Functions

The governing concept is the range of a constant function. A range consists of all output values that the function actually produces as the input runs through its domain. Here the domain is R, but f(x) is always equal to 2, regardless of whether x is positive, negative, or zero. Consequently, every real input has the same image, 2, and no other output can occur. Therefore Range(f) = {2}, so option A is correct. Option B confuses the codomain R with the actual range. Option C would be correct only for the constant function f(x) = 0, while option D incorrectly suggests that every positive real number is attained. The singleton set notation is essential because the range contains exactly one value.

Step 1: The expression inside the square root, (x+2), must be non-negative. Step 2: (x+2\ge 0) gives (x\ge -2). Step 3: Hence the real domain is ([-2,\infty)).

Step 1: The denominator (x+3) must not be zero. Step 2: (x+3=0) gives (x=-3). Step 3: So all real numbers except (-3) form the domain.

Step 1: The expression inside the square root must satisfy (5-x\ge 0). Step 2: This gives (x\le 5). Step 3: Therefore the domain is ((-\infty,5]).

Step 1: The denominator of the function is (x+1). Step 2: Setting (x+1=0) gives (x=-1). Step 3: A fraction is undefined when the denominator is zero.

Step 1: (A) has (2) elements and (B) gives (3) choices. Step 2: Each element has (3) possible images. Step 3: Total functions are (3^2=9).

Step 1: The number of functions is (n^m), where (m) is the number of domain elements and (n) is the number of codomain elements. Step 2: Here (m=2) and (n=4). Step 3: Therefore the number is (4^2=16).

Step 1: In a one-one function, the three inputs must have distinct images. Step 2: The first input has (3) choices, the second has (2), and the third has (1). Step 3: Total one-one functions are (3\cdot2\cdot1=6).

Step 1: The two inputs have different images, so the function is one-one. Step 2: Both elements (a) and (b) of (B) appear as images, so it is onto. Step 3: Being both one-one and onto, it is bijective.

Step 1: Both (1) and (2) map to (a), so the function is not one-one. Step 2: (c) is in the codomain but is not an image, so it is not onto. Step 3: Every input has one image, so it is a function, but neither one-one nor onto.

Step 1: Assume (f(a)=f(b)). Step 2: Then (3a+1=3b+1), so (3a=3b) and (a=b). Step 3: Hence different inputs have different images, so the function is one-one.

Step 1: In a one-one function, different inputs must have different images. Step 2: (2\ne -2), but (f(2)=0) and (f(-2)=0). Step 3: Therefore this function is not one-one.

Step 1: For onto, every real (y) must have a preimage. Step 2: From (x^3+1=y), we get (x=\sqrt[3]{y-1}). Step 3: This is real for every real (y), so the function is onto.

Step 1: (f(0)=|0-1|=1). Step 2: (f(2)=|2-1|=1). Step 3: Two different inputs have the same image, so the function is not one-one.

Step 1: An absolute value is always (0) or greater. Step 2: At (x=1), (|x-1|=0). Step 3: Therefore the range is ([0,\infty)).

Step 1: In an onto function, every element of the codomain becomes an image. Step 2: This means the range is equal to (B). Step 3: Therefore such a function is called onto.

Step 1: The main feature of a one-one function is that different inputs have different images. Step 2: This means two different elements do not map to the same image. Step 3: Therefore such a function is called one-one.

Step 1: (f(x)=x) is the identity function. Step 2: The identity function keeps every value unchanged. Step 3: Therefore its inverse is also (f^{-1}(x)=x).

The direct answer is A, \\(x\\). For \\(g\\circ f\\), the function on the right acts first, so \\( (g\\circ f)(x)=g(f(x)) \\). First calculate \\(f(x)=x+1\\). Now use this result as the input of \\(g\\). Since \\(g(t)=t-1\\), we obtain \\(g(x+1)=(x+1)-1=x\\). The addition of 1 and subtraction of 1 cancel exactly. Option A is correct because the output returns to the original input. Option B, \\(x+2\\), incorrectly adds the two changes. Option C, \\(x-2\\), incorrectly treats both operations as subtraction. Option D, \\(2x\\), assumes multiplication, although neither function multiplies by 2. Always substitute the complete expression, not just part of it. The useful exam cue is: composition is substitution, and the rightmost function is performed first.

Step 1: (f(g(x))=f(\frac{x}{2})=x). Step 2: (g(f(x))=g(2x)=x). Step 3: Both composites give the identity function, so they are inverses.

Step 1: A bijective function has a well-defined inverse. Step 2: If the original function maps from (A) to (B), the inverse reverses the direction. Step 3: Therefore (f^{-1}) maps from (B) to (A).

Step 1: Put (x=2) in the function. Step 2: (f(2)=6\cdot2+2=12+2=14). Step 3: In a linear function, multiply first and then add.

Step 1: Put (x=2). Step 2: (f(2)=2^2+3\cdot2=4+6=10). Step 3: Do the square and multiplication in the correct order.

Step 1: Put (x=-2). Step 2: (f(-2)=2(-2)^2-1=2\cdot4-1=7). Step 3: The square of a negative number is positive.

Step 1: In an ordered pair, the first component is the input and the second is the image. Step 2: Since ((2,6)) is given, the image of (2) is (6). Step 3: Do not reverse the order of components.