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

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Answer:

Correct Answer: 6. (a)

Solution:

  1. 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.



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