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: {Check injectivity} 
For \( x_1, x_2 \) such that: \[ f(x_1) = f(x_2), \] \[ \frac{x_1}{\sqrt{1 + x_1^2}} = \frac{x_2}{\sqrt{1 + x_2^2}}. \] Squaring both sides: \[ x_1^2(1 + x_2^2) = x_2^2(1 + x_1^2). \] Solving, we get: \[ x_1 = x_2. \] Thus, \( f(x) \) is injective. 
Step 2: {Check surjectivity} 
Solving for \( y \): \[ y = \frac{x}{\sqrt{1 + x^2}}. \] This implies: \[ y^2(1 + x^2) = x^2. \] Solving, we find \( y \in (-1,1) \), meaning \( f(x) \) is not surjective. 
Step 3: {Conclusion} 
\( f(x) \) is injective but not surjective.

Answer 2

Step 1: Check if the function is injective
We are given the function \( f(x) = \frac{x}{\sqrt{1 + x^2}} \). To check if it is injective, we need to verify if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). Assume \( f(x_1) = f(x_2) \): \[ \frac{x_1}{\sqrt{1 + x_1^2}} = \frac{x_2}{\sqrt{1 + x_2^2}}. \] Cross-multiply to get: \[ x_1 \sqrt{1 + x_2^2} = x_2 \sqrt{1 + x_1^2}. \] Square both sides: \[ x_1^2 (1 + x_2^2) = x_2^2 (1 + x_1^2). \] Simplify: \[ x_1^2 + x_1^2 x_2^2 = x_2^2 + x_1^2 x_2^2. \] Cancel out the common terms: \[ x_1^2 = x_2^2. \] Thus, \( x_1 = x_2 \) or \( x_1 = -x_2 \). Since the function is an odd function (\( f(-x) = -f(x) \)), it is injective when restricted to non-negative or non-positive values of \( x \). Therefore, \( f(x) \) is injective on \( \mathbb{R} \). 
Step 2: Check if the function is surjective
For surjectivity, we need to check if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \). We know that the function \( f(x) = \frac{x}{\sqrt{1 + x^2}} \) takes values between \( -1 \) and \( 1 \) (since \( \frac{x}{\sqrt{1 + x^2}} \) is always bounded within this range). 
Thus, \( f(x) \) cannot take any value outside of this range, making the function not surjective on \( \mathbb{R} \). 
Final Answer:
The function is:

injective but not surjective