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Question
Let the function
\[f : R - \left\{ - b \right\} \to R - \left\{ 1 \right\}\]
\[f\left( x \right) = \frac{x + a}{x + b}, a \neq b .\text{Then},\]
Options
f is one-one but not onto
f is onto but not one-one
f is both one-one and onto
None of these
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Solution
c) f is both one-one and onto
Injectivity:
Let x and y be two elements in the domain R- {-b}, such that
\[f\left( x \right) = f\left( y \right)\]
\[ \Rightarrow \frac{x + a}{x + b} = \frac{y + a}{y + b}\]
\[ \Rightarrow \left( x + a \right)\left( y + b \right) = \left( x + b \right)\left( y + a \right)\]
\[ \Rightarrow xy + bx + ay + ab = xy + ax + by + ab\]
\[ \Rightarrow bx + ay = ax + by\]
\[ \Rightarrow \left( a - b \right)x = \left( a - b \right)y\]
\[ \Rightarrow x = y\]
So, f is one-one.
Surjectivity:
Let y be an element in the co-domain of f, i.e. R-{1}, such that f (x)=y
\[f\left( x \right) = y\]
\[ \Rightarrow \frac{x + a}{x + b} = y\]
\[ \Rightarrow x + a = yx + yb\]
\[ \Rightarrow x - yx = yb - a\]
\[ \Rightarrow x\left( 1 - y \right) = yb - a\]
\[ \Rightarrow x = \frac{yb - a}{1 - y} \in R - \left\{ - b \right\}\]
So, f is onto.
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