightarrow \;\; -30 \le -5x \le 0

\Rightarrow \;\; 60-30 \le 60-5x \le 60

\Rightarrow \;\; 10 \le \frac{60-5x}{3} \le 20

and

7 \le x \le 10

\Rightarrow \;\; -30 \le -3x \le -21 \Rightarrow 1 \le 31-3x \le 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.

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