x ≥ sin(x) for ∀ x ≥ 0
True
There are multiple ways to solve this question. Here, we will look at how it can be solved using the concept of monotonicity.
Let’s define a function f(x) such that f(x) = x - sin(x)
f(x) = x - sin(x)
By simply putting x = 0 we can see that f(0) = 0
Now, if somehow we could prove that f(x) is a monotonically increasing function in the interval (0, ∞) then we can say that f(x) ≥ 0 ∀ x ε (0, ∞)
Or in other words we can say that x ≥ sin(x)
Now let’s see whether the function is monotonically increasing or not in the interval of (0, ∞)
f(x) = x - sin(x)
f’(x) = 1 - cos(x)
We know that cos function varies from [-1, 1] so the minimum value f’(x) can attain is zero.
So, f’(x) ≥ 0
And we know that if f’(x) ≥ 0 then f(x) is monotonically increasing.
Since the function f(x) is monotonically increasing after x =0 and at x =0 the value of the function is 0, we can say that after x=0 f(x) will be giving only nonnegative values.
Or x - sin(x) ≥ 0
Or x ≥ sin(x).
So the given statement is correct