CBSE Class 12 Mathematics Relations And Functions Class Test Set 16

Question 1. If \(f: R \to R\) is defined by \(f(x) = x^2 - 3x + 2\). Find \(f(f(x))\)
Answer: Given function is \(f(x) = x^2 - 3x + 2\)
We need to find \(f(f(x))\), which means substituting \(f(x)\) into itself.
\(f(f(x)) = f(x^2 - 3x + 2)\)
Substitute \(x^2 - 3x + 2\) in place of \(x\) in the given function:
\(= (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2\)
Expand the terms:
\(= (x^4 + 9x^2 + 4 - 6x^3 + 4x^2 - 12x) - (3x^2 - 9x + 6) + 2\)
\(= x^4 - 6x^3 + 13x^2 - 12x + 4 - 3x^2 + 9x - 6 + 2\)
Combine like terms:
\(= x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)\)
\(= x^4 - 6x^3 + 10x^2 - 3x\)
In simple words: Substitute the entire function expression into the function itself and simplify algebraically.

Exam Tip: Be careful with signs when expanding squares of trinomials and distributing negative coefficients.

Question 2. Show that the function \(f: R \to R\) given by \(f(x) = x^3\) is injective. Is it surjective also?
Answer: Injectivity (One-to-One):
Let \(x_1, x_2 \in R\) such that \(f(x_1) = f(x_2)\).
\(\Rightarrow x_1^3 = x_2^3\)
Taking the cube root on both sides gives:
\(x_1 = x_2\)
Since \(f(x_1) = f(x_2) \Rightarrow x_1 = x_2\), the function \(f\) is injective (one-to-one).

Surjectivity (Onto):
Let \(y \in R\) be any element in the co-domain \(R\).
For \(f\) to be surjective, there must exist an element \(x \in R\) such that \(f(x) = y\).
\(\Rightarrow x^3 = y\)
\(\Rightarrow x = y^{1/3}\)
For any real number \(y\), its cube root \(y^{1/3}\) is also a unique real number in the domain \(R\).
Thus, for every \(y \in R\), there exists \(x = y^{1/3} \in R\) such that \(f(x) = (y^{1/3})^3 = y\).
Therefore, the function \(f\) is surjective.
In simple words: A cubic function in real numbers produces unique outputs for unique inputs and covers all possible real numbers as outputs.

Exam Tip: To prove injectivity, start with \(f(x_1) = f(x_2)\) and show \(x_1 = x_2\). To prove surjectivity, express \(x\) in terms of \(y\) and verify it belongs to the domain.

Question 3. Let \(*\) be the binary operation on \(N\) defined by \(a * b = \text{H.C.F of } a \text{ and } b\). Is \(*\) commutative? Is \(*\) associative?
Answer: Commutativity:
For any \(a, b \in N\),
\(a * b = \text{H.C.F of } a \text{ and } b\)
\(b * a = \text{H.C.F of } b \text{ and } a\)
Since the Highest Common Factor of two numbers is independent of their order (i.e., H.C.F of \(a\) and \(b\) is equal to H.C.F of \(b\) and \(a\)), we have:
\(a * b = b * a\)
Therefore, the binary operation \(*\) is commutative.

Associativity:
For any \(a, b, c \in N\),
\((a * b) * c = (\text{H.C.F of } a \text{ and } b) * c = \text{H.C.F of } (a, b, \text{ and } c)\)
Also,
\(a * (b * c) = a * (\text{H.C.F of } b \text{ and } c) = \text{H.C.F of } (a, b, \text{ and } c)\)
Since \((a * b) * c = a * (b * c)\), the binary operation \(*\) is associative.
In simple words: The HCF of numbers does not depend on the order in which you take them or how you group them.

Exam Tip: Use fundamental properties of number theory (like HCF and LCM) directly when testing binary operations.

Question 4. Let \(L\) be the set of all lines in the xy-plane and \(R\) be the relation in \(L\) defined by \(R = \{(L_1, L_2) : L_1 \parallel L_2 \text{ for } L_1, L_2 \in L\}\). Show that \(R\) is an equivalence relation. Find the set of all lines related to the line \(y = 2x + 5\).
Answer: 1. Reflexivity:
For any line \(L_1 \in L\), every line is parallel to itself (i.e., \(L_1 \parallel L_1\)).
Therefore, \((L_1, L_1) \in R\) for all \(L_1 \in L\). Hence, \(R\) is reflexive.

2. Symmetry:
Let \((L_1, L_2) \in R\).
\(\Rightarrow L_1 \parallel L_2\)
\(\Rightarrow L_2 \parallel L_1\) (Since if a line is parallel to another, the second is parallel to the first)
\(\Rightarrow (L_2, L_1) \in R\). Hence, \(R\) is symmetric.

3. Transitivity:
Let \((L_1, L_2) \in R\) and \((L_2, L_3) \in R\).
\(\Rightarrow L_1 \parallel L_2\) and \(L_2 \parallel L_3\)
\(\Rightarrow L_1 \parallel L_3\) (Two distinct lines parallel to the same line are parallel to each other)
\(\Rightarrow (L_1, L_3) \in R\). Hence, \(R\) is transitive.

Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.

Set of all lines related to \(y = 2x + 5\):
The relation \(R\) consists of parallel lines. Two lines are parallel if and only if they have the same slope.
The given line is \(y = 2x + 5\), which has a slope of \(m = 2\).
All lines parallel to this line will have the same slope \(2\) but can have any \(y\)-intercept \(c\).
Therefore, the set of all lines related to \(y = 2x + 5\) is given by:
\(\{y = 2x + c : c \in R\}\)
In simple words: Parallel lines form equivalence classes where every line in the same class shares the exact same slope.

Exam Tip: For equivalence relations, verify all three conditions separately with clear statements. For the related set, identify the invariant property (slope).