Functions

Step 1: In a function, every element of the domain must have exactly one image. Step 2: If one element has two different images, the relation is not a function. Step 3: A function need not be one-one or onto, but each input must have exactly one output.

Step 1: Every element of (A) has exactly one image, so it is a function. Step 2: The images (1,2,3) are all distinct and cover the whole set (A). Step 3: On a finite set, if all images are distinct and the sizes match, the function is bijective.

Step 1: The relation contains both ((1,2)) and ((1,3)). Step 2: This means the same input (1) has two different outputs (2) and (3), which violates the definition of a function. Step 3: In ordered pairs, check the first component to test whether the relation is a function.

Step 1: If (a\neq0), the slope of (ax+b) is non-zero, so it is one-one. Step 2: For any (y), (x=\frac{y-b}{a}) exists, so it is onto. Step 3: For a linear function, the key condition is (a\neq0); (b) may be any real number.

Step 1: For any (t\neq0), (t) and (-t) are distinct. Step 2: But (a(t)^2=a(-t)^2), so they have the same image. Step 3: For even-power functions, compare (x) and (-x).

Step 1: (f(0)=0^3-3\cdot0=0). Step 2: (f(\sqrt{3})=(\sqrt{3})^3-3\sqrt{3}=3\sqrt{3}-3\sqrt{3}=0). Step 3: To disprove one-one, it is enough to show two distinct inputs with the same image.

Step 1: Suppose (f(a_1)=f(a_2)). Step 2: Applying (g) to both sides gives (g(f(a_1))=g(f(a_2))), so (a_1=a_2). Step 3: If a left inverse exists, the original function is one-one.

Step 1: Take any (b\in B). Step 2: Since (f(g(b))=b), (b) is the image of the element (g(b)\in A). Step 3: If a right inverse exists, the original function is onto.

Step 1: \((f\circ f)(x)=f(f(x))\). Step 2: \(f\left(\frac{1}{x}\right)=\frac{1}{\frac{1}{x}}=x\). Step 3: If applying a function twice gives the original input, the function is its own inverse.

Step 1: (g(f(x))=g(2x+3)=\frac{2x+3-3}{2}=x). Step 2: (f(g(x))=2\cdot\frac{x-3}{2}+3=x). Step 3: If both compositions give the identity function, the two functions are inverses.

Step 1: (f(2)=4) and (f(-2)=4), so two different inputs have the same image. Step 2: In the inverse, (4) would point to both (2) and (-2), violating the function rule. Step 3: For an inverse function to exist, the original function must be one-one.

Step 1: (x^2+2x=(x+1)^2-1). Step 2: Since ((x+1)^2\ge0), the minimum value is (-1). Step 3: When writing the range, include the minimum or maximum value correctly.

Step 1: Put (x=0) to get (f(1)=f(0)+2=5). Step 2: Put (x=1) to get (f(2)=f(1)+2=7). Step 3: In functional equations, move step by step from the given value.

Step 1: For an odd function, (f(-x)=-f(x)). Step 2: Putting (x=4), we get (f(-4)=-f(4)=-9). Step 3: Remembering even and odd function definitions helps solve value-based questions quickly.

Step 1: For an even function, (f(-x)=f(x)). Step 2: Since (f(-6)=f(6)), we get (f(6)=11). Step 3: For even functions, changing the sign of the input does not change the value.

Step 1: (f(-x)=(-x)^4+(-x)^2=x^4+x^2). Step 2: Hence (f(-x)=f(x)), which is the condition for an even function. Step 3: Polynomials containing only even powers are often even functions.

Step 1: (f(-x)=(-x)^5-3(-x)=-x^5+3x). Step 2: This equals (-(x^5-3x)=-f(x)). Step 3: If (f(-x)=-f(x)), the function is odd.

Step 1: For (x=-5), (|-5|=5). Step 2: Therefore (f(-5)=-5+5=0). Step 3: In modulus questions, first evaluate (|x|) according to the sign of (x).

Step 1: If (x<0), then (x+|x|=x-x=0). Step 2: If (x\ge0), then (x+|x|=2x), which gives all values in ([0,\infty)). Step 3: For modulus functions, split the function into cases to find the range.

Step 1: (f(1.2)=1) and (f(1.8)=1), so it is not one-one. Step 2: Its values are always integers, so a real number like (0.5) is not obtained. Step 3: The greatest integer function maps many values in an interval to the same integer.

Step 1: For every integer (n), taking (x=n) gives (\lfloor x\rfloor=n), so it is onto (\mathbb{Z}). Step 2: But (\lfloor 2.1\rfloor=\lfloor 2.9\rfloor=2), so it is not one-one. Step 3: Changing the codomain to (\mathbb{Z}) makes this function onto.

Step 1: For (x\ge0), (f(x)=\frac{x}{1+x}), which increases from (0) and approaches (1) but never equals (1). Step 2: For (x<0), the value stays greater than (-1) and less than (0). Step 3: Approaching an endpoint and attaining it are different things.

Step 1: \(f(2)=\frac{2}{1+4}=\frac{2}{5}\). Step 2: \(f\left(\frac{1}{2}\right)=\frac{\frac{1}{2}}{1+\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{5}{4}}=\frac{2}{5}\). Step 3: Once two distinct inputs have the same image, the function is not one-one.

Step 1: Both sets have (3) elements. Step 2: A function that is one-one and onto must pair every element uniquely, so the number is (3!). Step 3: In exams, for two finite sets of the same size, bijections are counted as (n!).

Step 1: Put (y=2x-5). Step 2: Solving for (x), we get (x=\frac{y+5}{2}). Step 3: While finding inverse, first write (y=f(x)), then isolate (x).