Functions 3 Question 6
6. If $f(x)=3 x-5$, then $f^{-1}(x)$
$(1998,2 M)$
(a) is given by $\frac{1}{3 x-5}$
(b) is given by $\frac{x+5}{3}$
(c) does not exist because $f$ is not one-one
(d) does not exist because $f$ is not onto
Show Answer
Answer:
Correct Answer: 6. (a)
Solution:
- We have a function $f: A \rightarrow R$ defined as, $f(x)=\frac{2 x}{x-1}$
One-one Let $x _1, x _2 \in A$ such that
$$ \begin{array}{rlrl} & & f\left(x _1\right) & =f\left(x _2\right) \\ \Rightarrow & & \frac{2 x _1}{x _1-1} & =\frac{2 x _2}{x _2-1} \\ \Rightarrow & 2 x _1 x _2-2 x _1 & =2 x _1 x _2-2 x _2 \\ \Rightarrow & x _1 & =x _2 \end{array} $$
Thus, $f\left(x _1\right)=f\left(x _2\right)$ has only one solution, $x _1=x _2$
$\therefore f(x)$ is one-one (injective)
Onto Let $x=2$, then $f(2)=\frac{2 \times 2}{2-1}=4$
But $x=2$ is not in the domain, and $f(x)$ is one-one function
$\therefore f(x)$ can never be 4
Similarly, $f(x)$ can not take many values.
Hence, $f(x)$ is into (not surjective).
$\therefore f(x)$ is injective but not surjective.
