Functions

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).

Step 1: For finite sets with equal size, a bijection is like a permutation. Step 2: There are (3!) ways to assign (3) distinct images. Step 3: Since (3!=6), there are (6) bijective functions.

Step 1: (\sin 0=0) and (\sin \pi=0), so it is not one-one. Step 2: The range of (\sin x) is ([-1,1]), so it is not onto (\mathbb{R}). Step 3: For trigonometric functions, check periodicity and range.

Step 1: On the given interval, (\sin x) is strictly increasing, so it is one-one. Step 2: Its range on this interval is exactly ([-1,1]). Step 3: Restricting the domain can make the same function bijective.

Step 1: (\cos 0=1) and (\cos 2\pi=1), so the function is not one-one. Step 2: The range of (\cos x) is ([-1,1]), so it is not onto (\mathbb{R}). Step 3: If codomain is (\mathbb{R}), a bounded range function is not onto.

Step 1: On \([0,\pi]\), (\cos x) is strictly decreasing, so it is one-one. Step 2: On this interval, it takes all values from (1) to (-1). Step 3: Hence it is also onto the codomain ([-1,1]).

Step 1: (f^{-1}(k)=4) means (f(4)=k). Step 2: (f(4)=2\cdot4+3=11). Step 3: In inverse function questions, translate the statement into the original function.

Step 1: First find (g(2)=2+1=3). Step 2: Now (f(g(2))=f(3)=3^2-1=8). Step 3: In composition, always evaluate the inner function first.