CBSE Class 12 Mathematics Relations And Functions Assignment Set 05

I. Multiple Choice Questions (MCQs) Choose the correct answer from the given options.

Question. R is a relation over the set N × N and it is defined by (a, b) R(c, d) ⇒ a + d = b + c, then R is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) equivalence relation

Answer: D

Question. Let R be a relation over the set of straight lines in a plane such that l₁ R l₂ such that l₁ ⊥ l₂, then R is
(a) symmetric
(b) reflexive
(c) transitive
(d) equivalence relation

Answer: A

Question. A relation R is defined on plane A of triangles in a given planes defined as R = {(T₁, T₂) : T₁ @ T₂ and T₁, T₂ ∈ A}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) equivalence

Answer: D

Question. Consider the non-empty set consisting of children in a family and a relation R defined as a Rb if a is a brother of b. Then R is
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) neither symmetric nor transitive
(d) both symmetric and transitive.

Answer: B

Question. Let R be the relation on the set R of all real numbers defined by a R b iff |a – b| ≤ 1. Then R is
(a) reflexive and symmetric
(b) symmetric only
(c) transitive only
(d) anti symmetric only

Answer: A

Question. A relation R = {(a, b) : a divides b} on the set N of all natural numbers is
(a) reflexive
(b) symmetric
(c) reflexive and transitive
(d) symmetric and transitive

Answer: C

Question. Let R be a relation on the set N defined by {(x, y) : x, y ∈ N, 2x + y = 41}. Then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) None of these.

Answer: D

Question. Let S be the set of all real numbers. Then the relation R = {(a, b) : 1 + ab > 0} on S is
(a) reflexive and symmetric but not transitive.
(b) reflexive and transitive but not symmetric.
(c) symmetric and transitive but not reflexive.
(d) reflexive, symmetric and transitive.

Answer: A

Question. Let W denote the words in the English dictionary. The relation R is defined by R = {(x, y) ∈ W × W : the words x and y have at least one letter in common}. Then R is
(a) not reflexive, symmetric and transitive
(b) reflexive, symmetric and not transitive
(c) reflexive, symmetric and transitive
(d) reflexive, transitive and not symmetric

Answer: B

Question. On the set N of all natural numbers define the relation R by aRb if and only if the G.C.D. of a and b is 2, then R is
(a) reflexive but not symmetric
(b) symmetric only
(c) reflexive and transitive
(d) reflexive, symmetric and transitive.

Answer: B

II. Very Short Answer Type Questions

Question. State the reason why the relation R = {(a, b) : a ≥ b²} defined in the set of real numbers is not reflexive.
Answer: R is not reflexive because a ≥ b² does not imply a ≥ a² ∀ a.

Question. State the reason of the relation R in the set {1, 2, 3} given by: R = {(1, 2), (2, 1)} not to be transitive.
Answer: Relation R is not transitive because: (1, 2), (2, 1) ∈ R but (1, 1) ∉ R.

Question. If R = {(x, y) : x + 2y = 8} is a relation on N. Write the domain of R.
Answer: Here R = {(2, 3), (4, 2), (6, 1)},
Domain of R = {2, 4, 6}

Question. Let R be the equivalence relation in the set A = {0, 1, 2, 3, 4, 5} given by R = {(a, b) : 2 divides (a – b); a, b ∈ A}. Write the equivalence class of [0].
Answer: Equivalence class of [0] = {0, 2, 4}

Question. Let A = {1, 2, 3, 4} and R be a relation in set A given by R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1), (1, 3)} then show that the relation R is reflexive and symmetric.
Answer: Relation R is reflexive and symmetric because (1, 1),
(2, 2), (3, 3) ∈ R
⇒ R is reflexive and (1, 2), (2, 1), (3, 1), (1, 3) ∈ R
⇒ R is symmetric.

Question. Show that the relation R less than in the set of natural numbers is transitive.
Answer: Reason: Take a, b, c ∈ N such that a < b and b < c
⇒ a < c \ (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R.
Hence, relation R is transitive.

Question. For real numbers x and y, we write xRy ⇒ x – y + √2 is an irrational number, then show that the relation R is reflexive.
Answer: Relation R = x – y + 2 is reflexive because
x R x = x –x + √2 = √2 .

Question. A relation R = {(x, y) : x + y < 10; x, y ∈ A} defined in the set A = {1, 2, 3, 4}. Is relation R symmetric? Give reason.
Answer: Relation R is symmetric because
x + y < 10 ⇒ y + x < 10
⇒ (x, y) ∈ R ⇒ (y, x) ∈ R

Question. Take a relation R = {(a, b) : a = b, a, b ∈ A} defined on a set A of real numbers. Is R a reflexive relation? Give reason.
Answer: Relation R is reflexive because a = a
⇒ (a, a) ∈ R for all a ∈ A.

III. Short Answer Type Questions-I

Question. If set A = {1, 2, 3, 4}, set B = {a, b, c, d} and relation R = {(1, a), (2, b), (3, c), (4, d)} is a relation defined from set A to set B. Write relation R–1.
Answer: R^–1 = {(a, 1), (b, 2), (c, 3), (d, 4)}.

Question. A relation R defined in set A = {1, 2, 3, ...., 10} as R = {(x, y) : 3x – y = 0, x, y ∈ A}. Show that the relation is not reflexive.
Answer: Given 3x – y = 0. Put y = x in 3x –y = 0
It gives 3x – x = 0            ⇒ 2x = 0.
It is not true.
∴ R is not reflexive.

Question. Let A = {a, b, c} and a relation R is defined in set A as: R = {(a, a), (b, c), (a, b)}. Write the minimum number of ordered pairs be included in relation R to make it reflexive and symmetric.
Answer: To make relation R reflexive, include (b, b) and (c, c). Now, (a, b) ∈ R and (b, c) ∈ R but (b, a) and (c, b) are not included in R. Therefore, include (b, a) and (c, b) to make R symmetric.
Hence, to make R reflexive and symmetric include (b, b), (c, c), (b, a) and (c, b).

Question. Check whether the relation R defined in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y = x + 1, x, y ∈ A} is reflexive or symmetric.
Answer: (i) Relation: y = x + 1.
If y = x makes it true, R will be reflexive
∴ Put y = x in y = x + 1.
It gives x = x + 1
⇒ 0 = 1 (False)
∴ R is not reflexive.
(ii) Relation: y = x + 1.
Put x = y and y = x in y = x + 1.
It gives x = y + 1.
It is not same as y = x + 1.
Hence, R is no symmetric.

Question. Show that the relation: R = {(1, 2), (2, 3), (3, 2), (1, 3), (2, 1), (2, 2), (3, 3), (3, 1), (1, 1)} is reflexive, symmetric and transitive.
Answer: (i) Reflexive: R is reflexive because
{(1, 1), (2, 2), (3, 3)} ∈ R
(ii) Symmetric: R is symmetric because
{(1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)} ∈ R
(iii) Transitive: R is transitive because (1, 2) ∈ R and (2, 3) ∈ R
⇒ (1, 3) ∈ R and so on.

Question. Show that the relation: R ={(1, 1), (1, 2), (2, 1), (2, 2)} is reflexive, symmetric and transitive.
Answer: (i) Reflexive: R is reflexive because
{(1, 1), (2, 2)} ∈ R
(ii) Symmetric: R is symmetric because
{(1, 2), (2, 1)} ∈ R
(iii) Transitive: R is transitive because {1, 2) ∈ R and (2, 1) ∈ R
⇒ (1, 1) ∈ R and so on.

Question. Show that the relation: R = {(1, 1), (3, 3), (5, 5), (1, 3), (3, 1), (3, 5)} is reflexive and is neither symmetric nor transitive.
Answer: (i) Reflexive: R is reflexive because
{(1, 1), (3, 3), (5, 5)} ∈ R
(ii) Symmetric: R is not symmetric because (3, 5) ∈ R but (5, 3) ∉ R.
(iii) Transitive: R is not transitive because (1, 3) ∈ R and (3, 5)
∈ R does not imply (1, 5) ∈ R.

Question. Take a set A = {1, 3, 7} and find a relation R = A × A, then show that it is reflexive, symmetric and transitive.
Answer: A = {1, 3, 7}
R = A × A = {(1, 1), (1, 3), (1, 7), (3, 1), (3, 3), (3, 7), (7, 1), (7, 3), (7, 7)}
(i) R is reflexive because {(1, 1), (3, 3), (7, 7)} ∈ R
(ii) R is symmetric because
{(1, 3), (3, 1) (1, 7), (7, 1), (3, 7), (7, 3)} ∈ R.
(iii) R is transitive because (1, 3) ∈ R and (3, 7) ∈ R
⇒ (1, 7) ∈ R and so on.

Question. Relation R defined in a set of natural numbers N as: R = {(x, y) : x ≥ y ∀ x, y ∈ N}. Show that relation R is reflexive and transitive but not symmetric.
Answer: R = {(x, y) : x ≥ y, x, y ∈ N}
(i) R is reflexive: Since x ≥ x            ⇒ (x, x) ∈ R
⇒ R is reflexive.
(ii) R is not symmetric: x ≥ y does not imply y ≥ x.
∴ R is not symmetric.
(iii) Transitive: Take three natural numbers x, y and z such that
x ≥ y and y ≥ z            ⇒ x ≥ z

Question. Relation R defined in a set of natural numbers N as: R = {(a, b) : (a + b) is an even natural number, where a, b ∈ N}. Show that R is reflexive, symmetric and transitive.
Answer: R = {(a, b) : (a + b) is an even natural number and a, b ∈ N}
(i) R is reflexive: Take a ∈ N. Since (a + a) is an even number ∀ a ∈ N.
(a, a) ∈ R ⇒ R is reflexive.
(ii) Symmetric: Let a, b ∈ N.
If (a + b) is an even number then (b + a) is also an even number.
∴ (a, b) ∈ R ⇒ (b, a) ∈ R
⇒ R is symmetric.
(iii) Transitive: Since sum of two even numbers is even or sum of two odd numbers is even.
∴ Let a, b, c ∈ N are either all even numbers or odd numbers.
∴ (a + b) is even and (b + c) is even
⇒ (a + c) is also even
∴ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R
⇒ R is transitive.

Question. Relation R defined in a set of natural numbers N as: R = {(a, b) : (a – b) is an even natural number and a, b ∈ N}. Show that R is neither reflexive nor symmetric but transitive.
Answer: R = {(a, b) : (a – b) is an even natural number and a, b ∈ N}
(i) Reflexive: Since (a – a) = 0 (not a natural no.)
⇒ (a, a) ∉ R            ∴ R is not reflexive.
(ii) Symmetric: If (a – b) is even natural number then (b – a) is not an even natural number.
∴ (a, b) ∈ R whereas (b, a) ∉ R.
Hence, R is not symmetric.
(iii) Transitive: a, b, c ∈ N such that (a – b) and (b – c) are even natural numbers then (a – c) will also be an even natural numbers.
∴ (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R.
Hence, R is transitive.

Question. Relation R defined in a set of real numbers R as: R = {(x, y) : (x + y) is divisible by 5 ∀ x, y ∈ R}. Show that R is neither reflexive nor transitive but symmetric. Note: In a set of real numbers R, if relation R = {(a, b) : (a + b) is a multiple of 3, 4, 5, 6, …, n ∀ a, b ∈ R} then R is neither reflexive nor transitive but symmetric.
Answer: R = {(x, y) : (x + y) is divisible by 5 ∀ x, y ∈ R}
(i) Reflexive: Take x ∈ R.
Since (x + x) is not divisible by 5 for any real number x.
⇒ R is not reflexive.
(ii) Symmetric: Let x, y ∈ R be such that (x + y) is divisible by 5, then (y + x) is also divisible by 5.
∴ (x, y) ∈ R            ⇒ (y, x) ∈ R
Hence, R is symmetric.
(iii) Transitive: Take x, y, z ∈ R such that (x + y) and (y + z) are divisible by 5, then (x + z) may or may not be divisible by 5.
∴ (a, b) ∈ R and (b, c ) ∈ R does not imply (a, c) ∈ R.
Hence, R is not transitive.

Question. Show that the relation R in set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.
Answer: in set A = {1, 2, 3} is
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
(i) Reflexive: Since {(1, 1), (2, 2), (3, 3)} ∈ R
⇒ R is reflexive.
(ii) Symmetric: Since (1, 2) ∈ R but (2, 1) ∉ R and (2, 3) ∈ R but (3, 2) ∉ R.
⇒ R is not symmetric.
(iii) Transitive: Since (1, 2) ∈ R and (2, 3) ∈ R does not imply (1, 3) ∈ R.
Hence R is not transitive.

Question. Let L be the set of lines in a plane and R be the relation in L defined as: R = {(L₁, L₂) : L₁ is perpendicular to L₂, L₁ and L₂ ∈ L}. Show that R is symmetric but neither reflexive nor transitive.
Answer: R = {(L₁, L₂) : L₁ ⊥ L₂ and L_1, L₂ ∈ L}
(i) Reflexive: Let L₁ ∈ L. Since line L₁ cannot be perpendicular to itself
∴ (L₁, L₁) ∉ R
Hence, R is not reflexive.
(ii) Symmetric: Let L1, L₂ ∈ L. If line L₁ ⊥ L₂ then L₂ ⊥ L₁
Hence (L₁, L₂) ∈ R and (L₂, L₁) ∈ R
⇒ R is symmetric.
(iii) Transitive: Let L₁, L₂, L3 ∈ L and L₁ ⊥ L₂ and L₂ ⊥ L3 then
L₁ is not perpendicular to L3
∴ (L₁, L₂) ∈ R and (L₂, L3) ∈ R does not imply that (L₁, L3) ∈ R.
Hence, R is not transitive

Question 1. If f(x) = |x| and g(x) = [x]. Evaluate - (fog) (-5/3) - (gof) (-5/3)
Answer: Given functions are \(f(x) = |x|\) and \(g(x) = [x]\) (greatest integer function).
First, evaluate \(g\left(-\frac{5}{3}\right)\):
\(-\frac{5}{3} = -1.666...\)
\(g\left(-\frac{5}{3}\right) = \left[-\frac{5}{3}\right] = -2\)
Now evaluate \((fog)\left(-\frac{5}{3}\right) = f\left(g\left(-\frac{5}{3}\right)\right) = f(-2) = |-2| = 2\)
Next, evaluate \(f\left(-\frac{5}{3}\right)\):
\(f\left(-\frac{5}{3}\right) = \left|-\frac{5}{3}\right| = \frac{5}{3}\)
Now evaluate \((gof)\left(-\frac{5}{3}\right) = g\left(f\left(-\frac{5}{3}\right)\right) = g\left(\frac{5}{3}\right) = \left[\frac{5}{3}\right] = [1.666...] = 1\)
Substitute these values into the expression:
\(- (fog) \left(-\frac{5}{3}\right) - (gof) \left(-\frac{5}{3}\right) = -(2) - (1) = -3\)
In simple words: We first find the output of the inner function and plug it into the outer function using absolute value and greatest integer rules.

Exam Tip: Be very careful with negative numbers in the greatest integer function, as \([-1.67] = -2\), not \(-1\).

Question 2. If f(x) = e^{2x} and g(x) = \log\sqrt{x}, x > 0, find (i) fog (ii) gof (iii) f + g (iv) fg
Answer: Given \(f(x) = e^{2x}\) and \(g(x) = \log\sqrt{x} = \frac{1}{2}\log x\) for \(x > 0\).
(i) \((fog)(x) = f(g(x)) = f(\log\sqrt{x}) = e^{2\log\sqrt{x}} = e^{\log(\sqrt{x})^2} = e^{\log x} = x\)
(ii) \((gof)(x) = g(f(x)) = g(e^{2x}) = \log\sqrt{e^{2x}} = \log(e^x) = x\)
(iii) \((f + g)(x) = f(x) + g(x) = e^{2x} + \log\sqrt{x}\)
(iv) \((fg)(x) = f(x) \cdot g(x) = e^{2x} \log\sqrt{x}\)
In simple words: Composite functions combine two functions sequentially, while algebraic operations add or multiply their function outputs directly.

Exam Tip: Use logarithmic properties like \(\log(a^b) = b\log a\) and \(e^{\log x} = x\) to simplify composite expressions.

Question 3. Let N be the set of all natural numbers and let R be the relation on N \times N, defined by (a, b) R (c, d) \rightarrow ad = bc for all (a, b), (c, d) \in N \times N. Show that R is an equivalence relation on N \times N.
Answer: To show \(R\) is an equivalence relation, we must prove it is reflexive, symmetric, and transitive.
1. Reflexivity: For any \((a, b) \in N \times N\), we have \(ab = ba\) (since multiplication in natural numbers is commutative). Thus, \((a, b) R (a, b)\). Hence, \(R\) is reflexive.
2. Symmetry: Let \((a, b) R (c, d)\). Then \(ad = bc\). By commutativity of multiplication, \(cb = da\), which means \(bc = ad\), so \((c, d) R (a, b)\). Hence, \(R\) is symmetric.
3. Transitivity: Let \((a, b) R (c, d)\) and \((c, d) R (e, f)\). Then \(ad = bc\) and \(cf = de\).
Multiply both equations: \((ad)(cf) = (bc)(de) \implies acdf = bcde\).
Since \(c \in N\) (\(c \neq 0\)), we can cancel \(c\) from both sides: \(adf = bde\).
Since \(d \in N\) (\(d \neq 0\)), we can cancel \(d\) from both sides: \(af = be\), which implies \((a, b) R (e, f)\). Hence, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: We check three conditions (reflexive, symmetric, transitive) using basic cross-multiplication properties of numbers.

Question 4. Let A = \{x \in R : -1 \leq x \leq 1\} = B. Show that f : A \to B given by f(x) = x|x| is a bijection.
Answer: The function can be rewritten in piecewise form as:
\(f(x) = \begin{cases} -x^2, & -1 \leq x < 0 \\ x^2, & 0 \leq x \leq 1 \end{cases}\)
1. Injectivity (One-One):
Case 1: If \(x_1, x_2 < 0\), then \(f(x_1) = f(x_2) \implies -x_1^2 = -x_2^2 \implies x_1^2 = x_2^2 \implies x_1 = x_2\) (since both are negative).
Case 2: If \(x_1, x_2 \geq 0\), then \(f(x_1) = f(x_2) \implies x_1^2 = x_2^2 \implies x_1 = x_2\).
Case 3: If \(x_1 < 0\) and \(x_2 \geq 0\), then \(f(x_1) < 0\) and \(f(x_2) \geq 0\), so they cannot be equal unless \(x_1 = x_2 = 0\).
Thus, \(f\) is one-one.
2. Surjectivity (Onto):
For any \(y \in B = [-1, 1]\):
If \(y \in [-1, 0)\), we choose \(x = -\sqrt{-y} \in [-1, 0)\), then \(f(x) = -(-\sqrt{-y})^2 = -(-y) = y\).
If \(y \in [0, 1]\), we choose \(x = \sqrt{y} \in [0, 1]\), then \(f(x) = (\sqrt{y})^2 = y\).
Since every element \(y\) in the codomain has a pre-image in domain \(A\), \(f\) is onto.
Therefore, \(f\) is a bijection.
In simple words: A function is bijective if every output has one and only one input, which we prove by checking negative and positive intervals.

Exam Tip: Split absolute value functions into piecewise definitions before testing for one-one and onto properties.

Question 5. Let I be the set of integers. Define a relation R on I by aRb \rightarrow a - b is divisible by 5. Show that R is an equivalence relation.
Answer:
1. Reflexivity: For any \(a \in I\), \(a - a = 0\), which is divisible by 5 (since \(0 = 5 \times 0\)). Thus, \(aRa\). Hence, \(R\) is reflexive.
2. Symmetry: Let \(aRb\). Then \(a - b\) is divisible by 5, meaning \(a - b = 5k\) for some integer \(k\).
Then \(b - a = -(a - b) = -5k = 5(-k)\), which means \(b - a\) is divisible by 5. Thus, \(bRa\). Hence, \(R\) is symmetric.
3. Transitivity: Let \(aRb\) and \(bRc\). Then \(a - b = 5k_1\) and \(b - c = 5k_2\) for some integers \(k_1, k_2\).
Adding both equations: \((a - b) + (b - c) = 5k_1 + 5k_2 \implies a - c = 5(k_1 + k_2)\), which means \(a - c\) is divisible by 5. Thus, \(aRc\). Hence, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Divisibility by 5 satisfies reflexivity (difference is 0), symmetry (reversing sign keeps divisibility), and transitivity (adding differences maintains divisibility).

Exam Tip: Express divisibility using multiples of the integer (like \(5k\)) to make algebraic manipulation clear.

Question 6. (a) Define a binary relation * on Q as follows : a * b = a + b - ab ; a, b \in Q. Find the identity element of (Q, *)
(b) Test * for commutativity.
(c) If the binary relation * on Z is defined by a * b = a + b + 2, then write the identity element.
Answer:
(a) Let \(e \in Q\) be the identity element. Then for all \(a \in Q\), \(a * e = a\).
\(a + e - ae = a \implies e(1 - a) = 0 \implies e = 0\) (since this must hold for all \(a \in Q\)).
Thus, the identity element is \(0$.
(b) For commutativity, consider \(a * b = a + b - ab\) and \(b * a = b + a - ba\).
Since addition and multiplication are commutative in \(Q\), \(a + b - ab = b + a - ba\), so \(a * b = b * a\) for all \(a, b \in Q\). Thus, * is commutative.
(c) Let \(e \in Z\) be the identity element for \(a * b = a + b + 2\). Then \(a * e = a \implies a + e + 2 = a \implies e + 2 = 0 \implies e = -2\).
Thus, the identity element is \(-2\).
In simple words: The identity element is the value you operate with that leaves any number completely unchanged.

Exam Tip: To find the identity element \(e\), set up the equation \(a * e = a\) and solve for \(e\).

Question 7. Prove that the function f: N \to N defined by f(n) = n^2 + n + 1 is one one but not onto.
Answer:
1. Injectivity (One-One):
Let \(n_1, n_2 \in N\) such that \(f(n_1) = f(n_2)\).
\(n_1^2 + n_1 + 1 = n_2^2 + n_2 + 1 \implies n_1^2 - n_2^2 + n_1 - n_2 = 0 \implies (n_1 - n_2)(n_1 + n_2 + 1) = 0\).
Since \(n_1, n_2 \in N\), \(n_1 + n_2 + 1 \neq 0\). Therefore, \(n_1 - n_2 = 0 \implies n_1 = n_2\).
Thus, \(f\) is one-one.
2. Surjectivity (Onto):
The minimum value in codomain \(N\) is 1. For \(n = 1\), \(f(1) = 1^2 + 1 + 1 = 3\).
For \(n = 2\), \(f(2) = 2^2 + 2 + 1 = 7\).
Notice that there is no natural number \(n \in N\) such that \(f(n) = 2\), because for the smallest natural number \(n = 1\), \(f(1) = 3 > 2\).
Thus, the element \(2 \in N\) (codomain) has no pre-image in domain \(N\).
Hence, \(f\) is not onto.
In simple words: Different inputs give different outputs (one-one), but small numbers like 2 are never hit by any natural number input (not onto).

Exam Tip: To disprove onto, pick a small integer in the codomain (like 2) and show no natural number produces it.

Question 8. Let R be the set of real numbers. Show that the function f: R \to R : f(x) = \cos x is neither one one nor onto.
Answer:
1. One-One Test:
Consider \(x_1 = 0\) and \(x_2 = 2\pi\). Both belong to \(R\).
\(f(0) = \cos(0) = 1\) and \(f(2\pi) = \cos(2\pi) = 1\).
Since \(f(0) = f(2\pi)\) but \(0 \neq 2\pi\), the function is not one-one.
2. Onto Test:
We know that the range of \(f(x) = \cos x\) is \([-1, 1]\).
The codomain is given as \(R\) (Real numbers).
Since Range \(\neq\) Codomain (for instance, \(2 \in R\) has no pre-image as \(\cos x = 2\) has no solution), the function is not onto.
Thus, \(f(x) = \cos x\) is neither one-one nor onto.
In simple words: Cosine repeats its values periodically (so not one-one) and only outputs values between -1 and 1 (so not onto).

Exam Tip: Provide a counter-example (like 0 and \(2\pi\) for one-one, and a number outside \([-1, 1]\) for onto).

Question 9. Let f, g : |x| + x and g(x) = |x| - x, for every x \in R. Then find fog and gof.
Answer: Given \(f(x) = |x| + x\) and \(g(x) = |x| - x\) for \(x \in R\).
First, let us define \(f(x)\) and \(g(x)\) piecewise:
If \(x \geq 0\): \(f(x) = x + x = 2x\) and \(g(x) = x - x = 0\).
If \(x < 0\): \(f(x) = -x + x = 0\) and \(g(x) = -x - x = -2x\).
Now, let us find \((fog)(x) = f(g(x))\):
Case 1: If \(x \geq 0\), \(g(x) = 0\), so \(f(g(x)) = f(0) = |0| + 0 = 0\).
Case 2: If \(x < 0\), \(g(x) = -2x > 0\), so \(f(g(x)) = f(-2x) = |-2x| + (-2x) = -2x - 2x = -4x\) (since \(-2x > 0\)).
Thus, \((fog)(x) = 0\) for \(x \geq 0\) and \(-4x\) for \(x < 0\).
Now, let us find \((gof)(x) = g(f(x))\):
Case 1: If \(x \geq 0\), \(f(x) = 2x \geq 0\), so \(g(f(x)) = g(2x) = |2x| - (2x) = 2x - 2x = 0\).
Case 2: If \(x < 0\), \(f(x) = 0\), so \(g(f(x)) = g(0) = |0| - 0 = 0\).
Thus, \((gof)(x) = 0\) for all \(x \in R\).
In simple words: Substitute the piecewise definition of the inner function into the outer function carefully across positive and negative domains.

Exam Tip: Always break absolute value functions into cases based on whether the input is positive or negative.

Question 10. Let f : N \to R be a function defined as f(x) = 4x^2 + 12x + 15. Show that f : N \to \text{Range } f is invertible. Find the inverse of f.
Answer: Let \(y \in \text{Range } f\). Then \(y = 4x^2 + 12x + 15\) for some \(x \in N\).
Complete the square for the quadratic expression in \(x\):
\(y = (2x)^2 + 2(2x)(3) + 3^2 - 9 + 15 \implies y = (2x + 3)^2 + 6\)
Rearranging to solve for \(x\):
\((2x + 3)^2 = y - 6 \implies 2x + 3 = \sqrt{y - 6}\) (taking positive square root since \(x \in N\))
\(2x = \sqrt{y - 6} - 3 \implies x = \frac{\sqrt{y - 6} - 3}{2}\)
Since each \(y\) produces a unique \(x \in N\), the function is one-one and onto onto its range, hence invertible.
The inverse function \(f^{-1} : \text{Range } f \to N\) is given by:
\(f^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2}\)
In simple words: To find the inverse, set \(y = f(x)\), solve the quadratic equation for \(x\) in terms of \(y\), and replace \(x\) with \(f^{-1}(y)\).

Exam Tip: Completing the square is the standard technique for inverting quadratic functions.

Question 11. Answer the following short answers:
(i) Let f : R \to R be defined as f(x) = \frac{x}{1 + x^2}. Then find (fofof)(x)
(ii) If f(x) = 4 - (x - 7)^3, then find f^{-1}(x)
(iii) Let the binary operation * on N is defined by m * n = \text{g.c.d}(m, n). Determine whether * is commutative, associative. Write the value of (22) * (4)
Answer:
(i) First find \(f(f(x))\):
\(f(f(x)) = \frac{\frac{x}{1+x^2}}{1 + \left(\frac{x}{1+x^2}\right)^2} = \frac{\frac{x}{1+x^2}}{\frac{(1+x^2)^2 + x^2}{(1+x^2)^2}} = \frac{x(1+x^2)}{1 + 2x^2 + x^4 + x^2} = \frac{x(1+x^2)}{1 + 3x^2 + x^4}\)
Now find \((fofof)(x) = f(f(f(x)))\):
Applying \(f\) to the above result gives \((fofof)(x) = \frac{x}{1 + 3x^2}\) (after algebraic simplification).
(ii) Let \(y = 4 - (x - 7)^3\). Solve for \(x\):
\((x - 7)^3 = 4 - y \implies x - 7 = (4 - y)^{1/3} \implies x = 7 + (4 - y)^{1/3}\).
Thus, \(f^{-1}(x) = 7 + (4 - x)^{1/3}\).
(iii) Commutativity: Since \(\text{g.c.d}(m, n) = \text{g.c.d}(n, m)\) for all \(m, n \in N\), the operation * is commutative.
Associativity: Since \(\text{g.c.d}(m, \text{g.c.d}(n, k)) = \text{g.c.d}(\text{g.c.d}(m, n), k)\) for all \(m, n, k \in N\), the operation * is associative.
Value of \((22) * (4)\): \(\text{g.c.d}(22, 4) = 2\).
In simple words: Triple composites require step-by-step substitution, inverse functions swap variables and solve, and GCD is always commutative and associative.

Exam Tip: For GCD and LCM operations, commutativity and associative properties always hold true on natural numbers.