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Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}, \mathrm{a}, \mathrm{b}, \mathrm{c}$ being fixed
non-zero real numbers. Which one of the following statements is correct, in general?
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non-zero real numbers. Which one of the following statements is correct, in general?
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The correct answer is:
If $b^{2}-4 a c>0$, then $f^{-1}(0)$ does not contain 0
As given $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$, Let $\mathrm{y}=\mathrm{f}(\mathrm{x})$. To get $\mathrm{f}^{-1}(\mathrm{x})$
we express $x$ is terms of y in equation $y=a x^{2}+b x+c$.
$\frac{y}{a}=x^{2}+\frac{b}{a} x+\frac{c}{a}=\left(x+\frac{b}{2 a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}$
$\Rightarrow x+\frac{b}{2 a}=\sqrt{\frac{y}{a}+\frac{b^{2}}{4 a^{2}}-\frac{c}{a}}=\frac{\pm \sqrt{4 a y+b^{2}-4 a c}}{2 a}$
$x=\frac{-b \pm \sqrt{4 a y+b^{2}-4 a c}}{2 a}$
or, $f^{1}(x)=\frac{-b \pm \sqrt{4 a x+b^{2}-4 a c}}{2 a}$
Putting $\mathrm{x}=0$
$\mathrm{f}^{-1}(0)=\frac{-\mathrm{b} \pm \sqrt{0+\mathrm{b}^{2}-4 \mathrm{ac}}}{2 \mathrm{a}} \neq 0 \quad\left[\right.$ if $\left.\mathrm{b}^{2}-4 \mathrm{ac}>0\right]$
so, $\mathrm{f}^{-1}(0) \neq 0$. i.e. if $b^{2}-4 a c>0$
$\mathrm{f}^{-1}(0)$ does not contain 0, if $b^{2}-4 a c>0 .$
we express $x$ is terms of y in equation $y=a x^{2}+b x+c$.
$\frac{y}{a}=x^{2}+\frac{b}{a} x+\frac{c}{a}=\left(x+\frac{b}{2 a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}$
$\Rightarrow x+\frac{b}{2 a}=\sqrt{\frac{y}{a}+\frac{b^{2}}{4 a^{2}}-\frac{c}{a}}=\frac{\pm \sqrt{4 a y+b^{2}-4 a c}}{2 a}$
$x=\frac{-b \pm \sqrt{4 a y+b^{2}-4 a c}}{2 a}$
or, $f^{1}(x)=\frac{-b \pm \sqrt{4 a x+b^{2}-4 a c}}{2 a}$
Putting $\mathrm{x}=0$
$\mathrm{f}^{-1}(0)=\frac{-\mathrm{b} \pm \sqrt{0+\mathrm{b}^{2}-4 \mathrm{ac}}}{2 \mathrm{a}} \neq 0 \quad\left[\right.$ if $\left.\mathrm{b}^{2}-4 \mathrm{ac}>0\right]$
so, $\mathrm{f}^{-1}(0) \neq 0$. i.e. if $b^{2}-4 a c>0$
$\mathrm{f}^{-1}(0)$ does not contain 0, if $b^{2}-4 a c>0 .$
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