One-one function

Step 1: When (x\ge 2), (|x-2|=x-2). Step 2: The expression (x-2) increases linearly, so different inputs give different outputs. Step 3: Restricting an absolute value function to one side can make it one-one.

Step 1: In even powers, (x) and (-x) can give the same value. Step 2: (f(2)=16) and (f(-2)=16), while (2\ne -2). Step 3: Even power functions are generally not one-one on the whole real domain.

Step 1: The domain contains only non-negative values. Step 2: On (x\ge 0), (x^4) is strictly increasing. Step 3: Restricting an even power function to non-negative values makes it one-one.

Step 1: Substitute two different values to find equal outputs. Step 2: \(f(0)=1\) and \(f\left(-\frac{3}{2}\right)=1\), while the inputs are different. Step 3: A quadratic function is not one-one on the entire real domain.

Step 1: (f(x)=(x+1)^2+4). Step 2: For (x\ge -1), ((x+1)^2) increases. Step 3: A quadratic function restricted to the right side of its vertex is one-one.

Step 1: In a one-one function, distinct elements of (A) must have distinct outputs. Step 2: Since (A) has (5) elements, (B) must have at least (5) possible distinct outputs. Step 3: For finite sets, the codomain size cannot be smaller than the domain size for a one-one function.

Step 1: A one-one function needs four distinct outputs for four elements of (A). Step 2: (B) has only three elements, so four distinct outputs are impossible. Step 3: If the domain is larger than the codomain, a one-one function cannot exist.

Step 1: The first element of (A) has (5) choices. Step 2: The next elements have (4,3,2) choices respectively. Step 3: Total one-one functions are (5\times4\times3\times2=120).

Step 1: The first element has (6) choices. Step 2: The second and third elements have (5) and (4) choices respectively. Step 3: The number of one-one functions is (6\times5\times4=120).

Step 1: Not one-one is the negation of the one-one definition. Step 2: It means two distinct inputs give the same output. Step 3: If proof is needed, showing such two inputs is enough.

Step 1: If (k\ne 0), then (f(1)=k) and (f(-1)=k). Step 2: If (k=0), the function becomes constant. Step 3: Therefore, (kx^2) is not one-one on the whole real domain for any value of (k).

Step 1: (f(x)=(x+k)^2-k^2). Step 2: Two different inputs at equal distance from the vertex give the same output. Step 3: No quadratic function is one-one on the whole real domain.

Step 1: Assume (f(a)=f(b)). Step 2: Then (g(f(a))=g(f(b))), so ((g\circ f)(a)=(g\circ f)(b)). Step 3: Since (g\circ f) is one-one, (a=b), hence (f) is one-one.

Step 1: ((g\circ f)(x)=(2x+1)^2). Step 2: At (x=0) and (x=-1), the output is (1) both times. Step 3: The outer function (x^2) is not one-one on the whole real domain, so this composition is also not one-one here.

Step 1: ((g\circ f)(x)=g(x+4)=3(x+4)-2). Step 2: This is (3x+10), a linear function with coefficient (3\ne 0). Step 3: Such a linear composition is one-one.

Step 1: The expression inside the root is (x^2+1), so (x) and (-x) give the same output. Step 2: (f(1)=\sqrt{2}) and (f(-1)=\sqrt{2}), while (1\ne -1). Step 3: Functions based on even expressions must be checked carefully on the whole real domain.

Step 1: Here (x\ge 0). Step 2: On (x\ge 0), (x^2+1) increases and the square root also increases. Step 3: On the non-negative domain, this function gives different outputs for different inputs.

Step 1: (\lfloor x\rfloor) sends many values in an interval to the same integer. Step 2: (\lfloor 1.2\rfloor=1) and (\lfloor 1.8\rfloor=1), while (1.2\ne 1.8). Step 3: Such step functions are not one-one on the whole real domain.

Step 1: If (m>n), then (m^3>n^3) and (m>n). Step 2: Therefore, (m^3+m>n^3+n), so the function is increasing. Step 3: An increasing function on integers is one-one.

Step 1: Check the rule separately for even and odd natural numbers. Step 2: (f(2k)=2k+1) and (f(2k+1)=2k), so the outputs alternate but do not repeat. Step 3: Since no two different natural numbers give the same output, the function is one-one.

Step 1: For (x\ge 0), outputs lie in ([1,\infty)). Step 2: For (x<0), outputs lie in ((-\infty,-1)). The output ranges of the two parts do not overlap. Step 3: If each part is one-one and their output ranges do not clash, the whole piecewise function is one-one.

Step 1: To disprove one-one, it is enough to show two different inputs with the same output. Step 2: (f(0)=0) and (f(-3)=0), while (0\ne -3). Step 3: Even if a function looks cubic, first check simple values that may give the same output.

Step 1: For (x\ge 0), both (x^3) and (3x) increase. Step 2: If (a>b\ge 0), then (a^3>b^3) and (3a>3b), so (f(a)>f(b)). Step 3: A strictly increasing function is always one-one.

Step 1: The first element has (7) choices. Step 2: Because the function is one-one, the next elements have (6,5,4,3) choices respectively. Step 3: Total number is (7\times6\times5\times4\times3=2520).

Step 1: Check whether the two pieces can produce the same output. Step 2: For \(x<1\), outputs are less than (3); for \(x\ge 1\), outputs are at least (3). Step 3: Each piece is increasing and their output ranges do not overlap, so (f) is one-one.