Question

Let f: R $\rightarrow$ R be defined as f(x) = 10x. Then (Where R is the set of real numbers)

• A. f is both one-one and onto
• B. f is onto but not one-one
• C. f is one-one but not onto
• D. f is neither one-one nor onto

Hint:

A linear function of the form f(x) = ax + b, where a $\neq$ 0, defined from R to R, is always both one-one and onto. Visualizing its graph, a straight line, shows that it passes both the horizontal line test (one-one) and covers the entire y-axis (onto).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To solve this problem, we need to check if the function f(x) = 10x satisfies the conditions for being one-one (injective) and onto (surjective).


One-one (Injective): A function f is one-one if for any two distinct elements x\textsubscript{1} and x\textsubscript{2} in the domain, their images f(x\textsubscript{1}) and f(x\textsubscript{2}) are also distinct. Mathematically, if f(x\textsubscript{1}) = f(x\textsubscript{2}), then x\textsubscript{1} = x\textsubscript{2}.

Onto (Surjective): A function f from a set R to a set R is onto if for every element y in the codomain R, there exists at least one element x in the domain R such that f(x) = y. In other words, the range of the function is equal to its codomain.

Step 3: Detailed Explanation:
Checking for One-one:
Let's assume f(x\textsubscript{1}) = f(x\textsubscript{2}) for some x\textsubscript{1}, x\textsubscript{2} $\in$ R.
\[ 10x_1 = 10x_2 \] Dividing both sides by 10, we get:
\[ x_1 = x_2 \] Since f(x\textsubscript{1}) = f(x\textsubscript{2}) implies x\textsubscript{1} = x\textsubscript{2}, the function is one-one.
Checking for Onto:
Let y be an arbitrary element in the codomain R. We need to check if there is an x in the domain R such that f(x) = y.
\[ f(x) = y \] \[ 10x = y \] \[ x = \frac{y}{10} \] For any real number y, the value \(x = \frac{y}{10}\) is also a real number. Thus, for any y in the codomain, there exists a pre-image x in the domain. Therefore, the function is onto.
Step 4: Final Answer:
Since the function f(x) = 10x is both one-one and onto, option (1) is correct.