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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS5 Marks
Question
Let $\text{A} = \text{R} - \left\{3\right\}, \text{B} = \text{R} - \left\{1\right\}. \text{Let f : A} \rightarrow \text{B}$ be defined by $\text{f(x)}\frac{\text{x - 2}}{\text{x - 3}}, \forall \text{ x} \in \text{A}.$ Show that f is bijective. Also, find
  1. $\text{x, if f}^{-1} \text{(x) = 4}$
  2. $\text{f}^{-1} (7)$
✓

Answer

$\text{x}_{1}, \text{x}_{2} \in \text{A and } \text{f(x}_{1}) = \text{f(x}_{2})$
$\Rightarrow \frac{\text{x}_{1} - 2}{\text{x}_{1} - 3} = \frac{\text{x}_{2} - 2}{\text{x}_{2} - 3} \Rightarrow \text{(x}_{1} - 2) \text{(x}_{2} - 3) = \text{(x}_{1} - 3) \text{(x}_{2} - 2)$
$\Rightarrow \text{x}_{1} \text{x}_{2} - \text{3x}_{1} - 2\text{x}_{2} + 6 = \text{x}_{1} \text{x}_{2} - \text{2x}_{1} - \text{3x}_{2} + 6$
$\Rightarrow \text{x}_{1} = \text{x}_{2}$
Hence f is a one-one function
Let $\text{y} = \frac{\text{x - 2}}{\text{x - 3}} \text{ for y } \in \text{R} - \left\{1\right\}$
$\Rightarrow \text{x} = \frac{\text{3y - 2}}{\text{y - 1}} ; \text{y} \neq 1$
$\therefore \forall \text{ y} \in \text{R} - \left\{1\right\}, \text{x} \in \text{R} - \left\{3\right\}$
i.e. Range of f = co-domain of f.
Hence f is onto and so bijective.
Also, $\text{f}^{-1} \text{(x)} = \frac{\text{3x - 2}}{\text{x - 1}} ; \text{x} \neq 1$
Now, $\text{f}^{-1} \text{(x)} = 4 \Rightarrow \frac{\text{3x - 2}}{\text{x - 1}} = 4 \Rightarrow \text{x = 2}$
and $ \text{f}^{-1}(7) = \frac{19}{6}$

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