Q41 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics

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

  • A
    bijective function
  • B
    one-one but not onto function
  • Correct answer
    C
    onto but not one-one function
  • D
    neither one-one nor onto function

Solution

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.

Hence, the correct answer is option: 'C'.

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