CBSE Class 12 Mathematics Relations And Functions Class Test Set 17

Relations and Functions

Q1. If \(f : R \to R\) is defined by \(f(x) = x^2 - 3x + 2\). Find \(f(f(x))\).
Answer: Given function is \(f(x) = x^2 - 3x + 2\).
To find \(f(f(x))\), substitute \(f(x)\) in place of \(x\) in the function \(f(x)\):
\(f(f(x)) = (f(x))^2 - 3(f(x)) + 2\)
Substitute \(f(x) = x^2 - 3x + 2\):
\(f(f(x)) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2\)
Expand each term:
\((x^2 - 3x + 2)^2 = x^4 + 9x^2 + 4 - 6x^3 + 4x^2 - 12x = x^4 - 6x^3 + 13x^2 - 12x + 4\)
\(-3(x^2 - 3x + 2) = -3x^2 + 9x - 6\)
Add all terms together including the constant \(+2\):
\(f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2\)
\(f(f(x)) = x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)\)
\(f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x\)
In simple words: Replace every \(x\) in the original function formula with the formula of \(f(x)\) itself and simplify the algebraic expression.

Exam Tip: Be very careful when expanding the square of a trinomial to avoid algebraic sign errors.

Q2. 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 \(f(x_1) = f(x_2)\).
Then, \(x_1^3 = x_2^3\)
Taking the cube root on both sides gives \(x_1 = x_2\).
Since \(f(x_1) = f(x_2) \implies x_1 = x_2\) for all \(x_1, x_2 \in R\), 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 \(f(x) = y\), which means \(x^3 = y\).
Solving for \(x\), we get \(x = y^{1/3}\).
Since the domain and co-domain are the set of all real numbers \(R\), for every real number \(y\), its cube root \(y^{1/3}\) is also a unique real number in \(R\).
Thus, for every \(y \in R\), there exists \(x = y^{1/3} \in R\) such that \(f(x) = (y^{1/3})^3 = y\).
Therefore, \(f\) is surjective (onto).
In simple words: Distinct real numbers have distinct cubes, making it injective, and every real number has a real cube root, making it surjective.

Exam Tip: For polynomial functions of odd degrees from \(R \to R\), the function is always both injective and surjective.

Q3. 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 the order of the numbers, \(\text{H.C.F}(a, b) = \text{H.C.F}(b, a)\).
Therefore, \(a * b = b * a\). Hence, \(*\) is commutative.

Associativity:
For any \(a, b, c \in N\),
\((a * b) * c = (\text{H.C.F of } a \text{ and } b) * c = \text{H.C.F}(a, b, c)\)
\(a * (b * c) = a * (\text{H.C.F of } b \text{ and } c) = \text{H.C.F}(a, b, c)\)
Since \((a * b) * c = a * (b * c)\) for all \(a, b, c \in N\), the operation \(*\) is associative.
In simple words: The HCF of numbers does not change if you swap them or group them differently, so the operation is both commutative and associative.

Exam Tip: State the symmetric properties of standard number-theoretic operations clearly to justify commutativity and associativity quickly.

Q4. Let \(L\) be the set of all lines in \(xy\) plane and \(R\) be the relation in \(L\) defined by \(R = \{(L_1, L_2) : L_1 \parallel L_2\}\) for all \(L_1, L_2 \in L\). Show \(R\) is an equivalence relation. Find the set of all lines related to the line \(y = 2x + 5\).
Answer: Reflexivity:
For any line \(L_1 \in L\), every line is parallel to itself, i.e., \(L_1 \parallel L_1\).
Thus, \((L_1, L_1) \in R\) for all \(L_1 \in L\). Hence, \(R\) is reflexive.

Symmetry:
Let \((L_1, L_2) \in R\).
Then \(L_1 \parallel L_2\), which implies that \(L_2 \parallel L_1\).
Thus, \((L_2, L_1) \in R\). Hence, \(R\) is symmetric.

Transitivity:
Let \((L_1, L_2) \in R\) and \((L_2, L_3) \in R\).
Then \(L_1 \parallel L_2\) and \(L_2 \parallel L_3\).
Since lines parallel to the same line are parallel to each other, \(L_1 \parallel L_3\).
Thus, \((L_1, L_3) \in R\). Hence, \(R\) is transitive.

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

Set of all lines related to \(y = 2x + 5\):
Two lines are parallel if and only if they have the same slope. The given line has a slope of \(2\).
Any line parallel to \(y = 2x + 5\) must be of the form \(y = 2x + c\), where \(c\) is a real constant.
Therefore, the set of all lines related to \(y = 2x + 5\) is given by \(\{y = 2x + c : c \in R\}\).
In simple words: Parallelism is reflexive, symmetric, and transitive, so it forms an equivalence relation. All lines related to a given line are simply other lines with the exact same slope.

Exam Tip: For equivalence relations involving lines, explicitly mention the geometric properties of slopes and parallelism.