One-one function

Step 1: The function can be written as (f(x)=(x+2)^3+1). Step 2: A cubic form is strictly increasing on the whole real domain, so two different (x)-values cannot have the same image. Step 3: In such questions, first try to identify a perfect cubic form.

Step 1: To show that a function is not one-one, it is enough to find two different inputs with the same image. Step 2: (f(-4)=-64+48+5=-11) and (f(2)=8-24+5=-11), while (-4\ne2). Step 3: Even a cubic function can repeat values if it has turning points.

Step 1: (f(x)=x^2-4x+6=(x-2)^2+2). Step 2: The domain ([2,\infty)) lies to the right of the vertex, where the function is strictly increasing. Step 3: A quadratic becomes one-one when restricted to one side of its vertex.

Step 1: (f(x)=(x-4)^2-3), so the vertex is at (x=4). Step 2: The interval ([2,6]) contains points (2) and (6), equally distant from the vertex, giving equal values. Step 3: If an interval extends on both sides of the vertex, a quadratic is not one-one there.

Step 1: In a one-one function, the (6) elements of the domain must be sent to (6) distinct elements of the (8)-element codomain. Step 2: The total number is (8\cdot7\cdot6\cdot5\cdot4\cdot3=,^{8}P_6). Step 3: For one-one functions, select ordered distinct images from the codomain.

Step 1: The image of (1) is already fixed as (5). Step 2: The remaining (3) elements must be mapped to distinct elements among the remaining (4) elements of (B), giving (4\cdot3\cdot2=24). Step 3: In fixed-image questions, remove the fixed pair first.

Step 1: Assume (f(a)=f(b)). Step 2: From (\frac{3a-2}{a+5}=\frac{3b-2}{b+5}), cross-multiplication gives (17a=17b), hence (a=b). Step 3: Cross-multiplication is very useful for fractional linear functions.

Step 1: For (x\ne2), (\frac{x^2-4}{x-2}=x+2). Step 2: (x+2) gives distinct values for distinct (x)-values, so it is one-one. Step 3: Remember the domain condition before simplifying.

Step 1: The function contains only even powers of (x), so values at (x) and (-x) can be equal. Step 2: (f(1)=1-2=-1) and (f(-1)=1-2=-1), while (1\ne-1). Step 3: An even function on a symmetric domain is generally not one-one.

Step 1: (f'(x)=1-\frac{1}{x^2}). Step 2: On ((0,1]), (f'(x)\le0), so the function is strictly decreasing and one-one. Step 3: For such functions, choose one side of the minimum point.

Step 1: Both (|x|) and (x^2) give the same value at (x) and (-x). Step 2: (f(1)=1+1=2) and (f(-1)=1+1=2), while (1\ne-1). Step 3: When an expression is symmetric, test (x) and (-x).

Step 1: When (x<0), (|x|=-x). Step 2: Hence (f(x)=x-x=0), so many negative numbers have the same image. Step 3: If a function is constant on an entire part of its domain, it is not one-one.

Step 1: When (x\ge0), (|x|=x). Step 2: So (f(x)=x-x=0), meaning all non-negative numbers have the same value. Step 3: When many inputs have the same image, one-one nature fails.

Step 1: In the domain ([-3,\infty)), (x+3\ge0). Step 2: As (x) increases, both (x+3) and (\sqrt{x+3}) increase, so the image does not repeat. Step 3: A square-root form is usually increasing on its natural domain.

Step 1: In the domain, (x\le5), so (5-x\ge0). Step 2: As (x) increases, (5-x) decreases and (\sqrt{5-x}) also decreases, so the function is one-one. Step 3: A strictly decreasing function is also one-one.

Step 1: (\sin x+\cos x) can be written as (\sqrt{2}\sin(x+\frac{\pi}{4})). Step 2: It is periodic, so the same value repeats at many different (x)-values. Step 3: A periodic function is generally not one-one on the whole real domain.

Step 1: (f'(x)=\cos x-\sin x). Step 2: On ([0,\frac{\pi}{4}]), (f'(x)\ge0), so the function is increasing and one-one. Step 3: When the derivative changes sign, choose a monotonic part.

Step 1: (\arctan x) is defined on the whole real domain. Step 2: As (x) increases, (\arctan x) also increases, so no value repeats at two different (x)-values. Step 3: Remember the standard one-one behaviour of inverse trigonometric functions.

Step 1: The domain of (\arccos x) is ([-1,1]). Step 2: On this domain, as (x) increases, (\arccos x) decreases, so two different inputs do not have the same value. Step 3: A decreasing inverse trigonometric function is also one-one.

Step 1: (f(a)=f(b)\Rightarrow2a-1=2b-1\Rightarrow a=b), so the function is one-one. Step 2: Even natural numbers in the codomain are not obtained as images, so it is not onto. Step 3: Test one-one and onto separately.

Step 1: To prove not one-one, find two different integers with the same image. Step 2: (f(0)=0) and (f(1)=1-1=0), while (0\ne1). Step 3: On an integer domain, checking small values is a quick method.

Step 1: Put (f(a)=f(b)), giving (2a+3=2b+3). Step 2: This gives (2a=2b), hence (a=b). Step 3: (f(a)=f(b)\Rightarrow a=b) is a direct proof of one-one nature.

Step 1: Suppose (f(a)=f(b)). Step 2: Then (g(f(a))=g(f(b))), meaning ((g\circ f)(a)=(g\circ f)(b)). Since (g\circ f) is one-one, (a=b). Step 3: In composition, this is how one concludes that the first function is one-one.

Step 1: Even if (g) is not one-one on all of (B), it may give distinct values on the smaller set (f(A)). Step 2: Therefore, (g\circ f) can sometimes be one-one. Step 3: In composition, check the behaviour of (g) on (f(A)), not only on all of (B).

Step 1: (2x+1) is a strictly increasing linear expression. Step 2: (e^u) is also strictly increasing in (u), so (e^{2x+1}) is increasing. Step 3: A composition of increasing one-one functions can preserve one-one nature.