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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS5 Marks
Question
Consider the function $\text{f}:\text{R}^{+}\rightarrow[-9,\infty]$ given by $f(x) = 5x2 + 6x - 9$. Prove that f is invertible with $\text{f}^{-1}\text{(y)}=\frac{\sqrt{54+5\text{y}}-3}{5}.$
✓

Answer

$\text{f}:\text{R}^{+}\rightarrow\ [-9,\infty)$ given by $f(x) = 5x^2 + 6x - 9F$ or any $\text{x, y}\in\text{R}^{+}$
$f(x) = f(y)$
$\Rightarrow 5x^2 + 6x - 9 = 5y^2 + 6y - 9$
$\Rightarrow 5(x^2 - y^2) + 6(x - y) = 0$
$\Rightarrow (x - y)[5(x + y) + 6] = 0$
$\Rightarrow (x - y) = 0 [\because5(\text{x}+\text{y})+6\neq0\text{ as x, y}\in\text{R}^{+}]$
$\Rightarrow x = y$
So, f is an injection.
Let y be an arbitrary element of $[-9,\infty).$
$f(x) = y$
$\Rightarrow 5x^2 + 6x - 9 = y$
$\Rightarrow 25x^2 + 30x - 45 = 5y$
$\Rightarrow 25x^2 + 30x + 9 - 54 = 5y$
$\Rightarrow (5x + 3)^2 = 5y + 54$
$\Rightarrow(5\text{x}+3)=\sqrt{5\text{y}+54}$
$\Rightarrow\ \text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}$
Now, $\text{y}\in[-9,\infty)$
$\Rightarrow\ \text{y}\geq-9$
$\Rightarrow\ 5\text{y}+54\geq9$
$\Rightarrow\ \sqrt{5\text{y}+54}\geq3$
$\Rightarrow\ \sqrt{5\text{y}+54}-3\geq0$
$\Rightarrow\ \frac{\sqrt{5\text{y}+54}-3}{5}\geq0$
$\Rightarrow\ \text{x}\geq0\Rightarrow\ \text{x}\in\text{R}^{+}$
Thus, for every $\text{y}\in[-9,\infty)$ there exist $\text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}\in\text{R}^{+} such that f(x) = y.$
So, $ \text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is onto.
Thus, $\text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is a bijection and hence invertible.
Let $f^{-1} $ denotes the inverse of f.
Then,
$(fof^{-1})(y) = y$ for all $\text{y}\in[-9,\infty)$
$f(f^{-1}(y)) = y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 5(f^{-1}(y))^2 + 6(f^{-1}(y)) - 9 = y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) - 45 = 5y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) + 9 = 5y + 54$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow {5f^{-1}(y) + 3}^2 = 5y + 54$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 5f^{-1}(y) + 3 =\sqrt{5\text{y}+54} for all \text{y}\in[-9,\infty)$
$\Rightarrow\ \text{f}^{-1}(\text{y})=\frac{\sqrt{5\text{y}+54}-3}{5}$

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