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Let F : R → R F ( X ) = X 2 − 8 X 2 + 2 Then, F is (A) One-one but Not onto (B) One-one and onto (C) onto but Not One-one (D) Neither One-one Nor onto

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Question

Let

\[f : R \to R\]
\[f\left( x \right) = \frac{x^2 - 8}{x^2 + 2}\]
Then,  f is

Options

  • one-one but not onto

  • one-one and onto

  • onto but not one-one

  • neither one-one nor onto

MCQ
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Solution

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).
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Chapter 2: Functions - Exercise 2.6 [Page 77]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 2 Functions
Exercise 2.6 | Q 23 | Page 77

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