The function f : R → R defined by f(x) = 4 + 3 cos x is 

We have a function f: ℝ → ℝ defined as

f(x) = 4 + 3 cos(x)

Whether the given function is bijective, one-one, onto, or neither can be determined.
Step 1 Check if the given function is one-one (injective).

A function is said to be one-one (injective) if the different elements in the domain of a function go to different elements in the range i.e. f(x₁) = f(x₂) => x₁ = x₂.

Let’s see if the function is injective. Assume that f(x₁) = f(x₂), that is:

4 + 3 cos(x₁) = 4 + 3 cos(x₂)

We simplify this to obtain:

3 cos(x₁) = 3 cos(x₂)
cos(x₁) = cos(x₂)

The cosine function is periodic and not injective. That means that there may be different values of x₁ and x₂, like x₁ = 0 and x₂ = 2π, for which cos(x₁) = cos(x₂). In other words, the function is not injective.

Step 2: Determine whether the function is onto (surjective).

A function is onto (surjective) if for every element y ∈ ℝ, there exists an x ∈ ℝ such that f(x) = y.

For f(x) = 4 + 3 cos(x), the range of the cosine function is between -1 and 1, so the range of f(x) will be:

f(x) = 4 + 3 cos(x)

Since cos(x) lies between -1 and 1, the values of f(x) will lie between:

f(x) = 4 + 3(-1) = 1 and f(x) = 4 + 3(1) = 7

So, f(x) can only be taken in the interval [1, 7]. Since f(x) cannot take any value outside that interval, it is not onto because no values of x map to values of y that are outside that interval.

Conclusion:
The function f(x) = 4 + 3 cos(x) is neither one-one nor onto. So the correct option is
(d) neither one-one nor onto.

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