One-one function

Step 1: To disprove injectivity it is enough to show two different inputs with the same output. Step 2: (f(0)=0) and (f(\sqrt{3})=3\sqrt{3}-3\sqrt{3}=0), while (0\neq\sqrt{3}). Step 3: Do not decide by degree alone; test special values.

Step 1: Both (x^3) and (x) contribute increasing behavior on real numbers. Step 2: If (x_1<x_2), then (x_1^3<x_2^3), so (x_1^3+x_1<x_2^3+x_2). Step 3: A strictly increasing function is one-one.

Step 1: A linear function (ax+b) is one-one when (a\neq0). Step 2: Here the coefficient of (x) is (k+1). Step 3: Hence (k+1\neq0), so (k\neq-1).

Step 1: Look for different inputs giving the same value. Step 2: \(f(2)=2+\frac{1}{2}=\frac{5}{2}\) and \(f\left(\frac{1}{2}\right)=\frac{1}{2}+2=\frac{5}{2}\). Step 3: Same value at different inputs makes the function not one-one.

If \(f(x_1)=f(x_2)\), then applying \(g\) gives \((g\circ f)(x_1)=(g\circ f)(x_2)\). Since \(g\circ f\) is one-one, we must have \(x_1=x_2\). Hence, \(f\) must be one-one. However, \(g\) need not be one-one because it may take equal values at elements outside the image of \(f\). Exam tip: injectivity of a composite function implies injectivity of the first function.

Step 1: A one-one function needs five distinct images from (B) for the five elements of (A). Step 2: The number is (^{7}P_{5}=7\cdot6\cdot5\cdot4\cdot3=2520). Step 3: Use permutations for counting one-one functions.

Step 1: A one-one function needs a distinct codomain element for each distinct domain element. Step 2: Here the domain has (6) elements but the codomain has only (4). Step 3: When the domain has more elements, a one-one function is impossible.

Step 1: Suppose (g(x_1)=g(x_2)). Step 2: Then (f(g(x_1))=f(g(x_2))), so ((f\circ g)(x_1)=(f\circ g)(x_2)). Step 3: Since (f\circ g) is one-one, (x_1=x_2), hence (g) is one-one.

Step 1: (x^2) gives the same value for (x) and (-x). Step 2: Hence (f(1)=\ln2) and (f(-1)=\ln2), while (1\neq-1). Step 3: When an even power appears inside, check injectivity carefully.

Step 1: In the given domain (x\ge0). Step 2: As (x) increases, (x^2+1) increases and (\ln(x^2+1)) also increases. Step 3: A suitable domain restriction can make the same rule one-one.

Step 1: In (e^x+e^{-x}), replacing (x) by (-x) gives the same value. Step 2: (f(1)=e+e^{-1}) and (f(-1)=e^{-1}+e), so they are equal. Step 3: Functions with even behavior are not one-one on the whole real domain.

Step 1: (e^x) increases and (e^{-x}) decreases. Step 2: Therefore (e^x-e^{-x}) keeps increasing as (x) increases. Step 3: A strictly increasing function is one-one.

Step 1: The greatest integer function maps many values from an interval to the same integer. Step 2: (\lfloor1.2\rfloor=1) and (\lfloor1.9\rfloor=1), while (1.2\neq1.9). Step 3: Step functions commonly repeat outputs.

Step 1: The cube function preserves order on integers. Step 2: If (n_1^3=n_2^3), then (n_1=n_2). Step 3: This simple odd-power rule remains one-one on integers.

Step 1: Absolute value gives the same value for opposite integers. Step 2: (2\neq-2), but (|2|=2) and (|-2|=2). Step 3: One repeated image is enough to break injectivity.

Step 1: Each element of the domain has exactly one image. Step 2: The images (a,c,b,d) are all distinct. Step 3: For finite sets a function is one-one when second components do not repeat.

Step 1: In a one-one function different inputs must have different images. Step 2: In the fourth option both (p) and (r) have image (3). Step 3: In such questions first look for repeated images.

Step 1: In the inverse relation each image is linked back to its original input. Step 2: If (f) is one-one, each image has only one original input. Step 3: Injectivity is the key condition for the inverse to be a function.

Step 1: Injectivity means the same (y)-value should not come from two different (x)-values. Step 2: A horizontal line represents a fixed (y)-value. Step 3: If any horizontal line cuts the graph twice, the function is not one-one.

Step 1: For a rational function check repeated values carefully. Step 2: \(f(2)=\frac{2}{5}\) and \(f\left(\frac{1}{2}\right)=\frac{2}{5}\), while \(2\neq\frac{1}{2}\). Step 3: Same value at two different inputs destroys injectivity.

Step 1: Assume equal images and solve. Step 2: From (\frac{x_1-1}{x_1+1}=\frac{x_2-1}{x_2+1}), simplification gives (x_1=x_2). Step 3: In linear fractional functions remember the excluded domain value while cross multiplying.

Step 1: The rule contains only (x^2), so (x) and (-x) can give the same value. Step 2: (f(1)=0) and (f(-1)=0), while (1\neq-1). Step 3: In rational expressions with even powers, test opposite inputs quickly.

Step 1: In a piecewise function check the ranges of different pieces. Step 2: \(f\left(-\frac{1}{2}\right)=\frac{1}{2}\) and \(f\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2}\), while the inputs are different. Step 3: Same value from different pieces breaks injectivity.

Step 1: For (x<0), (f(x)<-1). Step 2: For (x\ge0), (f(x)\ge1), so the ranges of the two pieces are disjoint. Step 3: Each piece is one-one and their ranges do not overlap, so the whole function is one-one.

Step 1: On \([0,\pi]\), (\sin x) first increases and then decreases. Step 2: \(\sin\frac{\pi}{6}=\frac{1}{2}\) and \(\sin\frac{5\pi}{6}=\frac{1}{2}\), while the angles are different. Step 3: For trigonometric functions check the interval carefully.