One-one function

Step 1: If (a>b), then (a^3>b^3) and (3a>3b). Step 2: Hence (a^3+3a+5>b^3+3b+5), so different inputs give different images. Step 3: Recognising an increasing function is very useful in hard one-one questions.

Step 1: To disprove one-one nature, find different inputs with the same image. Step 2: (f(0)=0) and (f(\sqrt{3})=0), while (0\neq \sqrt{3}). Step 3: Once a polynomial gives the same image for different inputs, it is not one-one.

Step 1: We can write (f(x)=(x-1)^2+3). Step 2: On ([1,\infty)), ((x-1)^2) increases, so equal images occur only for equal inputs. Step 3: For a quadratic function, carefully check whether the domain is restricted to one side.

Step 1: (f(x)=(x-1)^2+3). Step 2: On the domain (x\leq 1), the left side of the parabola moves in one direction. Step 3: Taking only one side of a parabola can make the function one-one.

Step 1: (f(x)=(x-1)^2+3). Step 2: (f(0)=4) and (f(2)=4), while (0\neq 2). Step 3: On the whole real domain, both sides of a quadratic may give the same image.

Step 1: (e^x) is increasing and (x) is also increasing. Step 2: The sum of two increasing parts is increasing, so two different inputs cannot have the same image. Step 3: In such questions, quickly identify the increasing nature of the function.

Step 1: (f'(x)=e^x-1), which is negative for (x<0) and positive for (x>0). Step 2: The function decreases first and then increases, so it is not one-one on all of (\mathbb{R}). Step 3: If a function changes direction, equal images may occur.

Step 1: (\ln x) is increasing on its domain ((0,\infty)). Step 2: If (\ln a=\ln b), then (a=b). Step 3: Remember the increasing nature of the logarithmic function.

Step 1: Look for two different positive values with the same image. Step 2: \(f(2)=\frac{5}{2}\) and \(f\left(\frac{1}{2}\right)=\frac{5}{2}\), while \(2\neq \frac{1}{2}\). Step 3: In functions like \(x+\frac{1}{x}\), notice the roles of (x) and \(\frac{1}{x}\).

Step 1: For (x\geq 1), as (x) increases, (x+\frac{1}{x}) also increases. Step 2: In this domain, two different values do not form the same image through (x) and (\frac{1}{x}). Step 3: The same formula can become one-one on a restricted domain.

Step 1: Assume (f(a)=f(b)). Step 2: From (\frac{2a+3}{a-1}=\frac{2b+3}{b-1}), expanding gives (2ab-2a+3b-3=2ab-2b+3a-3), hence (5b=5a). Step 3: Since (a=b), the function is one-one.

Step 1: For (x\neq 2), (\frac{x^2-4}{x-2}=x+2). Step 2: In (x+2), the coefficient of (x) is non-zero, so equal images give equal inputs. Step 3: While simplifying, keep the removed point outside the domain.

Step 1: The function contains only even powers. Step 2: (f(1)=2) and (f(-1)=2), while (1\neq -1). Step 3: An even function is generally not one-one on all of (\mathbb{R}).

Step 1: For (x\geq 0), both (x^4) and (x^2) increase. Step 2: Their sum also increases, so different non-negative inputs give different images. Step 3: A function with even powers can be one-one on half of the real domain.

Step 1: If (a>b), then (a^5>b^5) and (a>b). Step 2: Therefore (a^5+a>b^5+b), so the function is increasing. Step 3: An increasing function is one-one on its whole domain.

Step 1: Check different inputs giving the same image. Step 2: (f(-1)=0), (f(0)=0), and (f(1)=0). Step 3: Since three different inputs have the same image, the function is not one-one.

Step 1: (\arctan x) is increasing on the whole real domain. Step 2: If (\arctan a=\arctan b), taking (\tan) on both sides gives (a=b). Step 3: Remember the principal domain of inverse trigonometric functions.

Step 1: (\arcsin x) is increasing on its domain ([-1,1]). Step 2: Equal (\arcsin) values imply equal inputs. Step 3: To make an inverse function, the original function is restricted to a one-one domain.

Step 1: On ([0,\pi]), (\cos x) is strictly decreasing. Step 2: A decreasing function is also one-one because two different inputs do not give the same image. Step 3: For one-one nature, increasing is not necessary; strictly decreasing is also enough.

Step 1: (\cos x) is an even function. Step 2: \(\cos \frac{\pi}{3}=\frac{1}{2}\) and \(\cos \left(-\frac{\pi}{3}\right)=\frac{1}{2}\), while the inputs are different. Step 3: Always check symmetric values for an even trigonometric function.

Step 1: On the given interval, (\sin x) is increasing. Step 2: In this interval, the same (\sin) value comes only from the same angle. Step 3: Restricting a trigonometric function to a proper interval gives one-one nature.

Step 1: To check one-one nature, look for different angles with the same image. Step 2: (\sin 0=0) and (\sin \pi=0), while (0\neq \pi). Step 3: On a large trigonometric interval, periodic behaviour can break one-one nature.

Step 1: In a piecewise function, the images from different pieces must be compared. Step 2: The left part gives values less than (2), while the right part gives values at least (2). Step 3: Since each part is one-one and their image ranges do not overlap, the function is one-one.

Step 1: Check if different pieces can give the same image. Step 2: (f(-2)=4) and (f(4)=4), while (-2\neq 4). Step 3: In a piecewise function, it is important to check overlap of ranges from different pieces.

Step 1: For (x<1), the first part gives values less than (2). Step 2: For (x\geq1), the second part gives values at least (2). Step 3: Both parts are increasing and their ranges do not overlap, so the function is one-one.