Functions

Step 1: ((f\circ g)(x)=x^2+2). Step 2: ((g\circ f)(x)=(x+2)^2). Step 3: Changing the order of composition generally changes the result.

Step 1: If ((g\circ f)(a_1)=(g\circ f)(a_2)), then (g(f(a_1))=g(f(a_2))). Step 2: Since (g) is one-one, (f(a_1)=f(a_2)), and since (f) is one-one, (a_1=a_2). Step 3: The composition of two one-one functions is one-one.

Step 1: Take any element of (C). Step 2: Since (g) is onto, it is the image of some element of (B), and since (f) is onto, that element of (B) comes from some element of (A). Step 3: Therefore (g\circ f) reaches every element of (C).

Step 1: Every element of (A) has exactly one image, so it is a function. Step 2: The images (2,3,1) are distinct and cover all of (A). Step 3: Since (1) does not map to (1), it is not the identity function.

Step 1: In a function, each element of the domain must have exactly one image. Step 2: In the first option, both (1) and (2) have exactly one image. Step 3: If an element has two images or is missing, it is not a function.

Step 1: The basic condition for a function is that each input has exactly one output. Step 2: If one input has two different outputs, the rule is not well-defined. Step 3: Two inputs may have the same output, but then the function is not one-one.

Step 1: The expression inside the square root must be non-negative. Step 2: (x-4\geq0) gives (x\geq4). Step 3: For square-root functions, set the radicand greater than or equal to (0).

Step 1: For the square root, (x-2\geq0) is needed. Step 2: Since the square root is in the denominator, (x-2=0) is not allowed. Step 3: Combining both conditions gives (x>2).

Step 1: The expression inside the square root must satisfy (9-x^2\geq0). Step 2: This gives (x^2\leq9), so (-3\leq x\leq3). Step 3: In square inequalities, remember both lower and upper bounds.

Step 1: (f(1)=1) and (f(-1)=1), so the function is not one-one. Step 2: Since (x^4\geq0), negative real numbers are not in the range. Step 3: For even powers, check both sign symmetry and range.

Step 1: (f(2)=16) and (f(-2)=16), so it is not one-one. Step 2: Every (y\geq0) is obtained using (x=\sqrt[4]{y}). Step 3: Changing the codomain can make the same function onto.

Step 1: Cubes of different integers are different, so the function is one-one. Step 2: Not every integer is a cube; for example, (2) is not the cube of any integer. Step 3: For power functions on integers, examine the range carefully.

Step 1: Adding (1) to different rational numbers gives different rational numbers. Step 2: For any rational (y), (x=y-1) is also rational, so every (y) is obtained. Step 3: Rational numbers are closed under addition and subtraction.

Step 1: (f(1)=1) and (f(-1)=1), so it is not one-one. Step 2: Negative rational numbers cannot be squares of rational numbers. Step 3: In square functions, positive and negative inputs can give the same value.

Step 1: For an inverse function, every output must correspond back to exactly one input. Step 2: One-one prevents multiple inputs, and onto ensures every codomain element is covered. Step 3: Hence being bijective is necessary and sufficient for an inverse function.

Step 1: First, (f) sends an element of (A) to (B). Step 2: Then (f^{-1}) brings it back to the original element. Step 3: Therefore (f^{-1}\circ f) is the identity function on (A).

Step 1: A binary operation on (A) is a function from (A\times A) to (A). Step 2: (A\times A) has (n^2) ordered pairs, and each pair has (n) choices. Step 3: Therefore the total number is (n^{n^2}).

Step 1: Each element of the domain has (3) choices in the codomain. Step 2: Since there are (3) elements, the count is (3^3=27). Step 3: In counting all functions, repeated images are allowed.

Step 1: In a constant function, all elements of (A) go to the same element of (B). Step 2: That single element can be any of the (5) elements of (B). Step 3: Hence the number of constant functions equals the number of elements in the codomain.

Step 1: For a finite set, a one-one function from (A) to (A) gives distinct images. Step 2: Since the domain and codomain have the same number of elements, all codomain elements are used. Step 3: On finite equal sets, one-one implies onto.