Functions Ques 18
- The function $f:[0,3] \rightarrow[1,29]$, defined by
$ f(x)=2 x^3-15 x^2+36 x+1 \text {, is } $
(a) one-one and onto
(b) onto but not one-one
(c) one-one but not onto
(d) neither one-one nor onto
Show Answer
Answer:
Correct Answer: 18.(b)
Solution: PLAN To check nature of function.
(i) One-one To check one-one, we must check whether $f^{\prime}(x)>0$ or $f^{\prime}(x)<0$ in given domain.
(ii) Onto To check onto, we must check Range $=$ Codomain
Description of Situation To find range in given domain $[a, b]$, put $f^{\prime}(x)=0$ and find $x=\alpha_1, \alpha_2, \ldots$, $\alpha_n \in[a, b]$
Now, find $\left\{f(a), f\left(\alpha_1\right), f\left(\alpha_2\right), \ldots, f\left(\alpha_n\right), f(b)\right\}$
its greatest and least values gives you range.
Now, $f:[0,3] \rightarrow[1,29]$
$ \begin{aligned} f(x) & =2 x^3-15 x^2+36 x+1 \\ f^{\prime}(x) & =6 x^2-30 x+36=6\left(x^2-5 x+6\right) \\ & =6(x-2)(x-3) \\ \end{aligned} $
For given domain $[0,3], f(x)$ is increasing as well as decreasing $\Rightarrow$ many-one
Now, put $ f^{\prime}(x)=0 $
$\Rightarrow \quad$ $ x=2,3 $
Thus, for range $f(0)=1, f(2)=29, f(3)=28$
$\Rightarrow \quad$ Range $\in[1,29]$
$\therefore \quad$ Onto but not one-one.
BETA
