Functions

Step 1: In a function, every element of the domain must be used. Step 2: Each element must have exactly one image, not two different images. Step 3: In exams, check the image of every domain element carefully.

Step 1: Put (x=4) in the given function. Step 2: (f(4)=2\cdot4+3=8+3=11). Step 3: In such questions, substitute first and then simplify.

Step 1: In the notation (f:A\to B), the first set is (A). Step 2: Inputs of the function are taken from (A), so (A) is called the domain. Step 3: Remember that (B) is the codomain, not the domain.

Step 1: In (f:A\to B), the second set is (B). Step 2: The images of the function lie in (B), so (B) is called the codomain. Step 3: The range is always a subset of the codomain, so do not confuse them.

Step 1: In ordered pairs, the first component belongs to the domain. Step 2: The first components here are (1,2,3). Step 3: Therefore the domain is ({1,2,3}).

Step 1: The range is the set of actual images obtained from the function. Step 2: The second components of the ordered pairs are (2,4,6). Step 3: Hence the range is ({2,4,6}).

Step 1: Each element (1) and (2) of (A) must have exactly one image. Step 2: In the first option, (1) maps to (3) and (2) maps to (4). Step 3: In the other options, an element is missing or has two images.

Step 1: Square each given value of (x). Step 2: The values obtained are (4,1,0,1,4). Step 3: Write repeated values only once, so the range is ({0,1,4}).

Step 1: Put (x=0) in the function. Step 2: (f(0)=0+5=5). Step 3: In a simple linear function, pay attention to the sign and constant term.

The governing concept is evaluation of a function at specified inputs. Substitute x = 2 into f(x) = 3x − 1: f(2) = 3(2) − 1 = 6 − 1 = 5. Next substitute x = 0: f(0) = 3(0) − 1 = −1. Add the two function values: f(2) + f(0) = 5 + (−1) = 4. Hence option A is correct. Option B gives only f(2) and ignores f(0); option C can result from forgetting the subtraction of 1, and option D does not follow from either substitution. The crucial point is that each input must be placed into the entire rule before the results are added.

Step 1: Put (-2) in place of (x). Step 2: (f(-2)=(-2)^2+1=4+1=5). Step 3: Remember that the square of a negative number is positive.

Step 1: Different values of (x) give different values of (2x), so the function is one-one. Step 2: For any real (y), choosing (x=\frac{y}{2}) gives (f(x)=y). Step 3: Hence it is also onto and therefore bijective.

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

Step 1: For onto, every element of the codomain must be an image. Step 2: (-1) is in (R), but (x^2=-1) has no real solution. Step 3: Therefore the range is not all of (R), so the function is not onto.

Step 1: The value of the function is (7) for every (x). Step 2: A function that gives the same image for all inputs is called a constant function. Step 3: Do not confuse a constant function with the identity function.

Step 1: In (f(x)=x), every input maps to itself. Step 2: Such a function is called the identity function. Step 3: In the identity function, no value is changed.

Step 1: ((f\circ g)(3)=f(g(3))). Step 2: (g(3)=2\cdot3=6). Step 3: (f(6)=6+1=7), so the value is (7).

Step 1: ((g\circ f)(2)=g(f(2))). Step 2: (f(2)=2^2=4). Step 3: (g(4)=4+2=6), so the correct value is (6).

Step 1: Write (y=3x+2). Step 2: Solving for (x), we get (x=\frac{y-2}{3}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{x-2}{3}).

The governing concept is finding an inverse by interchanging the input and output and solving for the original input. Write y = x − 4. To express x in terms of y, add 4 to both sides: x = y + 4. Now interchange the variable names, replacing y by x, to obtain f⁻¹(x) = x + 4. Therefore option A is correct. The inverse must undo the subtraction of 4, so it adds 4. Option B is the original function rather than its inverse; option C changes the sign of the variable unnecessarily, and option D represents multiplication by 4, not reversal of the given operation. Composition also confirms the result: f(f⁻¹(x)) = (x + 4) − 4 = x.