Functions

Step 1: A bijection from (A) to (A) is a permutation of the elements. Step 2: The number of permutations of (n) elements is (n!). Step 3: For bijections on the same finite set, use (n!).

Step 1: Total functions are (3^3=27). Step 2: One-one functions are (3!=6). Step 3: Not one-one functions are (27-6=21).

Step 1: To be not onto, one element of (B) must be completely missed. Step 2: There are two possibilities: all elements map to (a), or all map to (b). Step 3: Therefore the number of not onto functions is (2).

Step 1: Put \(y=\frac{x}{1+x^2}\). Step 2: Treat \(yx^2-x+y=0\) as a quadratic in (x) and require its discriminant to be non-negative. Step 3: From \(1-4y^2\geq0\), we get \(-\frac{1}{2}\leq y\leq\frac{1}{2}\).

For \(f(x)=e^x\), the output is positive for every real \(x\), and the function is strictly increasing. Hence, distinct inputs give distinct outputs, so it is one-one. However, it never produces 0 or a negative value; its range is \((0,\infty)\). Therefore, it is not onto \(\mathbb{R}\). The function \(x^3\) is both one-one and onto, whereas \(x^2\) is not one-one. Exam tip: To test whether a function is onto, check whether every element of the codomain occurs in its range.

Step 1: (x^3-6x^2+12x+1=(x-2)^3+9). Step 2: The cubic ((x-2)^3) is strictly increasing on (\mathbb{R}) and takes all real values. Step 3: Shifting does not change bijectivity.

Step 1: For (x<1), (2x+1<3), giving the range part ((-\infty,3)). Step 2: For (x\geq1), (x^2+2\geq3), and (3) occurs at (x=1). Step 3: Combining both parts gives all of (\mathbb{R}).

Step 1: (f^{-1}(5)) means the input (x) for which (f(x)=5). Step 2: From (\frac{2x+3}{x-1}=5), we get (2x+3=5x-5), so (3x=8). Step 3: Hence (x=\frac{8}{3}); for inverse values, set the original function equal to the target value and solve.

Step 1: Each element of (A) must choose exactly one image in (B). Step 2: Since (A) has (3) elements and (B) has (2) elements, the number of functions is (2^3=8). Step 3: Remember that the number of functions from (A) to (B) is (|B|^{|A|}).

Step 1: In a one-one function, distinct elements must have distinct images. Step 2: Here (A) has (4) elements but (B) has only (3) elements, so (4) distinct images are impossible. Step 3: First compare the sizes of domain and codomain in such questions.

Step 1: (f(1)=2) and (f(-1)=2), so the function is not one-one. Step 2: Since (x^2+1\ge 1), negative numbers and (0) are not attained, so it is not onto. Step 3: Always test one-one and onto separately.

Step 1: Let (y=3x-5). Step 2: Solving for (x), we get (x=\frac{y+5}{3}). Step 3: Replace (y) by (x), so (f^{-1}(x)=\frac{x+5}{3}).

Step 1: The function (x^3) is strictly increasing on real numbers, so distinct inputs give distinct outputs. Step 2: For every real (y), (x=\sqrt[3]{y}) exists, so every (y) has a preimage. Step 3: The cube function from (\mathbb{R}) to (\mathbb{R}) is bijective.

Step 1: (f(2)=4) and (f(-2)=4), so the function is not one-one. Step 2: The codomain is ([0,\infty)), and for every (y\ge 0), (x=\sqrt{y}) exists. Step 3: Onto property depends on the codomain, so always check it carefully.

Step 1: (f(3)=3) and (f(-3)=3), so it is not one-one. Step 2: (|x|) is never negative, so negative real numbers are not attained. Step 3: In absolute value functions, opposite inputs may give the same output.

Step 1: A bijective function is both one-one and onto. Step 2: For finite sets, a bijection is possible only when both sets have the same number of elements. Step 3: Therefore (|B|=|A|=7).

Step 1: ((g\circ f)(x)) means (g(f(x))). Step 2: Put (f(x)=2x+1) into (g), giving (g(2x+1)=(2x+1)^2). Step 3: Expanding gives (4x^2+4x+1).

Step 1: (f(g(x))=f(3x-1)=3x+1). Step 2: (g(f(x))=g(x+2)=3x+5). Step 3: The difference is (3x+1-(3x+5)=-4), so order matters in composition.

Step 1: Rewrite (x^2-4x+7) as ((x-2)^2+3). Step 2: Since ((x-2)^2\ge 0), the minimum value is (3). Step 3: Hence the range is ([3,\infty)).

Step 1: (5-2x) is a linear function with non-zero slope, so it is one-one. Step 2: For any real (y), (x=\frac{5-y}{2}) is real, so it is onto. Step 3: A non-constant linear function from (\mathbb{R}) to (\mathbb{R}) is bijective.