Question

Let f: R→R be defined as f(x) = 3x. Choose the correct answer.

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

Answer 1

The Correct Option is A

\(f:R→R,f(x)=3x\)

Let \(x, y ∈ R\) such that \(f(x) = f(y).\)
\(⇒ 3x = 3y\)
\(⇒ x = y\)
∴f is one-one.

Also, for any real number (y) in co-domain R, there exists \(\frac{y}{3}\) in R such that \(f(\frac{y}{3})=3(\frac{y}{3})=y\).
∴f is onto.
Hence, function f is one-one and onto.
The correct answer is A.

Answer 2

Let's find out if the function is one-to-one.
For a function to be one-to-one, whenever \(f(x_1) = f(x_2),\) then \(x_1 = x_2​.\)

Given \(f(x)=3x,\)
For one-to-one, if \(f(x_1) = f(x_2), 3x_1 = 3x_2\)
\(x_1 = x_2\)
Therefore, if \(f(x_1) = f(x_2)\), then \(x_1 = x_2​.\)
So, the function f is one-to-one.

Now, let's check if the function is onto.
For a function to be onto, for every y in the range, there exists an x in the domain such that \(f(x) = y.\)

Let's consider \(f(x)=3x.\)
Let \(f(x)=y\), where y is any real number.
3x=y
\(x = \frac{y}{3}\)

Now, for \(y=f(x),\)
Putting the value of x in \(f(x),\)
\(f(x) = f\left(\frac{y}{3}\right)\)

\(f(x) = 3\left(\frac{y}{3}\right)\)

\(f(x)=y\)

Thus, for every y in the real numbers, there exists an xxx in the real numbers such that \(f(x)=y.\)
Hence is onto.
So, f is both one-to-one and onto.

Therefore, option A is correct.