RBSE Solutions Class 11 Maths Chapter 2 Relations and Functions Exercise 2.4

Question 1. Classify the following functions is the form of one-one, many one, into and onto, also give reason to support your answer.
(i) f : Q → Q, f(x) = 3x + 7
(ii) f : C → R, f(x + iy) = x
(iii) f : if R → [-1, 1], f(x) = sin x
(iv) f : N → Z, f(x) = |x|
Answer:
(i) Given the function \( f : Q \rightarrow Q \) where \( f(x) = 3x + 7 \). Here, Q represents the set of rational numbers.
To check if it's one-one, we assume \( f(x_1) = f(x_2) \).
This means \( 3x_1 + 7 = 3x_2 + 7 \).
Subtracting 7 from both sides gives \( 3x_1 = 3x_2 \).
Dividing by 3 gives \( x_1 = x_2 \).
Therefore, the function f is one-one.
To check if it's onto, we take an element \( y \) from the co-domain Q. We need to find an \( x \) in the domain Q such that \( f(x) = y \).
So, \( 3x + 7 = y \).
Subtracting 7 gives \( 3x = y - 7 \).
Dividing by 3 gives \( x = \frac{y-7}{3} \).
Since \( y \) is a rational number, \( y-7 \) is also rational. Dividing by 3, \( \frac{y-7}{3} \) is also rational. This means \( x \) is in Q.
Thus, every element in the co-domain has a pre-image in the domain.
So, the function f is onto.
Combining both, f is a one-one, onto function. This type of function creates a direct and complete pairing between elements of the domain and co-domain.
In simple words: This function pairs up each rational number to a unique rational number, and every rational number in the target set has someone from the starting set pointing to it. So, it's perfect match!

🎯 Exam Tip: For linear functions like \( f(x) = ax + b \) with \( a \neq 0 \) over sets like R or Q, they are generally both one-one and onto, as each input gives a unique output and every output can be traced back to a unique input.

Question 1. Classify the following functions is the form of one-one, many one, into and onto, also give reason to support your answer.
(i) f : Q → Q, f(x) = 3x + 7
(ii) f : C → R, f(x + iy) = x
(iii) f : if R → [-1, 1], f(x) = sin x
(iv) f : N → Z, f(x) = |x|
Answer:
(ii) Given the function \( f : C \rightarrow R \) where \( f(x + iy) = x \). Here, C is the set of complex numbers and R is the set of real numbers.
To check if it's one-one, we choose two different complex numbers.
Let \( z_1 = 1 + i \) and \( z_2 = 1 + 2i \). Both are in C.
Then, \( f(z_1) = f(1 + i) = 1 \).
And, \( f(z_2) = f(1 + 2i) = 1 \).
We see that \( f(z_1) = f(z_2) \) even though \( z_1 \neq z_2 \). For instance, different imaginary parts lead to the same real part output.
Therefore, the function f is many-one.
To check if it's onto, we take an element \( y \) from the co-domain R. We need to find a complex number \( z = a + ib \) in the domain C such that \( f(z) = y \).
We know \( f(a + ib) = a \).
So, we need \( a = y \). We can choose \( b \) to be any real number, for example, \( b = 0 \).
Then, \( z = y + 0i = y \) is a complex number (which is also a real number). For any real \( y \), we can find a complex number \( y \) whose real part is \( y \).
Thus, every element in the co-domain has a pre-image in the domain.
So, the function f is onto.
Combining both, f is a many-one, onto function.
In simple words: Many different complex numbers can give the same real number as an answer, but every real number has at least one complex number that maps to it.

🎯 Exam Tip: When checking for many-one, it's often easiest to find two distinct inputs that produce the same output. For complex numbers, varying the imaginary part while keeping the real part constant is a common strategy if the function only depends on the real part.

Question 1. Classify the following functions is the form of one-one, many one, into and onto, also give reason to support your answer.
(i) f : Q → Q, f(x) = 3x + 7
(ii) f : C → R, f(x + iy) = x
(iii) f : if R → [-1, 1], f(x) = sin x
(iv) f : N → Z, f(x) = |x|
Answer:
(iii) Given the function \( f : R \rightarrow [-1, 1] \) where \( f(x) = \sin x \). Here, R is the set of real numbers.
To check if it's one-one, we look for different inputs that give the same output.
Consider \( x_1 = 0 \) and \( x_2 = \pi \). Both \( 0 \) and \( \pi \) are real numbers.
Then, \( f(x_1) = f(0) = \sin 0 = 0 \).
And, \( f(x_2) = f(\pi) = \sin \pi = 0 \).
We see that \( f(x_1) = f(x_2) \) but \( x_1 \neq x_2 \). The sine function repeats its values for different inputs, which is characteristic of periodic functions.
Therefore, the function f is many-one.
To check if it's onto, we compare the range of the function with its co-domain.
The range of \( \sin x \) for all real numbers \( x \) is \( [-1, 1] \).
The given co-domain of the function is also \( [-1, 1] \).
Since the range of the function is equal to its co-domain, f is an onto function.
Combining both, f is a many-one, onto function.
In simple words: This sine function takes many different angles and gives the same result. But, it covers all possible values from -1 to 1 in the target set.

🎯 Exam Tip: Remember that trigonometric functions like \( \sin x \) and \( \cos x \) are typically many-one over their natural domain R due to their periodic nature. Their range is usually a closed interval like \( [-1, 1] \).

Question 1. Classify the following functions is the form of one-one, many one, into and onto, also give reason to support your answer.
(i) f : Q → Q, f(x) = 3x + 7
(ii) f : C → R, f(x + iy) = x
(iii) f : if R → [-1, 1], f(x) = sin x
(iv) f : N → Z, f(x) = |x|
Answer:
(iv) Given the function \( f : N \rightarrow Z \) where \( f(x) = |x| \). Here, N is the set of natural numbers \( \{1, 2, 3, ...\} \) and Z is the set of integers \( \{..., -2, -1, 0, 1, 2, ...\} \).
To check if it's one-one, we assume \( f(x_1) = f(x_2) \).
This means \( |x_1| = |x_2| \).
Since \( x_1 \) and \( x_2 \) are natural numbers, they are always positive.
So, \( |x_1| = x_1 \) and \( |x_2| = x_2 \).
Therefore, \( x_1 = x_2 \).
Thus, the function f is one-one.
To check if it's onto, we compare the range of the function with its co-domain.
The range of \( f(x) = |x| \) for \( x \in N \) is \( \{|1|, |2|, |3|, ...\} = \{1, 2, 3, ...\} \). This is the set of natural numbers.
The co-domain is Z, the set of all integers, which includes negative numbers and zero.
Since the range \( \{1, 2, 3, ...\} \) is a proper subset of the co-domain Z (as Z contains elements like 0, -1, -2, etc., which have no pre-images in N under this function), f is an into function (not onto).
Combining both, f is a one-one, into function. The absolute value function on natural numbers means each number maps to itself, keeping it unique, but it can't map to negative integers.
In simple words: Each natural number gives a unique positive integer as an answer. But, no natural number can give 0 or a negative integer, so the target set (all integers) is not fully covered.

🎯 Exam Tip: For functions involving absolute values, carefully consider the domain. If the domain consists only of positive numbers (like natural numbers), the absolute value function often behaves as a simple identity function for one-one check. For onto, always compare the actual range with the stated co-domain.

Question 2. If A = \( \{x : -1 \le x \le 1\} \) = B. then find out which function is one-one, into or one-one onto defined from A to B
(i) f(x) = \( \frac{x}{2} \)
(ii) g(x) = |x|
(iii) h(x) = \( x^2 \)
(iv) k(x) = sinx
Answer:
Given: The domain A is \( [-1, 1] \) and the co-domain B is \( [-1, 1] \). All functions are defined from A to B.
(i) Given the function \( f(x) = \frac{x}{2} \).
To check if it's one-one, let \( x_1, x_2 \in A \) such that \( f(x_1) = f(x_2) \).
This means \( \frac{x_1}{2} = \frac{x_2}{2} \).
Multiplying by 2 gives \( x_1 = x_2 \).
Thus, f is a one-one function.
To check if it's onto, we find the range of the function.
For \( x \in [-1, 1] \), the smallest value of \( \frac{x}{2} \) is \( \frac{-1}{2} \) and the largest is \( \frac{1}{2} \).
So, the range of f is \( [-\frac{1}{2}, \frac{1}{2}] \).
The co-domain B is \( [-1, 1] \).
Since the range \( [-\frac{1}{2}, \frac{1}{2}] \) is a proper subset of the co-domain \( [-1, 1] \) (e.g., \( -0.7 \in B \) has no pre-image), f is an into function.
Conclusion: f is a one-one, into function. This function scales down the input, ensuring unique outputs but not covering the full target range.
In simple words: Each input gives a unique answer, but the answers only cover a smaller part of the possible target values.

🎯 Exam Tip: For linear functions like \( f(x) = mx + c \), if the slope \( m \neq 0 \), the function will always be one-one. To determine into/onto, always compare the calculated range with the given co-domain.

Question 2. If A = \( \{x : -1 \le x \le 1\} \) = B. then find out which function is one-one, into or one-one onto defined from A to B
(i) f(x) = \( \frac{x}{2} \)
(ii) g(x) = |x|
(iii) h(x) = \( x^2 \)
(iv) k(x) = sinx
Answer:
Given: The domain A is \( [-1, 1] \) and the co-domain B is \( [-1, 1] \). All functions are defined from A to B.
(ii) Given the function \( g(x) = |x| \).
To check if it's one-one, we look for different inputs that give the same output.
Consider \( x_1 = -1 \) and \( x_2 = 1 \). Both \( -1 \) and \( 1 \) are in A.
Then, \( g(x_1) = g(-1) = |-1| = 1 \).
And, \( g(x_2) = g(1) = |1| = 1 \).
We see that \( g(x_1) = g(x_2) \) but \( x_1 \neq x_2 \).
Thus, g is a many-one function.
To check if it's onto, we find the range of the function.
For \( x \in [-1, 1] \), the absolute value \( |x| \) will always be non-negative. The minimum value is \( |0|=0 \) and the maximum is \( |-1|=1 \) or \( |1|=1 \).
So, the range of g is \( [0, 1] \).
The co-domain B is \( [-1, 1] \).
Since the range \( [0, 1] \) is a proper subset of the co-domain \( [-1, 1] \) (as negative numbers in B have no pre-image under \( |x| \)), g is an into function.
Conclusion: g is a many-one, into function. The absolute value function turns negative inputs positive, leading to repeated outputs for positive/negative pairs, and only covers non-negative values.
In simple words: Both positive and negative inputs can give the same answer. Also, you can never get a negative answer, so not all target values are hit.

🎯 Exam Tip: When dealing with \( |x| \), always test both positive and negative values from the domain. If the domain includes both, it's often a sign of a many-one function, especially if the range does not cover negative numbers but the co-domain does.

Question 2. If A = \( \{x : -1 \le x \le 1\} \) = B. then find out which function is one-one, into or one-one onto defined from A to B
(i) f(x) = \( \frac{x}{2} \)
(ii) g(x) = |x|
(iii) h(x) = \( x^2 \)
(iv) k(x) = sinx
Answer:
Given: The domain A is \( [-1, 1] \) and the co-domain B is \( [-1, 1] \). All functions are defined from A to B.
(iii) Given the function \( h(x) = x^2 \).
To check if it's one-one, we look for different inputs that give the same output.
Consider \( x_1 = -1 \) and \( x_2 = 1 \). Both \( -1 \) and \( 1 \) are in A.
Then, \( h(x_1) = h(-1) = (-1)^2 = 1 \).
And, \( h(x_2) = h(1) = (1)^2 = 1 \).
We see that \( h(x_1) = h(x_2) \) but \( x_1 \neq x_2 \).
Thus, h is a many-one function.
To check if it's onto, we find the range of the function.
For \( x \in [-1, 1] \), the square \( x^2 \) will always be non-negative. The minimum value is \( 0^2=0 \) and the maximum is \( (-1)^2=1 \) or \( (1)^2=1 \).
So, the range of h is \( [0, 1] \).
The co-domain B is \( [-1, 1] \).
Since the range \( [0, 1] \) is a proper subset of the co-domain \( [-1, 1] \) (as negative numbers in B have no pre-image under \( x^2 \)), h is an into function.
Conclusion: h is a many-one, into function. Squaring a number makes it positive, causing different inputs (like -1 and 1) to map to the same output (1), and only covers non-negative values in the range.
In simple words: This function gives the same output for positive and negative inputs, so it's not unique. Also, it can't produce negative answers, so it misses part of the target.

🎯 Exam Tip: Functions involving even powers (like \( x^2, x^4 \)) are typically many-one if their domain includes both positive and negative numbers. Their range will also generally be restricted to non-negative values.

Question 2. If A = \( \{x : -1 \le x \le 1\} \) = B. then find out which function is one-one, into or one-one onto defined from A to B
(i) f(x) = \( \frac{x}{2} \)
(ii) g(x) = |x|
(iii) h(x) = \( x^2 \)
(iv) k(x) = sinx
Answer:
Given: The domain A is \( [-1, 1] \) and the co-domain B is \( [-1, 1] \). All functions are defined from A to B.
(iv) Given the function \( k(x) = \sin x \).
To check if it's one-one, let \( x_1, x_2 \in A \) such that \( k(x_1) = k(x_2) \).
This means \( \sin x_1 = \sin x_2 \).
In the interval \( [-1, 1] \), the sine function is strictly increasing. This means if \( x_1 \neq x_2 \), then \( \sin x_1 \neq \sin x_2 \).
Therefore, for \( \sin x_1 = \sin x_2 \) to hold within this interval, it must be that \( x_1 = x_2 \).
Thus, k is a one-one function.
To check if it's onto, we find the range of the function.
For \( x \in [-1, 1] \), the smallest value of \( \sin x \) is \( \sin(-1) \) and the largest value is \( \sin(1) \).
Numerically, \( \sin(-1) \approx -0.841 \) and \( \sin(1) \approx 0.841 \).
So, the range of k is approximately \( [-0.841, 0.841] \).
The co-domain B is \( [-1, 1] \).
Since the range \( [-0.841, 0.841] \) is a proper subset of the co-domain \( [-1, 1] \) (e.g., \( -0.9 \in B \) has no pre-image in A), k is an into function.
Conclusion: k is a one-one, into function. Within this narrow domain, each angle has a unique sine value. However, the sine values do not reach the full range of -1 to 1 in the co-domain.
In simple words: This sine function gives a different answer for each unique input in its small range. But, it doesn't cover all the possible answers in the target set, especially the numbers very close to -1 and 1.

🎯 Exam Tip: The behavior of trigonometric functions changes drastically when their domain is restricted. Always check if the function is monotonic (strictly increasing or decreasing) over the given restricted domain to determine if it's one-one. Then, calculate the exact range for that restricted domain to determine into/onto.

Question 3. If f : C → C, f(x + iy) = (x – iy), then prove that f is an one-one onto function.
Answer:
Given the function \( f : C \rightarrow C \) where \( f(x + iy) = x - iy \). This function is the complex conjugate operation.
To prove it is one-one, let \( z_1 = x_1 + iy_1 \) and \( z_2 = x_2 + iy_2 \) be two complex numbers in the domain C.
Assume \( f(z_1) = f(z_2) \).
This means \( x_1 - iy_1 = x_2 - iy_2 \).
For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal.
From the real parts: \( x_1 = x_2 \).
From the imaginary parts: \( -y_1 = -y_2 \), which implies \( y_1 = y_2 \).
Since \( x_1 = x_2 \) and \( y_1 = y_2 \), it means \( x_1 + iy_1 = x_2 + iy_2 \), so \( z_1 = z_2 \).
Therefore, the function f is one-one.
To prove it is onto, let \( w = u + iv \) be any complex number in the co-domain C. We need to find a complex number \( z = x + iy \) in the domain C such that \( f(z) = w \).
We set \( f(x + iy) = u + iv \).
This means \( x - iy = u + iv \).
Equating the real and imaginary parts:
From the real parts: \( x = u \).
From the imaginary parts: \( -y = v \), which implies \( y = -v \).
So, the complex number \( z \) that maps to \( w \) is \( z = u + i(-v) = u - iv \).
Since \( u \) and \( v \) are real numbers, \( u - iv \) is also a complex number, so \( z \in C \).
Thus, every element \( w \) in the co-domain has a pre-image \( z \) in the domain.
Therefore, the function f is onto.
Combining both, f is a one-one, onto function. The complex conjugate operation uniquely maps each complex number and covers all complex numbers in its range.
In simple words: This function (finding the complex conjugate) gives a unique answer for every complex number, and every complex number in the target set can be reached. It's a perfect two-way match.

🎯 Exam Tip: For complex functions, proving one-one and onto often involves equating real and imaginary parts. Functions like the complex conjugate are invertible and thus are bijections (one-one and onto).

Question 5. Prove that f : R → R, f(x) = cos x is a many- one into function. Change the domain and co-domain of f such that I become:
(i) One-one into
(ii) Many-one onto
(iii) One-one onto
Answer:
Given the function \( f : R \rightarrow R \) where \( f(x) = \cos x \).
To prove it is many-one, we look for different inputs that give the same output.
Consider \( x_1 = 0 \) and \( x_2 = 2\pi \). Both \( 0 \) and \( 2\pi \) are real numbers in the domain R.
Then, \( f(x_1) = f(0) = \cos 0 = 1 \).
And, \( f(x_2) = f(2\pi) = \cos(2\pi) = 1 \).
We see that \( f(x_1) = f(x_2) \) but \( x_1 \neq x_2 \). The cosine function is periodic, meaning it takes the same values at different points.
Therefore, the function f is many-one.
To prove it is into, we compare the range of the function with its co-domain.
The range of \( f(x) = \cos x \) for all real numbers \( x \) is \( [-1, 1] \).
The co-domain is R, the set of all real numbers.
Since the range \( [-1, 1] \) is a proper subset of the co-domain R (as elements like 2 or -5 in R have no pre-images under \( \cos x \)), f is an into function.
Thus, \( f : R \rightarrow R, f(x) = \cos x \) is a many-one, into function. The periodic nature of cosine means it's not unique for each input, and its values only span a small part of all real numbers.
In simple words: Many different angles give the same cosine value, so it's not unique. Also, cosine values are only between -1 and 1, so it doesn't cover all real numbers.

🎯 Exam Tip: When a function's domain is unrestricted (like R), periodic functions will always be many-one. For range, remember that trigonometric functions like sine and cosine are bounded between -1 and 1. If the co-domain is larger than this interval, the function will be into.

Question 6. If N= {1, 2, 3, 4, ...), O = (1, 3, 5, 7, ...}, E = (2, 4, 6, 8,.....) and f1, f2 are function defined as f1 : N → 0, f1(x) = 2x – 1; f2 : N → E, f2(x) = 2x Then prove that f₁ and f2 are one-one onto.
Answer:
Given sets are N (natural numbers), O (odd natural numbers), and E (even natural numbers).
(i) Consider the function \( f_1 : N \rightarrow O \) defined by \( f_1(x) = 2x - 1 \).
To prove it is one-one, let \( x_1, x_2 \in N \) such that \( f_1(x_1) = f_1(x_2) \).
This means \( 2x_1 - 1 = 2x_2 - 1 \).
Adding 1 to both sides gives \( 2x_1 = 2x_2 \).
Dividing by 2 gives \( x_1 = x_2 \).
Therefore, \( f_1 \) is one-one.
To prove it is onto, let \( y \in O \) (co-domain). We need to find an \( x \in N \) (domain) such that \( f_1(x) = y \).
Set \( 2x - 1 = y \).
Adding 1 gives \( 2x = y + 1 \).
Dividing by 2 gives \( x = \frac{y+1}{2} \).
Since \( y \) is an odd natural number, \( y+1 \) will always be an even natural number. For example, if \( y=1 \), \( x=1 \); if \( y=3 \), \( x=2 \).
Thus, \( \frac{y+1}{2} \) will always be a natural number, meaning \( x \in N \).
So, every element in the co-domain O has a pre-image in N.
Therefore, \( f_1 \) is onto.
Hence, \( f_1 \) is a one-one, onto function. This function creates a unique mapping from natural numbers to all odd natural numbers.
(ii) Consider the function \( f_2 : N \rightarrow E \) defined by \( f_2(x) = 2x \).
To prove it is one-one, let \( x_1, x_2 \in N \) such that \( f_2(x_1) = f_2(x_2) \).
This means \( 2x_1 = 2x_2 \).
Dividing by 2 gives \( x_1 = x_2 \).
Therefore, \( f_2 \) is one-one.
To prove it is onto, let \( y \in E \) (co-domain). We need to find an \( x \in N \) (domain) such that \( f_2(x) = y \).
Set \( 2x = y \).
Dividing by 2 gives \( x = \frac{y}{2} \).
Since \( y \) is an even natural number, \( \frac{y}{2} \) will always be a natural number. For example, if \( y=2 \), \( x=1 \); if \( y=4 \), \( x=2 \).
Thus, \( x \in N \).
So, every element in the co-domain E has a pre-image in N.
Therefore, \( f_2 \) is onto.
Hence, \( f_2 \) is a one-one, onto function. This function uniquely maps each natural number to all even natural numbers.
In simple words: Both functions are perfect matches. \( f_1 \) takes natural numbers and gives unique odd numbers, covering all of them. \( f_2 \) takes natural numbers and gives unique even numbers, covering all of them.

🎯 Exam Tip: Functions from N to specific subsets of N (like odd or even numbers) can be one-one onto if they create a direct correspondence, often involving simple arithmetic expressions like \( 2x \) or \( 2x-1 \).

Question 7. If function f is defined from set of real numbers R to R in the following way then classify them in the form of one-one, many-one, into or onto.
(i) f(x) = \( x^2 \)
(ii) f(x) = \( x^3 \)
(iii) f(x) = \( x^3 + 3 \)
(iv) f(x) = \( x^3 - x \)
Answer:
All functions are defined from \( f : R \rightarrow R \).
(i) Given the function \( f(x) = x^2 \).
To check if it's one-one, we look for different inputs that give the same output.
Consider \( x_1 = -1 \) and \( x_2 = 1 \). Both are real numbers.
Then, \( f(x_1) = f(-1) = (-1)^2 = 1 \).
And, \( f(x_2) = f(1) = (1)^2 = 1 \).
We see that \( f(x_1) = f(x_2) \) but \( x_1 \neq x_2 \).
Therefore, f is a many-one function.
To check if it's onto, we find the range of the function.
For any real number \( x \), \( x^2 \) is always greater than or equal to 0. So, the range of \( f(x) = x^2 \) is \( [0, \infty) \).
The co-domain is R, the set of all real numbers.
Since the range \( [0, \infty) \) is a proper subset of the co-domain R (negative real numbers have no pre-image), f is an into function.
Conclusion: f is a many-one, into function.
In simple words: Positive and negative numbers give the same square, so it's not unique. Also, squares can never be negative, so it doesn't cover all real numbers.

🎯 Exam Tip: The function \( f(x) = x^2 \) is a classic example of a many-one, into function when its domain and co-domain are R. Always consider positive and negative inputs and the non-negativity of squares.

Question 7. If function f is defined from set of real numbers R to R in the following way then classify them in the form of one-one, many-one, into or onto.
(i) f(x) = \( x^2 \)
(ii) f(x) = \( x^3 \)
(iii) f(x) = \( x^3 + 3 \)
(iv) f(x) = \( x^3 - x \)
Answer:
All functions are defined from \( f : R \rightarrow R \).
(ii) Given the function \( f(x) = x^3 \).
To check if it's one-one, let \( x_1, x_2 \in R \) such that \( f(x_1) = f(x_2) \).
This means \( x_1^3 = x_2^3 \).
Taking the cube root of both sides gives \( x_1 = x_2 \). Unlike squares, odd powers preserve the sign and uniqueness.
Therefore, f is a one-one function.
To check if it's onto, we find the range of the function.
For any real number \( x \), \( x^3 \) can take any real value (from \( -\infty \) to \( \infty \)). So, the range of \( f(x) = x^3 \) is \( (-\infty, \infty) \), which is R.
The co-domain is also R.
Since the range is equal to the co-domain, f is an onto function.
Conclusion: f is a one-one, onto function.
In simple words: Each real number gives a unique cube as an answer. And, every real number can be found as the cube of some other real number.

🎯 Exam Tip: Functions involving odd powers (like \( x^3, x^5 \)) are generally one-one and onto over the domain of real numbers, because they are always strictly monotonic (either increasing or decreasing) and their range covers all real numbers.

Question 7. If function f is defined from set of real numbers R to R in the following way then classify them in the form of one-one, many-one, into or onto.
(i) f(x) = \( x^2 \)
(ii) f(x) = \( x^3 \)
(iii) f(x) = \( x^3 + 3 \)
(iv) f(x) = \( x^3 - x \)
Answer:
All functions are defined from \( f : R \rightarrow R \).
(iii) Given the function \( f(x) = x^3 + 3 \).
To check if it's one-one, let \( x_1, x_2 \in R \) such that \( f(x_1) = f(x_2) \).
This means \( x_1^3 + 3 = x_2^3 + 3 \).
Subtracting 3 from both sides gives \( x_1^3 = x_2^3 \).
Taking the cube root of both sides gives \( x_1 = x_2 \). The addition of a constant does not affect the one-one nature.
Therefore, f is a one-one function.
To check if it's onto, we find the range of the function.
For any real number \( x \), \( x^3 \) can take any real value. Adding 3 to it will also result in any real value (from \( -\infty \) to \( \infty \)). So, the range of \( f(x) = x^3 + 3 \) is \( (-\infty, \infty) \), which is R.
The co-domain is also R.
Since the range is equal to the co-domain, f is an onto function.
Conclusion: f is a one-one, onto function.
In simple words: Each unique input gives a unique answer. Also, every real number can be found as an answer from this function.

🎯 Exam Tip: Adding or subtracting a constant to a function does not change its one-one or onto property, as it just shifts the graph vertically without altering its shape or extent.

Question 7. If function f is defined from set of real numbers R to R in the following way then classify them in the form of one-one, many-one, into or onto.
(i) f(x) = \( x^2 \)
(ii) f(x) = \( x^3 \)
(iii) f(x) = \( x^3 + 3 \)
(iv) f(x) = \( x^3 - x \)
Answer:
All functions are defined from \( f : R \rightarrow R \).
(iv) Given the function \( f(x) = x^3 - x \).
To check if it's one-one, we look for different inputs that give the same output.
Consider \( x_1 = 0 \), \( x_2 = 1 \), and \( x_3 = -1 \). All are real numbers.
Then, \( f(x_1) = f(0) = 0^3 - 0 = 0 \).
And, \( f(x_2) = f(1) = 1^3 - 1 = 1 - 1 = 0 \).
And, \( f(x_3) = f(-1) = (-1)^3 - (-1) = -1 + 1 = 0 \).
We see that \( f(0) = f(1) = f(-1) = 0 \) but \( 0 \neq 1 \neq -1 \).
Therefore, f is a many-one function.
To check if it's onto, we find the range of the function.
The function is a polynomial of odd degree. For such functions, as \( x \rightarrow \infty \), \( f(x) \rightarrow \infty \), and as \( x \rightarrow -\infty \), \( f(x) \rightarrow -\infty \).
Since it is a continuous function and its limits extend to positive and negative infinity, it covers all real values. So, the range of \( f(x) = x^3 - x \) is \( (-\infty, \infty) \), which is R.
The co-domain is also R.
Since the range is equal to the co-domain, f is an onto function.
Conclusion: f is a many-one, onto function.
In simple words: This function can give the same answer for different inputs (like 0, 1, and -1 all give 0). However, it covers all possible real numbers in its answers.

🎯 Exam Tip: For polynomial functions, if the degree is odd, the range is always R, making it onto. To check for one-one, finding distinct inputs that yield the same output (especially simple integers or roots) is a quick method.