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Q.1 Let $f : R \rightarrow R$ be defined as $f (x) = 10x + 7$ . Find the function $g : R \rightarrow R$ such
that $g o f = f o g = I_R$ .

Answers (1)

best_answer

$f : R \rightarrow R$

$f (x) = 10x + 7$

$g : R \rightarrow R$ and $g o f = f o g = I_R$

For one-one:

$f(x)=f(y)$

$10x+7=10y+7$

$10x=10y$

$x=y$

Thus, f is one-one.

For onto:

For $y \in R$ , $y=10x+7$

$x= \frac{y-7}{10} \in R$

Thus, for $y \in R$ , there exists $x= \frac{y-7}{10} \in R$ such that

$f(x) = f(\frac{y-7}{10})=10(\frac{y-7}{10})+7=y-7+7=y$

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let $g : R \rightarrow R$ as $f(y)=\frac{y-7}{10}$

$gof(x)=g(f(x))= g(10x+7) =\frac{(10x+7)-7}{10}=\frac{10x}{10}=x$

$fog(x)=f(g(x))= f(\frac{y-7}{10}) =10\frac{y-7}{10}+7=y-7+7=y$

$\therefore$ $gof(x)=I_R$ and $fog(x)=I_R$

Hence, $g : R \rightarrow R$ defined as $g(y)=\frac{y-7}{10}$

Posted by

seema garhwal

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