Functions

Step 1: The value of (|x|+2) is always at least (2). Step 2: The codomain (R) contains (1), but it cannot be the image of any (x). Step 3: Therefore the function is not onto.

Step 1: In a bijective function, elements are matched exactly one-to-one. Step 2: Hence the domain and codomain have the same number of elements. Step 3: Since (A) has (5) elements, (B) also has (5).

Step 1: Both compositions give identity functions. Step 2: This means (f) and (g) undo each other's action. Step 3: Therefore (g=f^{-1}).

Step 1: (f^{-1}(13)) means the value of (x) for which (f(x)=13). Step 2: From (2x+5=13), we get (2x=8), so (x=4). Step 3: While finding an inverse value, equate the original function to the given value.

Step 1: Write (f(x+2)=(x+2)^2-3(x+2)). Step 2: Simplifying gives (x^2+x-2). Step 3: Now (x^2+x-2-(x^2-3x)=4x-2).

Step 1: The denominator of the fraction is (x-3). Step 2: The denominator cannot be zero, so (x-3\ne0). Step 3: Removing (x=3), all other real numbers remain in the domain.

Step 1: Write (x^2+4x+7=(x+2)^2+3). Step 2: Since ((x+2)^2\ge0), (f(x)\ge3). Step 3: The minimum value is (3), so the range is ([3,\infty)).

Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=x+5) into (f). Step 3: (f(x+5)=3(x+5)-2=3x+13).

Step 1: ((f\circ g)(x)=f(x-1)=(x-1)^2+2). Step 2: ((g\circ f)(x)=g(x^2+2)=x^2+1). Step 3: The expressions are generally different, so order is important in composition.

Step 1: In (7x+2), different (x) values give different outputs, so the function is one-one. Step 2: For any (y\in R), choose (x=\frac{y-2}{7}). Step 3: Thus every real (y) is an image, so the function is onto.

Step 1: (f(2)=f(-2)=-1), so the function is not one-one. Step 2: Since (x^2-5\ge-5), a number like (-6) cannot be an image. Step 3: Hence it is not onto either.

Step 1: For different (x), (x-1) is different, so the function is one-one. Step 2: For every (y\ge0), (x=y+1), which lies in the domain ([1,\infty)). Step 3: Hence the function is onto as well.

Step 1: For every (y\ge4), (x=\sqrt{y-4}) or (x=-\sqrt{y-4}) can be chosen, so it is onto. Step 2: (f(1)=f(-1)=5), so it is not one-one. Step 3: Changing the codomain can change onto status.

Step 1: The expression inside the square root, (2x-6), must be non-negative. Step 2: From (2x-6\ge0), we get (x\ge3). Step 3: At (x=3), the value is (0), so (3) is included.

Step 1: For the square root, (x+2\ge0) is needed. Step 2: But the same square root is in the denominator, so it cannot be zero. Step 3: Thus (x+2>0), so the domain is ((-2,\infty)).

Step 1: The denominator is (x^2-16=(x-4)(x+4)). Step 2: The denominator cannot be zero, so (x\ne4) and (x\ne-4). Step 3: Even if (x-4) cancels, (x=4) is not valid in the original function.

Step 1: To find the preimage, solve (|x+1|=4). Step 2: This gives (x+1=4) or (x+1=-4). Step 3: Thus (x=3) or (x=-5), so the preimage is ({-5,3}).

Step 1: The images of (1,2,3,4) are distinct, so the function is one-one. Step 2: (t) is in the codomain but is not an image. Step 3: Therefore the function is not onto.

Step 1: Total functions are (2^4=16). Step 2: There are two non-onto functions, where all elements go only to (a) or only to (b). Step 3: Therefore onto functions are (16-2=14).

Step 1: In a one-one function, the two inputs must have distinct images. Step 2: The first input has (5) choices and the second has (4) choices. Step 3: Total one-one functions are (5\cdot4=20).