One-one function

Step 1: On ([0,\pi]), (\cos x) is strictly decreasing. Step 2: In a strictly decreasing function two different inputs cannot give the same value. Step 3: A decreasing function can also be one-one.

The function \(f(x)=x^3\) is strictly increasing on the real numbers. Hence, if \(x_1\ne x_2\), then \(x_1^3\ne x_2^3\); distinct inputs give distinct outputs, so it is one-one. The function \(x^2\) is not one-one because \(f(2)=f(-2)\). Similarly, \(|x|\) gives the same output for positive and negative inputs of equal magnitude. Exam tip: To disprove one-one-ness, find two different inputs with the same output.

Step 1: Three elements need three distinct images. Step 2: The number is (3!=6). Step 3: For finite sets of equal size, a one-one function is just an arrangement of all elements distinctly.

Step 1: All powers in the rule are even. Step 2: (f(1)=1+1=2) and (f(-1)=1+1=2), while (1\neq-1). Step 3: Polynomials with even-power symmetry are usually not one-one on the whole real domain.

Step 1: In the given domain (x\ge0). Step 2: As (x) increases, both (x^2) and (x^4) increase. Step 3: Therefore different inputs give different outputs and the function is one-one.

Step 1: To disprove injectivity, show the same value at different inputs. Step 2: (f(-1)=0), (f(0)=0), and (f(1)=0). Step 3: Multiple roots giving the same output quickly break injectivity.

Step 1: (x^5), (x^3), and (x) are all increasing terms. Step 2: As (x) increases, their sum also increases. Step 3: A strictly increasing function is one-one, so this function is one-one.

Step 1: Injectivity means different inputs have different images. Step 2: Therefore if (a\neq b), then (f(a)\neq f(b)). Step 3: Remember both forms of the definition to handle proofs easily.

Step 1: Assume equal values and cross multiply. Step 2: Simplifying (\frac{ax_1+b}{cx_1+d}=\frac{ax_2+b}{cx_2+d}) gives ((ad-bc)(x_1-x_2)=0). Step 3: Since (ad-bc\neq0), (x_1=x_2), so the function is one-one.

Step 1: This is a linear fractional form. Step 2: Here (ad-bc=2\cdot3-1\cdot1=5\neq0). Step 3: Therefore the function is one-one on its given domain.

Step 1: In the given domain (x\ge2). Step 2: Therefore (|x-2|=x-2), which is an increasing linear function. Step 3: Restricting an absolute value function to one side can make it one-one.

Step 1: In the given domain (x\le2). Step 2: Therefore (|x-2|=2-x), which is a decreasing linear function. Step 3: A decreasing linear function is also one-one.

Step 1: Absolute value gives the same value to inputs equally distant from the center. Step 2: (1) and (3) are equally distant from (2), so (f(1)=1) and (f(3)=1). Step 3: On the whole real domain an absolute value function is generally not one-one.

Step 1: Because of (x^2), opposite inputs give the same inner value. Step 2: (f(1)=\sqrt2) and (f(-1)=\sqrt2), while (1\neq-1). Step 3: Before focusing on the square root, check the square inside it.

Step 1: In the given domain (x\ge0). Step 2: As (x) increases, (x^2+1) and then (\sqrt{x^2+1}) increase. Step 3: On the correct domain this square-root function becomes one-one.

Step 1: In (g\circ f), (g) is used only on the range of (f). Step 2: (g) may not be one-one on its whole domain but can behave distinctly on (f(A)). Step 3: In composition, distinguish the used part from the whole domain.

Step 1: Four elements of (A) need four distinct images from (B). Step 2: The number is (^{6}P_{4}=6\cdot5\cdot4\cdot3=360). Step 3: For one-one functions use permutations, not combinations, because domain elements are distinct.

Step 1: The given domain contains only zero and negative numbers. Step 2: On this interval, as (x) increases, (x^2) decreases, so the same value does not occur twice. Step 3: Restricting (x^2) to one side of zero makes it one-one.

Step 1: The domain ((-\infty,2]) contains both (1) and (-1). Step 2: (1\neq-1), but (1^2=(-1)^2=1). Step 3: If the domain spreads on both sides of zero, (x^2) is not one-one.

Step 1: (2x^3-5) is a transformation of the cube function. Step 2: From (2x_1^3-5=2x_2^3-5), we get (x_1^3=x_2^3), so (x_1=x_2). Step 3: Stretching and shifting the cube function does not destroy injectivity.

Step 1: (x^2-6x+10=(x-3)^2+1). Step 2: (0) and (6) are equally distant from (3), so (f(0)=10) and (f(6)=10). Step 3: For quadratics quickly test points equally distant from the vertex.

Step 1: (x^2) increases to the right of zero and can give the same value on both sides of zero. Step 2: If (a\ge0), the whole domain stays to the right of zero, so two different inputs do not give the same value. Step 3: For a quadratic function the domain should lie on one side of the vertex.

Step 1: Both (x^3) and (3x) contribute increasing behavior. Step 2: As (x) increases, (x^3+3x+2) keeps increasing, so the same value does not occur at two different inputs. Step 3: A strictly increasing function is always one-one.

Step 1: To disprove injectivity it is enough to show equal output for two different inputs. Step 2: (f(0)=0) and (f(3)=27-54+27=0), while (0\neq3). Step 3: In polynomial questions use simple values to quickly find repeated images.

Step 1: The logarithmic function increases on its proper domain ((0,\infty)). Step 2: If (\log x_1=\log x_2), then (x_1=x_2). Step 3: While checking injectivity of a logarithmic function, always check its domain first.