Question

Let f: R→R be defined by f(x) = cos x. Then

• A. f is one-one and odd
• B. f is odd but not one-one
• C. f is even and onto
• D. f is one-one and even
• E. f is even but not onto

Answer 1

Answer: (E)

We are given the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \cos(x) \).

Step 1: Check if \( f(x) \) is even.

A function \( f(x) \) is even if \( f(-x) = f(x) \) for all \( x \). For \( f(x) = \cos(x) \), we have:

\[ f(-x) = \cos(-x) = \cos(x) \] Therefore, \( f(x) = \cos(x) \) is an even function.

Step 2: Check if \( f(x) \) is one-one (injective).

A function is one-one if for every distinct pair \( x_1 \) and \( x_2 \), \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). For \( f(x) = \cos(x) \), this is not true because \( \cos(x) \) is periodic and takes the same value for different \( x \) values (for example, \( \cos(0) = \cos(2\pi) \)). Hence, \( f(x) \) is not one-one.

Step 3: Check if \( f(x) \) is onto (surjective).

A function is onto if for every element in the codomain, there is a corresponding element in the domain. The range of \( \cos(x) \) is limited to the interval \([-1, 1]\), so it does not cover all real numbers. Therefore, \( f(x) \) is not onto.

Thus, the function \( f(x) = \cos(x) \) is even but not onto, which matches option (E).

The correct option is (E) : f is even but not onto

Answer 2

We are given the function f(x) = cos(x) defined on R → R. Let's analyze its properties:

Even/Odd:

A function is even if f(-x) = f(x) for all x in its domain.

A function is odd if f(-x) = -f(x) for all x in its domain.

For f(x) = cos(x), we have f(-x) = cos(-x) = cos(x) = f(x). Therefore, f(x) = cos(x) is an even function.

One-to-one:

A function is one-to-one (injective) if for any x₁ ≠ x₂, f(x₁) ≠ f(x₂).

Consider x₁ = 0 and x₂ = 2π. Then f(0) = cos(0) = 1 and f(2π) = cos(2π) = 1. Since f(0) = f(2π) but 0 ≠ 2π, the function is not one-to-one.

Onto:

A function is onto (surjective) if for every y in the codomain (R), there exists an x in the domain (R) such that f(x) = y.

The range of the cosine function is [-1, 1]. Since the codomain is R (all real numbers), there are values in the codomain that are not in the range (e.g., y = 2). Therefore, f(x) = cos(x) is not onto.

Therefore:

• f is even.
• f is not one-to-one.
• f is not onto.

Thus, the correct answer is f is even but not onto.