Question

The function f: R\(\rightarrow\) R is defined by \[ f(x) = \frac{x}{\sqrt{1 + x^2}} \] is:

• A. surjective but not injective
• B. bijective
• C. injective but not surjective
• D. neither injective nor surjective

Hint:

A function is injective if distinct inputs yield distinct outputs. It is surjective if its range covers the entire codomain.

Answer 1

The Correct Option is C

Step 1: {Injectivity Check}
To determine if \( f(x) \) is injective, consider two values \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \). This condition is expressed as: \[ \frac{x_1}{\sqrt{1 + x_1^2}} = \frac{x_2}{\sqrt{1 + x_2^2}}. \] Squaring both sides yields: \[ x_1^2(1 + x_2^2) = x_2^2(1 + x_1^2). \] Algebraic manipulation leads to the conclusion: \[ x_1 = x_2. \] Therefore, \( f(x) \) is injective.
Step 2: {Surjectivity Check} 
To check for surjectivity, we attempt to solve for \( x \) in terms of \( y \) from the equation \( y = \frac{x}{\sqrt{1 + x^2}} \). This equation can be rewritten as: \[ y^2(1 + x^2) = x^2. \] Solving this equation for \( x \) reveals that \( y \) must be within the interval \( (-1,1) \). Consequently, \( f(x) \) is not surjective. 
Step 3: {Conclusion} 
\( f(x) \) is determined to be injective but not surjective.