Functions

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.

Step 1: Onto means every codomain element is used. Step 2: For finite equal sets, if two domain elements had the same image, some codomain element would be left unused. Step 3: Therefore an onto function from (A) to (A) is also one-one.

Step 1: ((f\circ f)(x)=f(f(x))). Step 2: (f(f(x))=\frac{1}{\frac{1}{x+1}+1}=\frac{x+1}{x+2}). Step 3: In fractional compositions, simplify the denominator carefully.

Step 1: Write (y=\sqrt{x-2}). Step 2: Squaring both sides gives (y^2=x-2), so (x=y^2+2). Step 3: When writing the inverse, the domain and range are interchanged.

Step 1: (f(0)=0), (f(\sqrt{3})=0), and (f(-\sqrt{3})=0), so the function is not one-one. Step 2: A cubic polynomial goes from very negative to very positive values, so every real value is attained. Step 3: A cubic function need not always be one-one.

Step 1: (A) has (5) elements and (B) has (3), so it is impossible to assign distinct images to all (5) elements. Step 2: But it is possible to cover all (3) elements of (B) using (5) elements of (A). Step 3: For finite sets, first compare cardinalities.

Step 1: (|x-2|\geq0) always. Step 2: Therefore (|x-2|+3\geq3), and at (x=2), the value is (3). Step 3: For modulus functions, the minimum is found by making the inside expression zero.

Step 1: If (x\neq0), then (\frac{1}{x}\neq0), so the function is well-defined. Step 2: (f(f(x))=\frac{1}{\frac{1}{x}}=x). Step 3: When (f\circ f) is the identity, the function is its own inverse.

Step 1: The function (x^5+x^3+x) is strictly increasing because its derivative (5x^4+3x^2+1) is always positive. Step 2: For very large negative (x), the value becomes very negative, and for very large positive (x), the value becomes very positive, so every real value is attained. Step 3: For increasing odd-power polynomials, check both one-one and onto carefully.