Functions

Step 1: In an onto function, every element of the codomain is an image. Step 2: Therefore the set of actual images equals the whole (B). Step 3: To identify onto, check whether any codomain element is left out.

Step 1: In a one-one function, two different inputs cannot have the same image. Step 2: Therefore different inputs have different images. Step 3: If two different inputs give the same image, the function is not one-one.

Step 1: A bijective function means two properties hold together. Step 2: It is both one-one and onto. Step 3: Such a function generally has a well-defined inverse.

Step 1: (A) has (3) elements and (B) gives (2) choices. Step 2: Each element of (A) has (2) possible images. Step 3: Total functions are (2^3=8).

Step 1: Each element of domain (A) must choose one image in (B). Step 2: Each element has (n) choices, and there are (m) elements. Step 3: Therefore the number of functions is (n^m).

Step 1: (f^{-1}(5)) means the (x) for which (f(x)=5). Step 2: From (x+2=5), we get (x=3). Step 3: Therefore (f^{-1}(5)=3).

Step 1: For (x^3), different real inputs give different outputs, so it is one-one. Step 2: For every real (y), (x=\sqrt[3]{y}) is real. Step 3: Hence every (y) has a preimage, so the function is onto as well.

Step 1: For one-one, different inputs must have different images. Step 2: (2\ne -2), but (|2|=|-2|=2). Step 3: Therefore this function is not one-one.

Step 1: To be onto, every element of (R) must be an image. Step 2: (-2\in R), but (|x|) is never negative. Step 3: Hence the range is not (R), and the function is not onto.

Step 1: The denominator of a fraction cannot be zero. Step 2: At (x=0), (\frac{1}{0}) is undefined. Step 3: So the domain is all real numbers except (0).

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

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

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

Step 1: For (f(a)), replace (x) by (a). Step 2: Replacing (x) by (a) in (2x-3) gives (2a-3). Step 3: In symbolic evaluation, change only the given variable.

Step 1: For (f(a+h)), replace (x) by the whole expression (a+h). Step 2: Putting (x=a+h) in (x^2) gives ((a+h)^2). Step 3: Keep the whole expression in brackets.

Step 1: For (f(x+2)), replace the original (x) by (x+2). Step 2: (f(x+2)=(x+2)+1=x+3). Step 3: Treat the entire inside expression as the input.

Step 1: ((f+g)(x)=f(x)+g(x)). Step 2: (2x+(x-1)=3x-1). Step 3: In the sum of functions, add their values.

Step 1: ((fg)(x)=f(x)g(x)). Step 2: (f(2)=4) and (g(2)=2). Step 3: Therefore ((fg)(2)=4\cdot2=8).

Step 1: This is a constant function, so its value is (5) for every input. Step 2: (f(3)=5) and (f(1)=5). Step 3: Hence (f(3)-f(1)=5-5=0).

Step 1: (f:A\to B) means the function maps from (A) into (B). Step 2: The image of any element of (A) lies in (B). Step 3: Therefore (f(a)\in B).