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Flashcards in GLORIOUS REVIEW MASTER RACE Deck (12)
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1

Let f, g : ℝ → ℝ be two functions.

If f is strictly monotone, then f is injective. 

True.

Without loss of generality, assume that f is strictly increasing and let x1 , x2 ∈ R be such that x1 < x2. Then f(x1) < f(x2) and so f(x1) ≠ f(x2). 

2

Let f, g : ℝ → ℝ be two functions.

If f is injective, then f is monotone. 

False.

Take for example f (x) = 2[x] − x + 1. 

3

Let f, g : ℝ → ℝ be two functions.

If f is bijective and croissante, then its inverse f−1 is decreasing. 

False.

Take for example f(x) = f−1(x) = x. 

4

Let f, g : ℝ → ℝ be two functions.

If f is bijective and odd, then its inverse f−1 is odd. 

True.

Let y ∈ R. There exists x ∈ R such that f(x) = y.

Then we have f−1(−y) = f−1(−f(x)) = f−1(f(−x)) = −x = −f−1(y). 

5

Let f, g : ℝ → ℝ be two functions.

f ◦ g = g ◦ f ⇔ f = g.

False.

Take for example f(x) = x and g(x) =x2 that satisfy (f◦g)(x)=x2 =(g◦f)(x) with f ≠ g. 

6

Let f, g : ℝ → ℝ be two functions.

If f and g are injective, then f ◦ g is injective. 

True.

 

7

Let f, g : ℝ → ℝ be two functions.

If f ◦ f is injective, then f is injective. 

True.

Let x1, x2 ∈ ℝ be such that f(x1) = f(x2). Then we have f(f(x1)) = f(f(x2)). Since f ◦ f is injective, we conclude that x1 = x2

8

Let f, g : ℝ → ℝ be two functions.

If f ◦ g is injective, then g is injective. 

True.

Let x1,x2 ∈ R be such that g(x1) = g(x2). Then we have f(g(x1)) = f(g(x2)). Since f ◦ g is injective, we conclude that x1 = x2. 

9

Let f, g : ℝ → ℝ be two functions.

If f ◦ g is injective, then f is injective. 

False.

Take for example f(x) = x2 and g(x) = ex defined from ℝ to ℝ. f is not injective but (f ◦ g)(x) = e2x is injective. 

10

Let f, g : ℝ → ℝ be two functions.

If f ◦ g is surjective, then f is surjective.

True.

Let y ∈ R, then there exists x ∈ R such that (f ◦ g)(x) = y. Then there exists z = g(x) such that f(z) = y, and therefore f is surjective. 

11

Let f, g : ℝ → ℝ be two functions.

 If f ◦ g is decreasing, then f and g are decreasing. 

False 

Take for example f(x) = x and g(x) = −x. 

12

Let f, g : ℝ → ℝ be two functions.

Let A, B be two subsets of R, then f([A∩B]) = f([A]) ∩ f([B]). 

False