One-one function

Step 1: Assuming (f(a)=f(b)), we get (3a-7=3b-7). Step 2: Simplifying gives (3a=3b), hence (a=b). Step 3: A linear function with a non-zero coefficient of (x) is one-one.

Step 1: (x^2-4x+6=(x-2)^2+2). Step 2: (f(1)=3) and (f(3)=3), while (1\neq3). Step 3: A quadratic function on the whole real domain is generally not one-one.

Step 1: (x^2-4x+5=(x-2)^2+1). Step 2: In the given domain, (x-2\geq0), so ((x-2)^2) increases. Step 3: A quadratic function can be one-one on one side of its axis.

Step 1: (f(4)=|4-5|=1). Step 2: (f(6)=|6-5|=1), while (4\neq6). Step 3: In a modulus function, inputs at equal distance from the center may give the same output.

Step 1: For (x\geq5), we have (x-5\geq0). Step 2: Hence (|x-5|=x-5), which is an increasing linear function. Step 3: A modulus function can become one-one on a suitable domain.

Step 1: Assuming (f(a)=f(b)), we get (\frac{a+1}{a-2}=\frac{b+1}{b-2}). Step 2: Cross-multiplication gives ((a+1)(b-2)=(b+1)(a-2)), and simplification gives (a=b). Step 3: For rational functions, remove denominator-zero values before testing one-one.

Step 1: From (f(a)=f(b)), we get (\frac{2a-1}{a+3}=\frac{2b-1}{b+3}). Step 2: Cross-multiplication gives (2ab+6a-b-3=2ab+6b-a-3), so (7a=7b). Step 3: Hence (a=b), so the function is one-one.

Step 1: (x^3+2x) is an increasing function. Step 2: If (a>b), then (a^3-b^3+2(a-b)>0), so (f(a)>f(b)). Step 3: A strictly increasing function is one-one.

Step 1: (f(0)=0). Step 2: (f(\sqrt{3})=(\sqrt{3})^3-3\sqrt{3}=3\sqrt{3}-3\sqrt{3}=0). Step 3: Do not assume every cubic-looking expression is one-one.

Step 1: \(f(2)=2+\frac{1}{2}=\frac{5}{2}\). Step 2: \(f\left(\frac{1}{2}\right)=\frac{1}{2}+2=\frac{5}{2}\), while \(2\neq\frac{1}{2}\). Step 3: In functions involving (x) and \(\frac{1}{x}\), check reciprocal pairs carefully.

Step 1: For (x\geq1), (x+\frac{1}{x}) increases as (x) increases. Step 2: Although (\frac{1}{x}) decreases, the total value increases on this domain. Step 3: Restricting the domain can turn a non-one-one function into a one-one function.

Step 1: (f(1)=2+3=5). Step 2: (f(-1)=2+3=5), while (1\neq-1). Step 3: A positive coefficient with a square does not make it one-one on all real numbers.

Step 1: On (x\geq0), (x^2) increases. Step 2: (2x^2+3) also increases on that domain, so different (x) values give different outputs. Step 3: A square function becomes one-one when restricted to the non-negative half-domain.

Step 1: (\sin 0=0). Step 2: (\sin \pi=0), while (0\neq\pi). Step 3: Periodic functions can give the same value for different inputs.

Step 1: On ([0,\pi]), (\cos x) is decreasing. Step 2: A decreasing function gives different outputs for different inputs. Step 3: A function need not increase to be one-one; strictly decreasing also works.

Step 1: (\cos 0=1). Step 2: (\cos 2\pi=1), while (0\neq2\pi). Step 3: If different inputs give the same value, the function is not one-one.

Step 1: (\lfloor x\rfloor) gives the greatest integer less than or equal to (x). Step 2: (f(2.1)=2) and (f(2.8)=2), while (2.1\neq2.8). Step 3: A step-like function may give the same value for many inputs.

Step 1: Cubes of different integers are different. Step 2: If (a^3=b^3), then (a=b). Step 3: The basic odd-power cubic function is one-one on integers.

Step 1: (1) and (-1) are different integers. Step 2: (f(1)=0) and (f(-1)=0). Step 3: A square-based function is not one-one on the full integer domain.

Step 1: From (f(n_1)=f(n_2)), we get (2n_1+5=2n_2+5). Step 2: Simplifying gives (n_1=n_2). Step 3: A linear rule with non-zero coefficient is one-one on natural numbers.

Step 1: In a one-one function, images are not repeated. Step 2: The first, second, and third elements have (6,5,4) choices respectively. Step 3: The total number is (6\cdot5\cdot4=120).

Step 1: A one-one function needs four distinct images for the four elements of (A). Step 2: (B) has only three elements, so this is impossible. Step 3: If the domain is larger than the codomain, no one-one function can exist.

Step 1: Every element of (A) has exactly one image. Step 2: The images (a,c,e,b) are all distinct. Step 3: For finite sets, no repetition of images means the function is one-one.

Step 1: (2) and (4) are different domain elements. Step 2: Both have image (q), so an image is repeated. Step 3: In a one-one function, two different domain elements cannot have the same image.

Step 1: In a one-one function, distinct domain elements have distinct images. Step 2: The (8) elements of (A) give (8) distinct images. Step 3: Therefore the range has as many elements as the domain.