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.

Answer:

\[f:{{R}^{+}}\to [-\,9,\,\,\infty ]\]

\[f(x)=5{{x}^{2}}+6x-9\]

\[f'(x)=10x+6\]     [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}.