OP Malhotra Class 12 Maths Solutions Chapter 2 Functions Exercise 2 (A)

S Chand Class 12 ICSE Maths Solutions Chapter 2 Functions Ex 2(a)

Question 1. If A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6} are two sets and function f : A → B is defined by f (x) = x + 2, ∀x ∈ A, then the function f is
(a) bijective
(b) onto
(c) one-one
Answer: (c) one-one
In simple words: The function takes each number from set A and adds 2 to it to get a number in set B. Because every number in A gives a different number in B, it is "one-one". However, some numbers in B (like 1 and 2) are not formed by any number from A, so it is not "onto".

🎯 Exam Tip: To check if a function is one-one, ensure each input maps to a unique output. To check if it's onto, make sure every element in the codomain has at least one pre-image in the domain.

Question 2. Show that a function f : R → R given by f (x) = ax + b, a, b ∈ R, a ≠ 0 is a bijective.
Answer:
We are given a function \( f : R \rightarrow R \) defined by \( f(x) = ax + b \), where \( a, b \in R \) and \( a \neq 0 \).

**To prove injectivity (one-one):**
Let \( x, y \in R \) such that \( f(x) = f(y) \).
\( \implies ax + b = ay + b \)
\( \implies ax = ay \)
\( \implies x = y \) (since \( a \neq 0 \), we can divide by \( a \)).
Since \( f(x) = f(y) \) implies \( x = y \), the function \( f \) is one-one (injective).

**To prove surjectivity (onto):**
Let \( y \) be any arbitrary element in the codomain \( R \). We need to find an \( x \in R \) such that \( f(x) = y \).
Set \( y = ax + b \).
\( \implies y - b = ax \)
\( \implies x = \frac{y - b}{a} \)
Since \( y, b, a \) are real numbers and \( a \neq 0 \), \( x = \frac{y - b}{a} \) is always a real number. This means for every \( y \) in the codomain, there exists a corresponding \( x \) in the domain such that \( f(x) = y \).
Therefore, the function \( f \) is onto (surjective).

Since \( f \) is both one-one and onto, it is a bijective function.
In simple words: A function is "bijective" if it is both "one-one" and "onto". We showed it is one-one because different inputs always give different outputs. We showed it is onto because every possible output value in the second set can be reached from some input value in the first set. This means the function creates a perfect match between all elements of the domain and codomain.

🎯 Exam Tip: For linear functions like \( f(x) = ax+b \) where \( a \neq 0 \), remember they are almost always bijective over real numbers. Focus on clearly showing the algebraic steps for both one-one and onto conditions.

Question 3. Let f : N → N given by f (x) = 2x ∀x ∈ N. Show that/is one-one and into.
Answer:
We are given a function \( f : N \rightarrow N \) defined by \( f(x) = 2x \), where \( N \) represents the set of natural numbers \( \{1, 2, 3, \dots\} \).

**To prove injectivity (one-one):**
Let \( x, y \in N \) such that \( f(x) = f(y) \).
\( \implies 2x = 2y \)
\( \implies x = y \)
Since \( f(x) = f(y) \) implies \( x = y \), the function \( f \) is one-one (injective).

**To prove "into" (not onto):**
A function is "into" if it is not surjective (onto). This means there is at least one element in the codomain \( N \) that does not have a pre-image in the domain \( N \).
Consider the element \( 3 \in N \) (codomain). We check if there exists an \( x \in N \) (domain) such that \( f(x) = 3 \).
Set \( f(x) = 3 \).
\( \implies 2x = 3 \)
\( \implies x = \frac{3}{2} \)
Since \( x = \frac{3}{2} \) is not a natural number, there is no element in the domain \( N \) that maps to \( 3 \). Thus, \( 3 \) has no pre-image.
Therefore, the function \( f \) is into (not surjective). The set of odd natural numbers will also not have pre-images.

Hence, the function \( f \) is one-one and into.

🎯 Exam Tip: When proving a function is "into" (not onto), it's sufficient to find just one element in the codomain that has no pre-image. For "one-one", you must show that `f(x)=f(y)` always implies `x=y`.

Question 4. Let f: R → R be defined a f(x) = 3x. Choose the correct answer.
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Answer: (a) f is one-one onto
In simple words: This function is "one-one" because every different input number gives a different output number. It is also "onto" because every number in the second set (R) can be made by putting some number from the first set (R) into the function. It perfectly matches inputs and outputs.

🎯 Exam Tip: Simple linear functions like \( f(x) = kx \) (where \( k \neq 0 \)) are generally bijective (one-one and onto) when mapping from \( R \) to \( R \). Quickly confirm by testing for injectivity and surjectivity.

Question 5. Show that f : R → R, defined by f(x) = x³ is a bijection.
Answer:
We are given a function \( f : R \rightarrow R \) defined by \( f(x) = x^3 \).

**To prove injectivity (one-one):**
Let \( x, y \in R \) such that \( f(x) = f(y) \).
\( \implies x^3 = y^3 \)
\( \implies x^3 - y^3 = 0 \)
\( \implies (x - y)(x^2 + xy + y^2) = 0 \)
For real numbers, the term \( x^2 + xy + y^2 \) can be written as \( (x + \frac{y}{2})^2 + \frac{3y^2}{4} \). This expression is zero only if \( x=0 \) and \( y=0 \). Otherwise, it is positive. Therefore, for \( x^3 - y^3 = 0 \), we must have \( x - y = 0 \).
\( \implies x = y \)
Since \( f(x) = f(y) \) implies \( x = y \), the function \( f \) is one-one (injective).

**To prove surjectivity (onto):**
Let \( y \) be any arbitrary element in the codomain \( R \). We need to find an \( x \in R \) such that \( f(x) = y \).
Set \( y = x^3 \).
\( \implies x = \sqrt[3]{y} \)
For any real number \( y \), its cube root \( \sqrt[3]{y} \) is always a real number. This means for every \( y \) in the codomain, there exists a corresponding \( x \) in the domain such that \( f(x) = y \).
Therefore, the function \( f \) is onto (surjective).

Since \( f \) is both one-one and onto, it is a bijective function.
In simple words: This function takes a number and cubes it. It is "one-one" because each unique starting number gives a unique cubed result. It is "onto" because any real number can be found by cubing some other real number (its cube root). So, it's a "bijective" function, meaning it perfectly pairs up every input with every output.

🎯 Exam Tip: When dealing with powers, remember that for odd powers (like \( x^3, x^5 \)), \( x^n = y^n \) implies \( x=y \) for real numbers, making them good candidates for one-one. For even powers (like \( x^2, x^4 \)), this is not true (e.g., \( (-2)^2 = 2^2 \)).

Question 6. Check the injectivity and surjectivity of the following functions:
(a) f : N → N given by f(x) = x²
(b) f : Z → Z given by f(x) = x²
(c) f : R → R given by f (x) = x²
(d) f : N → N given by f(x) = x³
(e) f : Z → Z given by f(x) = x³
Answer:
**(a) For f : N → N given by f(x) = x²**
*Injectivity (one-one):*
Let \( x, y \in N \) such that \( f(x) = f(y) \).
\( \implies x^2 = y^2 \)
\( \implies x^2 - y^2 = 0 \)
\( \implies (x - y)(x + y) = 0 \)
Since \( x, y \in N \) (natural numbers), both \( x \) and \( y \) are positive. Thus, \( x+y \) will always be positive and cannot be zero. So, we must have \( x - y = 0 \).
\( \implies x = y \)
Therefore, \( f \) is one-one (injective).
*Surjectivity (onto):*
Let's check for an element in the codomain that has no pre-image.
Consider \( 2 \in N \) (codomain). Is there an \( x \in N \) (domain) such that \( f(x) = 2 \)?
Set \( x^2 = 2 \).
\( \implies x = \pm\sqrt{2} \)
Since \( \pm\sqrt{2} \) are not natural numbers, there is no \( x \in N \) such that \( f(x) = 2 \).
Therefore, \( f \) is not onto (surjective).
**Conclusion for (a):** The function \( f(x) = x^2 \) from \( N \) to \( N \) is injective but not surjective.

**(b) For f : Z → Z given by f(x) = x²**
*Injectivity (one-one):*
Let's test with specific integer values.
\( f(1) = 1^2 = 1 \)
\( f(-1) = (-1)^2 = 1 \)
Here, different elements (1 and -1) from the domain \( Z \) map to the same image (1) in the codomain \( Z \).
Therefore, \( f \) is not one-one (it is many-one).
*Surjectivity (onto):*
Let's check for an element in the codomain that has no pre-image.
Consider \( 2 \in Z \) (codomain). Is there an \( x \in Z \) (domain) such that \( f(x) = 2 \)?
Set \( x^2 = 2 \).
\( \implies x = \pm\sqrt{2} \)
Since \( \pm\sqrt{2} \) are not integers, there is no \( x \in Z \) such that \( f(x) = 2 \).
Therefore, \( f \) is not onto (surjective).
**Conclusion for (b):** The function \( f(x) = x^2 \) from \( Z \) to \( Z \) is neither injective nor surjective.

**(c) For f : R → R given by f (x) = x²**
*Injectivity (one-one):*
Consider \( 1 \in R \) and \( -1 \in R \).
\( f(1) = 1^2 = 1 \)
\( f(-1) = (-1)^2 = 1 \)
Different elements (1 and -1) from the domain \( R \) map to the same image (1) in the codomain \( R \).
Therefore, \( f \) is not one-one (it is many-one).
*Surjectivity (onto):*
Consider \( -1 \in R \) (codomain). Is there an \( x \in R \) (domain) such that \( f(x) = -1 \)?
Set \( x^2 = -1 \).
There is no real number \( x \) whose square is \( -1 \).
Therefore, \( f \) is not onto (surjective).
**Conclusion for (c):** The function \( f(x) = x^2 \) from \( R \) to \( R \) is neither injective nor surjective.

**(d) For f : N → N given by f(x) = x³**
*Injectivity (one-one):*
Let \( x, y \in N \) such that \( f(x) = f(y) \).
\( \implies x^3 = y^3 \)
For natural numbers, if their cubes are equal, the numbers themselves must be equal.
\( \implies x = y \)
Therefore, \( f \) is one-one (injective).
*Surjectivity (onto):*
Consider \( 2 \in N \) (codomain). Is there an \( x \in N \) (domain) such that \( f(x) = 2 \)?
Set \( x^3 = 2 \).
\( \implies x = \sqrt[3]{2} \)
Since \( \sqrt[3]{2} \) is not a natural number, there is no \( x \in N \) such that \( f(x) = 2 \).
Therefore, \( f \) is not onto (surjective).
**Conclusion for (d):** The function \( f(x) = x^3 \) from \( N \) to \( N \) is injective but not surjective.

**(e) For f : Z → Z given by f(x) = x³**
*Injectivity (one-one):*
Let \( x, y \in Z \) such that \( f(x) = f(y) \).
\( \implies x^3 = y^3 \)
\( \implies x^3 - y^3 = 0 \)
\( \implies (x - y)(x^2 + xy + y^2) = 0 \)
For integers, the expression \( x^2 + xy + y^2 \) is zero only when \( x=0 \) and \( y=0 \). Otherwise, it is positive if \( x \ne y \). Therefore, to satisfy \( x^3 - y^3 = 0 \), we must have \( x - y = 0 \).
\( \implies x = y \)
Therefore, \( f \) is one-one (injective).
*Surjectivity (onto):*
Consider \( 2 \in Z \) (codomain). Is there an \( x \in Z \) (domain) such that \( f(x) = 2 \)?
Set \( x^3 = 2 \).
\( \implies x = \sqrt[3]{2} \)
Since \( \sqrt[3]{2} \) is not an integer, there is no \( x \in Z \) such that \( f(x) = 2 \).
Therefore, \( f \) is not onto (surjective).
**Conclusion for (e):** The function \( f(x) = x^3 \) from \( Z \) to \( Z \) is injective but not surjective.
In simple words: This question explores how changing the set of numbers (domain and codomain) affects if a function is one-one or onto. For \( x^2 \), it's often not one-one because negative and positive numbers can give the same square. For \( x^3 \), it is usually one-one. Whether they are "onto" depends on if all target numbers can be reached, which is not always true for natural numbers or integers when using powers.

🎯 Exam Tip: Pay close attention to the domain and codomain when analyzing injectivity and surjectivity. A function's properties can change dramatically when the sets \( N \), \( Z \), \( Q \), or \( R \) are used.

Question 7. Show that the function f : W → W defined by f(x) = { x+1, if x is even ; x-1, if x is odd } is a bijective function.
Answer:
We are given a function \( f : W \rightarrow W \) defined by: \[ f(x) = \begin{cases} x+1, & \text{if } x \text{ is even} \\ x-1, & \text{if } x \text{ is odd} \end{cases} \] Here, \( W \) represents the set of whole numbers \( \{0, 1, 2, 3, \dots\} \).

**To prove injectivity (one-one):**
We consider four cases for \( x, y \in W \):
*Case 1: \( x \) and \( y \) are both even.*
Let \( f(x) = f(y) \).
\( \implies x+1 = y+1 \)
\( \implies x = y \)
*Case 2: \( x \) and \( y \) are both odd.*
Let \( f(x) = f(y) \).
\( \implies x-1 = y-1 \)
\( \implies x = y \)
*Case 3: \( x \) is odd and \( y \) is even.*
In this case, \( f(x) = x-1 \) (which is an even number) and \( f(y) = y+1 \) (which is an odd number).
An even number can never be equal to an odd number. So, \( f(x) \neq f(y) \).
This means if \( x \neq y \), then \( f(x) \neq f(y) \).
*Case 4: \( x \) is even and \( y \) is odd.*
Similar to Case 3, \( f(x) = x+1 \) (an odd number) and \( f(y) = y-1 \) (an even number).
So, \( f(x) \neq f(y) \).

Combining all four cases, we see that if \( f(x) = f(y) \), then \( x \) must be equal to \( y \). Therefore, \( f \) is one-one (injective).

**To prove surjectivity (onto):**
Let \( y \) be any arbitrary element in the codomain \( W \).
*If \( y \) is an even number:*
Consider \( x = y+1 \). Since \( y \) is even, \( y+1 \) is an odd number. Also, if \( y \in W \), then \( y+1 \in W \).
Then \( f(x) = f(y+1) = (y+1)-1 = y \) (because \( y+1 \) is odd, we use the second rule).
*If \( y \) is an odd number:*
Consider \( x = y-1 \). Since \( y \) is odd, \( y-1 \) is an even number. Also, if \( y \in W \) (and \( y \ge 1 \)), then \( y-1 \in W \).
Then \( f(x) = f(y-1) = (y-1)+1 = y \) (because \( y-1 \) is even, we use the first rule).
In both cases, for any \( y \in W \), we found an \( x \in W \) such that \( f(x) = y \).
Therefore, \( f \) is onto (surjective).

Since \( f \) is both one-one and onto, it is a bijective function.
In simple words: This function swaps numbers: even numbers go up by one, and odd numbers go down by one. We proved it's "one-one" because different starting numbers always give different results. We also proved it's "onto" because every whole number can be reached as a result: if you want an even number, just pick the next odd number and subtract one; if you want an odd number, pick the previous even number and add one. This means all whole numbers are covered in both directions, making it "bijective".

🎯 Exam Tip: For piecewise-defined functions, you must analyze each case (e.g., even/odd, positive/negative) separately for both injectivity and surjectivity. Pay extra attention to boundary conditions (like 0) and the domains/codomains.

Question 8. Show that the function f : R → R given by f(x) = cos x for all x ∈ R is neither one-one nor onto.
Answer:
We are given a function \( f : R \rightarrow R \) defined by \( f(x) = \cos x \).

**To prove "not one-one" (many-one):**
A function is not one-one if different inputs can lead to the same output.
Consider \( x_1 = 0 \) and \( x_2 = 2\pi \) (both are in the domain \( R \)).
\( f(0) = \cos(0) = 1 \)
\( f(2\pi) = \cos(2\pi) = 1 \)
Since \( f(0) = f(2\pi) \) but \( 0 \neq 2\pi \), the function \( f \) is not one-one (it is many-one). This is true for all \( 2n\pi \) where \( n \) is an integer.

**To prove "not onto" (into):**
A function is not onto if there is at least one element in the codomain that has no pre-image.
The range of the cosine function is \( [-1, 1] \). This means that for any \( x \in R \), \( -1 \le \cos x \le 1 \).
Consider an element \( 2 \in R \) (codomain). Is there an \( x \in R \) (domain) such that \( f(x) = 2 \)?
Set \( \cos x = 2 \).
There is no real number \( x \) for which \( \cos x = 2 \) (or any value outside \( [-1, 1] \)).
Therefore, \( 2 \) has no pre-image in the domain \( R \).
Hence, the function \( f \) is not onto (surjective).

Since \( f \) is neither one-one nor onto, it is neither injective nor surjective.
In simple words: The cosine function takes real numbers and gives a result between -1 and 1. It's "not one-one" because different angles (like 0 and 360 degrees) give the same cosine value (1). It's also "not onto" because you can never get an output like 2 or -5 from a cosine function; the outputs are always limited to values between -1 and 1. So, it fails both conditions.

🎯 Exam Tip: For trigonometric functions, remember their periodic nature (which causes them to be many-one) and their bounded range (which causes them to be not onto unless the codomain is restricted to their range).

Question 9. Show that the function f: [0, ∞ ) → [0, ∞) defined by f(x) = \( \frac{2x}{1+2x} \) is
(a) one-one and onto
(b) one-one but not onto
(c) not one-one but onto
(d) neither one-one nor onto
Answer: (b) one-one but not onto
In simple words: This function takes any non-negative number and changes it. It is "one-one" because if two inputs give the same result, those inputs must have been the same number. However, it is "not onto" because you can never get the number 1 as an output from this function, even though 1 is in the set of allowed output numbers. So, some numbers in the target set are missed.

🎯 Exam Tip: For rational functions (fractions with variables), always check the denominator for values that would make the function undefined. This can give clues about the function's domain and range. Also, testing a specific value for the "not onto" part is often quicker than a full proof.

Question 10. If f : R → R be a function defined by f(x) = 2x³ – 5, show that the function f is a bijective function.
Answer:
We are given a function \( f : R \rightarrow R \) defined by \( f(x) = 2x^3 - 5 \).

**To prove injectivity (one-one):**
Let \( x, y \in R \) such that \( f(x) = f(y) \).
\( \implies 2x^3 - 5 = 2y^3 - 5 \)
\( \implies 2x^3 = 2y^3 \)
\( \implies x^3 = y^3 \)
For real numbers, if their cubes are equal, the numbers themselves must be equal.
\( \implies x = y \)
Since \( f(x) = f(y) \) implies \( x = y \), the function \( f \) is one-one (injective).

**To prove surjectivity (onto):**
Let \( y \) be any arbitrary element in the codomain \( R \). We need to find an \( x \in R \) such that \( f(x) = y \).
Set \( y = 2x^3 - 5 \).
\( \implies y + 5 = 2x^3 \)
\( \implies \frac{y + 5}{2} = x^3 \)
\( \implies x = \sqrt[3]{\frac{y + 5}{2}} \)
For any real number \( y \), \( \frac{y+5}{2} \) is a real number, and its cube root \( \sqrt[3]{\frac{y+5}{2}} \) is also always a real number. This means for every \( y \) in the codomain, there exists a corresponding \( x \) in the domain such that \( f(x) = y \).
Therefore, the function \( f \) is onto (surjective).

Since \( f \) is both one-one and onto, it is a bijective function.
In simple words: This function takes a number, cubes it, multiplies by two, and then subtracts five. It's "one-one" because if you get the same answer, you must have started with the same number. It's "onto" because every real number can be an output; you can always find a starting number that will give you any specific target number. This makes it a "bijective" function.

🎯 Exam Tip: A function involving an odd power of x (like \( x^3 \)) combined with linear operations (multiplication, addition, subtraction) tends to be bijective over real numbers because odd powers preserve uniqueness and cover all real values.

Question 11. Let f : R → R be defined as f(x) = x⁵. Show that it is a bijective function.
Answer:
We are given a function \( f : R \rightarrow R \) defined by \( f(x) = x^5 \).

**To prove injectivity (one-one):**
Let \( x, y \in R \) such that \( f(x) = f(y) \).
\( \implies x^5 = y^5 \)
\( \implies x^5 - y^5 = 0 \)
\( \implies (x - y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4) = 0 \)
For real numbers, the expression \( (x^4 + x^3y + x^2y^2 + xy^3 + y^4) \) is zero only if \( x=0 \) and \( y=0 \). Otherwise, it is positive. Therefore, for \( x^5 - y^5 = 0 \), we must have \( x - y = 0 \).
\( \implies x = y \)
Since \( f(x) = f(y) \) implies \( x = y \), the function \( f \) is one-one (injective).

**To prove surjectivity (onto):**
Let \( y \) be any arbitrary element in the codomain \( R \). We need to find an \( x \in R \) such that \( f(x) = y \).
Set \( y = x^5 \).
\( \implies x = \sqrt[5]{y} \)
For any real number \( y \), its fifth root \( \sqrt[5]{y} \) is always a real number. This means for every \( y \) in the codomain, there exists a corresponding \( x \) in the domain such that \( f(x) = y \).
Therefore, the function \( f \) is onto (surjective).

Since \( f \) is both one-one and onto, it is a bijective function.
In simple words: This function takes a number and raises it to the power of five. It's "one-one" because different starting numbers always give different results when raised to an odd power. It's "onto" because any real number can be created as an output by taking its fifth root as the input. This perfect pairing means it's a "bijective" function.

🎯 Exam Tip: Functions with odd powers (like \( x^3, x^5, \dots \)) are generally bijective when mapping real numbers to real numbers, as they are strictly increasing or decreasing and their inverse functions (odd roots) are also well-defined over real numbers.

Question 12. A mapping f : N → N, where N is the set of natural numbers is defined as f(n) = { n², for n is odd ; 2n+1, for n is even } for n ∈ N. Show that/is neither injective nor surjective.
Answer:
We are given a function \( f : N \rightarrow N \) defined by: \[ f(x) = \begin{cases} x^2, & \text{if } x \text{ is odd} \\ 2x+1, & \text{if } x \text{ is even} \end{cases} \] Here, \( N \) represents the set of natural numbers \( \{1, 2, 3, \dots\} \).

**To prove "not injective" (many-one):**
We need to find two different elements in the domain \( N \) that map to the same element in the codomain \( N \).
Consider \( x_1 = 3 \) (which is odd).
\( f(3) = 3^2 = 9 \)
Consider \( x_2 = 4 \) (which is even).
\( f(4) = 2(4) + 1 = 8 + 1 = 9 \)
Here, \( f(3) = f(4) \) but \( 3 \neq 4 \). Therefore, the function \( f \) is not one-one (it is many-one).

**To prove "not surjective" (into):**
We need to find an element in the codomain \( N \) that has no pre-image in the domain \( N \).
Consider \( 2 \in N \) (codomain). Let's see if there's any \( x \in N \) such that \( f(x) = 2 \).
*Case 1: If \( x \) is odd.*
Set \( f(x) = x^2 = 2 \).
\( \implies x = \pm\sqrt{2} \)
Since \( \pm\sqrt{2} \) are not natural numbers, there is no odd \( x \in N \) that maps to \( 2 \).
*Case 2: If \( x \) is even.*
Set \( f(x) = 2x+1 = 2 \).
\( \implies 2x = 1 \)
\( \implies x = \frac{1}{2} \)
Since \( \frac{1}{2} \) is not a natural number, there is no even \( x \in N \) that maps to \( 2 \).
Since in neither case did we find an \( x \in N \) that maps to \( 2 \), the element \( 2 \) in the codomain has no pre-image.
Therefore, the function \( f \) is not onto (surjective).

Since \( f \) is neither injective nor surjective, it is neither one-one nor onto.
In simple words: This function acts differently based on whether the input number is odd or even. It's "not one-one" because some different starting numbers (like 3 and 4) end up giving the same result (9). It's also "not onto" because some numbers in the target set (like 2) can never be produced as an output from this function. So, it fails both conditions for a "bijective" function.

🎯 Exam Tip: When testing for "not one-one", try a small set of numbers that could potentially map to the same value (e.g., consecutive odd/even numbers or numbers with similar properties). For "not onto", look for small integers in the codomain that are clearly outside the function's potential outputs.