One-one function

Step 1: Assuming (f(a)=f(b)), we get (4a+7=4b+7). Step 2: Simplifying gives (4a=4b), so (a=b). Step 3: A linear function with a non-zero coefficient of (x) is one-one.

Step 1: From (f(a)=f(b)), we get (-3a+8=-3b+8). Step 2: This gives (-3a=-3b), hence (a=b). Step 3: A negative slope does not stop a function from being one-one.

Step 1: This is a constant function because its value is (9) for every (x). Step 2: (f(2)=9) and (f(6)=9), while (2\neq6). Step 3: A constant function on a larger domain is not one-one.

Step 1: (x^2-6x+9=(x-3)^2). Step 2: (f(2)=1) and (f(4)=1), while (2\neq4). Step 3: A quadratic function is generally not one-one on the whole real domain.

Step 1: In the given domain, (x-3\geq0). Step 2: On ([3,\infty)), ((x-3)^2) increases, so different (x) values give different outputs. Step 3: A quadratic function can be one-one on one side of its axis.

Step 1: (f(0)=|1|=1). Step 2: (f(-2)=|-1|=1), while (0\neq-2). Step 3: In a modulus function, inputs equally distant from the center can give the same output.

Step 1: For (x\geq-1), (x+1\geq0). Step 2: Hence (|x+1|=x+1), which is linear and increasing. Step 3: Restricting the domain properly can make a modulus function one-one.

Step 1: From (f(n_1)=f(n_2)), we get (3n_1+2=3n_2+2). Step 2: This gives (n_1=n_2). Step 3: Such a linear rule on natural numbers gives different outputs for different inputs.

Step 1: (1) and (-1) are different integers. Step 2: (f(1)=2) and (f(-1)=2), so equal images occur. Step 3: Square-based rules are often not one-one on integers.

Step 1: Cubes of different integers are different. Step 2: Adding (5) preserves that difference. Step 3: A basic cubic function and its translation keep the one-one property.

Step 1: Each element of the domain has exactly one image. Step 2: The images (a,c,d) are distinct. Step 3: In a finite set, if images are not repeated, the function is one-one.

Step 1: Here (1) and (4) are different domain elements. Step 2: Both have image (p), so repetition occurs. Step 3: Repetition of an image is not allowed in a one-one function.

Step 1: The first element has (5) choices. Step 2: The second element has (4) choices because repetition is not allowed. Step 3: Total one-one functions are (5\cdot4=20).

Step 1: The first element has (4) choices. Step 2: The next two elements have (3) and (2) choices respectively. Step 3: The total number is (4\cdot3\cdot2=24).

Step 1: A one-one function needs (5) distinct images. Step 2: The codomain has only (4) elements, so this is impossible. Step 3: If the domain has more elements than the codomain, no one-one function can exist.

Step 1: The basic identity of a one-one function is distinct images for distinct inputs. Step 2: If the images become equal, one-one property fails. Step 3: Remember the definition: equal images only come from equal inputs.

Step 1: This is the standard algebraic condition for a one-one function. Step 2: Equal images must imply equal inputs. Step 3: This form is most commonly used in proofs.

Step 1: This condition says different inputs give different images. Step 2: That is the direct definition of a one-one function. Step 3: Remember that this statement and (f(a)=f(b)\Rightarrow a=b) have the same meaning.

Step 1: (5) and (9) are two different inputs. Step 2: Their images are equal, so the one-one condition fails. Step 3: A single counterexample is enough to disprove one-one.

Step 1: In (x^2+2), (x) and (-x) can give the same value. Step 2: (f(1)=3) and (f(-1)=3), while (1\neq-1). Step 3: In options, check square-based functions carefully.

Step 1: The cube-root function increases on real numbers. Step 2: Different real inputs have different cube roots. Step 3: Cubic and cube-root functions are simple examples of one-one behavior.

Step 1: (\sqrt{x}) increases on its domain ([0,\infty)). Step 2: If (\sqrt{a}=\sqrt{b}), then (a=b). Step 3: The square-root function is one-one on its proper domain.

Step 1: (f(0)=0). Step 2: (f(-1)=1-1=0), while (0\neq-1). Step 3: For a quadratic function, find two inputs with the same value to disprove one-one.

In a one-one function, distinct inputs must have distinct images. Therefore, if \(f(x_1)=f(x_2)\), then necessarily \(x_1=x_2\); hence option A is correct. Option C is the condition for an onto function, where every element of \(B\) must have at least one preimage in \(A\). Exam tip: for one-one functions, check whether equal outputs force the inputs to be equal.

Step 1: Because of modulus, (x) and (-x) can give the same value. Step 2: (f(3)=6) and (f(-3)=6), while (3\neq-3). Step 3: Multiplying a modulus function by a positive constant does not make it one-one.