Question

Let $f:[0,\infty )\to R$ be a function defined by $f\left(x\right)=9x^2+6x-5$. Prove that f is not invertible. Modify, only the codomain of f to make f invertible and then find its inverse.

OR

Let $\ast$ be a binary operation defined on $Q\times Q$ by $(a,b)\ast (c,d)=(ac,b+ad)$, where Q is the set of rational numbers. Determine whether $\ast$ is commutative and associative. Find the identity element for $\ast$ and the invertible elements of $Q\times Q$.

Answer 1

Let $y=f\left(x\right)\forall xϵ[0,\infty )$

$\therefore y=9x^2+6x-5=(3x+1{)}^2-6$

Clearly when $xϵ[0,\infty )$ then $y\ge -5\therefore$ Range of $f=[-5,\infty )\ne$ Codomain of f

Hence f is not onto and, therefore f(x) is not invertible.

Let us take the modified codomain of f as $[-5,\infty )$
Let's now check whether f is one-one : Let $x_1,x_2ϵ[0,\infty )$ such that $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow (3x_1+1{)}^2-6=(3x_2+1{)}^2-6\Rightarrow 3x_1+1=3x_2+1\Rightarrow x_1=x_2.$ So, f is one-one.

Since, with the modified codomain, we have range of f = new codomain of f. So, f is onto.

No for any $yϵ[-5,\infty ),y=(3x+1{)}^2-6$

$\Rightarrow y+6=(3x+1{)}^2\Rightarrow x=\frac{\sqrt{y+6}-1}{3}$

$\therefore f^{-1}:[-5,\infty )\to [0,\infty ),f^{-1}\left(y\right)=\frac{\sqrt{y+6}-1}{3}$

OR

Let $A=Q\times Q$, where Q is the set of all rational numbers, and $\ast$ be a binary operation on A defined by $(a,b)\ast (c,d)=(ac,b+ad)$ for $(a,b),(c,d)ϵA$

Note that $(1,2)\ast (2,3)=(2,5)$ and $(2,3)\ast (1,2)=(2,7)$ which implies $(2,5)\ne (2,7)$

That is $(a,b)\ast (c,d)\ne (c,d)\ast (a,b)\forall (a,b),(c,d)ϵQ\times Q$. So, $\ast$ isn't commutative.

Let $(a,b),(c,d),(e,f)\epsilon (Q\times Q)$.

Now $\left[\right(a,b)\ast (c,d\left)\right]\ast (e,f)=(ac,b+ad)\ast (e,f)=(ace,b+ad+acf)\dots \left(i\right)$

Also, $(a,b)\ast \left[\right(c,d)\ast (e,f\left)\right]=(a,b)\ast (ce,d+cf)=(ace,b+ad+acf)\dots \left(ii\right)$

By (i) and (ii), $\left[\right(a,b)\ast (c,d\left)\right]\ast (e,f)=(a,b)\ast \left[\right(c,d)\ast (e,f\left)\right]$.

Hence $\ast$ is associative

Let (e,e') be the identity element of $\ast$ in a. Then $(a,b)\ast (e,e^{\prime })=(a,b)=(e,e^{\prime })\ast (a,b)$

$\begin{array}{ccc}& & ax=1\Rightarrow x=\frac{1}{a}\\ \Rightarrow (ax,b+ay)=(1,0)& \Rightarrow & \\ & & b+ay=0\Rightarrow y=-\frac{b}{a}\end{array}]\therefore$ Inverse of $(a,b)=(\frac{1}{a}.-\frac{b}{a})$