Functions 3 Question 8
8. Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x}$, where $b$ is a constant such that $0<b<1$. Then,
(2011)
(a) $f$ is not invertible on $(0,1)$
(b) $f \neq f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$
(c) $f=f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$
(d) $f^{-1}$ is differentiable on $(0,1)$
Assertion and Reason
For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I.
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
Show Answer
Answer:
Correct Answer: 8. (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 ${f(a), f\left(\alpha _1\right), f\left(\alpha _2\right), \ldots, f\left(\alpha _n\right), f(b) }$
its greatest and least values gives you range.
Now, $f:[0,3] \rightarrow[1,29]$
$$ \begin{array}{rl} \quad f(x) & =2 x^{3}-15 x^{2}+36 x+1 \\ \therefore \quad f^{\prime}(x) & =6 x^{2}-30 x+36=6\left(x^{2}-5 x+6\right) \\ & =6(x-2)(x-3) \\ & +\quad-\quad++ \\ 2 & 3 \end{array} $$
For given domain $[0,3], f(x)$ is increasing as well as decreasing $\Rightarrow$ many-one
$$ \begin{array}{rlrl} \text { Now, put } & f^{\prime}(x) & =0 \\ \Rightarrow & & x & =2,3 \end{array} $$
Thus, for range $f(0)=1, f(2)=29, f(3)=28$
$\Rightarrow \quad$ Range $\in[1,29]$
$\therefore$ Onto but not one-one.
