Question

If \( f : \mathbb{R} \to \mathbb{R} \) is a function defined by \( f(x) = \sin x \), then which of the following is true?

• A. \( f \) is 1-1 but not onto
• B. \( f \) is onto but not 1-1
• C. \( f \) is both 1-1 and onto
• D. \( f \) is neither 1-1 nor onto
• E. \( f \) has finite number of zeros

Hint:

Trigonometric functions like sine are periodic, so they are never one-one over \( \mathbb{R} \), and their range is bounded.

Answer 1

The Correct Option is D

Concept: A function is:
• One-one (injective) if different inputs give different outputs.
• Onto (surjective) if every element of the codomain has a pre-image.

Step 1: Check if \( f(x) = \sin x \) is one-one. For a function to be one-one: \[ f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \] But for sine function: \[ \sin 0 = 0 \text{and} \sin \pi = 0 \] Here: \[ 0 \neq \pi \text{but} \sin 0 = \sin \pi \] Hence, different inputs give same output. \[ \Rightarrow f \text{ is not one-one} \]

Step 2: Check if \( f(x) = \sin x \) is onto. Range of sine function is: \[ -1 \leq \sin x \leq 1 \] So: \[ \text{Range} = [-1,1] \] But codomain is: \[ \mathbb{R} \] Since values like \(2, -5, 100\) are not attained by \( \sin x \), \[ \Rightarrow f \text{ is not onto} \]

Step 3: Check option (E). Zeros of sine function: \[ \sin x = 0 \Rightarrow x = n\pi, n \in \mathbb{Z} \] There are infinitely many zeros. So statement (E) is false.