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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS5 Marks
Question
$\text{Let f : N}\rightarrow\text{N}$ be a function defined as $\text{f(x)} = 9x^{2} + 6x -5.$ Show that $\text{f : N}\rightarrow\text{S},$ where S is the range of f, is invertible. Find the inverse of f and hence find $\text{f}^{-1}(43) \text{and f}^{-1}(163). $
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Answer

$\text{Let x}_{1},\text{x}_{2}\in \text{N and f (x}_{1}) =\text{f(x}_{2}) $
$\Rightarrow\text{9x}^{2}_{1} + 6\text{x}_{1} - 5 = \text{9x}^{2}_{2} + 6\text{x}_{2} - 5$
$^2_1 - \text {x}^{2}_{2}) + 6(\text{x}_{1} - \text{x}_{2}) = 0 \Rightarrow\text{(x}_{1} - \text{x}_{2}) \text{(9x}_{1} + \text{9x}_{2} + 6) = 0$
$\Rightarrow\text{x}_{1} - \text{x}_{2} = \text{0 or x}_{1} = \text{x}_{2}\text{ as}\text{(9x}_{1} + \text{9x}_{2} + 6\neq 0,\text{x}_{1}, \text{x}_{2} \in \text{N}$
$\therefore$ f is one one function
$\text{f : N}\rightarrow\text{S is ONTO as co-domain = Range}$
Hence f is invertible
$\text{y} = \text{9x}^{2} + \text{6x} - 5 = (\text{3x + 1)}^{2} - 6 \Rightarrow\text{x} = \frac{\sqrt{\text{y} + 6} -1}{3}$
$\therefore\text{f}^{-1}\text{(y)} = \frac{\sqrt{\text{y} + 6} - 1}{3}, \text{y}\in\text{S}$
$\text{f}^{-1}(43) = \frac{\sqrt{49}-1}{3} = 2$
$\text{f}^{-1}(163) = \frac{\sqrt{169} - 1}{3} = 4$

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