• # question_answer Consider $f:{{R}^{+}}\to [-\,9,\,\,\infty ]$ given   by $f(x)=5{{x}^{2}}+6x-9.$ Prove that f is invertible with ${{f}^{-1}}(y)=\left( \frac{\sqrt{54+5y}-3}{5} \right)$ (Where, ${{R}^{+}}$ is the set of all positive real numbers). OR Let '*' be a binary operation on the set {0, 1, 2, 3, 4, 5} defined as $a*b=\left\{ \begin{matrix} a+b, & \text{if}\,\,a+b<6 \\ a+b-6, & \text{if}\,\,a+b\ge 6 \\ \end{matrix} \right.$ Show that zero is the identity for this operation and each element a of the set is invertible with $b-a,$ being the inverse of a.

$f:{{R}^{+}}\to [-\,9,\,\,\infty ]$
$f(x)=5{{x}^{2}}+6x-9$
$f'(x)=10x+6$     [domain of $f(x)$ in ${{R}_{+}}$]
$\therefore$      $x>0,$$f'(x)>0,$ hence the function is increasing and we know increasing function is one-one.
$\therefore$$f(x)$ is one-one
Range of function
$\therefore$Function is increasing in$[0,\,\,\infty ]$.
$\therefore$Lowest value of function will be at$x=0$.
$\therefore$$f(x)=0+0-9=-\,9$
$\therefore$Range for ${{R}_{+}}$is $[-\,9,\,\,\infty )$.
$\therefore$$[-\,\,9,\,\,\infty )$ is codomain of given function.
$\therefore$Hence range = codomain
$\therefore$Function is onto.
Hence, the given function is invertible.
Now, $f(x)=5{{x}^{2}}+6x-9$
$y=5\left( {{x}^{2}}+\frac{6x}{5}-\frac{9}{5} \right)$ $[let\,f(x)=y]$
$=5\left( {{x}^{2}}+2x\frac{3}{5}+\frac{9}{25}-\frac{9}{5}-\frac{9}{25} \right)$
$\left[ \text{adding}\,\,\text{and}\,\,\text{subtracting}\frac{9}{25} \right]$
$=5\left[ {{x}^{2}}+2x\frac{3}{5}+{{\left( \frac{3}{5} \right)}^{2}}-\frac{54}{25} \right]$
$=5{{\left( x+\frac{3}{5} \right)}^{2}}-\frac{54}{5}$
$\Rightarrow$$y+\frac{54}{5}=5{{\left( x+\frac{3}{5} \right)}^{2}}$$\Rightarrow$$\frac{y}{5}+\frac{54}{25}={{\left( x+\frac{3}{5} \right)}^{2}}$
$\Rightarrow$   $x+\frac{3}{5}=\pm \sqrt{\frac{y}{5}+\frac{54}{25}}$
$\Rightarrow$   $x-\frac{3}{5}\pm \sqrt{\frac{y}{5}+\frac{54}{25}}$
$\therefore$      $x>0$
$\therefore$we will take only
$x=\frac{-3}{5}+\sqrt{\frac{y}{5}+\frac{54}{25}}\Rightarrow 5x=-\,3+\sqrt{5y+54}$
$\therefore$${{f}^{-1}}(y)=\frac{\sqrt{5y+54}-3}{5}$             Hence proved.
OR
a*b=\left\{ \begin{align} & \,\,\,a+b,\,\,\,\,\,\,\,\,if\,\,a+b<6 \\ & a+b-6,\,\,if\,\,a+b\ge 6 \\ \end{align} \right.
$2*3=2+3=5,$ since $2+3=5<6$
$4*5=4+5-6=9-6=3$, since $4+5=9\ge 6$
The composition table for $*$ is given as
 $*$ 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4
The first row coincides with the top most row and first column coincides with the left most column. They intersect at 0, so, 0 is the identity element.
$a*(6-a)=a+(6-a)-6=0,$$[\because a+(6-a)=6\ge 6]$
$\therefore$$6-a$is the inverse of a for which each a is
{0, 1, 2, 3, 4, 5}.

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