Question

Let

\[f : R \to R\]

\[f\left( x \right) = \frac{x^2 - 8}{x^2 + 2}\]

Then,  f is

पर्याय

• one-one but not onto

• one-one and onto

• onto but not one-one

• neither one-one nor onto

Answer 1

Injectivity:
Let x and y be two elements in the domain (R), such that

\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow \frac{x^2 - 8}{x^2 + 2} = \frac{y^2 - 8}{y^2 + 2}\] 
\[ \Rightarrow \left( x^2 - 8 \right)\left( y^2 + 2 \right) = \left( x^2 + 2 \right)\left( y^2 - 8 \right)\] 
\[ \Rightarrow x^2 y^2 + 2 x^2 - 8 y^2 - 16 = x^2 y^2 - 8 x^2 + 2 y^2 - 16\] 
\[ \Rightarrow 10 x^2 = 10 y^2 \] 
\[ \Rightarrow x^2 = y^2 \] 
\[ \Rightarrow x = \pm y \]

So, f is not one-one .

Surjectivity:

\[f\left( - 1 \right) = \frac{\left( - 1 \right)^2 - 8}{\left( - 1 \right)^2 + 2} = \frac{1 - 8}{1 + 2} = \frac{- 7}{3}\] 

\[ \text{ and} f\left( 1 \right) = \frac{\left( 1 \right)^2 - 8}{\left( 1 \right)^2 + 2} = \frac{1 - 8}{1 + 2} = \frac{- 7}{3}\]

\[\Rightarrow\] f is not onto.
The correct answer is (d).