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Maurinko [17]
3 years ago
13

Prove that the Greatest Integer Function f: R -> R given by f(x) = [x], is neither one-once nor onto, where [x] denotes the g

reatest integer less than or equal to x.
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0
F: R → R is given by, f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1
So, f(1.2) = f(1.9), but 1.2 ≠ 1.9
f is not one-one

Now, consider 0.7 ε R
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ε R such that f(x) = 0.7

So, f is not onto
Hence, the greatest integer function is neither one-one nor onto.

The answer was quite complicated but I hope it will help you.
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djverab [1.8K]

Answer:

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Step-by-step explanation:

We need to find sine, cosine, and tangent of theta for this triangle.

First, define the different trigonometric functions:

- sine is opposite side to the angle divided by the hypotenuse (longest side not adjacent to the 90 degree angle)

- cosine is the adjacent side next to the angle divided by the hypotenuse

- tangent is the opposite side to the angle divided by the adjacent side

Now, look at theta (θ):

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- cos(θ) = adjacent / hypotenuse = 5/13

- tan(θ) = opposite / adjacent = 12/5

Thus the answers are: sin(θ) = 12/13, cos(θ) = 5/13, tan(θ) = 12/5.

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