Functions

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.

Step 1: In a constant function, every input has the same image (c). Step 2: Different inputs give the same output, so the one-one condition fails. Step 3: A constant function on an infinite domain is never one-one.

Step 1: (e^x) is strictly increasing, so it is one-one. Step 2: Its values are positive, and for every (y>0), (x=\ln y) exists. Step 3: With codomain ((0,\infty)), the function is bijective.

Step 1: (\ln x) is strictly increasing on ((0,\infty)), so it is one-one. Step 2: For every real (y), (x=e^y>0), so it is onto. Step 3: The logarithmic function is the inverse of the exponential function.

Step 1: Let (y=\frac{x-1}{x+1}). Step 2: From (y(x+1)=x-1), we get (x(y-1)=-(1+y)). Step 3: Hence (x=\frac{1+y}{1-y}), so (f^{-1}(x)=\frac{1+x}{1-x}).

Step 1: (x^3+x) is strictly increasing because its value keeps increasing with (x). Step 2: As (x\to\infty), the value goes to (\infty), and as (x\to-\infty), it goes to (-\infty). Step 3: Hence every real value is attained and the function is bijective.

Step 1: Since (f) is one-one, distinct elements of (A) have distinct images in (B). Step 2: Since (g) is also one-one, those distinct images remain distinct in (C). Step 3: The composition of one-one functions is one-one.

Step 1: (f(1)=1) and (f(-1)=1), so it is not one-one. Step 2: Since (x^4\ge 0), negative real values are not attained. Step 3: For even power functions, check both repeated outputs and restricted range.

Step 1: The domain is ([0,\infty)), so (x) cannot be negative. Step 2: From (y=x^2), we get (x=\sqrt{y}). Step 3: (-\sqrt{y}) is not allowed because the domain has no negative numbers.

Step 1: \(f(x)=\frac{1}{x}\). Step 2: \(f(f(x))=f\left(\frac{1}{x}\right)=\frac{1}{\frac{1}{x}}=x\). Step 3: Such a function behaves like its own inverse.

Step 1: If (a=0), then (f(x)=b) becomes constant. Step 2: A constant function is neither one-one nor onto on (\mathbb{R}). Step 3: Therefore a linear function is bijective only when (a\ne 0).

Step 1: (x^2+2x+2=(x+1)^2+1). Step 2: Since ((x+1)^2\ge 0), the smallest value is (1). Step 3: Completing the square quickly gives range and minimum value.

Step 1: (f(1)=2) and (f(-1)=2), so it is not one-one. Step 2: Since (x^2+1\ge 1), and for every (y\ge 1), (x=\sqrt{y-1}) works. Step 3: Reading the codomain correctly is the key.

Step 1: The image of (1) can be chosen in (3) ways. Step 2: The image of (2) must be different, so (2) choices remain. Step 3: Total one-one functions are (3\times2=6).