One-one function

Step 1: For (x<0), the range of (x+3) is less than (3). Step 2: For (x\geq0), the range of (x^2+3) is at least (3). Step 3: The two ranges do not overlap and both pieces are one-one on their domains, so the whole function is one-one.

Step 1: In a one-one function, the four elements of (A) must get distinct images from the five elements of (B). Step 2: The number of ways is ({}^5P_4=5\times4\times3\times2=120). Step 3: In one-one counting, order matters because each input is different.

Step 1: In a one-one function, every distinct element of (A) needs a distinct image in (B). Step 2: (A) has (5) elements but (B) has only (3) elements. Step 3: When the domain has more elements than the codomain, a one-one function is impossible.

Step 1: The first element of (A) has (7) choices. Step 2: The second has (6) choices and the third has (5) choices left. Step 3: The total number is (7\times6\times5=210).

Step 1: For finite sets of equal size, a one-one function gives a complete arrangement of images. Step 2: The number is (4!=4\times3\times2\times1=24). Step 3: When (|A|=|B|), a one-one function is also onto.

Step 1: A quadratic function has a parabolic graph. Step 2: On both sides of the vertex, equal height values occur. Step 3: Hence a quadratic with (a\neq0) is not one-one on all of (\mathbb{R}).

Step 1: When (a>0), both (x^3) and (ax) are increasing parts. Step 2: Their sum is also increasing, so different inputs give different images. Step 3: In parameter questions, carefully observe the sign of the coefficient.

Step 1: (f(0)=0). Step 2: (f(2\sqrt{3})=(2\sqrt{3})^3-12(2\sqrt{3})=24\sqrt{3}-24\sqrt{3}=0). Step 3: Equal images for different inputs stop the function from being one-one.

Step 1: (f'(x)=1+\cos x), which is never negative. Step 2: The function does not decrease, and the (x) term keeps values separated as inputs move forward. Step 3: A derivative may become zero at points, but being non-negative can help in checking one-one nature.

Step 1: (f'(x)=1-\cos x), which is always zero or positive. Step 2: The function does not turn back, so two different inputs do not give the same image. Step 3: For mixed trigonometric functions, decide by checking the direction through the derivative.

Step 1: On the whole real domain, (x^2) introduces a symmetric growth effect. Step 2: The function changes direction because (f'(x)=\cos x+2x) is negative for some negative (x) and positive for large positive (x). Step 3: Since the function is not strictly in one direction on all of (\mathbb{R}), it is not one-one.

Step 1: (2x+1) is an increasing linear expression. Step 2: (e^t) is also increasing, so the composition is increasing. Step 3: A composition of increasing one-one functions remains one-one.

Step 1: In (x^2), (x) and (-x) give the same value. Step 2: Therefore (f(1)=e) and (f(-1)=e), while (1\neq -1). Step 3: Even if the outer function is one-one, the whole function may fail if the inner part is not one-one.

Step 1: Assume (f(a)=f(b)). Step 2: Then (g(2a-3)=g(2b-3)), and since (g) is one-one, (2a-3=2b-3). Step 3: This gives (a=b), so (f) is one-one.

Step 1: Assume ((g\circ f)(a_1)=(g\circ f)(a_2)). Step 2: From (g(f(a_1))=g(f(a_2))), we get (f(a_1)=f(a_2)) because (g) is one-one. Then (a_1=a_2) because (f) is one-one. Step 3: The composition of two one-one functions is one-one.

Step 1: Suppose (f(a)=f(b)). Step 2: Then (g(f(a))=g(f(b))), so ((g\circ f)(a)=(g\circ f)(b)). Step 3: Since (g\circ f) is one-one, (a=b); hence (f) must be one-one.

Step 1: Find different inputs with the same image. Step 2: (f(1)=| -1|+|3|=4) and (f(-1)=| -3|+|1|=4), while (1\neq -1). Step 3: In sums of absolute values, always check symmetric inputs.

Step 1: The domain is (x\geq2), so (|x-2|=x-2). Step 2: The function becomes (f(x)=x+1), which is one-one. Step 3: Before removing absolute value, understand the sign from the domain.

Step 1: In the greatest integer function, many values in the same interval have the same image. Step 2: (f(1.2)=1) and (f(1.8)=1), while (1.2\neq1.8). Step 3: A step function is not one-one on the whole real domain.

Step 1: On integers, check neighbouring opposite-side values. Step 2: (f(0)=0) and (f(-1)=0), while (0\neq -1). Step 3: Even on an integer domain, different values can have the same image.

Step 1: In natural numbers, as (n) increases, both (n^2) and (n) increase. Step 2: Hence (n^2+n) also increases and different (n) give different images. Step 3: The same formula may behave differently on integers, so read the domain carefully.

Step 1: For (x\geq0), (f(x)=x^2), and for (x<0), (f(x)=-x^2). Step 2: The negative part gives negative images and the positive part gives positive images, while each part is one-one on its domain. Step 3: If the ranges of pieces are separated, the whole function can be one-one.

Step 1: In (|x|x^2), (x) and (-x) give the same value. Step 2: (f(1)=1) and (f(-1)=1), while (1\neq -1). Step 3: If a function behaves evenly, one-one nature fails on all of (\mathbb{R}).

Step 1: For (x\geq0), (f(x)=\frac{x}{1+x}), which is increasing. Step 2: For (x<0), (f(x)=\frac{x}{1-x}), which is also increasing and gives negative images. Step 3: The two pieces have separated ranges and each piece is one-one, so the function is one-one.

Step 1: To check one-one nature, assume (f(a)=f(b)). Step 2: From (\frac{a-1}{a+1}=\frac{b-1}{b+1}), we get ((a-1)(b+1)=(b-1)(a+1)); simplifying gives (2a=2b), so (a=b). Step 3: In rational functions, remember to exclude the value that makes the denominator zero.