Question

Let \[f:R \to R\]be a function defined by \[f\left( x \right) = \dfrac{{x - m}}{{x - n}}\], where \[m \ne n\], then check whether \[f\] is onto or into.
A \[f\] is one-one onto
B \[f\] is one-one into
C\[f\] is many-one onto
D\[f\] is many-one into

Answer 1

Hint: The function is onto when each element of set one there exists at least one image in the second set.

Complete step by step solution: The given function is \[f:R \to R\]. It is defined as \[f\left( x \right) = \dfrac{{x - m}}{{x - n}}\], where \[m \ne n\].
First check if the function is one-one.
Consider two elements in domain \[R\].
Now for \[\left( {x,y} \right) \in R\] consider \[f\left( x \right) = f\left( y \right)\].
As \[f\left( x \right) = \dfrac{{x - m}}{{x - n}}\] then \[f\left( y \right) = \dfrac{{y - m}}{{y - n}}\]
Replace \[f\left( x \right) = \dfrac{{x - m}}{{x - n}}\]and \[f\left( y \right) = \dfrac{{y - m}}{{y - n}}\] in \[f\left( x \right) = f\left( y \right)\].
\[\dfrac{{x - m}}{{x - n}} = \dfrac{{y - m}}{{y - n}}\]
Perform cross multiplication.
\[ \Rightarrow \left( {x - m} \right)\left( {y - n} \right) = \left( {y - m} \right)\left( {x - n} \right)\]
Now remove the brackets and simplify the equation.
\[ \Rightarrow xy - xn - ym + mn = xy - yn - xm + mn\]
Further simplify by eliminating like terms.
\[\begin{array}{l} \Rightarrow - xn - ym = - yn - xm\\ \Rightarrow xm - xn = ym - yn\end{array}\]
Take common factors.
\[\begin{array}{l} \Rightarrow x\left( {m - n} \right) = y\left( {m - n} \right)\\ \Rightarrow x = y\end{array}\]
Hence, the function \[f\] is one-one.
Now check whether the function is onto.
Consider \[\beta \] be the element in domain \[R\].
So, \[\beta \in R\]such that \[f\left( x \right) = \beta \].
Hence, \[\beta = \dfrac{{x - m}}{{x - n}}\]
Now simplify the equation.
\[\beta \left( {x - n} \right) = x - m\]
Further simplify the equation as follows.
\[\begin{array}{l} \Rightarrow \beta x - \beta n = x - m\\ \Rightarrow \beta x - x = \beta n - m\\ \Rightarrow x\left( {\beta - 1} \right) = \beta n - m\end{array}\]
Now isolate x.
\[x = \dfrac{{\beta n - m}}{{\beta - 1}}\]
But for \[\beta = 1\] , \[\dfrac{{\beta n - m}}{{\beta - 1}}\] is undefined.
So, \[x \notin R\]. Hence function \[f\] is not onto.
So, the given function is one-one into.

Thus, Option (B) is correct.

Note: The common mistake done by students is that \[\dfrac{{\beta n - m}}{{\beta - 1}}\] is undefined so \[1 \in R\] and hence they consider as function is onto which is wrong.