Functions

Step 1: The domain is formed by the first components of ordered pairs. Step 2: The first components are (4,5,6). Step 3: Therefore the domain is ({4,5,6}).

Step 1: The range is the set of actual images. Step 2: The second components are (1,1,2). Step 3: Repetition is not written in a set, so the range is ({1,2}).

Step 1: In a function, one input cannot have two different images. Step 2: In the first option, each of (1,2,3) has exactly one image. Step 3: In the other options, some first component has two different images.

Step 1: For being a function, each input must have only one image. Step 2: In the first option, (0) has two images (1) and (2). Step 3: Therefore the first option is not a function.

Step 1: Apply the function to every element of (A). Step 2: (f(0)=1), (f(1)=3), and (f(2)=5). Step 3: Hence (f(A)={1,3,5}).

Step 1: To find (f(a)), put (a) in place of (x). Step 2: (3x+4) becomes (3a+4). Step 3: In symbolic evaluation, replace only the input.

Step 1: In (f(a-1)), replace (x) by the whole expression (a-1). Step 2: (x^2+1) becomes ((a-1)^2+1). Step 3: Put the whole expression in brackets.

Step 1: Replace (x) by (x-2). Step 2: (f(x-2)=2(x-2)+5=2x-4+5). Step 3: Simplifying gives (2x+1).

Step 1: (f(x-1)=(x-1)^2) and (f(x)=x^2). Step 2: The difference is ((x-1)^2-x^2). Step 3: Simplifying gives (x^2-2x+1-x^2=-2x+1).

The governing concept is the inverse of a one-to-one linear function. Start with y = f(x) = x + 6. To isolate the original input, subtract 6 from both sides, giving x = y − 6. Now interchange the variable names in the inverse relation: f⁻¹(x) = x − 6. Therefore option A is correct. Option B is the original function, not its inverse, because it adds 6 again. Option C reverses the order and sign incorrectly, and option D multiplies by 6 without any basis in the given rule. The result can be checked directly: f(f⁻¹(x)) = (x − 6) + 6 = x and f⁻¹(f(x)) = (x + 6) − 6 = x, confirming that the two functions undo each other.

Step 1: Let (y=5x). Step 2: Then (x=\frac{y}{5}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{x}{5}).

Step 1: (f^{-1}(0)) means solving (f(x)=0). Step 2: From (3x+6=0), we get (3x=-6). Step 3: Hence (x=-2).

Step 1: (4x+9) gives different values for different (x). Step 2: For every real (y), we can take (x=\frac{y-9}{4}). Step 3: Hence the function is both one-one and onto.

Step 1: For every real (x), (x^2\ge 0). Step 2: Therefore (x^2+9\ge 9). Step 3: The minimum value is (9), so the range is ([9,\infty)).

Step 1: The value of (x^2+9) is always at least (9). Step 2: The codomain (R) contains (0), but no real (x) maps to it. Step 3: Therefore the function is not onto.

The governing concept is that the range is determined by actual outputs, not merely by the stated codomain. Since f(x) = 9 for every real x in the domain R, the function is constant. Substituting any input, such as 0, 4, or −10, gives the same output 9. Thus the set of all attainable outputs contains exactly one element: Range(f) = {9}. Hence option A is correct. Option B is the codomain and is not the range of this constant function. Option C would describe a function constantly equal to zero. Option D includes infinitely many values greater than or equal to 9, but none of those values except 9 is produced by f. The braces in {9} show that this is a singleton set.

Step 1: In a one-one function, different inputs must have different images. Step 2: Here (1\ne2), but (f(1)=9) and (f(2)=9). Step 3: Therefore the constant function is not one-one.

Step 1: The expression under the square root, (x-4), must be non-negative. Step 2: (x-4\ge 0) gives (x\ge 4). Step 3: Hence the real domain is ([4,\infty)).

Step 1: For the square root, (3+x\ge 0) is needed. Step 2: This gives (x\ge -3). Step 3: Therefore the domain is ([-3,\infty)).

Step 1: The denominator of the fraction is (x-5). Step 2: (x-5=0) gives (x=5). Step 3: A function with zero denominator is not defined there.