Functions

Step 1: For (x=-5), (|-5|=5). Step 2: Therefore (f(-5)=-5+5=0). Step 3: In modulus questions, first evaluate (|x|) according to the sign of (x).

Step 1: If (x<0), then (x+|x|=x-x=0). Step 2: If (x\ge0), then (x+|x|=2x), which gives all values in ([0,\infty)). Step 3: For modulus functions, split the function into cases to find the range.

Step 1: (f(1.2)=1) and (f(1.8)=1), so it is not one-one. Step 2: Its values are always integers, so a real number like (0.5) is not obtained. Step 3: The greatest integer function maps many values in an interval to the same integer.

Step 1: For every integer (n), taking (x=n) gives (\lfloor x\rfloor=n), so it is onto (\mathbb{Z}). Step 2: But (\lfloor 2.1\rfloor=\lfloor 2.9\rfloor=2), so it is not one-one. Step 3: Changing the codomain to (\mathbb{Z}) makes this function onto.

Step 1: For (x\ge0), (f(x)=\frac{x}{1+x}), which increases from (0) and approaches (1) but never equals (1). Step 2: For (x<0), the value stays greater than (-1) and less than (0). Step 3: Approaching an endpoint and attaining it are different things.

Step 1: \(f(2)=\frac{2}{1+4}=\frac{2}{5}\). Step 2: \(f\left(\frac{1}{2}\right)=\frac{\frac{1}{2}}{1+\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{5}{4}}=\frac{2}{5}\). Step 3: Once two distinct inputs have the same image, the function is not one-one.

Step 1: Both sets have (3) elements. Step 2: A function that is one-one and onto must pair every element uniquely, so the number is (3!). Step 3: In exams, for two finite sets of the same size, bijections are counted as (n!).

Step 1: Put (y=2x-5). Step 2: Solving for (x), we get (x=\frac{y+5}{2}). Step 3: While finding inverse, first write (y=f(x)), then isolate (x).

Step 1: ((f\circ g)(x)) means (f(g(x))). Step 2: Substitute (g(x)=3x-2) into (f), giving ((3x-2)^2+1=9x^2-12x+5). Step 3: In composition, the order matters.

Step 1: Total functions from (A) to (B) are (2^4=16). Step 2: Not onto functions use only (a) or only (b), giving (2) such functions. Step 3: Hence onto functions are (16-2=14).

Step 1: In an onto function, every element of the codomain must be an image of some domain element. Step 2: The (5) elements of (A) can cover at most (5) distinct elements. Step 3: Hence the maximum possible size of (B) is (5).

Step 1: (f(1)=1) and (f(-1)=1), so distinct inputs have the same image; it is not one-one. Step 2: No negative real number is an output, so it is not onto (\mathbb{R}). Step 3: For absolute value functions, check both domain and codomain carefully.

Step 1: First find (f(-3)), which is (-1). Step 2: Then compute (g(-1)=(-1)^2=1). Step 3: In composition, apply the right-side function first.

Step 1: In a one-one function, the three elements of (A) must get distinct images. Step 2: The choices are (4), then (3), then (2). Step 3: Total number is (4\times3\times2=24).

Step 1: Write (\frac{x-1}{x+1}=2). Step 2: Then (x-1=2x+2), so (x=-3). Step 3: For rational functions, also check that the denominator is not zero.

Step 1: (f) is a bijective linear function, so its inverse exists. Step 2: The inverse reverses the effect of (f), hence (f^{-1}(f(10))=10). Step 3: Recognize the property first; full calculation is not always needed.

Step 1: In a one-one function, every distinct element of (A) needs a distinct image. Step 2: For (6) distinct images, (B) must have at least (6) elements. Step 3: For injective functions, the codomain cannot be smaller than the domain.

Step 1: Complete the square: (x^2-4x+7=(x-2)^2+3). Step 2: The least value of ((x-2)^2) is (0). Step 3: Therefore, the minimum value of the function is (3).

Step 1: In a one-one function, distinct inputs must have distinct images. Step 2: Here (1\neq -1), but (f(1)=1) and (f(-1)=1). Step 3: Once two different inputs give the same output, the function is not one-one.

Step 1: In a constant function, all elements of (A) have the same image. Step 2: This common image can be (1), (2), or (3). Step 3: The number of constant functions equals the number of elements in the codomain.