One-one function

Step 1: The cube root function is defined on all real numbers. Step 2: If (\sqrt[3]{a}=\sqrt[3]{b}), then (a=b). Step 3: The cube root function is increasing, so it is one-one.

Step 1: The square root function increases from (0) onward. Step 2: (\sqrt{a}=\sqrt{b}) gives (a=b), where (a,b\geq 0). Step 3: The domain starting from (0) is important for this one-one nature.

Step 1: (\sin x) is periodic. Step 2: (\sin 0=0) and (\sin \pi=0), while (0\neq \pi). Step 3: Periodic functions are generally not one-one on the whole real domain.

Step 1: (\cos x) is also periodic. Step 2: (\cos 0=1) and (\cos 2\pi=1), but (0\neq 2\pi). Step 3: Once different inputs give the same image, the function is not one-one.

Step 1: On the given interval, (\tan x) is increasing. Step 2: In this interval, no two different values give the same (\tan) value. Step 3: For trigonometric functions, restricting the domain can give one-one nature.

Step 1: (x^2-4x+7=(x-2)^2+3). Step 2: (f(1)=4) and (f(3)=4), while (1\neq 3). Step 3: Completing the square helps solve such questions quickly.

Step 1: (f(x)=(x-2)^2+3). Step 2: On the domain ([2,\infty)), ((x-2)^2) increases. Step 3: Hence in this restricted domain, equal images occur only for equal inputs.

Step 1: To test one-one nature, look for different inputs with the same image. Step 2: (f(-1)=0), (f(0)=0), and (f(1)=0). Step 3: Several different inputs give the same image, so the function is not one-one.

Step 1: The main identity of a one-one function is different images for different inputs. Step 2: If two different inputs give the same image, the condition fails. Step 3: Understanding the definition in simple words is very useful in exams.

Step 1: Assume (f(p)=f(q)). Step 2: From (ap+b=aq+b), we get (a(p-q)=0). Step 3: If (a\neq 0), then (p=q), so the function is one-one.

Step 1: (x^3) is one-one. Step 2: Multiplication by (-1) keeps distinct images distinct. Step 3: From (-a^3=-b^3), we get (a^3=b^3), hence (a=b).

Step 1: (x^2+2x=(x+1)^2-1). Step 2: (f(0)=0) and (f(-2)=0), while (0\neq -2). Step 3: Converting to square form helps find equal images easily.

Step 1: The linear functions have non-zero slope, so they are one-one. Step 2: (x^3+4) is also increasing. Step 3: For (x^2-1), (f(1)=0) and (f(-1)=0), so it is not one-one.

Step 1: Assume (f(a)=f(b)). Step 2: From (\frac{a+1}{2}=\frac{b+1}{2}), we get (a+1=b+1), so (a=b). Step 3: A linear form with non-zero coefficient remains one-one.

Step 1: Due to absolute value, two different inputs can give the same positive value. Step 2: (f(0)=1) and (f(-1)=1), while (0\neq -1). Step 3: Check absolute value functions carefully on all of (\mathbb{R}).

Step 1: In a one-one function, different elements of (A) must have different images in (B). Step 2: In option (B), the images are (4,5,6), with no repetition. Step 3: For finite sets, absence of repeated images is the key check.

Step 1: One counterexample is enough. Step 2: (f(1)=5) and (f(-1)=5), but the inputs are different. Step 3: In even-power functions, always test (x) and (-x).

Step 1: Assume (f(a)=f(b)). Step 2: From (9-3a=9-3b), we get (-3a=-3b), so (a=b). Step 3: The slope (-3) is non-zero, so the function is one-one.

Step 1: The square contains (x+5), so the center is (x=-5). Step 2: (f(-4)=1) and (f(-6)=1), while (-4\neq -6). Step 3: Values equally distant from the center give the same image.

Step 1: In the given domain, (x+5\geq 0). Step 2: On this domain, ((x+5)^2) increases and equal images give equal inputs. Step 3: A square function can become one-one when restricted to one side.

Step 1: Write (f(a)=f(b)). Step 2: From (\frac{3a-2}{5}=\frac{3b-2}{5}), we get (3a-2=3b-2), so (a=b). Step 3: A linear function written as a fraction is also one-one if the coefficient of (x) is non-zero.

Step 1: Look at the images in each option. Step 2: In option (C), both (1) and (2) have image (2). Step 3: Repetition of an image destroys one-one nature.

Step 1: (3) and (5) are different inputs. Step 2: Both have image (8). Step 3: The same image for different inputs is direct proof against one-one nature.

Step 1: Let (a>b), where (a,b) are integers. Step 2: Then (a^3>b^3) and (a>b), so (a^3+a>b^3+b). Step 3: The function increases, so different integers give different images.

Step 1: (x^2-6x+10=(x-3)^2+1). Step 2: (f(2)=2) and (f(4)=2), while (2\neq 4). Step 3: In square form, check values equally distant from the center.