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}.
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