Question

The function \(f(x)=2x\), \(x\in R\) is:

• A. one-one but not onto
• B. one-one and onto
• C. many-one and onto
• D. many-one but not onto

Hint:

Check if \(2x_1=2x_2\) forces \(x_1=x_2\), and if every real y has a pre-image \(y/2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Here \( f:R\to R \) is defined by \( f(x)=2x \). To classify it, check whether it is one-one (injective) and whether it is onto (surjective).

Step 2: Checking one-one:
Assume \( f(x_1)=f(x_2) \) for some \( x_1,x_2\in R \).
\[ 2x_1=2x_2 \quad\Rightarrow\quad x_1=x_2 \]
Since equal outputs force equal inputs, f is one-one.

Step 3: Checking onto:
Take any real number y in the codomain and try to find x in the domain with \( f(x)=y \).
\[ 2x=y \quad\Rightarrow\quad x=\dfrac{y}{2} \]
Since \( \dfrac{y}{2} \) is a real number for every real y, every y has a pre-image, so f is onto.

Step 4: Why option A is wrong:
Option A says one-one but not onto; f was shown to be onto in Step 3, so this option is incorrect.

Step 5: Why option C is wrong:
Option C says many-one and onto; f was shown to be one-one in Step 2, so calling it many-one is incorrect.

Step 6: Why option D is wrong:
Option D says many-one but not onto; f is neither many-one nor fails to be onto, so this option is incorrect on both counts.

Final Answer:
A linear function like \( 2x \) with nonzero slope from R to R is always a bijection. \[ \boxed{\text{one-one and onto}} \]